Tag Archives: chemistry notes form 1-4 pdf

	Solid	Liquids	Gases
Mass	Definite	Definite	Definite
Shape	Definite	Indefinite: they take the volume of the container in which they are in.	Indefinite: they take the shape of the container in which they are.
Volume	Definite	Definite	Indefinite: volume may increase with increase in temperature; and decrease with decrease in pressure

Note: Conductors and non-conductors:

– The flow of electric current through materials is called electrical conductivity.

– Solid substances which allow electric current to flow through them are called conductors.

– Solid substances that do not allow electric current to flow through them are called non-conductors.

Role of chemistry in a society.

– Chemistry has enabled extraction of chemicals from plants.

– It is used in the manufacture of substances such as soap, glass, plastics, medicine, rubber, textiles etc from naturally occurring substances.

– Purification of substances from natural raw materials.

– It forms a basis for entry into careers e.g. teaching, medicine, chemical engineering etc.

Laboratory rules and safety symbols.

Necessity of laboratory safety rules.

To avoid accidents and injuries during practical experiments in the laboratory.
To avoid damage to and breakage of apparatus and laboratory fittings.
To avoid wastage of laboratory chemicals.

The laboratory safety rules and regulations.

– Never run while in the laboratory;

Reason: You may injure others or yourself in the laboratory.

– Never taste or eat something in the laboratory;

Reason: to avoid poisoning.

– Always consult your teacher before trying out any experiment; so as to avoid accidents.

– Label all chemicals in use so as to avoid confusion.

– Always use a clean spatula for scooping a substance from a container to minimize contamination.

– Always hold test-tubes and boiling tubes using test tube holder when heating; to avoid being burned.

– When heating a substance never let the open end of the tube face yourself or anybody else, because the liquid may spurt out and cause injury.

– Never look directly into flasks and test tubes where reactions are taking place, because the chemicals may spurt into your eyes and cause injury.

– Never smell gases directly. Instead, waft the gaseous fumes near your nose with your hand.

– Experiments in which poisonous gases and vapours are produced must be carried out in a fume cupboard or an open space outdoors.

– Always keep flammable substances away from flames because they easily catch fire.

– Always report any accidents to the teacher or the laboratory technician immediately for necessary action.

– In case of an accident do not scramble for the same exit, because it may hinder easy escape.

– Always put off flames that are not in use in order to avoid accidents and minimize fuel wastage.

– If a chemical gets on your skin or mouth rinse it immediately with a lot of water.

– Always dispose off the chemicals already used safely to avoid explosions and contaminations.

– Always work on a clean bench. After completing your experiment, clean all the pieces of apparatus you have used and return them to their correct storage places.

– Always read the label of the reagents before using them.

Safety symbols.

– These are signs found on the labels of bottles or cartons containing dangerous chemicals.

– The common safety symbols are as follows:

Symbols	Meaning.
	Toxic: are very poisonous and can easily kill if swallowed, inhaled or on contact with the skin. Examples: Chlorine and mercury;
	Harmful: Less harmful (dangerous) than the toxic substances; are only likely to cause pain and discomfort. Examples: copper (II) sulphate, lead (II) oxide
	Highly flammable: are substances that catch fire easily and must not be handled near open fire. Examples: ethanol, hydrogen
	Oxidizing: rapidly provide oxygen and can cause fire to burn more fiercely. Examples: potassium manganate (VII), hydrogen peroxide
	Corrosive: are substances that cause burns to skin and fabric; and can also react with other substances such as metals Examples: nitric (V) acid, conc. sulphuric acid, bromine.
	Irritant: can cause blisters or reddening of the skin; usually irritate the respiratory tract, skin, eyes etc. Examples: calcium chloride and zinc sulphate

Common Chemistry laboratory apparatus and their uses.

Name of apparatus	Diagram.	Use.
Test tubes		– General laboratory experiments; like heating solids; qualitative analysis etc
Boiling tubes		– Mainly used for heating small amounts of solids and liquids.
Test tube holder		Used for holding test tubes and boiling tubes during heating experiments.
Measuring cylinder		– Measuring accurate volumes of liquids in the laboratory
Beaker – Lipped glass or plastic vessels of various capacities.		– Glass beakers are used for boiling liquid substances; – Holding solutions during chemistry experiments.
Filtering funnel		– Directing liquids into containers with small narrow mouths; – Holding filter papers during filtration;
Name of apparatus	Diagram	Use
Stirring rod
Watch glass
Thermometer		– Measuring temperatures during experiments.
Conical flask		– Normal laboratory experiments like titration. – May be used for measuring volumes if graduated.
Round-bottomed flask		– Used when heating liquid substances because heat is supplied uniformly.
Flat-bottomed flask		– Used for general laboratory experiments.
Evaporating dish		– Used when evaporating liquids.
Name of apparatus	Diagram	Use
Crucible		– Use when heating solid substances that require strong heating.
Pestle and mortar		– Crushing substances while the mortar holds the substances being crushed.
Pie clay (ceramic) triangle		– Supporting crucibles during heating.
Tripod stand		– Supporting beakers and flasks in which liquids are being heated.
Wire gauze		– Used when glass apparatus are being heated; to facilitate even distribution of heat when heating substances in beakers or flasks
Clamp		– Supporting and holding pieces of apparatus during experiments.
Name of apparatus	Diagram	Use
Deflagrating spoon		– Holding burning substances.
Spatula		– Scooping solid substances from containers
Crucible tongs		– Holding solid chemicals.
Condenser
Separating funnel		– Separating immiscible liquids.
Thistle funnel		– Delivering liquid substances into other containers like flasks during reactions.
Name of apparatus	Diagram	Use
Wash bottle		– Holding water for rinsing apparatus
Dropping funnel
Test tube rack		– Holding boiling tubes and test tubes.
Teat pipette (dropper)		– Sucking liquid chemicals and placing them in another container dropwise.
Burette – It consists of a long narrow tube with a tap and a jet at the bottom.		– Delivering accurate volumes of liquids
Name of apparatus	Diagram	Use
Pipette		– Delivering a specified volume of liquid accurately.
Gas jar		– Gas collection.
Trough		– Holding some amount of water for some experiments e.g. gas preparation.
Reagent bottles		– Storing chemicals in liquid state.
Desiccator		– Drying substances or keeping substances free from moisture.
Name of apparatus	Diagram	Use
Spirit lamp Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments		– Heating substances in the laboratory.
Kerosene stove Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments		– Heating substances in the laboratory.
Electric heater		– Heating substances in the laboratory.
Candles Note: less preferred for heating because their flames are not hot enough; and they deposit soot on apparatus making them dirty hence difficulty in observing changes during experiments		– Heating substances in the laboratory.
Bunsen burner		– Heating substances in the laboratory.
Name of apparatus	Diagram	Use
Stop watch (clock)		– Measuring time particularly in determination of reaction rates.
Beam balance		– To take accurate weight measurements
Electronic balance		– Take accurate weight measurements; and can take extremely low weight measurements.
Volumetric flask

Note: most chemistry apparatus are made of glass:

Advantages of glassware:

– It does not react with most chemicals

– Glass is transparent and hence reactions can easily be observed as they progress.

– Glass materials are easy to wash and rinse after experiments.

– They can be used comfortable in heating experiments.

Disadvantages:

– Have higher chances of breakages in case they fall during experiments.

– They are comparatively expensive to plastics

– Some materials like beakers may however be made of plastics.

Advantages of plastic apparatus.

– Have lower chances of breaking.

– They are relatively cheaper to buy.

Disadvantages:

– Plastics tend to react with some laboratory chemicals

– may not be transparent and hence reactions cannot easily be observed as they progress.

– Glass materials are difficult to wash and rinse after experiments.

– They cannot be used in heating experiments.

The Bunsen burner.

– is the most common heating apparatus in the laboratory.

– Was invented by a German scientist known as Wilhelm Bunsen hence the name.

– It uses natural (cooking) gas for heating.

Parts of a bunsen burner

– Chimney

– The air hole

– Collar

– The jet

– Base

– Gas inlet

Diagram: structure of a bunsen burner. Diagram: Bunsen burner-components detached

Functions of the various parts:

The chimney:

– It is a hollow metallic cylinder with an air hole near its lower end.

Function:

– allows air and the laboratory gas from the jet mix before they start burning at the upper end of the chimney.

– Its upper opened end provided a site where the gas burns.

The air hole:

– Is a small aperture found at the lower end of the chimney and smaller than the collar.

– Its diameter (size) is regulated by the collar.

Function:

– Allows air (oxygen) to enter and mix with the laboratory gas in the chimney.

The collar:

– Is a metallic ring with an air hole whose diameter is the same size as that of the air hole in the chimney.

– It fits into the lower part of the chimney; and can rotate around the chimney opening or closing the air hole.

Function:

– Regulates the amount of air entering the chimney.

The jet:

– It is a very tiny opening just below the air hole, that connects the gas inlet to the chimney.

Function:

– allows the laboratory gas (methane) into the chimney at high pressure.

The gas inlet:

– It is a hollow metal connected to the base and extending into the jet.

– Its external opening is usually connected to a rubber tubing that is connected to a gas tap.

Function:

– Allows laboratory gas from the cylinders (reservoirs) in the lab; through the gas taps into the Bunsen burner.

The base:

– A thick heavy metal, that is usually circular or oval.

Function:

– It supports the Bunsen burner on the bench.

The Bunsen burner flames.

– A flame is a mass of burning gases.

– A bunsen burner can produce two types of flames depending on the size of the air hole and hence amount of air entering the chimney.

Types of flames.

Luminous flame.
Non-luminous flame.

(a). The luminous flame.

– It is a large bright yellow flame produced when the air hole is closed and hence no air enters the chimney.

Characteristics of a luminous flame.

– It is large, quiet and bright yellow.

– Colour is not uniform and it ahs four zones.

– It produces less heat.

– It gives a lot of light to the glow of unburnt hot carbon particles

– It produces soot.

Diagram: the luminous flame.

Parts of a luminous flame.

The thin outer zone:

– Is a fairly visible, narrow zone on the outer surface of the flame.

– At this point methane (lab gas) mixes with sufficient air from the outside and burns completely to carbon (IV) oxide and water.

The inner bright yellow zone:

– It is a large bright yellow zone that lies beneath the thin outer zone.

– Here, air supply is insufficient resulting to incomplete combustion of the gas.

– Consequently the gas burns producing tiny carbon particles instead of carbon (IV) oxide.

– The white hot carbon particles glow brightly and are responsible for the yellow colour and the emission of light.

– On cooling the carbon particles form soot, which blackens the bottom of the apparatus being heated.

The almost colourless inner zone.

– Is found below the yellow inner zone; and consists mainly of unburnt gases.

The blue zone (region)

– Is found on the outer side of the base of the flame.

– Here, air near the flame rises rapidly due to convection currents and mixes with the burning gas.

– This makes burning more complete than in the two upper parts above it.

Advantages of the luminous flame:

– Can be used for lighting purposes; because it produces more light.

Disadvantages.

– Produces less heat hence inefficient in heating.

– Due to production of soot it blackens apparatus thus preventing better observations of experiments.

(b). The non-luminous flame.

– It is a small blue flame produced when the air hole is completely open and hence a lot of air enters the chimney.

Characteristics of a non-luminous flame.

– It is small, noisy and blue.

– Colour is uniform and it ahs three regions.

– It produces comparatively more (a lot of) heat.

– It does not produce soot, due to complete combustion hence no carbon particles remain.

– It produces less light due to lack of white-hot carbon particles.

Diagram: the luminous flame.

Parts of a non-luminous flame.

The outer pale blue region.

– It is a large light blue zone.

– Here, there is a lot of air coming up the chimney from the air hole and from the outside.

– The air gas mixture thus burns completely to carbon (IV) oxide and water.

– No soot formation because there are no carbon particles.

The middle greenish blue region.

– consists of partially burnt gas-air mixture, due to insufficient air supply.

– However as the mixture rises up the pale blue region, it undergoes complete combustion due to plenty of air (from outside)

The inner almost colourless region.

– Is located at the base of the flame.

– It consists of unburnt gas-air mixture.

Advantages of a non-luminous flame.

– Gives out a lot of heat hence very efficient in hating.

– It does not form soot hence will leave apparatus clean even after experiment (heating).

Disadvantages:

– It uses a lot of laboratory gas in burning.

– cannot be used for lighting purposes since it produces very little light.

Differences between a luminous and a non-luminous flame.

Luminous flam	Non-luminous flame
Bright yellow in colour	Blue in colour
Produces a lot of light	Produces a lot of light.
Large and unsteady	Small and steady
Produces soot	Does not produce soot
Has four zones	Has three zones
Burns quietly	Burns noisily
Moderately hot	Very hot

Experiments on Bunsen burner flames.

To investigate the heating effects of the luminous and non-luminous flames.

Apparatus:

– Bunsen burners, 250 ml beakers, lighter, stopwatch, tripod stand, wire gauze.

Procedure

– 100 cm³ of water is put into ach of the two 250 ml beakers.

– One beaker is put over a luminous flame while the other is simultaneously put over a luminous flame

– Time taken for water to boil is noted for each set up.

– The bottom of ach beaker is observed for any changes.

Apparatus

Observations.

– Water heated over the non-luminous flame boiled ion a shorter time than the same amount of water heated over a non-luminous flame.

– The bottom of the beaker heated over the non-luminous flame remained clear but the one heated over the luminous flame was covered with black deposits of soot.

Explanations.

– The non-luminous flame is hotter than the luminous flame; hence boils the water faster

– The hottest part of the luminous flam is the outer blue zone.

– Incomplete combustion in the luminous flame leads to production of carbon particles, which when hot glow yellow and on cooling forms black soot on the beaker;

– Incomplete combustion in a non-luminous flame leads to production of carbon (IV) oxide and steam only, hence no soot formation.

Conclusions.

– The non-luminous flame is hotter than the luminous flame.

– The non-luminous flame is cleaner than the luminous flame.

To investigate the hottest part of a non-luminous flame.

Requirements

– Bunsen burner, stiff white paper (cardboard), wooden splint.

Procedure

– A bunsen burner is ignited with the air hole open to get anon-luminous flame.

– A piece of white paper (cardboard) is slipped into the flame in region marked X as shown below.

– The piece of paper is removed quickly before it catches fire.

– A fresh piece of paper is then slipped into region marked Y as shown below; then again quickly removed before it catches fire.

– The experiment for each of the regions marked X and Y is then repeated using wooden splints.

– The splints should be held long enough for some of their parts to get charred

Apparatus

Observations.

Using pieces of paper.

– In region X, the part of the paper that was in contact with the flame was charred uniformly as shown below.

– In region Y, the part of the paper in contact with the flame had a charred ring with an unburnt part in the middle of the ring as shown below

Diagrams

Using wooden splints.

– In region X, the part of the splint in contact with the flame was charred uniformly as shown below.

– In region Y, the part of the splint in contact with the flame had an unburnt part in between two charred regions as shown below.

Diagrams:

Explanations.

– Regions which become charred indicate that they are the hottest part of the flame.

– Region X corresponds to the outermost blue region of a non-luminous flame.

– Region Y is the almost colourless region of the non-luminous flame, which is however surrounded by the middle greenish blue and the outer pale blue zones.

– Thus in region X, the uniform charring of the paper and splint indicate that the outer pale blue zone is the hottest pat of the flame.

– Similarly the charred ring for experiment in region Y show that the parts in contact with the outer pale blue zone gets burnt faster before the parts in contact with the almost colourless or the greenish blue zones.

Conclusions.

– The hottest part of the non-luminous flame is the outermost pale blue zone.

– During heating the object being heated should not be placed nearer the chimney; these parts are less hot.

– For efficient heating the object being heated should be placed at the outermost region of the flame.

To show the presence of unburnt gases in a Bunsen burner flame.

Apparatus:

– Bunsen burner, tongs, narrow hard glass tubing.

Procedure

– A bunsen burner is lit and adjusted to get a non-luminous flame.

– A narrow hard glass tubing is held with a pair of tongs and one of its end s is placed in the colourless zone of the flame.

– A match is lit and placed at the free end of the glass tubing.

Apparatus

Observations.

– A flame is obtained at the free end of the glass tubing.

Explanations.

– The tubing trapped unburnt gases at the almost colourless zone of the flame.

– The trapped gases combined with atmospheric air (oxygen) at the other (free) end of the tubing hence the flame.

Conclusions.

– The almost colourless region contains unburnt gases.

To show the hottest part of the flame.

Apparatus:

– Bunsen burner, match stick

Procedure

– A matchstick is placed at the top of the bunsen burner chimney using a pin.

– A bunsen burner is lit and adjusted to get a non-luminous flame.

– The match stick is observed fro sometime for any changes.

– If no observable changes are made, the matchstick is then slowly raised towards the blue zone and observed keenly.

Apparatus

Observations.

– The matchstick did not ignite while it was at the bottom of the flame (resting on top of the chimney).

– It ignited as it was being raised towards the outer pale blue zone.

Explanations.

– The bottom of the flame (just on top of chimney) corresponds to the almost colourless zone.

– This zone contains unburnt gases, hence no burning occurs and is thus least hot to cause ignition of the matchstick.

– As the matchstick is raised upwards it moves past the greenish blue zone (where there is partial combustion) then to the outer pale blue zone where there is complete combustion and hence most heat.

– The heat in this region is adequate to cause ignition of the matchstick.

Conclusions.

– The outer pale blue zone is the hottest part of the non-luminous flame, and is thus the correct position to place an object during heating.

Methods of gas collection.

– Various chemical reactions produce gases; some of which are colourless while others are coloured.

– Additionally some gases are poisonous to the human body, while others are major causes of environmental pollution.

Examples:

Coloured gases:

Chlorine (green-yellow); nitrogen (IV) oxide (brown); bromine (red-brown)

Colourless gases:

Oxygen; carbon (II) oxide; carbon (IV) oxide; sulphur (IV) oxide; hydrogen; ammonia etc.

Factors affecting method used in collecting a gas.

– Density

– Solubility in water.

– Colour

– Toxicity

Summary on collection methods.

Method	Apparatus	Characteristic of gas
Upward delivery – Also called downward displacement of air		– Must be less dense than air. Examples: Hydrogen, ammonia gas. Note: being lighter the gas is supported by the denser air from below; – When used for colourless gases, it is not possible to know when the gas jar is full;
Downward delivery (upward displacement of air)		– Must be denser than air. Examples: carbon (IV) oxide; nitrogen (IV) oxide; chlorine gas; Note: – The gas displaces air and settle at the bottom of the collecting vessel – Unless the gas is coloured, it is difficult to know when the container is full
Over water		– insoluble or only slightly soluble in water; – does not react with water Examples: carbon (IV) oxide; hydrogen; carbon (II) oxide; Note: – with this method it is easy to tell when the gas jar or collecting tube is full of gas; – This method cannot be used when the gas is required dry;
Collecting syringe – the gas produced is collected in a syringe;		– Mainly for poisonous gases; since the gases are confined and leakages are limited; Note: this method allows collection of small volumes of gases; – It also allows direct measurement of volume of gas produced;

Drying of gases.

– Is the process by which the moisture in a gas being prepared is removed prior to collection.

– This is done by passing the gas through chemicals that absorb moisture.

– Such chemicals are called drying agents.

– The drying agents should not react with the gases being dried.

Examples of drying agents.

– Anhydrous calcium chloride

– Concentrated sulphuric acid.

– Calcium oxide.

Apparatus and drying agents

Collection and drying of some gases

Gas	Collection method	Drying agent
Oxygen	Over water	Concentrated sulphuric (VI) acid; anhydrous calcium chloride.
Hydrogen	Over water, upward delivery	Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Nitrogen	Over water	Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Carbon (IV) oxide	Over water, downward delivery	Concentrated sulphuric (VI) acid; anhydrous calcium chloride
Ammonia	Upward delivery	Calcium oxide

Drugs and drug abuse.

Drug: is a chemical substance that alters the functioning of the body.

Types of drugs

(i). Medicinal drugs (medicines):

– Are drugs mainly used for treatment and prevention of diseases.

– Are also classified into two: over the counter drugs and prescription drugs.

Over-the-counter drugs.

– Are medicinal drugs that can be bought at a pharmacy or retail shop without written instructions from a doctor.

Examples: Mild painkillers like aspirin, panadol, paracetamol, drugs for flu etc.

Prescription drugs:

– Are strong medicines which should only be taken upon a doctors instruction (prescription).

– In this prescription, the doctors give a dosage, which indicates the amount and the rate at which it should be taken.

(ii). Leisure drugs.

– Are drugs that are usually taken for pleasure.

– Are classified into two:

Mild drugs: alcohol, tobacco;
Narcotic drugs: marijuana, cocaine, heroin, mandrax etc.

Drug abuse:

– Is the indiscriminate use of a drug for purposes which it is meant for; or administration of an overdose or underdose of a drug; as well as use of drugs for leisure purposes.

Note:

– The worst form of drug abuse is the taking of drugs for leisure purposes; and the most commonly abused drugs are the leisure drugs.

Effects of commonly abused drugs.

Alcohol:

– Affects the brain and the nervous system

– Damages the liver, and is a common cause of liver cirrhosis.

– Poor health due to loss of appetite.

– Time for working is wasted in drinking and hence less productivity and even lose of jobs; which results to poverty and family disintegrations.

– Bad breath, discoloured fingers and teeth

– Cause diseases such as bronchitis and tuberculosis.

– damages the lungs and is a common cause of lung cancer due to chemicals found in the cigarettes.

– Smoking during pregnancy is a common cause of miscarriages or still births.

– It is expensive: money used for other better uses is wasted in cigarette smoking.

Narcotic drugs:

– Interferes with the functioning of the brain.

– Results to addiction and drug dependency.

– Some are administered directly into the blood through syringes and hence common routes of transmission of HIV/AIDS.

General effects of drug abuse on the society.

– Drug abuses spend most of their money on drugs and hence neglect their family leading to misery and societal breakdown.

– Drunk drivers cause accidents.

– People who are drunk with a drug are unreasonable and cannot make logical decisions; and hence cannot be productive at that time.

– Drug abuse has resulted into loss of morals leading to higher rates of rapes, violent crimes, murders, prostitution etc.

– Drug abuse has fueled the spread of sexually transmitted diseases and HIV/AIDS.

Note:

Drug addiction:

– Is a situation in which an individual becomes dependent on a particular drug such that he cannot function normally without it; and lack of it result to some discomfort.

UNIT 2: SIMPLE CLASSIFICATION OF SUBSTANCES

Unit Checklist:

Elements compounds and mixtures.
Mixtures

Types of mixtures
Separation of mixtures
Basic concepts
Method of separation of mixtures.
- Decantation
- Evaporation
- Condensation
- Filtration
- Crystallization
- Separating funnel separation
- Distillation
- Sublimation
- Chromatography
- Solvent extraction

Criteria for purity

Effects of impurity on melting point
Effects of impurity on melting point

Nature of matter and kinetic theory of matter.

Effects of heat on matter
- Melting
- Evaporation
- Condensation
- Freezing
- Freezing
- Sublimation

Permanent and non-permanent changes
Constituents of matter

Atoms
Elements
Molecules
Compounds

Names and symbols of common elements
Simple word equations.

Elements compound and mixtures.

(a). Element:

– Is a pure substance that cannot be split up into simpler substances by chemical means.

Examples: copper, hydrogen, carbon.

(b). Compound:

– A pure substance that consists of two or more elements that are chemically combined.

Examples:

Compound	Elements in the compound
Calcium carbonate	Calcium, carbon and oxygen
Sodium chloride	Sodium and chlorine
Ammonium nitrate	Nitrogen, hydrogen, oxygen
Iron (II) sulphate	Iron, sulphur, oxygen

(c). Mixture:

– A substance that consists of two or more elements or compounds that are not chemically combined

– Some mixtures can be naturally occurring while some are artificial.

Examples

Naturally occurring mixtures.

Mixture	Components
Air	Nitrogen, oxygen, carbon (IV) oxide, water vapour, noble gases etc
Sea water	Water and various salts like chlorides of sodium, potassium and magnesium
Crude oil	A mixture of hydrocarbons like methane, petrol, bitumen, etc
Magadi soda	Sodium carbonate, sodium hydrogen carbonate and sodium chloride

Artificial mixtures.

Mixture	Components
Soft drinks	Water, citric acid, sugar, carbon (IV) oxide, stabilizers, sodium benzoate
Black ink	Blue, black, yellow dyes and solvent
Cement	Oxides of aluminium, iron, silicon, calcium and calcium carbonate.

Types of mixtures:

– There are two types of mixtures;

Homogenous mixtures
Heterogenous mixtures

(i). Homogenous mixtures.

– Is a mixture with a uniform composition and properties throughout its mass.

– The parts (components) of the mixture are uniformly distributed throughout the mixture

Examples:

Tea with sugar solution.

(ii). Heterogenous mixture:

Is a mixture without uniform composition throughout its mass.

Examples:

– Soil, rocks and sand mixture.

Separating mixtures.

A mixture can be separated into its various components (constituents) by appropriate physical means, depending on type of mixture.

Basic concepts:

Residue: solid that remains on the filter paper during filtration

Filtrate: liquid that passes past the filter paper during filtration

Solute: a solid that dissolves in a particular liquid

Solvent: the liquid in which a solute dissolves.

Saturated solution: a solution in which no more solute can dissolve at a particular temperature

Unsaturated solution: a solution that can take more of the solute (solute) at a particular temperature.

Miscible liquids: liquids that can mix together completely.

Immiscible liquids: liquids that cannot mix together completely.

There are various methods that can be used to separate mixtures.

These include:

Decantation
Evaporation
Condensation
Filtration
Crystallization
Separating funnel separation
Distillation (simple and fractional)
Sublimation
Chromatography
Solvent extraction

Decantation:

– Is a method used to separate insoluble solids from liquids; a heterogenous mixture.

Procedure:

– The solid-liquid mixture is allowed to stand in a container.

– The insoluble solid settles at the bottom and the upper liquid portion poured out with care.

Apparatus.

Examples:

– Separation of sand-water mixture

– Separation of maize flour-water mixture.

Limitations (disadvantages) of decantation.

– It is not efficient as some fine suspended solids may come long with the liquid during pouring.

Filtration.

– Is the separation of an insoluble solid from a heterogenous mixture (liquid) using a porous filter that does not allow the solids to pass through.

– Upon filtration the undissolved solid is left on the filter paper and is called the residue.

– The liquid that passes the filter paper is called filtrate.

Examples: separation of sand from water.

(i). Procedure.

– The filter paper is folded into ¼ and opened to from a cone.

Diagram: folding a filter paper.

– It is carefully placed inside a filter funnel.

– The apparatus are then arranged as shown blow.

– The sand-water mixture is then poured into the filter paper in the filter funnel.

– The collecting liquid is directed into a conical flask.

(ii). Apparatus.

Applications of filtration.

– Filtration of domestic water.

– Extraction of medicinal substances from plants.

– Extraction of sugar from sugarcane.

– Operation of a vacuum cleaner.

– Fuel filters in automobile engines.

Evaporation.

– Is used to separate a soluble solid from its solution.

– Such solutions are usually homogenous mixtures.

– The solid is called a solute while the liquid is called a solvent.

Example: separation of salt from salt solution.

(i). Procedure:

– The salt solution is poured in an evaporating dish.

– The set up is then arranged as in the apparatus shown below.

– The solution is boiled under steam or sand bath until all the water in the salt solution evaporates and salt crystals remain in the dish.

(ii). Apparatus.

(iii). Observations and explanations.

– Upon heating the solution, water evaporates because it has lower boiling point than the salt.

– The solution is boiled until salt crystals start appearing on a glass rid dipped into the solution.

– This shall indicate that the solution is saturated.

– The saturated solution is allowed to cool and crystallize.

– The mother liquor (liquid that remains with the crystals) is poured and the salt (solid) dried between absorbent papers.

Note:

A crystal: is a solid that consists of particles arranged in an orderly repetitive manner.

– It is advantageous to boil the solution under a steam or sand bath rather than directly.

Reason:

– The steam or sand bath prevents the mixture from splashing out (spitting) of the evaporating dish.

– It also reduces chances of the evaporating dish cracking.

Applications of evaporation:

– Extraction of soda ash from Lake Magadi.

Crystallization and recrystallization.

(a). Crystallization:

– is the process of formation of crystals from a solution.

– It involves evaporation of the solution to form a concentrated solution.

Example: crystallization of potassium nitrate from its solution.

(i). Procedure:

– About 5g of powdered potassium nitrate is added to 10cm³ of water in a boiling tube.

– The solution is heated until all the solid dissolves and then allowed to cool and crystallize.

Note:

– More potassium nitrate dissolves in hot water than in cold water.

– The resultant solution is then heated until crystals start appearing; and this can be confirmed by dipping a glass rod into the solution and feeling for crystals.

– This is called a saturated solution i.e. a solution that cannot take in any more of the solute at a given temperature.

– The saturated solution is then allowed to cool and crystallize.

(ii). Observations:

– The resultant solid particles have definite shapes.

– Some are needle-like while others are flat and sharp-edged.

– These are the potassium nitrate crystals.

(b). Recrystallization:

– Is used in obtaining pure crystals from a soluble solid containing impurities.

– Involves filtration and evaporation.

Examples:

Obtaining pure copper (II) sulphate crystals from impure copper (II) sulphate.

Purification of rock salt.

Note: The process can be enhanced by suspending a small piece of pure crystal into the saturated solution.

Diagram: recrystallization of copper (II) sulphate.

Applications of crystallization.

– Separation of Trona from sodium chloride in Lake Magadi.

Distillation.

– Is the vapourisation of a liquid from a mixture and then condensing the vapour.

– Is used in the purification of liquids and separation of liquids from a mixture.

– It utilizes the differences in boiling points of the components of the mixture.

– Are of two types:

Simple distillation
Fractional distillation.

(i). Simple distillation.

– Is mainly used for purification of liquids containing dissolved substances.

– It is also useful in separating two miscible liquids with widely differing boiling points

Note:

Miscible liquids: Liquids that mix to from a uniform a uniform homogenous solution

– The liquid with the lower boiling point usually distills over first, and is collected.

Example: To obtain pure water from sea water.

(i). Procedure:

– Salty sea water is poured into a distillation flask.

– A few pieces of pumice or porcelain is added to the solution.

Reason:

– To increase the surface area fro condensation and evaporation.

– The solution is heated until it starts boiling, then the burner removed so that the liquid boils gently.

– The boiling goes on until the liquid (distillate) starts collecting in the beaker.

(ii). Apparatus.

(iii). Observations and explanations:

– The water boils and the resultant steam is passed through the Liebig condenser.

– As the vapour passes through the condenser, it is cooled by circulating cold water through the jacket of the condenser.

– The cold water enters through the lower bottom and leaves through the top upper part.

Reason:

– To provide more time for the cold water to condense the vapours.

– The distillate is collected in the beaker while the residue remains in the distillation flask.

Applications of simple distillation.

– Manufacture of wins and spirits.

– Desalinization of sea water to obtain fresh water.

(ii). Fractional distillation:

– Is a method used fro the separation of miscible liquids with very close boiling points.

Examples:

Ethanol and water.

– It is a modification of simple distillation in which the fractionating column is inserted on top of the distillation flask.

– All the components must be volatile at different extents in order for separation to be possible.

The fractionating column.

– Is usually an elongated (glass) tube, packed with pieces of glass beads or pieces of broken glass.

Role of glass beads.

– To increase the surface area for vapourisation of the various components of the mixture and allow the separation of the vapours to occur.

– Thus the more the glass beads in the fractionating column, the higher the efficiency of separation.

Note:

– The efficiency of the fractional distillation so s to get more pure components can also be done by:

Increasing the length of the fractionating column (making it longer)
Making the fractionating column narrower (decreasing the diameter)

Volatile liquids:

– Are liquids with the ability to change into vapour.

– More volatile liquids vapourize and condense faster than the less volatile liquids.

Note:

– During fractional distillation, the components of a mixture are collected at intervals, one at a time with the most volatile (lowest boiling point) coming out first.

– Each component collected in the receiver is called a fraction.

Example: separation of ethanol and water.

(i). Apparatus.

(ii). Procedure:

– Water-ethanol mixture is poured into a round-bottomed flask.

– The apparatus is then connected and set up as shown below.

Note:

– The thermometer bulb must be at the vapour outlet to the condenser.

Reason:

– For accurate determination of the vapourisation temperature for each fraction.

– The mixture is then strongly heated until the first fraction comes out of the distillation flask into the conical flask.

– Collection of the fractions should be done in a conical flask other than in a beaker.

Reason:

– To reduce the rates of evaporation of the fractions, especially the highly volatile ones (in this case ethanol)

– For this particular separation the first temperatures recorded by the thermometer should not exceed 80^oC; to ensure that the first fraction is only ethanol.

(iii). Discussion.

– Ethanol boils at 78^oC and water boils at 100^oC.

– When the mixture is heated, ethanol and water evaporate and pass through the fractionating column which is filled with glass beads to offer a large surface area

– The large surface area encourages evaporation of ethanol and condensation of water vapour.

– Water can be seen dropping back into the distillation flask.

-Ethanol vapour passes through the condenser and warm liquid ethanol is collected in the conical flask.

Note:

– The first portion is almost pure ethanol (about 97%) and burns quietly with a blue flame.

– It also has the characteristic smell of alcohols.

Industrial applications of fractional distillation.

– Separation of air into various components in BOC gases Kenya limited.

– Separation of crude oil into paraffin, petrol, kerosene diesel and other components in the Kenya oil refinery.

– Distillation of ethanol from molasses at Muhoroni Agro-chemicals company.

Sublimation.

– Is the process by which a solid changes directly to gaseous state upon heating.

– It is used to separate a mixture in which one of the components sublimes on heating.

Note:

Solid Gas

Solids that sublime have very weak forces of attraction between the atoms and hence are easily broken on slight heating.

Examples of solids that sublime on heating.

– Iodine; sublimes to from a purple vapour.

– Ammonium chloride; sublimes to from dense white fumes;

– Solid carbon (IV) oxide (dry ice);

– Anhydrous iron (III) chloride; sublimes to give red brown fumes.

Examples: separation of iodine from sodium chloride.

(i). apparatus:

(ii). Procedure:

– The iodine-common salt mixture is poured into a beaker and placed in a tripod stand.

– A watch glass full of cold water is placed on the beaker.

– The beaker is heated gently until some dense purple fumes are observed.

(iii). Observations:

– A purple vapour appears in the beaker.

– A dark-grey shiny solid collects on the bottom of the watch-glass.

– White solid remains in the beaker.

(iv). Explanations:

– Upon heating the mixture iodine sublimes and condenses on the cold watch glass to form a sublimate of pure iodine.

Note:

– Solid carbon (IV) oxide (dry ice) is used a s refrigerant by ice cream and soft drink vendors.

Reason:

– It sublimes on heating; as it sublimes it takes latent heat from ice cream (soft drinks) thus leaving it cold.

– It is also advantageous as it does not turn into liquid, which could be cumbersome to carry and would mess up the ice cream.

Chromatography.

– Is the separation of coloured substances using an eluting solvent.

– It is also used to identify the components of a coloured substance.

– It involves the use of a moving liquid (eluting solvent) on a material that absorbs the solvent.

– It involves two major processes:

Solubility:

The tendency of a substance to dissolve in a solvent.

Adsorption:

The tendency of a substance to stick on an adsorbent material.

Examples:

Separation of components of black ink.

(i). Procedure:

– A filter paper is placed on the rim of an evaporating dish or a small beaker.

– A drop of the black ink is placed at the centre of the filter paper; allowed to spread out and dry.

– A drop of water (ethanol) is then added to the ink and allowed to spread.

– After complete spread of the drop, a second drop is added.

– Water drops are added continuously until the disc of coloured substances almost reaches the edge.

(ii). Observations:

Note: The dry filter paper showing the separated components of a mixture is called a chromatogram.

(iii). Explanations:

– Water is the eluting solvent since ink is soluble in it.

– The various dyes in the black ink move at different distances from the black spot hence the bands.

Reasons:

– The dyes have different solubilities in the solvent; the more soluble the dye, the further the distance it travels on the absorbent paper

– They have different rates of adsorption i.e. the tendency of the dyes to stick on the absorbent material; dyes with low rates of absorption travel far from the original spot.

Note:

Solvent front.

– Is the furthest distance reached by the eluting solvent on the filter paper.

Baseline:

– The point at which the dye to be separated is placed; i.e. it is the starting point of separation.

To verify contents of red, black and blue inks.

(i). apparatus:

(ii). Observations:

(iii). Explanations:

– The mixtures A to D have various components with varying solubilities in the solvent (ethanol)

– Mixture D is the most pure because it has only one spot.

– Mixture C is the least pure (most impure), as it has the highest number of spots indicating it is composed of so many dyes (four)

– Mixture C has the most soluble dye; its last component is the one nearest to the solvent front.

– Mixtures with similar dyes in their composition have spots at same levels; in this case A, B and D.

Applications of chromatography.

– Purification of natural products such as hormones, vitamins and natural pigments.

– Detection of food poisons e.g. in canned foods and soft drinks.

Solvent extraction.

– Is the extraction of a solute from its original solvent by using a second solvent in which it has a higher solubility

Example: extraction of oil from nuts.

(i). Apparatus.

(ii). Procedure:

– Some nuts are crushed in a mortar using a pestle; to increase the surface area for solubility.

– A suitable solvent such as hexane or propanone (acetone) is added.

– The nuts are further crushed in the solvent.

– The resultant solution is decanted in an evaporating dish, and left in the sun to evaporate.

– The liquid remaining in the evaporating dish is smeared onto a clean filter paper.

(iv). Observations;

– A permanent translucent mark appears on the filter paper.

(v). Explanations:

– The nuts are crushed when in contact with the solvent to bring more of the oil in the nuts closer to the solvent.

– Upon evaporation oil is left behind because it has a higher boiling point than the solvent.

– A permanent translucent mark verifies the presence of oils.

Applications of solvent extraction.

– Used by dry-cleaners to remove dirt (grease) and stains from dry-clean-only clothes such as sweaters, suits, dresses etc.

Separating funnel separation.

– Is used fro separating a mixture with two or more immiscible liquids.

– Such liquids do not mix but instead form layers based on their densities.

– The heaviest liquid layer is found at the bottom of the separating funnel; while the lightest liquid is found at the top of the separating funnel.

– The liquids are drained one after the other by opening and closing the tap of the separating funnel.

Example: Separation of oil from water

Apparatus:

Beakers, separating funnel, paraffin oil, distilled water, rubber stopper.

Apparatus.

Procedure.

– The tap of the separating funnel is closed.

– Equal volumes of water and paraffin are put in a separating funnel until it is half full.

– The mouth of the funnel is closed with a stopper and the mixture shaken.

– The mixture is allowed to stand until two distinct layers are formed.

– The stopper is removed and the tap opened to allow the bottom layer to drain into the beaker.

– The tap is closed after most of the bottom layer has drained off.

– The beaker is removed and the rest of the bottom layer is drained into a separate container and discarded; to ensure that no part of the top layer (paraffin) gets into the beaker containing the bottom (water) layer.

– The other (top) layer is then drained into another beaker.

Observations:

– After the mixture has settled oil and paraffin separate into two layers.

– The first beaker contains only water; while the second beaker contains only paraffin.

Conclusion.

– Paraffin and water are immiscible.

– The top layer contains water which is denser while the top layer contains oil (paraffin) which is lighter.

Practical application:

– Extraction of useful substances from complex mixtures.

Example: Separation of iodine from sodium chloride.

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Use of magnets.

– Is used to separate solid mixtures, one of which is magnetic (usually iron).

– The iron is picked with a magnet leaving the other components of the mixture behind.

Examples:

– Separation of powdered iron from iron powder-sulphur mixture.

Practical applications:

-In the extraction (mining of iron); where magnetic iron ore is separated from other materials in the crushed ore.

– Separation of scrap iron from non-magnetic materials like glass and plastics in recycling plants.

Criteria for purity.

Pure substance:

– Is a substance that contains only one type of compound or element.

Determination of purity

(a). Solids:

– Purity of solids is determined by measuring the melting point.

– Pure solids melt sharply over a narrow temperature range.

Examples:

– Naphthalene melts at 80^oC 81^oC.

– Water melts at 0^oC.

Effects of impurity on melting point.

– Impurities lower the melting point of a substance making it melt over a wide range of temperatures i.e. the melting point is not sharp.

Applications:

Extraction of metals.

– Impurities are added to purified metal ores to lower their melting points, hence save energy and extraction costs.

To prevent knocking of engines due to freezing of water in car radiators during cold seasons, impurities like ethylene glycol or salt (NaCl) is added.
Defrosting of frozen roads and sidewalks in temperate countries, by sprinkling a salt such as sodium chloride.

(b). Liquids:

– Purity of a liquid is determined by measuring its boiling point.

– A pure liquid has a sharp boiling point.

Examples:

Pure water boils at 100^oC at 1 atmospheric pressure.

Pure ethanol boils at 78^oC.

Effect of impurity on boiling point.

– Impurities raise the boiling point of a liquid.

Example:

– Sea water boils at a higher temperature than pure water due to the presence of dissolved salts.

States of matter.

Matter:

– Matter is anything that occupies space and has mass.

– It is composed of either pure substances or a mixture of substances.

States of matter:

Matter exists in three states:

Solid
Liquid

Kinetic theory of matter:

– States that matter is made up of small particles which are in continuous random motion.

– The continuous random motion of particles in matter is called Brownian motion.

– The rate of movement of particles in matter depends on the state

(a). Solid state:

– Have closely packed particles held by strong forces of attraction.

– Particles do not move from one point to another but vibrate about a fixed position.

– Upon heating, they start to vibrate vigorously.

Reason:

– Due to increase in their kinetic energy.

(b). Liquid state:

– Particles are further apart from one another than those in the solid state

– Forces of attraction between the particles are weaker than those in the solid.

– The particles remain close, but are free to move from one position to another.

– On heating they gain energy and move rapidly.

(c). Gaseous state:

– The particles are far apart and free to move randomly in all directions.

– Consequently they lack definite shape and volume, but occupy the whole space within a container.

– Forces of attraction between the particles are very weak.

The effect of heat on substances.

– A substance can change from one state to another upon heating or cooling.

– These are physical changes and can be reversed.

– There are five processes involved in changes of state:

Melting
Evaporation
Condensation
Freezing
Sublimation

The processes involved in change of states of matter.

(i). Melting:

– Is a change of state from solid to liquid.

Melting point:

– Is the constant temperature at which the melting takes place.

Examples:
– Ice melts at 0^oC.

– Sodium chloride melts at 800^oC.

– During melting the energy supplied to the particles is used to weaken the forces of attraction so that particles can move about.

(ii). Vapourisation (evaporation)

– Is change of state from liquid to gas.

Boiling point:

– Is the constant temperature at which a liquid changes from solid to a gas.

– During boiling, the energy supplied is used to break the forces of attraction in the liquid thus moving the particles far away from each other.

Examples:

Water boils at 100^oC, while ethanol boils at 78^oC at one atmospheric pressure.

Reason:

The forces of attraction between the water particles are stronger than those of ethanol.

(c). Condensation:

– Is the change of state from a gas to a liquid.

– Is a change due to decrease in temperature.

– When the temperature of gas is decreased, the particles lose kinetic energy to the surroundings to move slowly.

– The attractive forces become stronger, and the sample changes to a liquid.

Note:

– The temperature at which condensation occurs is the same as the boiling point.

(d). Freezing:

– Is the change from a liquid to a solid.

– It is also due to decrease in temperature.

– When a liquid is cooled, the particles lose energy and move very slowly.

– They attract one another strongly, and ultimately remain in fixed positions.

Note:

– The freezing point is the same as the melting point.

(e). Sublimation.

– Is the process whereby a solid does not melt when heated, but changes directly to the gaseous state.

Example:

– Iodine solid changes to purple vapour when heated to 70^oC.

– Dry ice (solid CO₂), used to cool ice cream, evaporates without leaving a liquid.

Note:

– The reverse of sublimation, whereby a gas changes directly to solid is called deposition.

Summary on changes of state.

Solid Liquid Gas

Key: A: melting; B: vapourisation; C: Freezing; D: Condensation; E: Sublimation; F: Deposition;

Experiment: Investigating changes in temperature when ice is heated.

(i). Procedure:

– A 250ml beaker is half-filled with dry ice, and the initial temperature recorded.

– The ice is heated, while stirring with a thermometer and the temperature recorded every 30 seconds.

– Heating and recording is done until the resultant water starts to boil.

– A graph of temperature against time is plotted.

(ii). Results:

Temperatures (^oC)	-10	0	30	60	90	120	150	180	210
Time (seconds)

(iii). Graph:

Effect of heat on pure ice

(iv). Explanations:

Point A-B:

– As the ice is heated the temperature rises steadily from -10^oC to 0^oC.

Reason:

– The heat supplied increases the kinetic energy of the ice (solid water) molecules; collisions between them hence increased temperature.

Point B-C:

-The temperature of the ice remains constant even as heat is applied.

Reason:

– Heat supplied is used to break the forces of attraction between the water molecules in ice.

– This is the melting point hence at B-C the ice melts.

Points C-D:

– At C, all the ice has already melted (turned to water).

– Between C and D, the temperature of the water increases as heating continues.

Reason:

– The heat supplied increases the kinetic energy of the water molecules; their rate of collision increases hence increased temperatures.

Points D-E:

– Temperature of the water remains constant even as heat is being supplied.

– Point D-E is the boiling point i.e. 100^oC.

Reason:

– The heat supplied is used to break the forces of attraction between water molecules in the liquid.

Points E-F:

– At point E, all the liquid water has turned into vapour.

– Thus between E and F, the temperature of the vapour rises as heat is applied.

Note: Heating curve for an impure solid.

Effect of heat on impure ice

The cooling curve.

– Is a curve that shows how the temperature of a substance changes with time as it is cooled from a gas into a solid.

– It is the opposite of a heating curve.

Example: the cooling curve of water

Explanations:

Points A-B:

– Gaseous state; temperature is declining.

– Heat loss results into decrease in kinetic energy of the gaseous molecules.

Points B-C:

– This is the condensation point;

– The water vapour condenses to the liquid state; bonds are formed as the hest is lost.

Points C-D;

– Temperature of the liquid water is declining;

– Heat loss results into decrease in kinetic energy of the liquid molecules.

Points D-E;

– This is the freezing point;

– The temperature remains constant as the heat is being lost;

– The heat being lost results into bond formation; as the liquid forms a solid.

Points E-F;

– The water is now in solid state.

– The temperature of the solid declines as heat is being lost.

Permanent and temporary changes.

– Heat causes matter to change.

– Changes due to heat can either be permanent or non-permanent (temporary).

(a). Permanent changes.

– Are also called chemical changes.

– Involves substances that are relatively less stable to heat.

Characteristics of permanent (chemical) changes.

(i). New substances are formed.

(ii). Involves considerable heat changes; energy is either given out or absorbed.

(iii). The mass of the substance changes.

Types of permanent (chemical) changes.

– Chemical changes are of two main types:

Reversible permanent changes.
Irreversible (non-reversible) permanent changes.

(i). Reversible permanent changes.

– Are chemical changes in which the final new products can recombine to form the original substance, under certain conditions.

Generally:

A + B C + D

Reactants Products

Examples:

Effect of heat on hydrated blue copper (II) sulphate.

(i). Apparatus:

– Test tubes, Bunsen burner, test tube holder

– Hydrated copper (II) sulphate

(ii). Apparatus:

(ii). Procedure:

Dry crystals of hydrated blue copper (II) sulphate are put in a clean dry test tube.

– The apparatus are arranged as above.

– The copper (II) sulphate is heated until no further change.

– The delivery tube is removed from the collected liquid while heating continues.

Reason:

– To avoid sucking back of the condensing liquid which would otherwise rehydrate the anhydrous copper (II) sulphate

– The test tube is allowed to cool and the remaining solid is divided into two portions.

– To one portion of the powder, add distilled water, while to the other potion add the condensed liquid.

(iii). Observations:
– A white solid/ powder remains in the test tube after heating.

– A colourless liquid condenses in the test tube dipped into the ice cold water.

– The colourless liquid turns the white solid into blue.

(iv). Explanations:
– Hydrated blue copper (II) sulphate have water of crystallization, giving it the characteristic blue colour.

– During heating, the heat energy supplied is used to drive out the water molecules (particles) out of the crystals;

– Without water, the copper (II) sulphate turns white and thus called anhydrous copper (II) sulphate;

– The water driven out of the crystals condenses in the test tube immersed in the ice cold water.

Heat

Equation

Cool

Hydrated copper (II) sulphate anhydrous copper (II) sulphate + Water

Blue White

Heat

Cool

Chemically:
CuSO₄.5H₂O_(s) CuSO_4(s) + 5H₂O_(g);

Confirmatory test for water.

It turns white anhydrous copper (II) sulphate to blue;
It changes blue cobalt chloride paper pink;

Confirmatory test that the water is pure:

It boils at 100^oC;
It melts at 0^oC;
It has a density of 1g cm^-3;
It has a refractive index of 1.33;

Note:
– The delivery tube is removed from the collecting liquid while heating is continued; to ensure that no water condenses back into the copper (II) sulphate as this would cause rehydration;

– Addition of water to the anhydrous copper (III) sulphate changes its colour from white to blue;

Conclusion:

The effect of heat on copper (II) sulphate is a reversible chemical change;

Effect of heat on ammonium chloride.

– When ammonium chloride is heated, it produces ammonia gas and hydrogen chloride gas.

– These are seen as dense white fumes.

– Reversible, when ammonia and hydrogen chloride are gases are reacted or allowed to cool, they produce ammonium chloride;

Heat

In summary:

Cool

Ammonium chloride solid ammonium chloride + hydrogen chloride;

White Dense white fumes

Heat

Cool

Chemically:
NH₄Cl_(s) NH_3(g) + HCl_(g);

Ammonium chloride Ammonia gas hydrogen chloride gas

Decomposition of calcium carbonate.

Heat

In summary:

Cool

Calcium carbonate solid Calcium oxide + Carbon (IV) oxide;

(ii). Irreversible chemical change.

– Are chemical changes in which the resultant products cannot recombine to form the original substance (reactants);

– Majority of the chemical changes are irreversible;

Generally:
A + B C + D

Reactants Products

Examples:
1. Action of heat on potassium manganate (VII)

(i). Apparatus and chemicals.

– Bunsen burner, test tube, trough, wooden splint;

– Potassium managnate (VII);

(ii). Apparatus set up.

Heat

Potassium manganate (VII)

(iii). Procedure:
– 2 end-fulls of a spatula of potassium manganate (VII) are put in a hard glass test tube;

– The set up is assembled as shown above;

– The solid potassium manganate (VII) is heated, and the resultant gas collected over water;

– The resultant gas(es) is tested with a glowing splint;

(iv). Observations:

– The purple solid turns black;

– A colourless gas collects over water;

– The colourless relights a glowing splint;

(v). Explanations:
– Potassium manganate (VII), a purple solid was decomposed (splint up) on heating to yield (give potassium manganate (III) and oxygen.

– The potassium manganate (III) is the black residue;

– The colourless gas is oxygen; and relighting a glowing splint is the confirmatory test;

In summary:
Potassium manganate (VII) → Potassium manganate (III) + oxygen

Purple solid Black solid Colourless gas

Note: It is not possible for oxygen and potassium manganate (III) to recombine back to potassium manganate (VII); hence the change is irreversible;

Thermal decomposition of copper (II) nitrate.

– The blue solid decomposes to form a black solid; copper (II) oxide, red-brown fumes of nitrogen (IV) oxide and a colourless gas, oxygen;

In summary:
Copper (II) nitrate → Copper (II) oxide + Nitrogen (IV) oxide + oxygen

Blue solid Black solid Brown (red-brown) fumes Colourless gas

Note: Further examples of chemical changes

– The burning of any substance (except platinum);

– The rusting of iron;

– Addition of water to calcium oxide;

– Explosion of natural gas or hydrogen with air;

– Reacting of sodium in water;

Note: Exothermic and endothermic reactions:

(i). Exothermic reactions.

– Are reactions in which heat is released // given out to the surrounding;

– Usually the final temperature of the reaction vessel // mixture (e.g. beaker is higher than initially;

Examples:
– Freezing;

– Condensation;

– Deposition // sublimation of fumes to solid;

(ii). Endothermic reactions.

– Are chemical reactions in which heat is absorbed from the surrounding;

– The final temperature of the reaction vessel or reaction mixture is usually lower than the initial i.e. they are accompanied by a drop in temperature;

Examples.

– Melting;

– Vapourization;

– Sublimation (of solid to gas)

Summary on chemical changes.

Reaction	Appearance of substance	Changes during reaction	New substance(s)	Type of change
Heating hydrated copper (II) sulphate	Blue	Blue crystals turn into a white powder; colourless liquid condenses on cooling;	Anhydrous copper (II) sulphate and water	Chemical
Heating potassium manganate (VII)	Shiny purple crystals	The purple solid turns black; evolution of a colourless gas;	Potassium manganate (III) and oxygen;	Chemical;
Heating ammonium chloride	White solid // powder;	Dense white fumed that cools to a white solid;	Ammonia gas and hydrogen chloride gas;	Chemical;
Heating lead (II) nitrate;	White solid	The white solid turns into a red solid during heating which on cooling turns yellow; – Decrepitating sound; – Brown fumes; – colourless gas;	– Lead (II) oxide; nitrogen (IV) oxide and oxygen gas;	Chemical;
Heating lead (II) nitrate;	White solid	The white solid turns into a yellow solid during heating which on cooling turns white; – Decrepitating sound; – Brown fumes; colourless gas;	– Zinc (II) oxide; nitrogen (IV) oxide and oxygen gas;	Chemical;
Heating copper turnings	Brown turnings;	– Brown turnings // solid turn black;	– Copper (II) oxide;	Chemical;
Rusting of iron;	Grey solid	– Grey solid turns into a red brown solid;	– Hydrated iron (III) oxide;	Chemical;
Heating Copper (II) nitrate;	Blue solid	The blue solid turns into a black solid; – Brown fumes; – colourless gas;	– Copper (II) oxide; nitrogen (IV) oxide and oxygen gas;	Chemical;
Heating copper (II) carbonate	Green solid	The green solid turns into a black solid; – colourless gas;	– copper (II) oxide and carbon (IV) oxide;	Chemical;

(b). Temporary (non-permanent) changes.

– Are also called physical changes;

– They are changes that involve substances that are more stable to heat;

– On heating they do not decompose hence no new substances are formed;

Characteristics of permanent changes.

All are reversible upon changes in temperature;
No new substance is formed (instead there are only changes of state);
The mass of the substances do not change;

Examples:

Solid	Original appearance	Observations during heating then cooling
1. Candle wax	White sticky solid;	– The solid melts into a colourless viscous liquid; and on cooling solidifies to the original solid wax again;
2. Iodine solid.	Shiny dark-grey crystals;	– The solid turns directly to purple vapour (sublimation); – On cooling the purple iodine vapour (gas) changes directly to solid iodine (deposition); ie. Iodine solid ═ iodine vapour; *Dark grey Purple*
3. Zinc oxide	White solid;	– The white solid turns yellow on heating and upon cooling changes back to the original white colour; ie. Zinc oxide ═ Zinc oxide; *White (cold) Yellow (hot)*
4. Ice	White	– The solid water melts into liquid and on further heating the liquid vapourizes and turns into gas; – On cooling the gas condenses to liquid which then freezes back into solid; i.e Ice ═ Water ═ Gas
5. Platinum wire;		– A white glow of the metal is seen on heating, but on cooling the metal changes back to its original grey colour;
6. Lead (II) oxide;	Yellow	– The yellow solid turns red on heating and upon cooling changes back to the original yellow colour; ie. Lead (II) oxide ═ Lead (II) oxide; *White (cold) Yellow (hot)*

Differences between Physical and Chemical changes.

Physical change	Chemical change
1. Produces no new kind of substance;	– Always produces a new kind of substance;
2. UIs usually (generally) irreversible;	Are generally irreversible; with only few exceptions (i.e. most are irreversible);
3. The mass of the substance does not change;	– The mass of the substance changes;
4. No energy is given out or absorbed i.e. are not accompanied by great heat changes;	– Energy is usually given out or absorbed i.e are usually accompanied by great heat changes;

Constituents of matter

– A detailed examination of matter reveals that it is built of very tiny units called toms;

– Presently about 115 atoms have been identified;

– The arrangement and number of atoms in a substance will result into other much larger constituents of matter;

– These are:

Elements;
Molecules;
Compounds;
Mixtures;

The atom;

– Is the smallest particle of matter that can take part in a chemical reaction;

– It is the smallest particle into which an element can be divided without losing the properties of the element;

– Atoms of various elements all differ from one another;

Examples:
– Copper is made up of many copper atoms;

– Sodium element is made up of many sodium atoms;

Elements.

– An element is a substance that cannot be split into anything simple by any known chemical means;

– An element consists of a single type of atom;

– There are about 155 known elements, 90 of which occur naturally.

– Elements are classified into two main groups;

Metals:

– All are solids at room temperature (except mercury); and are good conductors of electricity;

Non-metals;

– Exists as solids and gases;

– All are poor electric conductors except graphite;

Examples of elements.

(i). Metals:

– Sodium, magnesium, potassium, aluminium, lead, iron, zinc, silver, gold, tin, platinum, uranium, calcium, manganese etc.

(ii). Non-elements.

– Carbon, nitrogen, sulphur, oxygen, chlorine, fluorine, argon, neon, bromine, iodine, silicon, boron, xenon, krypton.

Molecule.

– Is the smallest particle of a substance that can exist independently;

– It is made when 2 or more atoms (similar or dissimilar) are chemically combined together;

– However atoms of noble/ inert gases exist as single atoms;

Note:

– Depending on number of atoms molecules can be categorized into:

(i). Monoatomic molecule;

– made up of only one atom;

Examples:
– Argon;

→

Argon atom Argon molecule;

– Neon;

– Helium;

(ii). Diatomic molecules.

– Made up of 2 similar atoms; chemically combines;

Examples

– Oxygen gas;

→

Oxygen atom Oxygen molecule;

– Nitrogen gas;

– Hydrogen gas;

– Chlorine gas;

(ii). Triatomic molecules.

– Made up of 3 similar atoms; chemically combines;

Examples

– Ozone molecule;

→

Oxygen atom Ozone molecule;

Note:

– Other molecules are also made from atoms of different elements chemically combined together;

Examples:
(i). Hydrogen chloride;

+ →

Hydrogen Chlorine Hydrogen chloride molecule;

Atom. Atom;

(ii). Water molecule;

+ + →

Hydrogen atoms oxygen atom; Water molecule;

Compounds.

– A compound is a pure substance consisting of two or more elements that are chemically combined.

– Compounds usually have different properties from those of its constituent elements;

– Properties of a compound are uniform throughout any given sample and from one sample to another;

Examples of compounds and their constituent elements.

Compound	Constituent elements
Colourless water liquid;	Oxygen and hydrogen;
Green copper (II) carbonate;	Copper, carbon and oxygen;
White sodium nitrate;	Sodium, nitrogen and oxygen;
Black copper (II) oxide;	Copper and oxygen;
Blue copper (II) nitrate;	Copper, nitrogen and oxygen;
Blue copper (II) sulphate;	Copper, sulphur, oxygen and hydrogen;
Ammonium chloride;	Nitrogen, hydrogen and chlorine;

Note:

– Carbonates are derivatives of (derived from or made of) carbon and oxygen;

– Nitrates are derivatives of nitrogen and oxygen;

– Sulphates are derivatives of sulphur and oxygen;

– Hydrogen carbonates are derivatives of hydrogen, carbon and oxygen;

Mixtures

– A substance that consists of two or more elements or compounds that are not chemically combined.

Characteristics of mixtures.

– It properties are the average of the properties of its elements;

– Its components can be separated by physical means e.g. filtration, magnetism, distillation etc.

– Its components are not necessarily if fixed positions;

– Are formed by physical means; i.e. there is usually no heat change during its formation.

Examples of mixtures:
1. Air:

– A mixture of oxygen, nitrogen, carbon (IV) oxide, water vapour, and noble gases.

Sugar solution.

– A mixture of sugar and water.

Sea water.

– Water, dissolved salts;

Experiment: To distinguish between an element and a compound.

(i). Apparatus.

– Watch glass, test tube, wooden splint, magnet, iron fillings, sulphur powder, dilute hydrochloric acid.

(ii). Procedure.

– Approximately 7g of iron fillings and 4g of sulphur are mixed in a test tube and the mixture strongly heated;

Observations:
– A red glow starts and spreads throughout the mixture forming a black solid.

– The black solid is iron (II) sulphide.

– The two products // substances in steps 1 and 2 are subjected to the following tests;

Test // Analysis	Observations
Test // Analysis	Iron-sulphur mixture.	Iron (II) sulphide
1. Colour: The colour of the substance is noted;	– The resultant substance is yellow-grey due to the yellow sulphur and the grey iron powder;	– The yellow-grey mixture changes to a black solid; iron (II) sulphide on heating;
2. Separation: A magnet is passed over the substances separately; – Alternatively, water was added to each substance;	– Before heating the iron could be separated from sulphur by use of a magnet or sedimentation; Note: These are physical methods;	– Magnetism and sedimentation have no effect on iron (II) sulphide;
3. Reaction with dilute hydrochloric acid: To each of the substances, a few drops of hydrochloric acid is added;	– Iron reacted with dilute hydrochloric acid to form a colourless gas that burns with a pop sound. – This is hydrogen gas; – Sulphur is not affected;	– Iron (II) sulphide reacted with hydrochloric acid to produce a colourless gas with a characteristic pungent (rotten egg) smell; – The gas is hydrogen sulphide;
4. Heat change.	No heat was produced or applied in mixing iron and sulphur;	– After heating the mixture, the formation of the new substance, iron (II) sulphide produced enough heat hence the bright red glow;

Explanations:
– These four experiments summarize the four main differences between compounds and mixtures.

– From the results, iron and sulphur powder is a mixture; while iron (II) sulphide is a compound.

Differences between a mixture and a compound.

Compounds	Mixtures
1. Components are in fixed positions;	– Components are in any positions;
2. Components can only be separated by chemical means; which require large amounts of energy;	– Components ca be separated by physical means;
3. The properties are different from those of the constituent substances;	– The properties are the average of the properties of the constituent elements;
4. Are formed by chemical means // methods; i.e. a new substance is formed and there is evolution of heat;	– Are formed by physical mans; no new substance is formed and there is no // negligible heat change;
5. Formation involves heat changes; either liberation or absorption;	– No heat change in the formation of a mixture;

Names and symbols of common elements.

Chemical symbols.

– Are chemical short hands, written to refer to elements.

– They are usually based on the letters of the element;

– Chemical symbols consist of one or two letters which are usually derivatives of the Latin or English name of the element;

Rules in writing chemical symbols.

The first letter must always be a capital letter;
The letters should not be joined with each other, like in handwriting; they must be printed.
If present, the second letter of a symbol must be a small letter;

Note: The abbreviations of the chemical symbols are mainly derivatives of English, Latin or German names.

Examples:

Copper is ymbolised as Cu; derived from Cuprum which is ltin;

Iron; → Fe (Ferrum –Latin)
Potassium; → K (Kalium –Latin)
Sodium; → Na (Natrium –Latin)
Lead; → Pb (Plumbum –Latin);

– The symbol of each element represents one atom of that element.

Example:
– Ag represents one atom of silver;

– 2Ag represents 2 atoms of silver;

Some common elements and their symbols.

Element

Latin // Greek // German name

Symbol

Carbon

Fluorine

Hydrogen

Iodine

Nitrogen

Oxygen

Phosphorus

Sulphur

Aluminium

Argon

Barium

Calcium

Chlorine

Helium

Magnesium

Neon

Silicon

Zinc

Copper

Iron

Lead

Mercury

Potassium

Silver

Sodium

Gold

Tin

Manganese

–

Cuprum

Ferrum

Plumbum

Hydagyrum

Kalium

Argentum

Natrium

Aurum

–

F
H
I

S
Al

Ca
Cl

Importance of chemical names and symbols over common names.

Chemical names and symbols indicate the names in the compounds.

Examples:
– Compounds whose names end in -ide; contains only two elements;

Iron (II) sulphide: iron and sulphur;
Magnesium nitride: magnesium and nitrogen;

– Compounds whose names end in ate contain three elements and oe of them is oxygen;

Chemical names are universally known and accepted; hence provide mean of easy communication among chemists all over the world;

Simple word equations.

Equations:

– Is a linear summary of a chemical reaction, showing the reactants and products.

Examples:
Copper (II) oxide + hydrogen → copper + water;

Explanations:
– Substances on the left hand side are called reactants;

– Substances on the right hand side are called products;

– The addition sign (+) on the left hand side means “reacts with”;

– The arrow (→) means to form;

– The addition sign (+) on the right hand side (products side) means “and”.

Conclusion:

Copper (II) oxide reacts with hydrogen to form copper and water;

Note:
– Some chemical reactions are reversible and hence have two opposite arrows ( ) between reactants and products

– The arrows ( ) in chemistry means a reversible chemical reaction;

Further examples:
1. Copper (II) carbonate → Copper (II) oxide + carbon (IV) oxide;

Magnesium + oxygen → magnesium oxide;

UNIT 3: ACIDS BASES AND INDICATORS.

Unit checklist

Acids:

Meaning of acids;
Organic acids
Mineral acids;

Bases.

Meaning;

Indicators.

Meaning;
Preparation of acid-base indicators;
Commercial acid-base indicators;
Colour of indicators in acids and bases;
Classifications of substances as acids or bases using acid base indicators;
The universal indicator;
The pH scale;
pH values of various solutions in universal indiactor;

Properties of acids.

Physical properties;
- Taste;
- Effect on litmus papers;
- Electrical conductivity;
Chemical properties.
- Reaction with alkalis and bases;
- Reaction with metals;
- Reaction with carbonates and hydrogen carbonates;

Properties of bases

Physical properties;
- Taste;
- Texture;
- Effect on litmus papers;
- Electrical conductivity;
Chemical properties.
- Reaction with acids;
- Precipitation of some hydroxides;
- Effect of heat

Uses of some acids and bases.

Acids:

– Are substances that dissolve in water to release hydrogen ions.

– Acids can either be organic acids or mineral acids;

(i). Organic acids:

– Are acids found in plants and animals;

Examples:
– lactic acid in sour milk;

– Citric acid in citrus fruits like oranges;

– Ethanoic acid in vinegar;

– Tartaric acid in baking powder;

– Methanoic acid in bee and ant stings;

– Tannic acid in tea;

(ii). Mineral acids.

– Are acids made from minerals containing elements such as sulphur, chlorine, nitrogen etc.

– Are formed from reactions of chemicals;

– Main examples include:

Sulphuric (VI) acid (H₂SO₄); contains hydrogen, sulphur and oxygen;
Hydrochloric acid; contains hydrogen and chlorine;
Nitric (V) acid (HNO₃); contains nitrogen, oxygen and hydrogen;

Note:
– Mineral acids are more powerful than organic acids; because they yield // release more hydrogen ions in water

– They are thus more corrosive.

Bases.

– Are substances that dissolve in water to yield // release hydroxyl ions;

– Just like acids they are bitter to taste;

Examples:
– Sodium hydroxide;

– Ammonium hydroxide;

– Calcium hydroxide;

Note:
– Some bases insoluble in water while some are soluble in water;

– Soluble bases are called alkalis;

Indicators.

– Are substances which give definite colours in acidic or basic solutions;

– Are substances which can be used to determine whether a substance is an acid or a base;

– Consequently they are called acid-base indicators;

– The determination is based on colour changes, where each indicator have particular colourations in acids and bases.

– Indicators can be commercially or locally prepared in the laboratory;

Indicators:

Experiment: preparation of simple acid-base indicators from flower extracts.

(i). Apparatus and chemicals.

– Test tubes;

– Pestle and mortar;

– Flower petals;

– Ethanol // propanone;

– Water;

– Various test solutions: sulphuric (VI) acid, hydrochloric acid, Ethanoic acid, sodium hydroxide, magadi soda, ammonia solution.

(ii). Procedure:
– Flowers from selected plants are collected and assembled e.g. bougainvillea, hibiscus etc;

– They are crushed in a mortar using a pestle and some ethanol added with continued crushing;

– The resultant liquid is decanted into a small beaker; and its colour recorded.

– Using a dropper, two to three drops of the resultant indicator are added to the test solutions.

(iii). Observations:
(a). Colour of extract in acids and bases

Plant extract	Colour in hydrochloric acid	Colour in dilute sodium hydroxide
1
2

(b). Result with various test solutions:

Test substance

Colour (change)

Type of substance (acid/base)

Lemon juice

Wood ash

Ammonia

Sour milk

Vinegar

Nitric (V) acid

Toothpaste

Lime water

Baking powder;

Sugar

Potassium hydroxide

Note:

– Plant extracts acid-base indicators are not normally preferred in Chemistry experiments.

Reason:

– They dont give consistent (reproducible results because they are impure.

– Commercial indicators give more distinctive and reproducible results.

Commercial indicators.

– Are commercially prepared indicators which are sold in already purified forms.

Advantages of commercial indicators.

– They are relatively pure hence give consistent and reproducible results;

– They are readily available and easy to store in a Chemistry laboratory;

Main examples:
– Phenolpthalein;

– Methyl orange;

– Bromothymol blue;

– Litmus paper;

Note:
– Litmus is a blue vegetable compound which is extracted from plants called lichens;

– Litmus paper is an adsorbent paper which has been dipped in litmus indicator solution then dried;

Colours of various commercial indicators in acids and bases.

Indicator

Colour in.

Neutral

Base

Acid

1. Litmus;

2. Phenolphthalein;

3. Methyl orange;

4. Bromothymol blue

Purple

Colourless;

Orange

Blue

Blue;

Pink;

Yellow

Blue

Red;

Colourless;

Pink;

Yellow;

Classification of various substances as acids or bases using indicators.

Substance

Colour in

Classification

Litmus

Phenolphthalein

Methyl orange

Bromothymol blue

Hydrochloric acid

Sodium hydroxide

Omo (detergent)

Soda

Actal tablets

Lemon juice

Sour milk

Bleach (jik)

Fresh milk

Wood ash

The universal indicator.

– Is a full range indicator which gives range of colours depending on the strength of the acid or alkali.

– It is prepared by suitable mixing certain indicators;

– It gives a range of colour depending on the strength of acids and bases;

– Each universal indicator is supplied with a chart, to facilitate this classification.

The pH scale.

– Is a scale of numbers which shows the strength of acids or bases.

– It refers to the potential (power) of hydrogen;

– It ranges from 0 14;

– To determine the strength of an acid or base, the colour it gives in universal indicators solution is compared to the shades on the pH chart of the indicator;

Diagram: The pH scale.

Note:
– The strongest acid has a pH of 1;

– The strongest alkali has a pH of 14;

– Neutral substances have a pH of 7;

– Any pH less than 7 is acidic solution; while any pH above 7 is for a alkaline / basic solution;

Colour and pH of various solutions in universal indicator;

Substance

Colour

pH on chart

Classification

Hydrochloric acid

Sodium hydroxide

Omo (detergent)

Soda

Actal tablets

Lemon juice

Sour milk

Bleach (jik)

Fresh milk

Wood ash

Properties of acids.

(a). Physical properties.

They have a sour taste.

Examples:
– The sour taste of citric fruits is due to the citric acid in them.

– The sour taste in sour milk is due to lactic acid;

They turn blue litmus to red;

– Red litmus will remain red in acidic solution; blue litmus will turn red;

Electrical conductivity;

– Acids conduct electric current when dissolved in water;

– This is because they dissolve in water to release hydrogen ions; which are the ones that conduct electric current;

Thy destroy clothing when strong; i.e. they at away clothing material leaving holes in it;

Strong acids are corrosive; hence able to burn plant and animal tissues;

Chemical properties.

Reaction with alkalis ad bases.

– Acids react with alkalis to form salt and water only;

– These types of reactions are called neutralization reactions;

– The hydrogen ions of the acid react with the hydroxyl ion of the alkali to form water;

– The name of the salt is usually derived from the acid;

Examples:

Acid

Derivative salt

Sulphuric (VI) acid

Hydrochloric acid

Nitric (V) acid

Phosphoric acid

Sulphates;

Chlorides

Nitrates;

Phosphates;

Summary:

Acid + base (alkali) → salt + water; (a neutralization reaction);

Examples:
1. Sodium hydroxide + Hydrochloric acid → Sodium chloride + water;

Calcium oxide + Sulphuric (VI) acid → calcium sulphate + water;

Reaction with metals;

– Acids react with some metals to produce hydrogen;

Examples: Reaction with dilute hydrochloric acid and zinc metal;

Procedure:

– 2 cm3 of hydrochloric acid is put in a test tube;

– A spatula end-full of zinc powder is added.

– A burning splint is lowered in the test tube.

Observations.

– Effervescence of a colourless gas;

– The colourless gas burns with a pop sound;

Explanations.

– Zinc metal displaces the hydrogen ions in the acid which form the hydrogen gas;

– When a glowing splint is introduced into the hydrogen gas; it burns with a pop sound;

– This is the chemical test to confirm that a gas is hydrogen;

Conclusion;

– The gas produced is hydrogen gas;

– Thus, acids react with some metals to produce hydrogen gas, and a salt;

General equation:

Metal + Dilute acid → salt + Hydrogen gas;

Reaction equation:

Zinc + Hydrochloric acid → Zinc chloride + Hydrogen chloride;

Further examples:
i. Magnesium + Dilute sulphuric (VI) acid → magnesium sulphate + hydrogen gas;

Magnesium + Dilute Hydrochloric acid → magnesium chloride + Hydrogen gas;

Reaction with carbonates and hydrogen carbonates.

– Metal carbonates and hydrogen carbonates react with acids to form carbon (IV) oxide, water and a salt;

General equation:
Metal carbonate + Dilute acid → A salt + water + carbon (IV) oxide;

Metal hydrogen carbonate + Dilute acid → A salt + water + carbon (IV) oxide;

Examples: Reaction of sodium carbonate with dilute hydrochloric acid.

Procedure:

– About 2 cm³ of dilute hydrochloric acid is put in a test tube;

– A spatula end-full of sodium carbonate powder is then added;

– A burning splint is carefully lowered into the test tube.

Apparatus.

Observations.

– An effervescence occurs (bubbles); and a colourless gas is produced;

– The colourless gas does not relight a glowing splint; showing that it is carbon (IV) oxide;

Properties of bases.

Note:

– Bases are substances that release hydroxyl ions when added to water;

– Soluble bases are called alkalis;

Examples:
Sodium hydroxide + Water → Sodium ions + hydroxyl ions;

(a). Physical properties of bases

They are bitter to taste;
They are slippery or soapy to feel;
They turn litmus blue;
They conduct electricity / electric current. This is because when they are added to water they release hydroxyl ions which are the ones that conduct electricity;

(b). Chemical properties.

Reaction with acids.

– They react with acids to form a salt and water as the only products;

– This is a neutralization reaction; and is used to cure indigestion;

Example:
– Actal tablets contain a base that neutralizes the stomach acid.

They precipitate some metal hydroxides.

– Addition of some alkalis to salt solutions results in formation of solids;

– Most of these are normally hydroxides;

– A solid that is formed when two solutions are mixed is called a precipitate;

Example:

– Copper (II) sulphate + Sodium hydroxide → Copper (II) hydroxide + Sodium sulphate solution;

Blue solid;

Effects of heat.

– Most metal hydroxides are decomposed by heat to form their oxides and water;

Heat

General equation:
Metal hydroxide Metal oxide + Water;

Example:
Zinc hydroxide → Zinc oxide + water;

Applications of acids and bases.

Application of acids

Manufacture of aerated drinks;
Cleaning metal surfaces to remove oxide layer;
Sulphuric (VI) acid is used in car batteries to store and produce electricity;
Treatment of some insect bites.

Examples:
– Wasp and bee stings can be treated by applying vinegar (Ethanoic acid) or lemon juice;

– These acidic substances neutralize the alkaline insect stings;

Uses of bases.

Manufacture of anti-acid tablets to neutralize acid indigestions e.g. actal;
Calcium oxide ad calcium chloride are used to dry gases in the laboratory;

UNIT 4: AIR AND COMBUSTION.

Checklist.

Components
Determination of percentage of the active part of air.

Burning candle.
Heating copper turnings;
Heating magnesium turnings.
Smouldering of white phosphorus.
Rusting of iron;

Determination of presence of water and carbon (IV) oxide in water.
Fractional distillation of liquid air.
Rusting

Meaning and formula;
Conditions necessary for rusting;
Prevention of rusting;

Oxygen

Laboratory preparation of oxygen gas;
Chemical test for oxygen gas;
Alternative methods of oxygen preparation;
- Addition of water to solid sodium peroxide;
- Heating potassium manganate (VII) solid.
Use of oxygen

Burning substances in air;

Changes in mass;
Burning substances in oxygen;
- Metals;
- Non-metals;
Reactivity series;

Competition for oxygen among metals;
Applications for the competition for oxygen;
Atmospheric pollution.

Introduction:

– Air is a gaseous mixture constituted of several gases, water vapour and pollutants.

Combustion:

– Is the burning of substances, usually in presence of air // oxygen;

– During combustion only the oxygen component of air is used; .e the active part of air.

Percentage composition of air.

Component	Percentage volume.
Nitrogen	78%
Oxygen	21%
Carbon (IV) oxide	0.03%
Noble gases (argon)	About 1%
Water vapour	Variable
Smoke/dust particles	Variable;
Others	Trace

Note:

– From the noble gases argon is the most abundant, constituting about 0.93% of the entire 1%

Oxygen and combustion.

– When substances burn in air they consume oxygen.

– Thus the process of combustion utilizes mainly oxygen;

– The reactions in combustion are normally exothermic (give out heat) and often involve flame.

Note:

Combustion in which a flame is used is called burning;

– In combustion if all the oxygen in a given volume of air is used, the final volume of air reduces by about 21.0%;

– Since oxygen is the only constituent of air participating in combustion its termed the active part of air.

Experiments: Determination of the active part of air.

Burning candle in air.

Apparatus and requirements.

– Candle;

– Cork / evaporating dish;

– Sodium hydroxide solution;

Procedure:

– A candle about 3cm long is put on a wide cork/ evaporating dish;

– It is then floated in a dilute solution of sodium hydroxide solution just above the beehive shelf;

– It is carefully covered with a dry 100cm³ measuring cylinder, during which the level of solution in the cylinder is noted and marked;

– The measuring cylinder is removed and the candle lit;

– The lighting candle is then covered with a measuring cylinder;

– The experiment is allowed to proceed until the candle goes off;

Observations:

– The candle went off after sometime;

– The sodium hydroxide level inside the gas jar rises;

– The sodium hydroxide level in the trough goes down;

Diagrams:

Explanations:
– The candle wax is made up of hydrogen and carbon, hence called a hydrocarbon;

– During burning it melts in air consuming oxygen and producing carbon (IV) oxide and water vapour;

– The water vapour condenses giving a negligible volume of water;

– The resultant carbon (IV) oxide is absorbed by the sodium hydroxide;

– Absorption of carbon (IV) oxide in the gas jar creates a partial vacuum within it;

– The sodium hydroxide in the trough rises to fill the resultant space; and hence a drop in the sodium hydroxide level in the tough;

Conclusion.

– Oxygen is the active part of air that is utilized during burning;

– Air is basically made up of 2 parts; an active part that supports burning and an inactive part that does not support burning;

Quantitative determination of percentage of oxygen in air.

Apparatus and chemicals.

– Tough;

– Beehive;

– Candle and gas jar;

– A 30ml ruler;

– Sodium hydroxide solution;

Procedure:

– The entire apparatus is arranged as shown below;

– An empty gas jar is inverted over the candle before lighting it;

– The initial height A, is measured and recorded;

– The gas jar is then removed; the candle lit and covered with the gas jar again;

– The set up is allowed to run till the candle extinguishes (goes off); and the final height (B) of the air column measured.

Diagrams

Calculations:
– Amount of air in the gas jar initially = A cm³;

– Final amount of air remaining after burning; = Bcm³;

– Amount of oxygen used; = (A B)

Thus;

Percentage of oxygen in air: = (A B) x 10

= C%

Sample data:
Volume of air in the gas jar before burning =

Volume of air in the gas jar after burning =

Volume of air used during burning =

Percentage of air (by volume) used up = Volume used in burning x 100

Original (initial) volume

Substituting:

Conclusion:
– When candle burns in air, about 20% of air, which is oxygen used up;

Determination of percentage of air used up in rusting.

Apparatus:
– Gas jar;

– Trough;

– Beehive;

– Iron fillings//powder

Procedure:
– The gas jar is divided into five equal portions by marking around it using a waterproof marker;

– The gas jar is wet near the bottom and some iron fillings sprinkled on it;

– Some water is put in a trough and the jar with iron fillings // wool // powder inverted over it;

– The initial colour of iron fillings is noted;

– The set up is left undisturbed for a few days until the water shows no further change in rising;

Diagrams of apparatus set up:

Observations:
– The iron fillings change colour from a grey to form a brown solid;

– Water level in the gas jar rises // increases until the first mark in the gas jar;

– The water level in the trough decreases;

Explanations:
– The gas jar is moistened to make the iron fillings stick onto its surface so that the fillings do not fall in the water when the gas jar is inverted;

– The brown substance formed is called rust and its chemical name is hydrated iron (III) oxide;

– During rusting, oxygen is utilized, thus creating a partial vacuum in the gas jar;

– This causes the water level in the gas jar to rise up and the water level in the trough to go down;

– The rise in water level is equivalent to about 1/5 of the original air volume, which translates to about 20%;

Conclusion:
– When rusting occurs about 20% of air, which is oxygen, is used up;

Determination of percentage of air used up when air is passed over heated copper.

Apparatus and chemicals.

– Two 100cm³ syringes labeled Y and Y;

– Hard glass test tube;

– Glass wool;

– Bunsen burner;

– Copper turnings;

Procedure:

– A small amount of copper turnings is put in a hard glass tube and glass wool put at both ends of the tube;

– All the air in the syringe Y is removed by pushing the plunger inside, upon which the syringe is tightly fixed at one end of the tube// glass tube;

– The plunger of syringe Z is pulled out to the 100cm3 mark; to fill it with air.

The apparatus is arranged as below.

Diagram:

– The tube containing copper turnings is strongly heated;

– Air is then passed over the hot copper turnings by slowly pushing the plunger Z to and fro for several times.

Reason:

– Ensure complete reaction between the hot copper turnings and oxygen (air);

– When no further air change in volume of air in the syringe occurs, the apparatus is allowed to cool;

– The volume of air left in syringe Z is recorded;

Observations:
– The brown solid (copper) turns into a black solid (copper (II) oxide);

– The plunger of syringe Z moves inwards to approximately 80 cm3 mark;

Explanations:
– The heated copper reacted with oxygen in air to form black copper (II) oxide;

– The percentage of oxygen that was in the air is approximately 20%, causing the plunger to move inwards to the 80 cm³ mark;

Equation:

Copper + Oxygen → Copper (II) oxide;

(Brown) (Colourless) (Black)

2Cu_(s) + O_2(g) → 2CuO_(s)

Brown (Colourless) (Black)

Conclusion:

– Burning of copper in air utilizes oxygen and produces black copper (II) oxide.

Sample results and calculations:

– Initial air volume in syringe Z = cm³;

– Final air volume in syringe Z = cm³;

– Volume of air used = (100 80) = 20 cm³;

Percentage of oxygen in air = 20 x 100 = 20%;

100

Determination of percentage of air used up when air is passed over heated Magnesium.

– When the same set up is used to investigate the percentage of air used up in combustion of magnesium the volume of air used up is relatively higher than the 20%.

Reason:

Magnesium produces a lot of heat during combustion and thus reacts with both oxygen and nitrogen to form two products; magnesium oxide and magnesium nitride respectively;

Observations:

– Magnesium glows giving a bright blinding flame;

– Formation of a mixture of two white powders.

Equations:

Reaction with oxygen:

Magnesium + Oxygen → Magnesium oxide;

(Grey) (Colourless) (White)

2Mg_(s) + O_2(g) → 2MgO_(s)

(Grey) (Colourless) (White)

Reaction with nitrogen:

Magnesium + Nitrogen → Magnesium nitride;

(Grey) (Colourless) (White)

3Mg_(s) + N_2(g) → Mg₃N_2(s)

(Grey) (Colourless) (White)

Note:

– Sodium metal will also react with both oxygen and nitrogen during combustion; forming sodium oxide and sodium nitride respectively;

Smouldering of white phosphorus.

Apparatus and requirements.

– Graduated measuring cylinder;

– Water tough // pneumatic trough;

– Copper wire;

– White phosphorus;

Procedure:
– An empty measuring cylinder is inverted in a water trough and the water level noted;

– A small piece of white phosphorus is attached to the end of a piece of copper wire then put // inserted into the inverted measuring cylinder ensuring it is above the water;

– The set up is left undisturbed for 24 hours;

Precaution:

– Avoid contact with the phosphorus;

– Avoid inhalation of the fumes;

Observations:

– White fumes inside the cylinder at the start of the experiment;

– After 24 hours:
– water level inside the measuring cylinder rises;

– Water level in the trough drops;

Explanations:
– Yellow or white phosphorus smoulders in air; due to the fact that phosphorus reacts with oxygen to form phosphorus oxides;

-The phosphorus oxides are the white fumes;

– The phosphorus oxides then dissolves in water; forming acidic solutions of phosphoric acids;

– The water level rises inside the cylinder to occupy the volume of oxygen used up in reaction with phosphorus;

Equations:

Phosphorus + Oxygen → Phosphorus (V) oxide

White // yellow Colourless White fumes

P_4(s) + 5O_2(g)→ 2P₂O_5(g);

White Colourless White fumes

Yellow

Phosphorus + Oxygen → Phosphorus (III) oxide

White // yellow Colourless White fumes

P_4(s) + 3O_2(g)→ 2P₂O_3(g);

White Colourless White fumes

Yellow

Conclusion:

– Phosphorus smolders easily in air, reacting with oxygen (active part of air) to form phosphorus (III) or phosphorus (V) oxide;

– For this reason phosphorus is stored under water; to prevent it from reacting with atmospheric oxygen;

Note:
– This reaction can be made much faster by heating the copper wire; which will transmit heat to the piece of phosphorus at the tip, causing rapid burning of phosphorus to give dense white fumes of phosphorus (V) oxide // phosphorus (III) oxides;

Test for evidence of some components of air.

Water vapour.

(i). Formation of dew;

(ii). When white anhydrous copper (II) sulphate is left in the open overnight; it forms a blue solid of hydrated copper (II) sulphate;

Reason:
– The white anhydrous copper (II) sulphate absorbs atmospheric water vapour;

– Upon hydration the copper (II) sulphate turns blue;

(iii). Sodium hydroxide pellets form a colourless solution when left in the open air overnight.

Reason:

– The sodium hydroxide pellets absorbs atmospheric water vapour and dissolves in it forming sodium hydroxide solution;

(iv). When air is passed through anhydrous calcium hydroxide solid in a U-tube for sometime; there is formation of a colourless solution in the U-tube.

Apparatus:

Reason:

– The anhydrous calcium chloride absorbs atmospheric water vapour forming a colourless solution of calcium chloride;

Equation:

Calcium chloride + water → calcium chloride solution;

White Colourless solution;

Note:
– Substances that absorb moisture from the air to form a colourless solution are called deliquescent substances.

– Other examples of deliquescent substances include: iron (III) chloride, magnesium chloride and zinc chloride;

Carbon (IV) oxide.

(i). Glass stoppers of reagent bottles containing sodium hydroxide solution tend to stick when left on for sometime in the laboratory;

Reason:

– The sodium hydroxide solution at the edges of the stopper is exposed to air; and thus reacts with atmospheric carbon (IV) oxide forming white sodium carbonate solid.

Equation:
Sodium hydroxide + Carbon (IV) oxide → Sodium carbonate + Water;

2NaOH_(aq) + CO_2(g) → Na₂CO_3(s) + H₂O_(l);

(ii). Bubbling atmospheric air through lime water (calcium hydroxide) to form a white insoluble salt of calcium carbonate.

Apparatus:

Reason:

– Atmospheric carbon (IV) oxide reacts with limewater (calcium hydroxide) to form a white insoluble precipitate of calcium carbonate salt;

Equation:
Calcium hydroxide + Carbon (IV) oxide → Calcium carbonate + Water;

Ca(OH)_2(aq) + CO_2(g) → CaCO_3(s) + H₂O_(l);

Note:

– When the air is bubbled on even after the formation of the white precipitate; the white precipitate dissolves after sometime to form a colourless solution;

Reason:

– Excess carbon (IV) oxide reacts with the calcium carbonate to form soluble calcium hydrogen carbonate.

Equation:

Calcium carbonate + Water + Carbon (IV) oxide → Calcium hydroxide solution;

CaCO_3(s) + H₂O_(g) + CO_2(g) → Ca(HCO₃)_2(aq);

Fractional distillation of Liquefied air.

– Air is a mixture of gases;

– It can be separated into its constituents by fractional distillation of liquid air.

– During the process air is passed through a series of steps during which it is purified, some components eliminated then it is compressed into liquid prior to fractional distillation.

– The process can be divided into two main stages;

Purification and liquefaction;

Fractional distillation of air;

(a). Purification and liquefaction.

Step 1: Purification:
– The air is purified by removal of dust particles ;

– This is done through the following ways:

Passage through filters; during which air is passed through a series of filters; the dust particles remain within the filters while dust free air passes on to the next stage;
Electrostatic precipitation where air is passed through charged electrodes which trap oppositely charged dust particles;

Step 2: Removal of carbon (IV) oxide.

– The dust-free air is passed through a chamber containing calcium hydroxide solution;

– The sodium hydroxide solution dissolves the carbon (IV) oxide present in the air;

– During the reaction, sodium carbonate and water are formed;

– Over a prolonged time; the sodium carbonate absorbs more (excess) carbon (IV) oxide forming sodium hydrogen carbonate;

Equations:

Sodium hydroxide + carbon (IV) oxide → Sodium carbonate + Water;

2NaOH_(aq) + CO_2(g) → Na₂CO_3(aq) + H₂O_(l);

In excess;

Sodium carbonate + Water + Carbon (IV) oxide → Sodium hydrogen carbonate;

Na₂CO_3(aq) + H₂O_(l) + CO_2(g) → 2NaHCO_3(aq);

Step 3: Removal of water vapour;

– The dust-free, CO₂ – free air is then cooled to -25^oC;

– This process solidifies the water vapour out as ice;

– This cooling process may be done at temperatures a s low as -80; so as to solidify any carbon (IV) oxide (freezing point -78^oC) that may have escaped absorption by the sodium hydroxide;

– The removal of water vapour and carbon (IV) oxide are important because it prevents blockage of the pipes in the rest of the system;

Step 4: Liquefaction of air;

– The dry, dust-free and carbon (IV) oxide-free air is compressed to about 100 atmospheres of pressure; causing it to warm;

– The compressed air is cooled by refrigeration;

– The cold compressed air is made to expand rapidly by passage through a nozzle which cools it further;

– The repeated compression, cooling and expansion of air causes it to liquefy at about -200^oC:

Note:
At this temperature only neon and helium whose boiling points re lower than -200^oC remain in gaseous states;

(b). Fractional distillation;

– The liquid air now consists only of nitrogen, oxygen and noble gases (especially argon);

– The liquid air is fed at the bottom of a fractionating column;

– It is warmed to a temperature of -192^oC;

-Nitrogen distils over fast at -196^oC because it has a lower boiling point; and is collected at the top of the fractionating column;

Note:

– Any vapours of oxygen and argon which rise together with nitrogen vapour condense in the column and fall back as liquids;

– The nitrogen collected is 99% pure;

The small amounts of impurities include neon and helium;

– The liquids remaining at the bottom of the fractionating column after vaporization of all nitrogen is mainly oxygen and argon; with traces of krypton and xenon;

– The liquid is again warmed further to a temperature of -185^oC; causing the vapourization of argon whose boiling point is -186^oC;

– This is collected as a gas at the top of the fractionating column;

– The residue liquid is mainly oxygen with minute quantities of krypton and xenon which have even high boiling points;

– The oxygen is drained off and stored as pressurized oxygen in steel cylinders;

Uses of the products:

Oxygen;

– Used in hospitals with patients with breathing difficulties;

– It is used by mountain climbers and deep-sea divers for breathing;

– It is used to burn fuels;

– It is combined with acetylene to form oxy-acetylene flame which is used in welding;

– During steel making oxygen is used to remove carbon impurities;

Nitrogen:

– Manufacture of ammonia;

– Used in light bulbs; because of its inert nature it dies not react with the filament;

– As a refrigerant e.g. storage of semen for artificial insemination;

Rusting.

– Is the corrosion of iron in presence of oxygen and moisture to form brown hydrated iron (III) oxide;

– The chemical name rust is therefore hydrated iron (III) oxide with the formula Fe₂O₃.2H₂O;

– Rust itself is a brown porous substance;

Disadvantage of rusting:

– It weakens the structure of the metal (iron) and hence eventually destroys them.

Experiment: To show the conditions necessary for rusting.

Experiment	Procedure	Observation	Explanation
1	– Two clean iron nails are put inside the test tube; – 10 cm³ of tap water are then added; – Examine for two days;	– Iron nails turn brown implying there is rusting;	– There is presence of both oxygen and water;
2	– Two clean iron nails are added to the test tube; – 10 cm³ of boiled hot water is added followed by about 3 cm³ of oil; – Examine for two days;	– No rusting occurs;	– There is water but no oxygen so no rusting occur; – Boiling the water removes any dissolved oxygen; – Addition of the oil on top prevents dissolution // entry of any air containing oxygen into the water;
3	– Two clean iron nails are added to the test tube; – Push a piece of cotton wool half way the test tube; – Place some anhydrous calcium chloride on it and cork the tube tightly; – Examine for two days;	– No rusting occurs;	– There is no air // oxygen but no water; – Anhydrous calcium chloride absorbs any moisture form the air in the test tube; – Corking the tube tightly prevents more moisture from the atmosphere from getting into the tube as the calcium chloride may get saturated and allow moisture into the nails;
4	– Two clean iron nails are added to the test tube; – Examine for two days;	– Some little rusting occurs;	– Air contains and oxygen and some moisture that will facilitate rusting;
5	– Two clean iron nails are added to the test tube; – Add salty water; – Examine for two days;	Rusting occurs; and at a faster rate than the rest;	– Rusting occurs due to presence of both water and oxygen in the salty water; – Rusting is faster because the salty water contains ions which gain electrons hence facilitate faster oxidation of iron;

Summary diagrams

Further explanations:
– During rusting the first step is the oxidation of iron b7y xygen (in the air) to form anhydrous iron (III) oxide;

Equation:

Iron + Oxygen → Iron (III) oxide;

4Fe_(s) + 3O_2(g)→ 2Fe₂O_3(s);

– The anhydrous ion (III) oxide then undergoes hydration with water to form brown hydrated iron (III) oxide;

Equation:
Anhydrous iron (III) oxide + Water → Hydrated iron (III) oxide.

Black Brown;

– Rusting occurs faster in salty conditions;

Reason:

– The initial step is the oxidation of iron, from iron (II) ions (Fe²⁺) to iron (III) ions (Fe³⁺);

– During oxidation iron (II) ions give out electrons to undergo oxidation and form iron (III) ions;

– Salty water contains several dissolved salts whose ions easily accept electrons from the iron (II) ions and thus accelerating the oxidation of iron and hence rusting;

Prevention of rusting.

Note:
– Rusting destroys materials; equipment and roofs made of iron;

– Rust is porous and thus allows air and water to reach the iron beneath.

– Thus if not removed iron will continue corroding until it is all eaten up.

Methods of preventing rusting.

Galvanizing.

– Is the coating of iron with a small layer of zinc;

– Can be done by either dipping the iron object in molten zinc, spraying with a spray of molten zinc, or by electroplating (electrolytic deposition);

– On exposure to air the zinc acquires an inert layer of zinc oxide that is impervious to both air and water;

– The iron beneath is thus prevented from air and water and thus rusting.

Note:
– The iron is protected even if the zinc coating is scratched.

Reason:
– Upon scratching both the iron and zinc get into contact with air and water;

– Since zinc is more reactive than iron, air and water reacts with zinc at the expense (instead) of iron;

Electroplating:
– Refers to electrolytic coating of metals (iron) with less reactive (less corrosive metals);

– This is done through the process of electrolysis where floe of electric current causes the less reactive metal to coat the metal being protected from rusting;

Example:

– Most tin cans are in fact made yup of steel coated with a thin layer of tin.

– Other than being non-toxic tin is unreactive and rarely reacts with the contents of the can or air;

Note:

– Unlike in galvanizing, when an electroplated material gets scratched, the metal underneath (iron) rusts, and very fast;

Reason:

– Both iron and the electroplating metal (tin) are exposed to air and water;

– Since iron is more reactive than tin (the less reactive electroplating material) it reacts with oxygen and water in preference to tin;

– This explain why galvanization is more durable than electroplating;

– Other less reactive metals that are used to coat iron objects include chromium, silver and gold;

– Some electroplating metals such as gold and silver also increase the aesthetic value of the electroplated object;

Sacrificial protection.

– Blocks of a more reactive metal such as zinc or magnesium are attached to the iron structure;

– The more reactive metal will be corroded in preference to iron;

– To keep the iron structure from rusting, the block of reactive metal has to be replaced regularly;

– This metal is used for the protection of underground water pipes as well as ship hulls;

– The blocks of reactive metal are either attached directly to the iron structure or connected to it by a wire.

Diagrams:

Painting.

– The paint coats the metal surface and thus prevents contact with air and water hence no rusting;

– However if the paint is scratched, rusting occurs quickly;

– It is used mainly in ion railings, gates, bridges, roofs, ships ad cars;

Alloying.

– Alloys are mixtures of two or more metals;

– Thus to prevent iron from rusting it may be mixed with one o more metals resulting into a substance that does not rust;

Example:

– Stainless steel is an alloy of iron with chromium, nickel and manganese and it resistant to rusting.

Oiling and greasing;

– Oil is used in moving engine parts while grease is used I other movable metal joints;

– The oil // grease forms a barrier that prevents water and air from coming into contact with the metal surface and hence preventing rusting;

– Oiling and greasing are unique in the sense that they are the only methods that can be used to prevent rusting in movable car parts;

Oxygen.

– A very important constituent of air;

– Lavoisier (1743 1794), A French Chemist showed that it is the component of air used in respiration and also in burning fuels;

– It is the most abundant of all elements; occurring both freely as well as in combination with other elements;

– Freely it constitutes about 21% by volume of atmospheric air;

Laboratory preparation of oxygen gas.

Apparatus.

– Zinc; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, hydrogen peroxide (20% by volume), manganese (IV) oxide.

Diagram;

Procedure.

– Some manganese (IV) oxide is placed into a flat-bottomed flask;

– The apparatus is set up as shown in the diagram above;

– Add hydrogen peroxide from a thistle funnel into the flask dropwise;

– The gas is collected as shown;

Observations:

– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;

– The colourless gas collects on top of the water;

Explanations:
– Hydrogen peroxide decomposes slowly to oxygen and water under normal conditions;

– This process is however slow to collect enough volumes of oxygen;

– On addition of manganese (IV) oxide the decomposition is speeded up;

– Thus manganese (IV) oxide speeds up the decomposition of hydrogen peroxide and thus acts a s a catalyst;

Equation:

Without a catalyst:

Hydrogen peroxide → Water + oxygen;

2H₂O_2(l) → 2H₂O_(l) + O_2(g) (slow process)

With manganese (IV) oxide catalyst:

Manganese (IV) oxide

Hydrogen peroxide Water + oxygen;

2H₂O_2(l) → 2H₂O_(l) + O_2(g) (faster process)

Note:

The first few bubbles of oxygen gas are not collected.

Reason: The gas is mixed with air which was originally in air and hence impure.

Method of collection;

– Over water collection.

Reason:

– It is insoluble in water and less dense than water

Physical properties of oxygen gas.

– It is colourless;

– It is odourless;

– Has a low boiling point of about -183^oC;

– Almost insoluble in water (hence collected over water);

Chemical test for oxygen gas.

– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;

Drying of oxygen gas.

– The resultant oxygen is usually moist due to the fact that it is collected over water;

– If required dry the gas ca be died using either of the two methods:

(i). Using sulphuric (VI) acid.

– Bubbling the gas through a wash bottle containing concentrated sulphuric (VI) acid;

– The concentrated sulphuric (VI) acid absorbs moisture from the gas leaving it dry;

– The dry gas is then draw into collection syringe;

Diagram;

(ii). Using anhydrous calcium chloride.

– From the flask the gas is passed through a U-tube containing anhydrous calcium chloride;

– The anhydrous calcium chloride also absorbs moisture from the gas leaving it dry;

– The dry gas is then drawn into a collection syringe;

Diagram:

Alternative methods of oxygen preparation.

Addition of water to sodium peroxide.

Apparatus:
– Sodium peroxide; round-bottomed // flask bottomed flask; thistle // dropping funnel; rubber stopper, deliver tubes, rubber tubings, beehive shelf, trough, gas jars, wooden splint, water;

Diagram of apparatus.

Procedure.

– Some sodium peroxide is placed into a flat-bottomed flask;

– The apparatus is set up as shown in the diagram above;

– Add water from a thistle funnel into the flask dropwise;

– The gas is collected as shown;

Observations:

– Bubbles of a colourless gas are released from the flask through the water then into the gas jar;

– The colourless gas collects on top of the water;

Explanations:
– Sodium peroxide reacts with water to liberate oxygen;

– A solution of sodium hydroxide remains in the flask;

– This solution will turn litmus paper blue showing it is alkaline.

Equation:

Sodium peroxide + water → Sodium hydroxide + oxygen;

2Na₂O_2(l) + 2H₂O_(l)→ 4NaOH_(aq) + O_2(g)

Note:

The first few bubbles of oxygen gas are not collected.

Reason: The gas is mixed with air which was originally in air and hence impure.

Method of collection;

– Over water collection;

Reason:

– It is insoluble in water and less dense than water;

Chemical test for oxygen gas.

– On inserting a glowing splint on a gas jar full of oxygen gas; it relights a glowing splint;

Heating potassium manganate (VII) solid.

Apparatus:
– Ignition tube // boiling tube; means of heating; solid potassium manganate (VII); rubber stopper, deliver tubes, beehive shelf, trough, gas jars, wooden splint, water;

Diagram of apparatus.

Procedure:

– The apparatus is set up as shown above.

– Some solid potassium manganate (VII) is put in a hard ignition// combustion tube and strongly heated as shown above.

– The resultant gas is collected over water as shown above.

Observations;

– The purple solid forms a black solid (potassium manganate (II) solid);

– Bubbles of a colourless gas are evolved and collect over water;

Explanations:
– Upon heating potassium manganate (VII) decompose to manganese (VI) oxide; potassium

Equation:
Potassium manganate (VII) → Potassium manganate (II) + Oxygen gas

KMnO_4(s) → KMnO_2(s) + O_2(g);

Uses of oxygen.

Used in hospitals for breathing by patients with breathing difficulties;
It is used by mountain climbers and deep sea divers for breathing;
It is used to burn fuels e.g. burning fuels for propelling rockets;
Used in welding and cutting metals

Examples:
– It combines with hydrogen to form a very hot oxy-hydrogen flame that is used in welding and cutting metals;

– It combines with acetylene to form oxy-acetylene flame which is also used in welding and cutting metals;

During steel making, oxygen is used to remove iron impurities.

– During this process oxygen is blown over hot impure iron.

– The oxygen react with carbon impurities forming carbon (IV) oxide which escapes laving pure iron which is steel due to its higher purity;

Burning substances in air.

– When substances burn in air they mainly react mainly with oxygen (the active part of air);

– Some metals however also react with nitrogen;

– During burning there is usually change in mass;

Experiment: To investigate burning substances in air.

Requirements:
– Metal (magnesium ribbon); crucible; tripod stand; pipe clay triangle; means of heating;

Apparatus.

Procedure:

– About 1g of magnesium is put in the crucible;

– The crucible (with the magnesium is then weighed)

– The apparatus is set up as above;

– The crucible is heated with the lid lifted occasionally; so as to allow in air;

– No content of the crucible is allowed to escape; to ensure all products of the burning are retained;

– After all the magnesium has burned the crucible is allowed to cool;

– The crucible and its contents are weighed again;

Observations

Mass of crucible + magnesium before burning = xg

Mass of crucible + contents after burning = yg

Change in mass = (x y) g;

Mass of product before burning is lower // less than the mass of the product after burning;

Explanations:

– When the magnesium is burned in a closed crucible in a closed container, most of the air is consumed;

– It is therefore necessary to allow in air so that the burning can continue;

– During burning the magnesium combines with air to form a new product;

– Magnesium combines with both oxygen and nitrogen in air to form magnesium nitride and magnesium oxide;

Equations

With oxygen:
Magnesium + oxygen → Magnesium oxide;

Mg_(s) + O_2(g) → MgO_(s);

With nitrogen:

Magnesium + Nitrogen →Magnesium nitride;

3Mg_(s) + N_2(g) → Mg₃N_2(s);

Conclusion:

– Generally when metals burn in air, there is increase in mass;

– All metals react with oxygen to form metal oxides;

– Only more reactive metals react with nitrogen in air;

Note:

– During burning if the product(s) of the burning is gaseous, then there would be decrease in mass.

Examples:
Phosphorus → Phosphorus (V) oxide;

Lead (II) nitrate → Phosphorus (V) oxide + Nitrogen (IV) oxide + Oxygen gas;

Calcium carbonate → Calcium oxide + carbon (IV) oxide;

Burning metals in air and in oxygen.

Requirements:

– Metals; deflagrating spoon; gas jar; source of heat;

Diagram of apparatus;

Procedure:

– A piece of sodium is warmed on a deflagrating spoon until it begins to burn;

– It is then lowered into a gas jar of air as shown above;

– The flame colour is noted;

– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;

– Any gases produced are tested by smell and also with litmus papers;

– The experiment is then repeated with pure oxygen;

– The whole procedure is repeated with other metals;

Observations;

– When substances burn in oxygen they form only oxides; as opposed to burning substances in air where some react with both air and nitrogen;

– Different substances produce different flame colours;

– Many metals burn in air and in oxygen at different speeds; with more reactive metals burning more vigorously than the less reactive metals;

– Burning is faster in oxygen than in air;

Reason:

– Oxygen is pure but in air there are other constituents such as nitrogen, carbon (IV) oxide and noble gases which slow down the burning;

– In air products are generally oxides and in some few cases (magnesium and sodium) nitrides as well;

– Metals that tend to be more reactive are the ones that react with both oxygen and nitrogen;

– In oxygen products are strictly oxides;

– Some of then products are soluble in water while others are not.

Sample equations:

Magnesium:

With oxygen:
Magnesium + oxygen → Magnesium oxide;

Mg_(s) + O_2(g) → MgO_(s);

With nitrogen:

Magnesium + Nitrogen →Magnesium nitride;

3Mg_(s) + N_2(g) → Mg₃N_2(s);

Sodium:

With oxygen:
Sodium + oxygen → Sodium oxide;

4Na_(s) + O_2(g) → 2Na₂O_(s);

With nitrogen:

Sodium + Nitrogen →Magnesium nitride;

6Na_(s) + N_2(g) → 2Na₃N_(s);

Summary: burning metals in air.

Metal	How it burns	Appearance of product	Name of products	Solubility of product in water	Effect of solution on litmus paper
Magnesium	Burns with a bright white flame;	White powder	Magnesium oxide and magnesium nitride;	Slightly soluble; alkaline gas (ammonia) is produced during the process;	Turns blue;
Copper	Burns with a blue flame; surface turns black;	Black solid;	Copper (II) oxide;	Insoluble;	No effect;
Iron.	Glows to red hot; produces sparks;	Brown black (dark brown) solid;	Iron (II) oxide	Insoluble;	No effect;
Sodium	Buns very vigorously with a golden yellow flame;	White solid;	Sodium oxide and sodium nitride	Soluble; alkaline gas (ammonia) is produced in the process;	Turns litmus blue;
Calcium	Vigorous with a red flame;	White solid;	Calcium oxide and calcium nitride;	Slightly soluble; alkaline gas evolved in the process;	Turns blue;
Zinc		Yellow solid which cools to white;	Zinc oxide	Insoluble;	No effect
Lead		Red solid which cools to yellow;	Lead (II) oxide	Insoluble;	No effect;
Potassium	Very vigorously with a lilac flame;	White solid;	Potassium oxide and potassium nitride;	Soluble; alkaline gas evolved in the process;	Turns blue;

Note:
– When metals combine with oxygen, it forms metal oxides. In these reactions oxygen is added to the metals; hence the reaction is called oxidation.

– Oxidation refers to the addition of oxygen to a substance;

– The reactivity of various metals with oxygen differs.

– The arrangement of the metals in order of their activity forms the reactivity series;

– Metallic oxides generally turn litmus paper blue and are thus said to be basic oxides;

– Some metallic oxides however have both acidic and basic properties and are thus termed amphoteric oxides e.g. aluminium oxides;

The Reactivity series of metals;

Potassium; Most reactive;

Sodium;

Calcium;

Aluminium;

Zinc; increasing reactivity;

Iron;

Lead;

Copper;

Mercury;

Silver;

Gold; Least reactive

Burning non-metals in oxygen

Requirements:

– Non-metals; deflagrating spoon; gas jar; source of heat;

Diagram of apparatus;

Procedure:

– A piece of sulphur is heated on a deflagrating spoon until it begins to burn;

– It is then lowered into a gas jar of oxygen as shown above;

– The flame colour is noted;

– The gas jar is allowed to cool; some water added to the product(s) in the gas jar and shook well;

– Any resultant solution is tested with litmus papers;

– Any gases produced are tested by smell and also with litmus papers;

– The experiment is then repeated with pure oxygen;

– The whole procedure is repeated with other non-metals such as carbon and phosphorus;

Explanations:
1. Sulphur.

– Burns in oxygen with a blue flame to give a colourless gas with a choking irritating smell;

– The gas is sulphur (IV) oxide;

Equation:
Sulphur + Oxygen → Sulphur (IV) oxide;

S + O_2(g) → SO_2(g);

– The sulphur (IV) oxide dissolves in water to form sulphurous acid, which turns litmus rd;

SO_2(g) + H₂O_(l) → H₂SO_3(l);

Carbon

– Glows red to give a colourless gas that forms a white precipitate in lime water;

– The gas is Carbon (IV) oxide;

Equation:
Carbon + Oxygen → Carbon (IV) oxide;

C + O_2(g) → CO_2(g);

– The Carbon (IV) oxide dissolves in water to form weak carbonic acid, which turns litmus rd;

CO_2(g) + H₂O_(l) → H₂CO_3(l);

Note:
– In limited oxygen the carbon undergoes partial oxidation forming carbon (II) oxide;

Equation:
2C_(s) + O_2(g) → 2CO_(g)

Phosphorus.

– Burns in oxygen with a white flame to give dense white fumes;

– The white fumes are either phosphorus (V) oxide or phosphorus (III) oxide;

– Both solids // fumes dissolve in water to form phosphoric acid;

Equations:

With limited supply of oxygen:
Phosphorus + Oxygen → Phosphorus (III) oxide;

P_4(s) + 3O_2(g) →2P₂O_3(g);

With excess oxygen.

Phosphorus + Oxygen → Phosphorus (V) oxide;

P_4(s) + 5O_2(g) →P₂O_10(g);

Summary: effects of burning non-metals in air.

Non-metal	How it burns in oxygen	Name of products formed	Appearance of the product	Effect of solution on litmus paper
Sulphur	Burns with a blue flame;	Sulphur (IV) oxide	White fumes;	Turns ed
Carbon	Glows red	Carbon (IV) oxide	Colourless gas;	Turns red;
Phosphorus	Burns with a white flame	Phosphorus (V) oxide and phosphorus (III) oxide	White fumes	Turns red;

Precautions:
The experiment should be done in a fume cupboard.

Reason:
Products of burning sulphur and phosphorus in air are poisonous.

Note:
– Most non-metallic oxides are acidic in nature and therefore turn litmus red and are thus referred to as acidic oxides;

– Some non-metallic oxides form oxides which are neither acidic nor basic and are thus termed neutral oxides; e.g. carbon (II) oxide and water (hydrogen oxide);

Competition for combined oxygen among elements.

Apparatus:

– Metal oxides, source of heat, metals.

Procedure:

– A spatula end full of copper (II) oxide in a bottle top.

– Magnesium powder and mixed well;

– Record the observations;

– The experiment is repeated using other metal oxides with various other metals like zinc, iron etc.

Observations:

Metal Metal oxide	Magnesium	Zinc	Iron	Lead	Copper
Magnesium oxide (white)	No reaction	No reaction	No reaction	No reaction	No reaction
Zinc oxide (white)	White magnesium oxide and grey zinc metal	No reaction	No reaction	No reaction	No reaction
Iron (III) oxide	White magnesium oxide and grey iron metal;	White zinc oxide and iron;	No reaction	No reaction	No reaction
Lead (II) oxide (yellow when	White magnesium oxide and	White zinc oxide and lead;	Iron (III) oxide and lead;	No reaction	No reaction
Copper (II) oxide (Black	White magnesium oxide and brown copper metal;	White zinc oxide and brown copper metal;	Brown iron (III) oxide and brown copper metal	Yellow lead (II) oxide and brown copper metal	No reaction

Explanations:

– A more reactive metal takes away oxygen from a less reactive metal;

– This is because a more reactive metal reacts more readily with a less reactive metal;

– These reactions are called displacement reactions;

– Some metals can displace other metals from their oxides upon heating;

– Metals which are higher in the reactivity series can displace metals which are lower in the reactivity series from their oxides;

– From the table none of the metals can displace magnesium from its oxide, while copper can be displaced from its oxides by all the metals.

– Thus from the list magnesium is the most reactive while copper is the least reactive.

– Such results of displacement reactions can also be used to develop a reactivity series of the metals (elements) concerned.

Selected equations:

Copper (II) oxide + Magnesium → Magnesium oxide + Copper

CuO_(s) + Mg_(s) → MgO_(s) + Cu_(s);

Black Grey White Brown.

Zinc (II) oxide + Magnesium → Magnesium oxide + Zinc

White when cold Grey White Grey.

Yellow when hot

ZnO_(s) + Mg_(s) → MgO_(s) + Cu_(s);

Copper (II) oxide + Zinc → Zinc oxide + Copper

Black Grey White when cold Brown.

Yellow when hot

CuO_(s) + Mg_(s) → MgO_(s) + Cu_(s);

Lead (II) oxide + Magnesium → Magnesium oxide + Lead

Yellow when cold Grey White Grey.

Redwhen hot

CuO_(s) + Mg_(s) → MgO_(s) + Cu_(s);

Copper (II) oxide + Zinc → Magnesium oxide + Copper

CuO_(s) + Zn_(s) → MgO_(s) + Zn_(s);

Black Grey White Brown.

Typical reactivity series from the results above:

Magnesium ↑ Most reactive

Zinc

Iron

Lead

Copper ↓ Least reactive

Note:

– Removal of oxygen is called reduction;

– Addition of oxygen is called oxidation;

– A substance that loses oxygen during a reaction is said to be reduced while a substance that removes oxygen from another is called reducing agent;

– A substance that gains oxygen during a reaction is said to be oxidized while a substance that loses / donates oxygen to another is called an oxidizing agent;

Examples:

Copper (II) oxide + Magnesium → Magnesium oxide + Copper

CuO_(s) + Mg_(s) → MgO_(s) + Cu_(s);

Black Grey White Brown.

Reducing agent: magnesium

Oxidizing agent: Copper (II) oxide

Oxidized species: Magnesium

Reduced species: Copper

Copper (II) oxide + Zinc → Magnesium oxide + Copper

CuO_(s) + Zn_(s) → MgO_(s) + Zn_(s);

Black Grey White Brown.

Reducing agent: Zinc

Oxidizing agent: Copper (II) oxide

Oxidized species: Zinc

Reduced species: Copper

– In the above reactions both reduction and oxidation take place at the same time;

– A reaction in which both reduction and oxidation occur at the same time is called a redox reaction; red from reduction and ox from oxidation;

Application of Redox reactions:

Extraction of metals;

– Ores of metals such as zinc, iron lead etc are roasted in air to form corresponding metal oxides;

– The metal oxides are then reduced to corresponding metals using common reducing agents like carbon and carbon (II) oxide.

Examples:

Zinc (II) oxide + Carbon (II) oxide → Zinc + carbon (IV) oxide;

Atmospheric pollution and percentage composition of air.

– Human activities have changes the normal composition of air in some places;

– This has not only altered the percentage composition of the main components but also added other components into the air.

Examples:

– Mining increases the amount of dust particles in the air;

– Geothermal power drilling may result into emission of gases like hydrogen sulphide, sulphur (IV) oxide into the air;

– Industrial processes like manufacture of nitric (V) acid, contact process etc may add gases ilke sulphur (IV) oxide, nitrogen oxides into the air;

These gases and emissions cause atmospheric pollution:

– Gases like sulphur (IV) oxide and nitrogen (IV) oxide dissolve I rain water to form acidic rain that causes corrosion of buildings, iron sheet roofing, bleaching of plants; irritation in bodies and respiratory surfaces in animals etc;

– Dust particles may block stomata in plants; cause smog formation hence reducing visibility (leading to more cases of road accidents).

Uses of oxygen:
1. Used for breathing in hospitals fro patients with breathing difficulties.

Used for breathing by mountain climbers and deep sea divers.
Used to burn fuels e.g. burning fuels to propel rockets.
Manufacture of the oxy-acetylene flame that is used in welding and cutting of metals;
Removal of iron impurities during steel making i.e. oxygen is blown through impure iron; the oxygen then reacts with carbon impurities forming carbon (IV) oxide which escapes leaving behind pure iron (steel).

UNIT 5: WATER AND HYDROGEN.

Checklist:
1. Introduction

Burning candle wax in air.
Reaction of water with metals

Sodium
Potassium
Calcium

Reaction of metals with steam.

Calcium
Magnesium
Zinc
Iron
Aluminium

Hydrogen

Hydrogen;
Reduction property of hydrogen
Burning hydrogen in air;

Uses of hydrogen

Introduction:

– Water is the most abundant substances on earth;

– It covers about 71% of the earths surface;

– Main sources of water include seas, lakes, rivers, oceans.

Burning candle wax in air.

Apparatus:

Procedure.

– The candle is lit under the funnel and the suction pump turned on.

– The set up is left undisturbed for about 15 minutes.

Observations;

– The candle continues to burn.

– Droplets of a colourless liquid in the tube A;

– The colourless liquid turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt (II) chloride into pink;

– A white precipitate forms in the calcium hydroxide in tube B;

– Deposits of a black solid on the inner sides of the funnel;

Explanations;

– The suction pump ensures continuous supply of air hence the candle continues to burn;

– Candle wax buns in oxygen to form carbon (IV) oxide and steam;

– The carbon (IV) oxide is sucked out through the apparatus by the suction pump;

– Carbon (IV) oxide forms a white precipitate of calcium carbonate when bubbled through lime water (calcium hydroxide)

– Incomplete combustion of the carbon in the candle wax produces carbon particles which cools and deposits as black solids;

Equations:

As the candle burns:
Carbon + Oxygen → Carbon (IV) oxide;

C_(s) + O_2(g) → CO_2(g);

Hydrogen + Oxygen → Carbon (IV) oxide;

2H_2(s) + O_2(g) → 2H₂O_(g);

For the formation of the black deposits (soot)

Carbon + Oxygen → Carbon + Carbon (IV) oxide;

2C_(s) + O_2(g) → C_(s) + CO_2(g);

In the calcium hydroxide:

Carbon (IV) oxide + calcium hydroxide → Calcium carbonate + Water

Colourless Colourless White precipitate Colourless

CO_2(g) + Ca(OH)_2(aq) → CaCO_3(s) + H₂O_(l);

– The steam condenses into water in the boiling tube;

– Water turns white anhydrous copper (II) sulphate to blue and blue anhydrous cobalt chloride paper into pink;

General equation:
Hydrocarbon + Oxygen → Water + Carbon (IV) oxide;

Conclusion:

– Candle wax is a compound of carbon and hydrogen only; and such compounds are defined as hydrocarbons;

– When burned in air (oxygen) hydrocarbons produce carbon (IV) oxide ad steam (water);

– Other examples of hydrocarbons include: petrol; diesel; kerosene etc;

Note: Effects of repeating the same experiment without a suction pump.

Apparatus:

Observations:
– The candle went off;

– Deposition of black solid on the inner sides of the funnel;

– No colourless liquid in tube A;

– No white precipitate in tube B;

Explanations:
– The carbon (IV) oxide and steam produced would accumulate in the filter funnel hence making the flame to go off;

– Incomplete combustion of the candle would produce carbon particles which cool as soot;

– Only negligible amount of water and carbon (IV) oxide would pass through the apparatus;

Reactions of water with metals:
1. Potassium.

Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It explosively melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;

– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;

(b). Reaction equations.

Equation I

2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g);

Equation II

4K_(s) + O_{2 (g)} → 2K₂O_(s);

Equation III:

K₂O_(s) + H₂O_(l) → 2KOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Sodium.

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

(b). Reaction equations.

Equation I

2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g);

Equation II

4Na_(s) + O_{2 (g)} → 2Na₂O_(s);

Equation III:

Na₂O_(s) + H₂O_(l) → 2NaOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Calcium.

Procedure:
– A small piece of calcium metal is cut and dropped into a trough containing water;

– A filter funnel is inverted over it;

– A test tube filled with water is then inverted over the funnel;

– The gas given out is collected as shown in the apparatus below.

– The resultant gas is then tested with a burning splint;

– The resultant solution in the trough is tested with litmus paper.

Diagram of apparatus:

Observations and explanations:

– Calcium sinks to the bottom of the beaker; because it is denser than water;

– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;

– Soapy solution formed; due to formation of alkaline calcium hydroxide;

– A white suspension is formed; because calcium hydroxide is slightly soluble in water;

Reaction equation:

Ca_(s) + H₂O_(l) → Ca (OH)_2(aq) + H_{2 (g)};

Effect of resultant solution on litmus paper;

– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;

Magnesium and other metals.

– Magnesium reacts with atmospheric oxygen to form magnesium oxide that coast the metal surface;

– Thus before reacting it with water this oxide layer has to be removed e.g. by polishing metal surface using sand paper;

– Reaction between magnesium and cold water is generally very slow; with very slow evolution of hydrogen gas;

– Zinc and iron metals do not react with cold water;

Reaction of metals with steam.

Note:

– Metals that react with cold water would react very explosively with steam and thus their reactions with steam should not be attempted in the laboratory;

– However some metals which react only sparingly with cold water or do not react with cold water at all react with steam to produce respective metal oxide and hydrogen gas;

Magnesium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;

– The magnesium ribbon is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

Diagram of apparatus:

Observations and explanations.

– Magnesium burns with a bright blinding flame;

– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;

– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

Reaction equation.

Magnesium + Steam → Magnesium oxide + Hydrogen gas;

Mg_(s) + H₂O_(g) → MgO_(s) + H_2(g);

Zinc

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of zinc put in the middle of the combustion tube;

– The zinc is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

Diagram of apparatus:

Observations and explanations.

– Zinc metal does not burn but rather glows;

– Grey solid (zinc) forms a yellow solid which cools to a white solid (zinc oxide);

– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;

Reaction equation.

Zinc + Steam → Zinc oxide + Hydrogen gas;

Grey Colourless Yellow when hot Colourless

White on cooling

Zn_(s) + H₂O_(g) → ZnO_(s) + H_2(g);

Iron

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of iron put in the middle of the combustion tube;

– The iron is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

Diagram of apparatus:

Observations and explanations.

– Iron metal does not burn but rather glows;

– Grey solid (zinc) forms a black solid; due to formation of tri-iron tetra-oxide;

– Evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

Reaction equation.

Iron + Steam → Tri-iron tetra-oxide + Hydrogen gas;

Grey Colourless Black Colourless

3Fe_(s) + 4H₂O_(g) → Fe₃O_4(s) + 4H_2(g);

Aluminium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of aluminium put in the middle of the combustion tube;

– The aluminium is heated strongly then the wet sand is warmed gently;

– The delivery tube is removed before heating stops; and the gas produced is tested using a burning splint;

Diagram of apparatus:

Observations and explanations.

– Aluminium burns in steam but the reaction quickly stops; because the reaction forms a layer of aluminium oxide that coats the metal surface preventing further reaction;

– Grey solid (aluminium) forms a white solid of aluminium oxide;

– Slight evolution of a colourless gas that burns with a pop sound; confirming it is hydrogen;

– The production of the gas however stops soon after the reaction starts because the oxide layer stops further reaction;

Reaction equation.

Aluminium + Steam → Zinc oxide + Hydrogen gas;

Grey Colourless White Colourless

2Al_(s) + 3H₂O_(g) → Al₂O_3(s) + 3H_2(g);

Other metals.

– Lead and copper do not react with steam;

Summary of the reaction between metals with cold water and steam

Metal

Action of metal on water

Action of metal on steam

Potassium

Sodium

Calcium

Magnesium

Aluminium

Zinc

Iron

Lead

Copper

Violent

Moderate

Very slow

No reaction

Explosive

Violent

Rapid

Slow

No reaction

Note:

– Metals can thus be arranged in order of their reactivities with water; resulting to a reactivity series similar to that obtained form reaction between metals with oxygen;

Reactivity series of metals:

Potassium; Most reactive;

Sodium;

Calcium;

Magnesium

Aluminium;

Zinc; increasing reactivity;

Iron;

Lead;

Copper; Least reactive;

Hydrogen

– An element that does not exist freely in nature;

– Generally exists in compounds such as water, sugars, fuels etc;

Laboratory preparation of hydrogen gas.

Note:

– Hydrogen gas is generally prepared by the reaction between dilute acids and metals;

– Most suitable acids are dilute hydrochloric acid and dilute sulphuric (VI) acid;

– Most suitable metal is zinc metal;

Apparatus:

Procedure:
– Zinc granules are added to dilute sulphuric (VI) acid;

– Small amounts of copper (II) sulphate are added to the zinc – acid mixture;

Reason: To act as a catalyst hence speed up the reaction;

– The resultant colourless gas is collected over water;

Reason: The gas is insoluble in water;

– If the gas is required dry, the gas is passed through concentrated sulphuric (VI) acid or a U-tube containing calcium chloride;

Diagrams for drying the gas:

Zinc granules

– The dry gas is collected by upward delivery (downward displacement of air);

Reason: It is less dense than air (note that hydrogen is the lightest gas known);

Note:

– Nitric (V) acid is not used in preparation of hydrogen gas; except very dilute nitric (V) acid and magnesium

Reason: Nitric acid is a strong oxidizing agent hence the hydrogen formed is immediately oxidized to water
– Potassium, sodium, lithium and calcium are not used in laboratory preparation of hydrogen gas;
Reason: They react explosively with acids;

– Magnesium is not usually used for laboratory preparation of hydrogen;

Reason: It is expensive;

– Prior to using it for preparation of hydrogen; aluminium should be washed with concentrated hydrochloric acid;

Reason: To remove the protective oxide layer that usually forms on the aluminium surface on its exposure to air;

– Hydrogen gas produced from iron metal tends to have a foul smell;

Reason: Iron gives a mixture of gases due to impurities in the iron; the foul smell is usually due to production of hydrogen sulphide that results from sulphide impurities in the iron metal;

Reaction equations:

Iron + Sulphuric (VI) acid → Iron (II) sulphate + Hydrogen gas;

Fe_(s) + H₂SO_4(aq) → FeSO_4(aq) + H_2(g)

Iron (II) sulphide + Sulphuric (VI) acid → Iron (II) sulphate + Hydrogen sulphide gas;

FeS_(s) + H₂SO_4(aq) → FeSO_4(aq) + H₂S_(g);

Other sources of hydrogen.

– Cracking of alkanes;

– Fractional distillation of petroleum;

Properties of Hydrogen gas;

Physical properties.

Colourless and odourless;
It is insoluble in water;
It is less dense than air (it is the lightest gas known);

Chemical properties.

It has no effect on litmus paper implying that it is neutral;
It burns with a pop sound when mixed with air;
Pure hydrogen burns quietly with a blue flame producing water;
Hydrogen does not support combustion but it burns in air;
It is a reducing agent;

Chemical test for hydrogen;

– When a burning splint is introduced into a gas jar full of hydrogen gas; the gas buns with a “pop sound”

Note:
– The intensity of the “pop sound” diminishes as the purity of hydrogen increases;

Reactions of hydrogen:

Passing hydrogen over heated copper (II) oxide;

Requirements:

Copper (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

The mixture of hydrogen and oxygen is explosive when ignited;

Observations and explanations

– The black solid turns into a brown solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

Reason:

The hot black copper (II) oxide is reduced by hydrogen gas into brown copper metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

Reaction equations:

In the combustion tube:

Copper (II) oxide + Hydrogen → Copper + Water;

Black solid Colourless gas Brown solid Colourless liquid

CuO_(s) + H_2(g) → Cu_(s) + H₂O_(l);

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H_2(g) + O_2(g) → 2H₂O_(l);

Conclusion:

– Hydrogen is a reducing agent and thus reduces the copper (II) oxide to copper metals while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Copper (II) oxide is the oxidizing agent;

Passing hydrogen over heated Lead (II) oxide;

Requirements:

Lead (II) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

The mixture of hydrogen and oxygen is explosive when ignited;

Observations and explanations

– The yellow solid turns red on heating then finally into a grey solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

Reason:

The hot red lead (II) oxide is reduced by hydrogen gas into grey lead metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

Reaction equations:

In the combustion tube:

Lead (II) oxide + Hydrogen → Lead + Water;

Yellow-cold; red – hot Colourless gas Grey solid Colourless liquid

PbO_(s) + H_2(g) → Pb_(s) + H₂O_(l);

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H_2(g) + O_2(g) → 2H₂O_(l);

Conclusion:

– Hydrogen is a reducing agent and thus reduces the lead (II) oxide to lead metals while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Lead (II) oxide is the oxidizing agent;

Passing hydrogen over heated Iron (III) oxide;

Requirements:

Iron (III) oxide; porcelain boat; mean of heating; combustion tube; dry hydrogen gas;

Diagram of apparatus.

Procedure:

– Apparatus are arranged as shown above;

– Dry hydrogen gas is passed through the combustion tube for sometime prior to heating the oxide;

Reason:

To drive out all the air from the apparatus which may otherwise re-oxidize the metal after reduction.
A mixture of hydrogen and air will explode when the combustion tube is heated;

– The gas is continuously collected at the jet and tested; until the gas burns smoothly without a “pop” sound;

– The gas is then lit at the jet and the copper oxide heated;

– This is done until no further change;

– The apparatus is allowed to cool as hydrogen is still continuously allowed to pass through;

Reason:

To prevent re-entry of atmospheric oxygen which will re-oxidize the hot metal back to the metal oxide;

– Excess hydrogen must be burnt at the jet so that excess gas is not allowed to escape into the air;

Reason:

The mixture of hydrogen and oxygen is explosive when ignited;

Observations and explanations

– The Brown solid turns into a grey solid;

– Droplets of a colourless liquid on the cooler parts of the combustion tube;

Reason:

The hot brown iron (III) oxide is reduced by hydrogen gas into grey iron metal while hydrogen gas is oxidized to water;

– The colourless liquid is confirmed to be water by:

Adding drops of it to anhydrous copper (II) sulphate which forms a blue solid;
Adding drops of it onto blue anhydrous cobalt (II) chloride which turns pink;

Reaction equations:

In the combustion tube:

Iron (III) oxide + Hydrogen → Iron + Water;

Brown solid Colourless gas Grey solid Colourless liquid

Fe₂O_3(s) + 3H_2(g) → 2Fe_(s) + 3H₂O_(l);

At the jet:

– Hydrogen burns with a blue flame producing water;

Equation:

2H_2(g) + O_2(g) → 2H₂O_(l);

Conclusion:

– Hydrogen is a reducing agent and thus reduces the iron (III) oxide to iron metal while hydrogen itself undergoes oxidation to form water;

– Hydrogen is the reducing agent;

– Iron (III) oxide is the oxidizing agent;

Note:

– Hydrogen does not reduce (remove oxygen) from oxides of metals above it in the reactivity series;

Products of burning hydrogen gas in air.

Apparatus and requirements:

– The apparatus is arranged as shown below.

Procedure:

– Apparatus is arranged as shown below.

– A stream of hydrogen is passed through anhydrous calcium chloride;

– The gas is tested for purity by collecting samples over the jet and testing with a burning splint.

Note: pure hydrogen gas should burn smoothly without the typical “pop sound”;

– The gas is then lit and the pump tuned on;

– The products of burning hydrogen is drown in through the apparatus using the pump for about 15 minutes;

– The product condensing in the test tube in cold water is tested with white anhydrous copper (II) sulphate and blue cobalt chloride paper;

Observations:
– Pure hydrogen burns with a blue flame;

– A colourless liquid condenses in the test tube immersed in cold water;

– The liquid turns white anhydrous copper (II) sulphate blue;

– The colourless liquid turns blue anhydrous cobalt chloride pink;

Explanations:

– The calcium chloride in the U-tube is used to dry the gas;

– The pure dry hydrogen gas burns with a blue flame to form steam which condenses into liquid water;

– Water turns anhydrous copper (II) chloride from white to blue; and turns blue anhydrous cobalt chloride paper into pink;

Reaction equations:

At the jet:

2H_2(g) + O_2(g) → 2H₂O_(g);

Uses of Hydrogen

Large scale manufacture of ammonia in the Haber process;
Hydrogenation for the manufacture of margarine.

– This refers to hardening of oils into fats.

– In this reaction Hydrogen gas is bubbled into liquid oil in presence o0f nickel catalyst;

– The oil takes up hydrogen and is converted into fat;

Hydrogen is used in weather balloons because it is lighter than air;

– Usually a radio transmitter is connected to a weather balloon filled with air; as the balloon floats in air the transmitter collects information which is conveyed to weather stations for interpretation by meteorologists;

A mixture of hydrogen and oxygen forms the very hot oxy-hydrogen flame (with temperatures up to 2000^oC that is used in welding and cutting metals;
It is used in rocket fuels for propulsion of rockets;
Manufacture of hydrochloric acid; during which hydrogen is burnt in chlorine;

Teachers' Resources

CHEMISTRY FORM ONE TEACHING NOTES

October 5, 2025 Hillary Kangwana Leave a comment

CHEMISTRY FORM ONE NOTES

INTRODUCTION TO CHEMISTRY

Chemistry is a branch of Science. Science is basically the study of living and non-living things. The branch of science that study living things is called Biology. The branch of science that study non-living things is called Physical Science. Physical Science is made up of:

Physics- the study of matter in relation to energy
Chemistry- the study of the composition of matter.

Chemistry is thus defined as the branch of science that deals with the structure composition, properties and behavior of matter.

Basic Chemistry involves studying:

States/phases of matter

Matter is anything that has weight/mass and occupies space/volume. Naturally, there are basically three states of matter.

(i) Solid-e.g. soil, sand, copper metal, bucket, ice.

(ii)Liquid- e.g. water, Petrol, ethanol/alcohol, Mercury (liquid metal).

(iii) gas- e.g. Oxygen, Nitrogen ,Water vapour.

A solid is made up of particles which are very closely packed. It thus has a definite/fixed shape and fixed/definite volume /occupies definite space. It has a very high density.

A liquid is made up of particles which have some degree of freedom. It thus has no definite/fixed shape. It takes the shape of the container it is put. A liquid has fixed/definite volume/occupies definite space.

A gas is made up of particles free from each other. It thus has no definite/fixed shape. It takes the shape of the container it is put. It has no fixed/definite volume/occupies every space in a container.

(b) Separation of mixture

A mixture is a combination of two or more substances that can be separated by physical means. Simple methods of separating mixtures at basic chemistry level include:

i) Sorting/picking-this involve physically picking one pure substance from a mixture with another/other. e. g. sorting maize from maize beans mixture.
ii) Decantation-this involve pouring out a liquid from a solid that has settled /sinking solid in it. e. g. Decanting water forms sand.

iii)Filtration-this involves sieving /passing particles of a mixture through a filter containing small holes that allow smaller particle to pass through but do not allow bigger particle to pass through.

iv) Skimming-this involve scooping floating particles. E.g. cream from milk

(c) Metals and non-metals

Metals are shiny, ductile(able to form wires), malleable(able to form sheet) and coil without breaking. E.g. Iron, gold, silver, copper. Mercury is the only liquid metal known.

Non-metals are dull, not ductile (do not form wires), not malleable (do not form sheet) and break on coiling/brittle. E.g. Charcoal, Sulphur, pla-stics.

(d) Conductors and non-conductors

A conductor is a solid that allow electric current to pass through. A non-conductor is a solid that do not allow electric current to pass through.

All metals conduct electricity. All non-metals do not conduct electricity except carbon graphite.

(e) Drugs

A drug is a natural or synthetic/man-made substance that when taken changes/alter the body functioning. A natural or synthetic/man-made substance that when taken changes/alter the abnormal body functioning to normal is called medicine. Medicines are thus drugs intended to correct abnormal body functions. . Medicines should therefore be taken on prescription and dosage.

A prescription is a medical instruction to a patient/sick on the correct type of medicine to take and period/time between one intake to the other.

A dosage is the correct quantity of drug required to alter the abnormal body function back to normal. This is called treatment. It is the professional work of qualified doctors/pharmacists to administer correct prescription and dosage of drugs/medicine to the sick. Prescription and dosage of drugs/medicine to the sick use medical language.

Example

(i) 2 x 4 ; means “2” tablets for solid drugs/spoonfuls for liquid drugs taken “4” times for a duration of one day/24 hours and then repeated and continued until all the drug given is finished.

(ii) 1 x 2 ; means “1” tablets for solid drugs/spoonfuls for liquid drugs taken “2” times for a duration of one day/24 hours and then repeated and continued until all the drug given is finished.

Some drugs need minimal prescription and thus are available without pharmacist/ doctor’s prescription. They are called Over The Counter (OTC) drugs. OTC drugs used to treat mild headaches, stomach upsets, common cold include:

(i) Painkillers

(ii) Anti-acids

(iii) cold/flu drugs.

All medicine requires correct intake dosage. When a prescription dosage is not followed, this is called drug misuse/abuse. Some drugs are used for other purposes other than that intended. This is called drug abuse.

Drug abuse is when a drug is intentionally used to alter the normal functioning of the body. The intentional abnormal function of the drug is to make the victim have false feeling of well being. The victim lack both mental and physical coordination.

Some drugs that induce a false feeling of well being are illegal. They include heroin, cocaine, bhang, Mandrax and morphine.

Some abused drugs which are not illegal include: Miraa, alcohol, tobacco, sleeping pills.

The role of chemistry in society

(a) Chemistry is used in the following:

(i) Washing/cleaning with soap:

Washing/cleaning is a chemical process that involves interaction of water, soap and dirt so as to remove the dirt from a garment.

(ii) Understanding chemicals of life

Living thing grow, respire and feed. The formation and growth of cells involve chemical processes in living things using carbohydrates, proteins and vitamins.

(iii) Baking:

Adding baking powder to dough and then heating in an oven involves interactions that require understanding of chemistry.

(iv) Medicine:

Discovery, test, prescription and dosage of drugs to be used for medicinal purposes require advanced understanding of chemistry

(v) Fractional distillation of crude oil:

Crude oil is fractional distilled to useful portions like petrol, diesel, kerosene by applying chemistry.

(vi) Manufacture of synthetic compounds/substances

Large amounts of plastics, glass, fertilizers, insecticides, soaps, cements, are manufactured worldwide. Advanced understanding of the chemical processes involved is a requirement.

(vii) Diagnosis/test for abnormal body functions.

If the body is not functioning normally, it is said to be sick/ill. Laboratory test are done to diagnose the illness/sickness.

(b) The following career fields require Chemistry as one of subject areas of advanced/specialized study:

(i) Chemical engineering/chemical engineer

(ii) Veterinary medicine/Veterinary doctor

(iii) Medicine/Medical doctor/pharmacist/nurse

(iv) Beauty/Beautician

(v) Teaching/Chemistry teacher.

The School Chemistry Laboratory

Chemistry is studied mainly in a science room called a school chemistry laboratory. The room is better ventilated than normal classroom. It has electricity, gas and water taps. A school chemistry laboratory has a qualified professional whose called Laboratory technician/assistant.

All students user in a school chemistry laboratory must consult the Laboratory technician/assistant for all their laboratory work. A school chemistry laboratory has chemicals and apparatus.

A chemical is a substance whose composition is known. All chemical are thus labeled as they are. This is because whereas physically a substance may appear similar, chemically they may be different.

All Chemicals which are not labeled should never be used. Some chemicals are toxic/poisonous, explosive, corrosive, caustic, irritants, flammable, oxidizing, carcinogenic, or radioactive.

Care should always be taken when handling any chemical which have any of the above characteristic properties.

Common school chemistry laboratory chemicals include:

(i) Distilled water

(ii) Concentrated mineral acid which are very corrosive (on contact with skin they cause painful open wounds)

(iii) Concentrated alkali/bases which are caustic (on contact with skin they cause painful blisters)

(iv) Very many types of salts

The following safety guideline rules should be followed by chemistry laboratory users:

(i) Enter the laboratory with permission in an orderly manner without rushing/pushing/scrabbling.

(ii) Do not try unauthorized experiments. They may produce flammable, explosive or toxic substances that affect your health.

(iii) Do not taste any chemical in the laboratory. They may be poisonous.

(iv) Waft gas fumes to your nose with your palm. Do not inhale/smell gases directly. They may be highly poisonous/toxic.

(v) Boil substances with mouth of the test tube facing away from others and yourself. Boiling liquids spurt out portions of the hot liquid. Products of heating solids may be a highly poisonous/toxic gas.

(vi) Wash with lots of water any skin contact with chemicals immediately. Report immediately to teacher/laboratory technician any irritation, cut, burn, bruise or feelings arising from laboratory work.

(vii) Read and follow safety instruction. All experiments that evolve/produce poisonous gases should be done in the open or in a fume chamber.

(viii )Clean your laboratory work station after use. Wash your hand before leaving the chemistry laboratory.

(ix) In case of fire, remain calm, switch of the source of fuel-gas tap. Leave the laboratory through the emergency door. Use fire extinguishers near the chemistry laboratory to put of medium fires. Leave strong fires wholly to professional fire fighters.

(x) Do not carry unauthorized item from a chemistry laboratory.

An apparator /apparatus are scientific tools/equipment used in performing scientific experiments. The conventional apparator used in performing scientific experiments is called standard apparator/apparatus. If the conventional standard apparator/apparatus is not available, an improvised apparator/apparatus may be used in performing scientific experiments. An improvised apparator/apparatus is one used in performing a scientific experiment for a standard apparator/apparatus. Most standard apparatus in a school chemistry laboratory are made of glass because:

(i)Glass is transparent and thus reactions /interactions inside are clearly visible from outside

(ii) Glass is comparatively cheaper which reduces cost of equipping the school chemistry laboratory

(iii) Glass is comparatively easy to clean/wash after use.
(iv) Glass is comparatively unreactive to many chemicals.

Apparatus are designed for the purpose they are intended in a school chemistry laboratory:

Apparatus for measuring volume

Measuring cylinder

Measuring cylinders are apparatus used to measure volume of liquid/ solutions. They are calibrated/ graduated to measure any volume required to the maximum. Measuring cylinders are named according to the maximum calibrated/graduated volume e.g.

“10ml” measuring cylinder is can hold maximum calibrated/graduated volume of “10mililitres” /“10 cubic centimetres”

“50ml” measuring cylinder is can hold maximum calibrated/graduated volume of “50mililitres” /“50 cubic centimetres”

“250ml” measuring cylinder is can hold maximum calibrated/graduated volume of “250mililitres” /“250 cubic centimetres”

“1000ml” measuring cylinder is can hold maximum calibrated/graduated volume of “1000mililitres” /“1000 cubic centimetres”

Burette

Burette is a long and narrow/thin apparatus used to measure small accurate and exact volumes of a liquid solution. It must be clamped first on a stand before being used. It has a tap to run out the required amount out. They are calibrated/ graduated to run out small volume required to the maximum 50ml/50cm3.

The maximum 50ml/50cm3 calibration/ graduation reading is at the bottom .This ensure the amount run out from a tap below can be determined directly from burette reading before and after during volumetric analysis.

Burettes are expensive and care should be taken when using them.

(i) Pipette

Pipette is a long and narrow/thin apparatus that widens at the middle used to measure and transfer small very accurate/exact volumes of a liquid solution.

It is open on either ends.

The maximum 25ml/25cm3 calibration/ graduation mark is a visible ring on one thin end.

To fill a pipette to this mark, the user must suck up a liquid solution upto a level above the mark then adjust to the mark using a finger.

This requires practice.

(ii) Pipette filler

Pipette filler is used to suck in a liquid solution into a pipette instead of using the mouth. It has a suck, adjust and eject button for ensuring the exact volume is attained. This requires practice.

Volumetric flask.

A volumetric flask is thin /narrow but widens at the base/bottom. It is used to measure very accurate/exact volumes of a liquid solution.

The maximum calibration / graduation mark is a visible ring.

Volumetric flasks are named according to the maximum calibrated/graduated volume e.g.

“250ml” volumetric flask has a calibrated/graduated mark at exact volume of “250mililitres” /“250centimetres”

“1l” volumetric flask has a calibrated/graduated mark at exact volume of “one litre” /“1000 cubic centimeters”

“2l” volumetric flask has a calibrated/graduated mark at exact volume of “two litres” /“2000 cubic centimeters”

Dropper/teat pipette

A dropper/teat pipette is a long thin/narrow glass/rubber apparatus that has a flexible rubber head.

A dropper/teat pipette is used to measure very small amount/ drops of liquid solution by pressing the flexible rubber head. The numbers of drops needed are counted by pressing the rubber gently at a time

(b)Apparatus for measuring mass

Beam balance

A beam balance has a pan where a substance of unknown mass is placed. The scales on the opposite end are adjusted to “balance” with the mass of the unknown substance. The mass from a beam balance is in grams.

Electronic/electric balance.

An electronic/electric balance has a pan where a substance of unknown mass is placed. The mass of the unknown substance in grams is available immediately on the screen.

(c)Apparatus for measuring temperature

A thermometer has alcohol or mercury trapped in a bulb with a thin enclosed outlet for the alcohol/mercury in the bulb.

If temperature rises in the bulb, the alcohol /mercury expand along the thin narrow enclosed outlet.

The higher the temperature, the more the expansion

Outside, a calibration /graduation correspond to this expansion and thus changes in temperature.

A thermometer therefore determines the temperature when the bulb is fully dipped in to the substance being tested. To determine the temperature of solid is thus very difficult.

(d)Apparatus for measuring time

The stop watch/clock is the standard apparatus for measuring time. Time is measured using hours, minutes and second.

Common school stop watch/clock has start, stop and reset button for determining time for a chemical reaction. This requires practice.

(e) Apparatus for scooping

Spatula

A spatula is used to scoop solids which do not require accurate measurement. Both ends of the spatula can be used at a time.

A solid scooped to the brim is “one spatula end full” A solid scooped to halfbrim is “half spatula end full”.

Deflagrating spoon

A deflagrating spoon is used to scoop solids which do not require accurate measurement mainly for heating. Unlike a spatula, a deflagrating spoon is longer.

(f) Apparatus for putting liquids/solid for heating.

Test tube.

A test tube is a narrow/thin glass apparatus open on one side. The end of the opening is commonly called the “the mouth of the test tube”.

Boiling/ignition tube.

A boiling/ignition tube is a wide glass apparatus than a test tube open on one side. The end of the opening is commonly called the “the mouth of the boiling/ignition tube”.

Beaker.

Beaker is a wide calibrated/graduated lipped glass/plastic apparatus used for transferring liquid solution which do not normally require very accurate measurements

Beakers are named according to the maximum calibrated/graduated volume they can hold e.g.

“250ml” beaker has a maximum calibrated/graduated volume of “250mililitres” /“250 cubic centimeters”

“1l” beaker has a maximum calibrated/graduated volume of “one litre” /“1000 cubic centimeters”

“5 l” beaker has a maximum calibrated/graduated volume of “two litres” /“2000 cubic centimeters”

Conical flask.

A conical flask is a moderately narrow glass apparatus with a wide base and no calibration/graduation. Conical flasks thus carry/hold exact volumes of liquids that have been measured using other apparatus. It can also be put some solids. The narrow mouth ensures no spillage.

Conical flasks are named according to the maximum volume they can hold e.g. “250ml” Conical flasks hold a maximum volume of “250mililitres” /“250 cubic centimeters”

“500ml” Conical flasks hold a maximum volume of “500ml” /“1000 cubic centimeters”

Round bottomed flask

A round bottomed flask is a moderately narrow glass apparatus with a wide round base and no calibration/graduation. Round bottomed flask thus carry/hold exact volumes of liquids that have been measured using other apparatus. The narrow/thin mouth prevents spillage. The flask can also hold (weighed) solids. A round bottomed flask must be held/ clamped when in use because of its wide narrow base.

Flat bottomed flask

A flat bottomed flask is a moderately narrow glass apparatus with a wide round base with a small flat bottom. It has no calibration/graduation.

Flat bottomed flasks thus carry/hold exact volumes of liquids that have been measured using other apparatus. The narrow/thin mouth prevents spirage. They can also hold (weighed) solids. A flat bottomed flask must be held/ clamped when in use because it’s flat narrow base is not stable.

(g) Apparatus for holding unstable apparatus (during heating).

Tripod stand

A tripod stand is a three legged metallic apparatus which unstable apparatus are placed on (during heating).Beakers. Conical flasks, round bottomed flask and flat bottomed flasks are placed on top of tripod stand (during heating).

Wire gauze/mesh

Wire gauze/mesh is a metallic/iron plate of wires crossings. It is placed on top of a tripod stand:

(i) Ensure even distribution of heat to prevent cracking glass apparatus

(ii) Hold smaller apparatus that cannot reach the edges of tripod stand

3 Clamp stand

A clamp stand is a metallic apparatus which tightly hold apparatus at their “neck” firmly.

A clamp stand has a wide metallic base that ensures maximum stability. The height and position of clamping is variable. This require practice

Test tube holder

A test tube holder is a hand held metallic apparatus which tightly hold test/boiling/ignition tube at their “neck” firmly on the other end.

Some test tube holders have wooden handle that prevent heat conduction to the hand during heating.

Pair of tong.

A pair of tong is a scissor-like hand held metallic apparatus which tightly hold firmly a small solid sample on the other end.

Gas jar

A gas jar is a long wide glass apparatus with a wide base.

It is open on one end. It is used to collect/put gases.

This requires practice.

(h) Apparatus for holding/directing liquid solutions/funnels (to avoid spillage).

Filter funnel

A filter funnel is a wide mouthed (mainly plastic) apparatus that narrow drastically at the bottom to a long extension.

When the long extension is placed on top of another apparatus, a liquid solution can safely be directed through the wide mouth of the filter funnel into the apparatus without spirage.

Filter funnel is also used to place a filter paper during filtration.

Thistle funnel

A thistle funnel is a wide mouthed glass apparatus that narrow drastically at the bottom to a very long extension.

The long extension is usually drilled through a stopper/cork.

A liquid solution can thus be directed into a stoppered container without spillage

Dropping funnel

A dropping funnel is a wide mouthed glass apparatus with a tap that narrow drastically at the bottom to a very long extension.

The long extension is usually drilled through a stopper/cork.

A liquid solution can thus be directed into a stoppered container without spillage at the rate determined by adjusting the tap.

Separating funnel

A separating funnel is a wide mouthed glass apparatus with a tap at the bottom narrow extension.

A liquid solution can thus be directed into a separating funnel without spillage. It can also safely be removed from the funnel by opening the tap.

It is used to separate two or more liquid solution mixtures that form layers/immiscible. This requires practice.

(h) Apparatus for heating/Burners

Candle, spirit burner, kerosene stove, charcoal burner/jiko are some apparatus that can be used for heating.

Any flammable fuel when put in a container and ignited can produce some heat.

Bunsen burner

The Bunsen burner is the standard apparatus for heating in a Chemistry school laboratory.

It was discovered by the German Scientist Robert Wilhelm Bunsen in1854.

(a)Diagram of a Bunsen burner

A Bunsen burner uses butane/laboratory gas as the fuel. The butane/laboratory gas is highly flammable and thus usually stored safely in a secure chamber outside Chemistry school laboratory. It is tapped and distributed into the laboratory through gas pipes.

The gas pipes end at the gas tap on a chemistry laboratory bench .If opened the gas tap releases butane/laboratory gas. Butane/laboratory gas has a characteristic odor/smell that alerts leakages/open gas tap.

The Bunsen burner is fixed to the gas tap using a strong rubber tube.

The Bunsen burner is made up of the following parts:

(i) Base plate –to ensure the burner can stand on its own

(ii)Jet-a hole through which laboratory gas enters the burner

(iii)Collar/sleeve-adjustable circular metal attached to the main chimney/burell with a side hole/entry. It controls the amount of air entering used during burning.

(iv)Air hole- a hole/entry formed when the collar side hole is in line with chimney side hole. If the collar side hole is not in line with chimney side hole, the air hole is said to be “closed” If the collar side hole is in line with chimney side hole, the air hole is said to be “open”

(v)Chimney- tall round metallic rod attached to the base plate.

(b)Procedure for lighting/igniting a Bunsen burner

Adjust the collar to ensure the air holes are closed.
Connect the burner to the gas tap using a rubber tubing. Ensure the rubber tubing has no side leaks.
Turn on the gas tap.
Ignite the top of the chimney using a lighted match stick/gas lighter/wooden splint.
Do not delay excessively procedure (iv) from (iii) to prevent highly flammable laboratory gas from escaping/leaking.

(c)Bunsen burner flames

A Bunsen burner produces two types of flames depending on the amount of air entering through the air holes.

If the air holes are fully open, a non luminous flame is produced. If the air holes are fully closed, a luminous flame is produced. If the air holes are partially open/ closed, a hybrid of non luminous and luminous flames is produced.

Characteristic differences between luminous and non-luminous flame

Luminous flame	Non-luminous flame
1. Produced when the air holes are fully/completely closed.	1. Produced when the air holes are fully/completely open.
2. when the air holes are fully/ completely closed there is incomplete burning/ combustion of the laboratory gas	2.when the air holes are fully/ completely open there is complete burning/ combustion of the laboratory gas
3. Incomplete burning/ combustion of the laboratory gas produces fine unburnt carbon particles which make the flame sooty/smoky	3. Complete burning/ combustion of the laboratory gas does not produce carbon particles. This makes the flame non-sooty /non- smoky.
4. Some carbon particles become white hot and emit light. This flame is thus bright yellow in colour producing light. This makes luminous flame useful for lighting	4. Is mainly blue in colour and is hotter than luminous flame. This makes non-luminous flame useful for heating
5. Is larger, quiet and wavy/easily swayed by wind	5.Is smaller, noisy and steady
Luminous flame has three main regions: (i)the top yellow region where there is incomplete combustion/burning (ii)the region of unburnt gas below the yellow region where the gas does not burn (iii) blue region on the sides of region of unburnt gas where there is complete burning	Non-luminous flame has four main regions: (i)the top colourless region (ii) Blue region just below where there is complete burning. It is the hottest region (iii) green region surrounded by the blue region where there is complete burning (Ii) The region of unburnt gas at the innermost surrounded by green and blue regions. No burning takes place here

Scientific apparatus are drawn:

(i) Using a proportional two dimension (2D) cross-sections. Three dimensions (3D) are not recommended.

(ii) Straight edges of the apparatus on a scientific diagram should be drawn using ruler.

(iii) Curved edges of the apparatus on a scientific diagram should be drawn using free hand.

(iv)The bench, tripod or clamp to support apparatus which cannot stand on their own should be shown.

CLASSIFICATION OF SUBSTANCES

Substances are either pure or impure. A pure substance is one which contains only one substance.

An impure substance is one which contains two or more substances. A pure substance is made up of a pure solid, pure liquid or pure gas.

A mixture is a combination of two or more pure substances which can be separated by physical means. The three states of matter in nature appear mainly as mixtures of one with the other.

Common mixtures include:

(a)Solutions/solid-liquid dissolved mixture

Experiment:

To make a solution of copper (II) sulphate (VI)/Potassium magnate(VII) /sodium chloride

Procedure

Put about 100 cm3 of water in three separate beakers. Separately place a half spatula end full of copper (II) sulphate (VI), Potassium manganate (VII) and sodium chloride crystals to each beaker. Stir for about two minutes.

Observation

Copper (II) sulphate (VI) crystals dissolve to form a blue solution

Potassium manganate (VII) crystals dissolve to form a purple solution

Sodium chloride crystals dissolve to form a colourless solution

Explanation

Some solids, liquids and gases dissolve in some other liquids.

A substance/liquid in which another substance dissolves is called solvent.

A substance /solid /gas which dissolves in a solvent is called solute.

When a solute dissolves in a solvent it forms a uniform mixture called solution.

A solute dissolved in water as the solvent exists in another state of matter called aqueous state. Water is referred as the universal solvent because it dissolves many solutes. A solute that dissolves in a solvent is said to be soluble. Soluble particles uniformly spread between the particles of water/solvent and cannot be seen.

Solute + Solvent -> solution

Solute + Water -> aqueous solution of solute

The solute dissolved in water gives the name of the solution e. g.

Sodium chloride solution is a solution formed after dissolving sodium chloride crystals/solid in water. Sodium chloride exists in aqueous state after dissolving.

Sodium chloride + Water -> Sodium chloride solution

NaCl(s) + (aq) -> NaCl(aq)

Ammonia solution is a solution formed after dissolving ammonia gas in water. Ammonia exists in aqueous state after dissolving.

Ammonia gas + Water -> aqueous ammonia

NH₃ (g) + (aq) -> NH₃ (aq)

Copper (II) sulphate (VI) solution is a solution formed after dissolving Copper (II) sulphate (VI) crystals/solid in water. Copper (II) sulphate (VI) exists in aqueous state after dissolving.

Copper (II) sulphate (VI) + Water -> Copper (II) sulphate (VI) solution

CuSO₄(s) + (aq) -> CuSO₄ (aq)

Potassium manganate(VII) solution is a solution formed after dissolving Potassium manganate(VII) crystals/solid in water.

Potassium manganate(VII)exist in aqueous state after dissolving.

Potassium manganate(VII) + Water -> Potassium manganate(VII) solution

KMnO₄(s) + (aq) -> KMnO₄ (aq)

(b)Suspension/ precipitates/solid-liquid mixture which do not dissolve

Experiment: To make soil, flour and Lead (II) Iodide suspension/precipitate

Procedure

Put about 100 cm3 of water in three separate beakers. Separately place a half spatula end full of soil, maize and lead (II) Iodide to each beaker. Stir for about two minutes.

Observation

Some soil, maize and lead (II) Iodide float in the water

A brown suspension/precipitate/particles suspended in water containing soil

A white suspension/precipitate/particles suspended in water containing flour

A yellow suspension/precipitate/particles suspended in water containing Lead (II) iodide. Some soil, maize and lead (II) Iodide settle at the bottom after some time.

Explanation

Some solid substances do not dissolve in a liquid. They are said to be insoluble in the solvent .When an insoluble solid is put in liquid:

(i) Some particles remain suspended/floating in the liquid to form a suspension /precipitate.

(ii) Some particles sink/settle to the bottom to form sediments after being allowed to stand.

An insoluble solid acquire the colour of the suspension/precipitate .e.g.

A white suspension /precipitate have some fine white particles suspended /floating in the liquid. Not “white solution”
A blue suspension /precipitate has some fine blue particles suspended /floating in the liquid.
A green suspension /precipitate has some fine green particles suspended /floating in the liquid.
A brown suspension /precipitate has some fine brown particles suspended /floating in the liquid.
A yellow suspension /precipitate has some fine yellow particles suspended /floating in the liquid.

To form water-ethanol and Kerosene-turpentine miscibles

Procedure

(i)Measure 50cm3 of ethanol into 100cm3 beaker. Measure 50cm3 of water. Place the water into the beaker containing ethanol. Swirl for about one minute.

(ii)Measure 50cm3 of kerosene into 100cm3 beaker. Measure 50cm3 of turpentine oil. Place the turpentine oil into the beaker containing kerosene. Swirl for about one minute.

Observation

Two liquids do not form layers.

Ethanol and water form a uniform mixture.

Kerosene and turpentine oil form uniform mixture

Explanation

Ethanol is miscible in Water. Kerosene is miscible in turpentine oil. Miscible mixture form uniform mixture. They do not form layers. The particles of one liquid are smaller than the particles of the other. The smaller particles occupy the spaces between the bigger particles.

Immiscibles /Liquid-liquid mixtures

To form water-turpentine oil and Kerosene-water miscibles

Procedure

(i)Measure 50cm3 of water into 100cm3 beaker. Measure 50cm3 of turpentine oil. Place the oil into the beaker containing water. Swirl for about one minute.

(ii) Measure 50cm3 of water into 100cm3 beaker. Measure 50cm3 of kerosene. Place the kerosene into the beaker containing water. Swirl for about one minute.

Observation

Two liquids form layers.

Turpentine and water do not form a uniform mixture.

Water and kerosene do not form uniform mixture

Explanation

Kerosene is immiscible in Water. Water is immiscible in turpentine oil. Immiscible mixtures do not form uniform mixtures. They form layers. The size of the particles of one liquid is almost equal to the particles of the other. The particles of one liquid cannot occupy the spaces between the particles of the other. The heavier particles settle at the bottom. The less dense particles settle on top.

(d)Solid-solid mixtures/Alloys

Before solidifying, some heated molten/liquid metals dissolve in another metal to form a uniform mixture of the two. On solidifying, a uniform mixture of the metals is formed. A uniform mixture of two metals on solidifying is called alloy. In the alloy, one metallic particle occupies the spaces between the metallic particles of the other.

c) Common alloys of metal.

Alloy name	Constituents of the alloy	Uses of the alloy
Brass	Copper and Zinc	Making screws and bulb caps
Bronze	Copper and Tin	Making clock springs, electrical contacts and copper coins
Soldier	Lead and Tin	Soldering, joining electrical contacts because of its low melting points and high thermal conductivity
Duralumin	Aluminum, Copper and Magnesium	Making aircraft, utensils, and windows frames because of its light weight and corrosion resistant.
Steel	Iron, Carbon ,Manganese and other metals	Railway lines, car bodies girders and utensils.
Nichrome	Nichrome and Chromium	Provide resistance in electric heaters and ovens
German silver	Copper, Zinc and Nickel	Making coins

METHODS OF SEPARATING MIXTURES

Mixtures can be separated from applying the following methods:

(a) Decantation

Sediments can be separated from a liquid by pouring out the liquid. This process is called decantation.

Experiment

Put some sand in a beaker. Add about 200cm3 of water. Allow sand to settle. Pour off water carefully into another beaker.

Observation

Sand settles at the bottom as sediments.

Less clean water is poured out.

Explanation

Sand does not dissolve in water. Sand is denser than water and thus settles at the bottom as sediment. When poured out, the less dense water flows out.

(b)Filtration

Decantation leaves suspended particles in the liquid after separation. Filtration is thus improved decantation.Filtration is the method of separating insoluble mixtures/particles/solids from a liquid.

Experiment: To separate soil and water using filtration

Fold a filter paper to fit well into a filter funnel. Place the funnel in an empty 250 cm3 beaker.

Put one spatula end full of soil into 50cm3 of water. Stir. Put the soil/water mixture into the filter funnel.

Observations

Clean water is collected below the filter funnel.

Soil remains above the filter paper.

Explanation

A filter paper is porous which act like a fine sieve with very small holes. The holes allow smaller water particles to pass through but do not allow bigger soil particles. The liquid which passes through is called filtrate. The solid which do not pass through is called residue.

Set up of apparatus

In industries, filtration is used in engine filters to clean up air.

(c)Evaporation

Evaporation is a method of separating a solute/solid from its solution. This involves heating a solution (solvent and solute)to vapourize the solvent out of the solution mixture leaving pure solute/solid. If a mixture contain insoluble solid, they are filtered out.

Experiment: To separate a mixture of soil and salt (sodium chloride).

Procedure:

Put one spatula end full of soil on a filter paper.

Put one spatula full of common salt/sodium chloride into the same filter paper. Mix well using the spatula,.

Place about 200cm3 of water into a beaker.

Put the contents of the filter paper into the water. Stir thoroughly using a glass/stirring rod for about one minute.

Fold a filter paper into a filter funnel.

Pour half portion of the contents in the beaker into the filter funnel.

Put the filtrate into an evaporating dish. Heat on a water bath.

Observation

(i)On mixing

Colourless crystals and brown soil particles appear on the filter paper.

(ii)On adding water

Common soil dissolves in water. Soil particles do not dissolve in water.

(iii)On filtration

Colourless liquid collected as filtrate below the filter funnel/paper.

Brown residue collected above the filter funnel/paper.

(iv)On evaporation

Colourless crystals collected after evaporation

Explanation

Solid mixture of sand and common salt take the colors of the two.

On adding water, common salt dissolves to form a solution.

Soil does not because it is insoluble in water and thus forms a suspension.

On filtration, a residue of insoluble soil does not pass through the filter paper.

It is collected as residue.

Common salt solution is collected as filtrate.

On heating the filtrate, the solvent/water evaporate/vaporize out of the evaporating dish leaving common salt crystals.

Vapourization/evaporation can take place even without heating.

This is the principle/process of drying wet clothes on the hanging line.

Set up of apparatus

(d) Distillation

Distillation is an improved evaporation where both the solute and the solvent in the solution are separated /collected. Distillation therefore is the process of separating a solution into constituent solid solute and the solvent. It involves heating the solution to evaporate/vaporize the solvent out. The solvent vapour is then condensed back to a liquid.

Experiment: To obtain copper (II) sulphate (VI) crystals and water from copper (II) sulphate (VI) solution.

Procedure:

Put one spatula end full of copper (II) sulphate (VI) crystals into a 250cm3 beaker.

Place about 200cm3 of water into the beaker.

Stir thoroughly using a glass/stirring rod for about one minute.

Pour half portion of the contents in the beaker into a round bottomed/flat/conical flask broken porcelain/sand/glass into the flask.

Put a few pieces of b Stopper the flask.

Connect the flask to a Liebig condenser using delivery tube.

Place a 200cm3 clean empty beaker/conical flask as a receiver at the end of the Liebig condenser.

Circulate water in the Liebig condenser.

Heat the flask strongly on a tripod stand with wire mesh/gauze until there is no more visible boiling bubbles in the flask.

Observation

Copper (II) sulphate (VI) crystals dissolve in water to form a blue solution.
On heating, colourless liquid is collected in the receiver.

Blue crystals are left in the flask.

(If gently heated further, the blue crystals turn to white powder)

Explanation

On heating blue Copper (II) sulphate (VI) solution, the colourless liquid solvents evaporate/vaporize.

The liquid vapour/gas passes through the delivery tube to the Liebig condenser.

The Liebig condenser has a cold water inlet near the receiver and cold water out let.

This ensures efficient cooling. If the cold water outlet/inlet is reversed, the water circulation would be less efficient.

The water in the receiver would be warm. In the Liebig condenser, the cold water condenses the liquid vapour into liquid.

The condensed liquid collects in the receiver as distillate.

The solute of blue Copper (II) sulphate (VI) crystals is left in the flask as residue.

During simple distillation, therefore, the solution is heated to vaporize /evaporate the solvent/one component which is condensed at a different part of the apparatus.

The purpose of pieces of broken porcelain/porous pot/glass/sand/ is to:

(i) Prevent bumping of the solution during boiling.

(ii) Ensure smooth and even boiling.

Salty sea water can be made pure through simple distillation.

Any mixture with a large difference /40^oC in boiling point can be separated using simple distillation.

Set up of apparatus

e)Fractional distillation

Fractional distillation is an improved simple distillation used specifically to separate miscible mixtures with very close /near boiling points.

Fractional distillation involves:

(i) Heating the mixture in a conical/round bottomed /flat bottomed flask.

The pure substance with a lower boiling point and thus more volatile evaporates/boils/vaporize first.e.g. Pure ethanol has a boiling point of 78^oC.Pure water has a boiling point of 100^oC at sea level/one atmosphere pressure.

When a miscible mixture of ethanol and water is heated, ethanol vaporizes /boils/ evaporates first because it is more volatile.

(ii)The conical/round bottomed /flat bottomed flask is connected to a long glass tube called fractionating column.

The purpose of the fractionating column is to offer areas of condensation for the less volatile pure mixture.

The fractionating column is packed with glass beads/broken glass/ porcelain/ shelves to increase the surface area of condensation of the less volatile pure mixture.

(iii)When the vapors rise they condense on the glass beads/broken glass /porcelain / shelves which become hot.

When the temperature of the glass beads/broken glass/porcelain/shelves is beyond the boiling point of the less volatile pure substance, the pure substance rise and condensation take place on the glass beads/broken glass/porcelain/shelves at a higher level on the fractionating column.

The less volatile pure substance trickles/drips back down the fractionating column or back into the conical/round bottomed /flat bottomed flask to be heated again. e.g.

If the temperature on glass beads/broken glass/porcelain/shelves is beyond 78^oC, the more volatile pure ethanol rise to condense on the glass beads/broken glass /porcelain/shelves higher in the fractionating column.

Water condenses and then drip/trickle to the glass beads/broken glass /porcelain /shelves lower in the fractionating column because it is less volatile.

(iv) The fractionating column is connected to a Liebig condenser. The Liebig condenser has a cold water inlet and outlet circulation.

The more volatile mixture that reach the top of the fractionating column is condenses by the Liebig condenser into a receiver. It is collected as the first fraction.

(v)At the top of the fractionating column, a thermometer is placed to note/monitor the temperature of the boiling mixtures.

Pure substances have constant/fixed boiling point. When one mixture is completely separated, the thermometer reading rises.

E.g. the thermometer reading remains at78^oC when ethanol is being separated. When no more ethanol is being separated, the mercury/alcohol level in the thermometer rises.

(vi)The second /subsequent fractions are collected in the receiver after noting a rise the mercury/alcohol level in the thermometer.

E.g. the thermometer reading rises to 100^oC when water is being separated. It is passed through the Liebig condenser with the cold water inlet and outlet circulation. It is collected different receiver as the second/subsequent fraction.

(vii)Each fraction collected should be confirmed from known physical/chemical properties/characteristic.

Example

Ethanol

Ethanol is a colourless liquid that has a characteristic smell .When it is put in a watch glass then ignited, it catches fire and burn with a blue flame.

Water

Water is a colourless liquid that has no smell/odour .When it is put in a watch glass then ignited, it does not catch fire.

Set up of apparatus

Industrial application of Fractional distillation

On a large scale,fractional distillation is used:

(i)In fractional distillation of crude oil in an oil refinery.

Crude oil is a mixture of many fractions. When heated in a furnace, the different fractions separate out according to their boiling point. In Kenya,fractional distillation takes place at Changamwe in Mombasa.

(ii)In fractional distillation of air.

Air contain a mixture of three main useful gases which are condensed by cooling to very low temperature (-200^oC) to form a liquid. The liquid is then heated. Nitrogen is the most volatile (-196^oC) and thus comes out as the first fraction. Argon (at -186^oC) is the second fraction. Oxygen ( at -183^oC) is the last fraction. The three gases are very useful industrial gases.

(f)Separation of immiscibles (Using a separating funnel)

Two or more liquids that form layers on mixing are immiscible. Immiscible mixture arrange themselves according to their densities

i.e. The denser liquid sink to the bottom. The less dense liquid floats on the denser one. Immicible mixtures can be separated from each other by using a separating funnel.

Experiment: To separate an immiscible mixture of paraffin and water.

Procedure

Place about 100cm3 of water into a 250cm3 beaker. Add about 100cm3 of paraffin into the beaker. Stir.

Transfer the mixture into a separating funnel. Allow to settle for about one minute. Open the tap, run out the lower layer out slowly into a clean beaker. Close the tap when the upper layer is very close to the tap.

Run out the intermediate small amount of the mixture near the tap into a beaker. Discard it.

Run out the remaining upper layer into a fresh beaker.

Place a portion of upper and lower layer into a watch glass separately after separating each. Ignite.

Observation

Water and paraffin are both colourless liquids.

Two layers are formed on mixing.

Colourless odorless liquid collected first. It does not catch fire.

A colourless liquid with characteristic smell collected later/second. It catches fire and burn with a yellow smoky flame.

Explanation

Water and paraffin are immiscible. Water is denser than paraffin. When put in a separating funnel, paraffin float on water. On opening the tap, water runs out. A mixture of water and paraffin at the junction of the two is discarded. It is not pure.

Set up of apparatus

(g)Sublimation/deposition

Some solids on heating do not melt to a liquid but change directly to a gas. The process by which a solid changes to a gas is called sublimation. The gas cools back and changes directly to a solid. The process by which a gas changes to a solid is called deposition. Sublimation and deposition therefore are the same but opposite processes.

GAS

Sublimation Deposition

SOLID

Some common substances that undergo sublimation/ deposition include:

(i)Iodine (ii)Carbon(IV)oxide (iii)Camphor (iv) ammonium chloride (v)Iron(III)chloride (vi)Aluminum(III)chloride

(vii) benzoic acid

If a mixture has any of the above as a component, then on heating it will change to a gas and be deposited away from the source of heating.

Procedure

Place about one spatula full of ammonium chloride crystals into a clean dry 100cm3 beaker. Add equal amount of sodium chloride crystals into the beaker. Swirl to mix.

Place the beaker on a tripod stand.

Put about 100cm3 of water into another beaker. Place carefully the beaker containing water on top of the beaker containing the solid mixture. Light/ignite a burner and heat the solid.

Set up of apparatus:

Observation

(i)With ammonium chloride/common salt mixture

White fumes produced.

White sublimate deposited

Colourless residue left

(ii)With Iodine/common salt mixture

Purple fumes produced.

Dark grey sublimate deposited

Colourless residue left

Explanation

(i)On heating a mixture of ammonium chloride and common salt, a white fume of ammonium chloride is produced. The white fumes solidify as white sublimate on the cooler parts. Common salt remains as residue.

Chemical equation:

Ammonium chloride solid Ammonium chloride gas

NH₄Cl(s) NH₄Cl(g)

(ii)On heating a mixture of Iodine and common salt, a purple fume of Iodine vapour is produced. The purple fumes solidify as dark grey sublimate on the cooler parts. Common salt remains as residue.

Chemical equation:

Iodine solid Iodine gas

I₂(s) I₂ (g)

(h)Chromatography

Chromatography is a method of separating components of a solution mixture by passing it through a medium where the different components move at different rates. The medium through which the solution mixture is passed is called absorbent material.

Paper chromatography is a method of separating colored dyes by using paper as the absorbent material.

Since dyes are insoluble/do not dissolve in water, ethanol and propanone are used as suitable solvents for dissolving the dye.

Practically, a simple paper chromatography involve placing a dye/material on the absorbent material, adding slowly a suitable soluble solvent on the dye/material using a dropper, the solvent spread out on the absorbent material carrying the soluble dye away from the origin.

The spot on which the dye is initially/originally placed is called baseline. The farthest point the solvent spread is called solvent front.

The farthest a dye can be spread by the solvent depend on:

(i) Density of the dye-the denser the dye, the less it spread from the basely ne by the solvent.

(ii) Stickiness of the dye-some dyes sticks on the absorbent material more than other thus do not spread far from baseline.

Experiment: To investigate the colors in ink

Procedure

Method 1

Place a filter paper on an empty beaker. Put a drop of black/blue ink in the centre of the filter paper. Wait for about one minute for the ink drop to spread. Using a clean teat pipette/dropper add one drop of ethanol/propanone. Wait for about one minute for the ink drop to spread further. Add about twenty other drops of ethanol waiting for about one minute before each addition. Allow the filter paper to dry.

Experiment: To investigate the colors in ink

Procedure

Method 2

Cut an 8 centimeter thin strip of a filter paper. At about 3cm on the strip, place a drop of ink. Place the filter paper in a 10cm length boiling tube containing 5cm3 of ethanol. Ensure the cut strip of the filter paper just dips into the ethanol towards the ink mark. Cover the boiling tube. Wait for about twenty minutes. Remove the boiling tube and allow the filter paper to dry.

Set up of apparatus

Method 1

Set up of apparatus

Method 2

Explanation

When a drop of ink is placed on an absorbent material it sticks. On adding an eluting solvent, it dissolves the dye spread out with it. The denser and sticky pure dye move least. The least dense/sticky pure dye move farthest. A pure dye will produce the same chromatogram/spot if the same eluting solvent is used on the same absorbent material. Comparing the distance moved by a pure dye with a mixture, the coloured dyes in a mixture can be deduced as below:

Example 1

The chromatogram of pure dyes A, B ,C and a dye mixture D is shown below Determine the pure dyes present in D. On the diagram show:

(i)the solvent front

(ii) Baseline

(Iii) the most soluble pure dye

(i) Solvent extraction

Solvent extraction is a method of separating oil from nuts/seeds. Most nuts contain oil. First the nuts are crushed to reduce their size and increase the surface area. A suitable volatile solvent is added. The mixture is filtered. The filtrate solvent is then allowed to crystallize leaving the oil/fat. If a filter paper is rubbed/smeared with the oil/fat, it becomes translucent. This is the test for the presence of oil/fat.

Experiment: To extract oil from Macadamia nut seeds

Procedure

Crush Macadamia nut seeds form the hard outer cover .Place the inner soft seed into a mortar. Crush (add a little sand to assist in crushing).

Add a little propanone and continue crushing. Continue crushing and adding a little propanone until there is more liquid mixture than the solid. Decant/filter. Put the filtrate into an evaporating dish. Vapourize the solvent using solar energy /sunlight. Smear/rub a portion of the residue left after evaporation on a clean dry filter paper.

Observation /Explanation

Propanone dissolve fat/oil in the macadamia nuts. Propanone is more volatile (lower boiling point) than oil/fat. In sunlight/solar energy, propanone evaporate/vaporize leaving oil/fat(has a higher boiling point).Any seed like corn, wheat , rice, soya bean may be used instead of macadamia seed. When oil/fat is rubbed/ smeared on an opaque paper, it becomes translucent.

(j) Crystallization

Crystallization is the process of using solubility of a solute/solid to obtain the solute/solid crystals from a saturated solution by cooling or heating the solution.

A crystal is the smallest regular shaped particle of a solute. Every solute has unique shape of its crystals.

Some solutions form crystals when heated. This is because less solute dissolves at higher temperature. Some other solutions form crystals when cooled. This is because less solute dissolves at lower temperature.

Experiment; To crystallize copper (II) sulphate (VI) solution

Procedure:

Place about one spatula full of hydrated copper sulphate (VI) crystals into 200cm3 of distilled water in a beaker. Stir. Continue adding a little more of the hydrated copper sulphate (VI) crystals and stirring until no more dissolve. Decant/filter. Cover the filtrate with a filter paper. Pierce and make small holes on the filter paper cover. Preserve the experiment for about seven days.

Observation/Explanation

Large blue crystals formed

When hydrated copper (II) sulphate crystals are placed in water, they dissolve to form copper (II) sulphate solution. After some days water slowly evaporate leaving large crystals of copper (II) sulphate. If the mixture is heated to dryness, small crystals are formed.

Physical/Temporary and Chemical changes

A physical/temporary change is one which no new substance is formed and is reversible back to original.

A chemical/permanent change is one which a new substance is formed and is irreversible back to original.

The following experiments illustrates physical and chemical changes

(a)Heating ice

Place about 10g of pure ice in a beaker. Determine its temperature. Record it at time “0.0” in the table below. Heat the ice on a strong Bunsen flame and determine its temperature after every 60seconds/1minute to complete the table below:

Time/minutes	0	1	2	3	4	5	6	7	8
Temperature (^oC)	-2	0	0	40	80	90	95	95	96

Plot a graph of time against Temperature (y-axes)

Explain the shape of your graph

Melting/freezing/fusion/solidification and boiling /vaporization /evaporation are the two physical processes.

Melting /freezing point of pure substances is fixed /constant.

The boiling point of pure substance depends on external atmospheric pressure.

Melting/fusion is the physical change of a solid to liquid.

Freezing is the physical change of a liquid to solid.

Melting/freezing/fusion/solidification is therefore two opposite but same reversible physical processes i.e.

A (s) A (l)

Boiling/vaporization/evaporation is the physical change of a liquid to gas.

Condensation/ liquidification is the physical change of gas to liquid.

Boiling/vaporization/evaporation and condensation/ liquidification are therefore two opposite but same reversible physical processes i.e.

B (l) B(g)

Practically

(i) Melting/liquidification/fusion involves heating a solid to weaken the strong bonds holding the solid particles together.

Solids are made up of very strong bonds holding the particles very close to each other (Kinetic Theory of matter).

On heating these particles gain energy/heat from the surrounding heat source to form a liquid with weaker bonds holding the particles close together but with some degree of freedom.

(ii)Freezing/fusion/solidification involves cooling a liquid to reform /rejoin the very strong bonds to hold the particles very close to each other as solid and thus lose their degree of freedom (Kinetic Theory of matter).

Freezing /fusion / solidification is an exothermic (–∆H) process that require particles holding the liquid together to lose energy to the surrounding.

(iii)Boiling/vaporization/evaporation involves heating a liquid to completely break/free the bonds holding the liquid particles together.

Gaseous particles have high degree of freedom (Kinetic Theory of matter).

Boiling /vaporization / evaporation is an endothermic (+∆H) process that require/absorb energy from the surrounding.

(iv)Condensation/liquidification is reverse process of boiling /vaporization / evaporation.

It involves gaseous particles losing energy to the surrounding to form a liquid.

AIR OXYGEN AND COMBUSTION

A.THE ATMOSPHERE

The atmosphere is made up of air. Air is a mixture of colourless, odorless gases which is felt as wind (air in motion).All living things breath in air for respiration. Plants use air for respiration and photosynthesis.
The main gases present in the atmosphere/air:

Gas	Approximate % composition by volume
Nitrogen	78.0
Oxygen	21.0
Carbon(IV)oxide	0.03
Noble gases	1.0
Water vapour	Vary from region

The following experiments below shows the presence and composition of the gases in air/atmosphere

(a)To find the composition of air supporting combustion using a candle stick

Procedure

Measure the length of an empty gas jar M₁. Place a candle stick on a Petri dish. Float it on water in basin/trough. Cover it with the gas jar. Mark the level of the water in the gas jar M₂. Remove the gas jar. Light the candle sick. Carefully cover it with the gas jar. Observe for two minutes. Mark the new level of the water M₃.

Set up of apparatus

Sample observations

Candle continues to burn then extinguished/goes off

Level of water in the gas jar rises after igniting the candle

Length of empty gas jar = M₁= 14cm

Length of gas jar without water before igniting candle = M₂= 10 cm

Length of gas jar with water before igniting candle = M_{1 –} M₂= 14- 10 = 4 cm

Length of gas jar with water after igniting candle = M₃= 8 cm

Length of gas jar without water after igniting candle = M_{1 –} M₃= 10 -8 = 2 cm

Explanation

Candle burns in air. In a closed system (vessel), the candle continues to burn using the part of air that support burning/combustion. This is called the active part of air. The candle goes off/extinguished when all theactive part of air is used up. The level of the water rises to occupy the space /volume occupied by the usedactive part of air.

The experiment is better when very dilute sodium/potassium hydroxide is used instead of water. Dilute Potassium/ sodium hydroxide absorb Carbon (IV) oxide gas that comes out from burning/combustion of candle stick.

From the experiment above the % composition of the:

(i) Active part of air can be calculated:

M_{2 –} M₃ x 100% => 10- 8 x 100% = 20%

M₂ 10cm

(ii) Inactive part of air can be calculated:

100% –20% = 80% // M₃=> 8 x 100% = 80%

M₂10cm

(b)To find the composition of active part of air using heated copper turnings.

Procedure

Clamp a completely packed/filled open ended glass tube with copper turnings. Seal the ends with glass/cotton wool.

Label two graduated syringes as “A” and “B” Push out air from syringe “A”. Pull in air into syringe “B”.

Attach both syringe “A” and “B” on opposite ends of the glass tube.

Determine and record the volume of air in syringe “B” V₁.

Heat the glass tube strongly for about three minutes.

Push all the air slowly from syringe “B” to syringe “A” as heating continues. Push all the air slowly from syringe “A” back to syringe “B” and repeatedly back and forth.

After about ten minutes, determine the new volume of air in syringe “B” V₂

Set up of apparatus

Sample observations

Colour change from brown to black

Volume of air in syringe “B” before heating V₁ = 158.0cm3

Volume of air in syringe “B” after heating V₂ = 127.2cm3

Volume of air in syringe “B” used by copper V₁– V₂= 30.8cm3

Sample questions

What is the purpose of:

(i) glass/cotton wool

To prevent/stop copper turnings from being blown into the syringe/out of the glass tube

(ii) Passing air through the glass tube repeatedly

To ensure all the active part of air is used up

(iii) Passing air through the glass tube slowly

To allow enough time of contact between the active part of and the heated copper turnings

State and explain the observations made in the glass tube.

Colour change from brown to black

Brown copper metal reacts with the active part of air/oxygen to form black copper (II) oxide.

Chemical equation

Copper + Oxygen -> Copper (II) oxide

2Cu(s) + O₂ (g) -> 2CuO(s)

The reaction reduces the amount/volume of oxygen in syringe “B” leaving the inactive part of air. Copper only react with oxygen when heated.

Calculate the % of

(i) Active part of air

% active part of air = V₁– V₂ x 100% => 30.8cm3 x 100% = 19.493%

V₁158.0cm3

(ii) Inactive part of air

Method 1

% inactive part of air = V₂ x 100% =>127.2cm3 x 100% = 80.506%

V₁158.0cm3

Method 2

% inactive part of air = 100% -% active part of air

=> 100 % – 19.493 % = 80.507%

The % of active part of air is theoretically higher than the above while % of inactive part of air is theoretically lower than the above. Explain.

Not all the active part of air reacted with copper

State the main gases that constitute:

(a )active part of air.

Oxygen

(b) Inactive part of air

Nitrogen, carbon (IV) oxide and noble gases

If the copper turnings are replaced with magnesium shavings the % of active part of air obtained is extraordinary very high. Explain.

Magnesium is more reactive than copper. The reaction is highly exothermic. It generates enough heat for magnesium to react with both oxygen and nitrogen in the air.

A white solid/ash mixture of Magnesium oxide and Magnesium nitride is formed. This considerably reduces the volume of air left after the experiment.

Chemical equation

Magnesium + Oxygen -> magnesium (II) oxide

2Mg(s) + O₂ (g) -> 2MgO(s)

Magnesium + Nitrogen -> magnesium (II) nitride

3Mg(s) + N₂ (g) -> Mg₃N₂ (s)

(c)To find the composition of active part of air using alkaline pyrogallol

Procedure

Measure about 2cm3 of dilute sodium hydroxide into a graduated gas jar. Record the volume of the graduated cylinder V₁.

Place about two spatula end full of pyrogallol/1, 2, 3-trihydroxobenzene into the gas jar. Immediately place a cover slip firmly on the mouth of the gas jar. Swirl thoroughly for about two minutes.

Invert the gas jar in a trough/basin containing water. Measure the volume of air in the gas jar V₂

Sample observations

Colour of pyrogallol/1, 2, 3-trihydroxobenzene change to brown.

Level of water in gas jar rises when inverted in basin/trough.

Volume of gas jar /air in gas jar V1= 800cm3

Volume of gas jar /air in gas jar after shaking with alkaline pyrogallol/1, 2, 3-trihydroxobenzene V₂= 640 cm3

Sample questions

Which gas is absorbed by alkaline pyrogallol/1,2,3-trihydroxobenzene

Oxygen

Calculate the

(i) % of active part of air

V₁-V₂ x 100% => (800cm3 – 640 cm3) x 100% = 20%

V₁ 800cm3

(ii) % of inactive part of air

V₂ x 100% => 640 cm3 x 100% = 80%

V₁ 800cm3

(d)To establish the presence of carbon (IV) oxide in air using lime water

Pass tap water slowly into an empty flask as in the set up below

Sample observation questions

What is the purpose of paper cover?

To ensure no air enters into the lime water.

What happens when water enters the flask?

It forces the air from the flask into the lime water.

What is observed when the air is bubbled in the lime water?

A white precipitate is formed. The white precipitate dissolves on prolonged bubbling of air.

(a) Identify the compound that form:

(i)lime water

Calcium hydroxide / Ca(OH)₂

(ii) White precipitate

Calcium carbonate/ CaCO₃

(iii) When the white precipitate dissolves

Calcium hydrogen carbonate/ CaHCO₃

(b)Write the chemical equation for the reaction that tale place when:

(i) White precipitate is formed

Calcium hydroxide + carbon (IV) oxide -> Calcium carbonate + water

Ca (OH)₂(aq) + CO₂ (g) -> CaCO₃(s) + H₂O (l)

(ii) White precipitate dissolves

Calcium carbonate + water+ carbon (IV) oxide -> Calcium hydrogen carbonate

CaCO₃(s) + H₂O (l) + CO₂ (g) -> CaHCO₃ (aq)

State the chemical test for the presence of carbon (IV) oxide gas based on 4(a) and (b) above:

Carbon (IV) oxide forms a white precipitate with lime water that dissolves in excess of the gas.

State the composition of carbon (IV) oxide gas by volume in the air.

About 0.03% by volume

B.OXYGEN

a) Occurrence.
Fifty 50% of the earth’s crust consist of Oxygen combined with other elements e.g. oxides of metals
About 70% of the earth is water made up of Hydrogen and Oxygen.
About 20% by volume of the atmospheric gases is Oxygen that form the active part of air.
b) School laboratory preparation.

Oxygen was first prepared in 1772 by Karl Scheele and later in 1774 by Joseph Priestly. It was Antony Lavoisier who gave it the name “Oxygen”

Procedure

Method 1: Using Hydrogen peroxide

Half fill a trough/basin with tap water. Place a bee hive shelf/stand into the water.

Completely fill the gas jar with water and invert in onto the bee hive shelf/stand.

Clamp a round bottomed flask and set up the apparatus as below.

Collect several gas jars of Oxygen covering each sample.

Sample observation questions

What is observed when the hydrogen peroxide is added into the flask?

Rapid effervescence/bubbling/fizzing

Describe the colour and smell of the gas

Colourless and odorless

(a)Name the method of gas collection used.

–Over water

-Upward delivery

-Down ward displacement of water

(b)What property of Oxygen makes it to be collected using the method above?

-Slightly soluble in water

What is the purpose of manganese (IV) oxide?

Manganese (IV) oxide is catalyst.

A catalyst is a substance that speeds up the rate of a chemical reaction but remain chemically unchanged at the end of the reaction.

Hydrogen peroxide decomposes slowly to form water and Oxygen gas.

A little Manganese (IV) oxide speeds up the rate of decomposition by reducing the time taken for a given volume of Oxygen to be produced.

Write the equation for the reaction.

Hydrogen peroxide -> Water + Oxygen

2H₂O₂ (aq) -> 2H₂O (l) + O₂ (g)

Lower a glowing splint slowly into a gas jar containing Oxygen gas. State what is observed.

The glowing splint relights/rekindles

Oxygen relights/rekindles a glowing splint. This is the confirmatory test for the presence of Oxygen gas

Method 1: Using Sodium peroxide

Half fill a trough/basin with tap water. Add four drops of phenolphthalein indicator.

Place a bee hive shelf/stand into the water.

Completely fill a gas jar with water and invert in onto the bee hive shelf/stand.

Clamp a round bottomed flask and set up the apparatus as below.

Collect several gas jars of Oxygen covering each sample.

Sample observation questions

What is observed when water is added?

(i) Into the flask containing sodium peroxide

Rapid effervescence/bubbling/fizzing

(ii) Phenolphththalein

Remains colourless /Phenolphthalein indicator is colourless in neutral solution

Describe the colour and smell of the gas

Colourless and odorless

3.(a)Name the method of gas collection used.

–Over water. Oxygen is slightly soluble in water.

Test the gas by lowering a glowing splint slowly into a gas jar containing the prepared sample.

The glowing splint relights/rekindles. This confirms the presence of Oxygen gas

Write the equation for the reaction.

Sodium peroxide + Water -> Sodium hydroxide + Oxygen

2Na₂O₂ (aq) + 2H2O (l) -> 4NaOH (aq) + O₂ (g)

Test the gas by lowering a glowing splint slowly into a gas jar containing the prepared sample.

The glowing splint relights/rekindles.

This confirms the presence of Oxygen gas

Write the equation for the reaction.

Potassium Chlorate (V) -> Potassium Chloride + Oxygen

2KClO₃ (aq) -> 2KCl (aq) + 3O₂ (g)

What is the purpose of manganese (IV) oxide?

Manganese (IV) oxide is catalyst.

A catalyst is a substance that speeds up the rate of a chemical reaction but remain chemically unchanged at the end of the reaction.

Potassium Chlorate (V) decomposes slowly to form potassium chloride and Oxygen gas.

A little Manganese (IV) oxide speeds up the rate of decomposition by reducing the time taken for a given volume of Oxygen to be produced.

(c)Uses of Oxygen

Oxygen is put in cylinders for use where natural supply is not sufficiently enough. This is mainly in:

(i)Mountain climbing/Mountaineering-at high altitudes, the concentration of air/oxygen is low. Mountain climbers must therefore carry their own supply of oxygen for breathing.

(ii) Deep sea diving-Deep sea divers carry their own supply of Oxygen.

(iii) Saving life in hospitals for patients with breathing problems and during anesthesia.

A mixture of oxygen and some other gases produces a flame that is very hot.

(i) Oxy-acetylene/ethyne flame is produced when Ethyne/acetylene gas is burnt in pure oxygen. The flame has a temperature of about 3000^oC.It is used for welding /cuttingmetals.

(ii)Oxy-hydrogen flame is produced when Hydrogen is burn in pure oxygen. The flame has a temperature of about 2000^oC.It is used also for welding /cuttingmetals.

Oxy-hydrogen mixture is used as rocket fuel
A mixture of charcoal, petrol and liquid Oxygen is an explosive.

(d) Chemical properties of Oxygen /combustion.

Oxygen is a very reactive non metal. Many elements react with oxygen through burning to form a group of compounds called Oxides.

Burning/combustion is the reaction of Oxygen with an element/substances.

Reaction in which a substance is added oxygen is called Oxidation reaction. Burning/combustion are an example of an oxidation reaction.

Most non metals burn in Oxygen/air to form an Oxide which in solution / dissolved in water is acidic in nature. They turn blue litmus red.e.g. Carbon (IV) oxide/CO₂, Nitrogen (IV) oxide/ NO₂, Sulphur (IV) oxide/ SO₂

Some non metals burn in Oxygen/air to form an Oxide which in solution / dissolved in water is neutral in nature. They don’t turn blue or red litmus. E.g. Carbon (II) oxide/CO, Water/ H₂O

All metals burns in Oxygen/air to form an Oxide which in solution/dissolved in water is basic/alkaline in nature. They turn red litmus blue.e.g.

Magnesium oxide/MgO, Sodium Oxide/ Na₂O, Copper (II) oxide/CuO
Elements/substances burn faster in pure Oxygen than in air

Air contains the inactive part of air that slows the rate of burning of substances/elements.

(i)Reaction of metals with Oxygen/air

The following experiments show the reaction of metals with Oxygen and air.

Burning Magnesium

Procedure

(a)Cut a 2cm length piece of magnesium ribbon. Using a pair of tongs introduce it to a Bunsen flame. Remove it when it catches fire. Observe.

Place the products in a beaker containing about 5cm3 of water. Test the solution/mixture using litmus papers

(b)Cut another 2cm length piece of magnesium ribbon. Using a pair of tongs introduce it to a Bunsen flame. When it catches fire, lower it slowly into a gas jar containing Oxygen.

Place about 5cm3 of water into the gas jar. Test the solution/mixture using litmus papers. Test the solution/mixture using litmus papers

Observations

(a)In air

Magnesium burns with a bright blindening flame in air forming white solid/ash /powder. Effervescence/bubbles/ fizzing Pungent smell of urine. Blue litmus paper remains blue. Red litmus paper turns blue

(b) In pure Oxygen

Magnesium burns faster with a very bright blindening flame pure oxygen forming white solid/ash /powder. No effervescence/bubbles/ fizzing. No pungent smell of urine. Blue litmus paper remains blue. Red litmus paper turns blue

Explanation

Magnesium burns in air producing enough heat energy to react with both Oxygen and Nitrogen to form Magnesium Oxide and Magnesium nitride. Both Magnesium Oxide and Magnesium nitride are white solid/ash /powder.

Chemical equations

Magnesium + Oxygen -> Magnesium Oxide

2Mg(s) + O₂(g) -> 2MgO(s)

Magnesium + Nitrogen -> Magnesium Nitride

3Mg(s) + N₂(g) -> Mg₃N₂ (s)

Magnesium Oxide dissolves in water to form a basic/alkaline solution of Magnesium hydroxide

Chemical equations

Magnesium Oxide + Water -> Magnesium hydroxide

2Mg(s) + O₂ (l) -> 2MgO(s)

Magnesium Nitride dissolves in water to form a basic/alkaline solution of Magnesium hydroxide and producing Ammonia gas. Ammonia is also an alkaline/basic gas that has a pungent smell of urine.

Chemical equations

Magnesium Nitride + Water -> Magnesium hydroxide + Ammonia gas

Mg₃N₂ (s) + 6H₂O (l) -> 3Mg (OH)₂ (aq) + 2NH₃(g)

Burning Sodium

Procedure

(a)Carefully cut a very small piece of sodium. Using a deflagrating spoon introduce it to a Bunsen flame. Remove it when it catches fire. Observe.

Place the products in a beaker containing about 20cm3 of water. Test the solution/mixture using litmus papers

(b) Carefully cut another very small piece of sodium. Using a deflagrating spoon introduce it to a Bunsen flame. When it catches fire, lower it slowly into a gas jar containing Oxygen.

Place about 20 cm3 of water into the gas jar. Test the solution/mixture using litmus papers. Test the solution/mixture using litmus papers

Observations

(a)In air

Sodium burns with a yellow flame in air forming a black solid. Blue litmus paper remains blue. Red litmus paper turns blue

(b) In pure Oxygen

Sodium burns faster with a golden yellow flame in pure oxygen forming a yellow solid. Effervescence/bubbles/ fizzing. Gas produced relights glowing splint. Blue litmus paper remains blue. Red litmus paper turns blue.

Explanation

(a)Sodium burns in air forming black Sodium Oxide

Chemical equations

Sodium + Oxygen/air -> Sodium Oxide

4Na(s) + O₂ (g) -> 2Na₂O(s)

Sodium Oxide dissolves in water to form a basic/alkaline solution of Sodium hydroxide

Chemical equations

Sodium Oxide + Water -> Sodium hydroxide

Na₂O(s) + H₂O (l) -> 2NaOH (aq)

(b)Sodium burns in pure oxygen forming yellow Sodium peroxide

Chemical equations

Sodium + Oxygen -> Sodium peroxide

2Na(s) + O₂ (g) -> Na₂O₂ (s)

Sodium peroxide dissolves in water to form a basic/alkaline solution of Sodium hydroxide. Oxygen is produced.

Chemical equations

Sodium Oxide + Water -> Sodium hydroxide + Oxygen

2Na₂O₂ (s) + 2H₂O (l) -> 4NaOH (aq) + O₂ (l)

III. Burning Calcium

Procedure

(a)Using a pair of tongs hold the piece of calcium on a bunsen flame.

Observe.

Place the products in a beaker containing about 2cm3 of water. Test the solution/mixture using litmus papers

(b)Using a pair of tongs hold another piece of calcium on a Bunsen flame. Quickly lower it into a gas jar containing Oxygen gas .Observe.

Place about 2cm3 of water. Swirl.

Test the solution/mixture using litmus papers

Observations

(a)In air

Calcium burns with difficulty producing a faint red flame in air forming a white solid. Blue litmus paper remains blue. Red litmus paper turns blue

(b) In pure Oxygen

Calcium burns with difficulty producing a less faint red flame Oxygen forming a white solid. Blue litmus paper remains blue. Red litmus paper turns blue

Explanation

(a)Calcium burns in air forming white calcium Oxide. Calcium Oxide coat/cover the calcium preventing further burning.

Chemical equations

Calcium + Oxygen/air -> calcium Oxide

2Ca(s) + O₂(g) -> 2CaO(s)

Small amount of Calcium Oxide dissolves in water to form a basic/alkaline solution of Calcium hydroxide. The common name of Calcium hydroxide is lime water.

Chemical equations

Calcium Oxide + Water -> Calcium hydroxide

CaO(s) + H₂O (l) -> Ca (OH)₂ (aq)

Burning Iron

Procedure

(a)Using a pair of tongs hold the piece of Iron wool/steel wire on a Bunsen flame.

Observe.

Place the products in a beaker containing about 2cm3 of water. Test the solution/mixture using litmus papers

(b)Using a pair of tongs hold another piece of Iron wool/steel wire on a Bunsen flame.

Quickly lower it into a gas jar containing Oxygen gas .Observe.

Place about 2cm3 of water. Swirl. Test the solution/mixture using litmus papers

Observations

(a)In air

Iron wool/steel wire burns producing an Orange flame in air forming a brown solid. Blue litmus paper remains blue. Red litmus paper turns faint blue

(b) In pure Oxygen

Iron wool/steel wire burns producing a golden Orange flame in Oxygen forming a Brown solid. Blue litmus paper remains blue. Red litmus paper turns faint blue

Explanation

(a)Iron burns in air forming brown Iron (III) Oxide

Chemical equations

Iron + Oxygen/air -> Iron (III) Oxide

4Fe(s) + 3O₂ (g) -> 2Fe₂O₃(s)

Very small amount of Iron (III) Oxide dissolves in water to form a weakly basic/alkaline brown solution of Iron (III) hydroxide.

Chemical equations

Calcium Oxide + Water -> Iron (III) hydroxide

Fe₂O₃(s) + 3H₂O (l) -> 2Fe (OH)₃ (s)

Burning Copper

Procedure

(a)Using a pair of tongs hold the piece of copper turnings/shavings on a Bunsen flame.

Observe.

Place the products in a beaker containing about 2cm3 of water. Test the solution/mixture using litmus papers

(b)Using a pair of tongs hold another piece of Copper turnings/shavings on a Bunsen flame. Quickly lower it into a gas jar containing Oxygen gas .Observe.

Place about 2cm3 of water. Swirl. Test the solution/mixture using litmus papers

Observations

(a)In air

Copper turnings/shavings burns with difficulty producing a green flame in air forming a black solid. Blue litmus paper remains blue. Red litmus paper turns faint blue

(b) In pure Oxygen

Copper turnings/shavings burns less difficulty producing a green flame in Oxygen forming a Brown solid. Blue litmus paper remains blue. Red litmus paper turns faint blue

Explanation

(a)Copper burns in air forming black Copper (II) Oxide

Chemical equations

Copper + Oxygen/air -> Copper (II) Oxide

2 Cu(s) + O₂ (g) -> 2CuO(s)

Very small amount of Copper (II) Oxide dissolves in water to form a weakly basic/alkaline blue solution of Copper (II) hydroxide.

Chemical equations

Copper (II) Oxide + Water -> Copper (II) hydroxide

CuO(s) + H₂O (l) -> Cu (OH)₂ (s)

(i)Reaction of non metals with Oxygen/air

The following experiments show the reaction of non metals with Oxygen and air.

Burning Carbon

Procedure

(a)Using a pair of tongs hold a dry piece of charcoal on a Bunsen flame.

Observe.

Place the products in a beaker containing about 2cm3 of water. Test the solution/mixture using litmus papers

(b)Using a pair of tongs hold another piece of dry charcoal on a Bunsen flame. Quickly lower it into a gas jar containing Oxygen gas .Observe.

Place about 2cm3 of water. Swirl. Test the solution/mixture using litmus papers

Observations

-Carbon chars then burns with a blue flame

-Colourless and odorless gas produced

-Solution formed turn blue litmus paper faint red.

Red litmus paper remains red.

Explanation

Carbon burns in air and faster in Oxygen with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas.

Carbon burns in limited supply of air with a blue non-sooty/non-smoky flame forming Carbon (IV) oxide gas.

Carbon (IV) oxide gas dissolves in water to form weak acidic solution of Carbonic (IV) acid.

Chemical Equation

Carbon + Oxygen -> Carbon (IV) oxide

(excess air/oxygen)

C(s) + O₂ (g) -> CO₂ (g) (in excess air)

Carbon + Oxygen -> Carbon (II) oxide

(limited air/oxygen)

2C(s) + O₂ (g) -> 2CO (g) (in limited air)

Carbon (IV) oxide + Water -> Carbonic (IV) acid

CO₂ (g) + H₂O (l) -> H₂CO₃ (aq) (very weak acid)

Burning Sulphur

Procedure

(a)Using a deflagrating spoon place sulphur powder on a Bunsen flame.

Observe.

Place the products in a beaker containing about 3cm3 of water. Test the solution/mixture using litmus papers

(b) Using a deflagrating spoon place sulphur powder on a Bunsen flame. Slowly lower it into a gas jar containing Oxygen gas. Observe.

Place about 5cm3 of water. Swirl. Test the solution/mixture using litmus papers.

Observations

-Sulphur burns with a blue flame

-Gas produced that has pungent choking smell

-Solution formed turn blue litmus paper faint red.

Red litmus paper remains red.

Explanation

Sulphur burns in air and faster in Oxygen with a blue non-sooty/non-smoky flame forming Sulphur (IV) oxide gas.

Sulphur (IV) oxide gas dissolves in water to form weak acidic solution of Sulphuric (IV) acid.

Chemical Equation

Sulphur + Oxygen -> Sulphur (IV) oxide

S(s) + O₂ (g) -> SO₂ (g) (in excess air)

Sulphur (IV) oxide + Water -> Sulphuric (IV) acid

SO₂ (g) + H₂O (l) -> H₂SO₃ (aq) (very weak acid)

III. Burning Phosphorus

Procedure

(a)Remove a small piece of phosphorus from water and using a deflagrating spoon (with a lid cover) places it on a Bunsen flame.

Observe.

Carefully put the burning phosphorus to cover gas jar containing about 3cm3 of water. Test the solution/mixture using litmus papers

(b) Remove another small piece of phosphorus from water and using a deflagrating spoon (with a lid cover) place it on a Bunsen flame.

Slowly lower it into a gas jar containing Oxygen gas with about 5 cm3 of water. Observe.

Swirl. Test the solution/mixture using litmus papers.

Observations

-Phosphorus catches fire before heating on Bunsen flame

-Dense white fumes of a gas produced that has pungent choking poisonous smell

-Solution formed turn blue litmus paper faint red.

Red litmus paper remains red.

Explanation

Phosphorus is stored in water. On exposure to air it instantaneously fumes then catch fire to burn in air and faster in Oxygen with a yellow flame producing dense white acidic fumes of Phosphorus (V) oxide gas.

Phosphoric (V) oxide gas dissolves in water to form weak acidic solution of Phosphoric (V) acid.

Chemical Equation

Phosphorus + Oxygen -> Phosphorous (V) oxide

4P(s) + 5O₂ (g) -> 2P₂O₅(s)

Phosphorous (V) oxide + Water -> Phosphoric (V) acid

P₂O₅(s) + 3H₂O (l) -> 2H₃PO₄ (aq) (very weak acid)

(e) Reactivity series/competition for combined Oxygen.

The reactivity series is a list of elements/metals according to their affinity for oxygen.

Some metals have higher affinity for Oxygen than others.

A metal/element with higher affinity for oxygen is placed higher/on top of the one less affinity.

The complete reactivity series of metals/elements

Most reactive

Element/Metal

Symbol

Potassium

Sodium

Calcium

Magnesium

Aluminum

Carbon

Zinc

Iron

Tin

Lead

Hydrogen

Copper

Mercury

Silver

Gold

Platinum

Least reactive

Metals compete for combined Oxygen. A metal/element with higher affinity for oxygen removes Oxygen from a metal lower in the reactivity series/less affinity for Oxygen.

When a metal/element gains/acquire Oxygen, the process is called Oxidation.

When metal/element donate/lose Oxygen, the process is called Reduction.

An element/metal/compound that undergoes Oxidation is called Reducing agent.

An element/metal/compound that undergoes Reduction is called Oxidizing agent.

A reaction in which both Oxidation and Reduction take place is called a Redox reaction.

Redox reaction between Magnesium and copper (II) Oxide

Procedure

Place about 2g of copper (II) oxide in a crucible with a lid. Place another 2g of Magnesium powder into the crucible. Mix thoroughly.

Cover the crucible with lid. Heat strongly for five minutes.

Allow the mixture to cool. Open the lid. Observe.

Observation

Colour change from black to brown. White solid power formed.

Explanation

Magnesium is higher in the reactivity series than Copper. It has therefore higher affinity for Oxygen than copper.

When a mixture of copper (II) oxide and Magnesium is heated, Magnesium reduces copper (II) oxide to brown copper metal and itself oxidized to Magnesium oxide. Magnesium is the reducing agent because it undergoes oxidation process.

Copper (II) oxide is the oxidizing agent because it undergoes redox reduction process.

The mixture should be cooled before opening the lid to prevent hot brown copper from being reoxidized back to black copper (II) oxide.

The reaction of Magnesium and Copper (II) oxide is a reaction

Chemical equation

Copper (II) oxide + Magnesium -> Magnesium oxide + Copper

(black) (white ash/solid) (brown)

CuO(s) + Mg(s) -> MgO(s) + Cu(s)

(Oxidizing Agent) (Reducing Agent)

Zinc (II) oxide + Magnesium -> Magnesium oxide + Zinc

(yellow when hot) (white ash/solid) (grey)

ZnO(s) + Mg(s) -> MgO(s) + Zn(s)

(Oxidizing agent) (Reducing agent)

Zinc (II) oxide + Carbon -> Carbon (IV) oxide gas + Zinc

(yellow when hot) (colourless gas) (grey)

ZnO(s) + C(s) -> CO₂ (g) + Zn(s)

(Oxidizing agent) (Reducing agent)

The reactivity series is used during extraction of metals from their ore. An ore is a rock containing mineral element which can be extracted for commercial purposes. Most metallic ores occur naturally as:

(i) oxides combined with Oxygen

(ii) sulphides combined with Sulphur

(iii) carbonates combined with carbon and Oxygen.

Metallic ores that naturally occur as metallic sulphides are first roasted in air to form the corresponding oxide. Sulphur (IV) oxide gas is produced. e.g.

Copper (I) sulphide + Oxygen -> Copper (I) Oxide + Sulphur (IV) oxide

Cu₂S(s) + O₂ (g) -> 2Cu(s) + SO₂ (g)

Zinc (II) sulphide + Oxygen -> Zinc (II) Oxide + Sulphur (IV) oxide

ZnS(s) + O₂ (g) -> Zn(s) + SO₂ (g)

Lead (II) sulphide + Oxygen -> Lead (II) Oxide + Sulphur (IV) oxide

PbS(s) + O₂ (g) -> Pb(s) + SO₂ (g)

Iron (II) sulphide + Oxygen -> Iron (II) Oxide + Sulphur (IV) oxide

FeS(s) + O₂ (g) -> Fe(s) + SO₂ (g)

Metallic ores that naturally occur as metallic carbonates are first heated in air. They decompose/split to form the corresponding oxide and produce Carbon (IV) oxide gas. .e.g.

Copper (II) carbonate -> Copper (II) oxide + Carbon (IV) oxide

CuCO₃(s) -> CuO(s) + CO₂ (g)

Zinc (II) carbonate -> Zinc (II) oxide + Carbon (IV) oxide

ZnCO₃(s) -> ZnO(s) + CO₂ (g)

Lead (II) carbonate -> Lead (II) oxide + Carbon (IV) oxide

PbCO₃(s) -> PbO(s) + CO₂ (g)

Iron (II) carbonate -> Iron (II) oxide + Carbon (IV) oxide

FeCO₃(s) -> FeO(s) + CO₂ (g)

Metallic ores

WATER AND HYDROGEN

A.WATER

Pure water is a colourless, odorless, tasteless, neutral liquid. Pure water does not exist in nature but naturally in varying degree of purity. The main sources of water include rain, springs, borehole, lakes, seas and oceans:

Water is generally used for the following purposes:

(i) Drinking by animals and plants.

(ii) Washing clothes.

(iii) Bleaching and dyeing.

(iv) Generating hydroelectric power.

(v) Cooling industrial processes.

Water dissolves many substances/solutes.

It is therefore called universal solvent.

It contains about 35% dissolved Oxygen which support aquatic fauna and flora.

Water naturally exists in three phases/states solid ice, liquid water and gaseous water vapour.

The three states of water are naturally interconvertible.

The natural interconvertion of the three phases/states of water forms the water cycle.

Precipitation

Evaporation(Water in gaseous state)

Liquid water in land, lakes, seas and oceans use the solar/sun energy to evaporate/vapourize to form water vapour/gas. Solar/sun energy is also used during transpiration by plants and respiration by animals.

During evaporation, the water vapour rises up the earth’s surface. Temperatures decrease with height above the earth surface increase. Water vapour therefore cools as it rises up. At a height where it is cold enough to below 373Kelvin/100^oC Water vapour looses enough energy to form tiny droplets of liquid.

The process by which a gas/water vapour changes to a liquid is called condensation/liquidification.

On further cooling, the liquid looses more energy to form ice/solid. The process by which a liquid/water changes to a ice/solid is called freezing/solidification. Minute/tiny ice/solid particles float in the atmosphere and coalesce/join together to form clouds. When the clouds become too heavy they fall to the earth’s surface as rain/snow as the temperature increase with the fall.

Interconversion of the three phases/states water

Solid/Ice

Liquid/Water

Gas/water vapour

Evaporation Liquidification/

/boiling/Vapourization condensation

Melting Freezing liquidification

Solidification

Pure water has:

(i) fixed/constant/sharp freezing point/melting point of 273K/0^oC

(ii) fixed/constant/sharp boiling point of 373K/100^oC at sea level/1 atmosphere pressure

(iii) fixed density of 1gcm^-3

This is the criteria of identifying pure/purity of water.

Whether a substance is water can be determined by using the following methods:

a) To test for presence of water using anhydrous copper (II) suphate (VI)

Procedure

Put about 2g of anhydrous copper (II) sulphate (VI) crystals into a clean test tube. Add three drops of tap water. Repeat the procedure using distilled water.

Observation

Colour changes from white to blue

Explanation

Anhydrous copper (II) sulphate (VI) is white. On adding water, anhydrous copper (II) sulphate (VI) gains/reacts with water to form hydrated copper (II) sulphate (VI).

Hydrated copper (II) sulphate (VI) is blue.Hydrated copper (II) sulphate (VI) contains water of crystallization.

The change of white anhydrous copper (II) sulphate (VI) to bluehydrated copper (II) sulphate (VI) is a confirmatory test for the presence of water

Chemical equation

Anhydrous Hydrated copper (II) sulphate (VI) + Water -> copper (II) sulphate (VI)

(white) (blue)

CuSO₄(s) + 5H2O (l) ->CuSO₄.5H₂O(s)

b) To test for presence of water using anhydrous cobalt (II) chloride

Procedure

Put about 5cm3 of water into a clean test tube.

Dip a dry anhydrouscobalt (II) chloride paper into the test tube.

Repeat the procedure using distilled water.

Observation

Colour changes from blue to pink

Explanation

Anhydrous cobalt (II) chloride is blue. On adding water, anhydrous cobalt (II) chloride gains/reacts with water to form hydrated cobalt (II) chloride.

Hydrated cobalt (II) chloride is pink.

Hydrated cobalt (II) chloride contains water of crystallization.

The change of blue anhydrous cobalt (II) chloride to pinkhydrated cobalt (II) chloride is a confirmatory test for the presence of water Chemical equation.

Anhydrous Hydrated cobalt (II) chloride + Water -> cobalt (II) chloride

(Blue) (pink)

CoCl₂ (s) + 5H2O (l) ->CoCl₂.5H₂O(s)

Burning a candle in air

Most organic substances/fuels burn in air to produce water. Carbon (IV) oxide gas is also produced if the air is sufficient/excess.

Procedure

Put about 2g of anhydrous copper (II) sulphate (VI) crystals in a boiling tube.

Put about 5cm3 of lime water in a boiling tube.

Light a small candle stick. Place it below an inverted thistle/filter funnel

Collect the products of the burning candle by setting the apparatus as below

Set up of apparatus

Observation

The sanction pump pulls the products of burning into the inverted funnel. Colour of anhydrous copper (II) sulphate (VI) changes from white to blue. A white precipitate is formed in the lime water/calcium hydroxide.

Explanation

When a candle burn it forms a water and carbon (IV) oxide.

Water turns anhydrous copper (II) sulphate (VI) changes from white to blue.

Carbon (IV) oxide gasforms white precipitate when bubbled in lime water/calcium hydroxide.

Since:

(i) hydrogen in the wax burn to form water

Hydrogen + Oxygen -> Water

(from candle) (from the air)

2H₂ (g) + O₂ (g) -> 2H₂O (g/l)

(ii) carbon in the wax burn to form carbon (IV) oxide

Hydrogen + Oxygen -> Water

(from candle) (from the air)

C(s) + O₂(g) -> CO₂ (g)

The candle before burning therefore contained only Carbon and Hydrogenonly. A compound made up of hydrogen and carbon is called Hydrocarbon.

A candle is a hydrocarbon.

Other hydrocarbons include: Petrol, diesel, Kerosene, and Laboratory gas. Hydrocarbons burn in air to form water and carbon (IV) oxide gas.

Hydrocarbons + Oxygen -> Water + Oxygen

Water pollution

Water pollution takes place when undesirable substances are added into the water. Sources of water pollution include:

(i)Industrial chemicals being disposed into water bodies like rivers, lakes and oceans.

(ii)Discharging untreated /raw sewage into water bodies.

(iii)Leaching of insecticides/herbicides form agricultural activities into water bodies.

(iv)Discharging non-biodegradable detergents after domestic and industrial use into water bodies.

(v)Petroleum oil spilling by ships and oil refineries

(vi)Toxic/poisonous gases from industries dissolving in rain.

(vii) Acidic gases from industries dissolving in rain to form “acid rain”

(viii)Discharging hot water into water bodies. This reduces the quantity of dissolved Oxygen in the water killing the aquatic fauna and flora.

Water pollution can be reduced by:

(i) Reducing the use of agricultural fertilizers and chemicals in agricultural activities.

(ii) Use of biological control method instead of insecticides and herbicides

(iii) Using biodegradable detergents

REACTION OF WATER WITH METALS.

Some metals react with water while others do not. The reaction of metals with water depends on the reactivity series. The higher the metal in the reactivity series the more reactive the metal with water .The following experiments shows the reaction of metals with cold water and water vapour/steam.

(a)Reaction of sodium/ potassium with cold water:

Procedure

Put about 500cm3 of water in a beaker. Add three drops of phenolphthalein indicator/litmus solution/universal indicator solution/methyl orange indicator into the water.

Cut a very small piece of sodium .Using a pair of forceps put the metal into the water.

Observation

Sodium melts to a silvery ball that floats and darts on the surface decreasing in size. Effervescence/fizzing/ bubbles of colourless gas produced.

Colour of phenolphthalein turns pink

Colour of litmus solution turns blue

Colour of methyl orange solution turns Orange

Colour of universal indicator solution turns blue

Explanation

Sodium is less dense than water. Sodium floats on water and vigorously reacts to form an alkaline solution of sodium hydroxide and producing hydrogen gas. Sodium is thus stored in paraffin to prevent contact with water.

Chemical equation

Sodium + Water -> Sodium hydroxide + Hydrogen gas

2Na(s) + 2H₂O (l) -> 2NaOH (aq) + H₂(g)

To collect hydrogen gas, Sodium metal is forced to sink to the bottom of the trough/beaker by wrapping it in wire gauze/mesh.

Potassium is more reactive than Sodium. On contact with water it explodes/burst into flames. An alkaline solution of potassium hydroxide is formed and hydrogen gas

Chemical equation

Potassium + Water -> Potassium hydroxide + Hydrogen gas

2K(s) + 2H₂O (l) -> 2KOH (aq) + H₂(g)

Caution: Reaction of Potassium with water is very risky to try in a school laboratory.

(b)Reaction of Lithium/ Calcium with cold water:

Procedure

Put about 200cm3 of water in a beaker. Add three drops of phenolphthalein indicator/litmus solution/universal indicator solution/methyl orange indicator into the water.

Cut a small piece of Lithium .Using a pair of forceps put the metal into the water.

Repeat with a piece Calcium metal

Observation

Lithium sinks to the bottom of the water. Rapid effervescence/fizzing/ bubbles of colourless gas produced.

Colour of phenolphthalein turns pink

Colour of litmus solution turns blue

Colour of methyl orange solution turns Orange

Colour of universal indicator solution turns blue

Explanation

Lithium and calcium are denser than water. Both sink in water and vigorously react to form an alkaline solution of Lithium hydroxide / calcium hydroxide and producing hydrogen gas. Lithium is more reactive than calcium. It is also stored in paraffin like Sodium to prevent contact with water.

Chemical equation

Lithium + Water -> Lithium hydroxide + Hydrogen gas

2Li(s) + 2H₂O (l) -> 2LiOH (aq) + H₂(g)

Calcium + Water -> Calcium hydroxide + Hydrogen gas

Ca(s) + 2H₂O (l) -> Ca (OH)₂(aq) + H₂ (g)

(c) Reaction of Magnesium/Zinc/ Iron with Steam/water vapour:

Procedure method1

Place some wet sand or cotton/glass wool soaked in water at the bottom of an ignition/hard glass boiling tube.

Polish magnesium ribbon using sand paper.

Coil it at the centre of the ignition/hard glass boiling tube.

Set up the apparatus as below.

Heat the wet sand or cotton/glass wool soaked in water gently to:

(i) Drive away air in the ignition/hard glass boiling tube.

(ii) Generate steam

Heat the coiled ribbon strongly using another burner. Repeat the experiment using Zinc powder and fresh Iron filings.

Set up of apparatus

Observations

(i)With Magnesium ribbon:

The Magnesium glows with a bright flame (and continues to burn even if heating is stopped)

White solid /ash formed

White solid /ash formed dissolve in water to form a colourless solution

Colourless gas produced/collected that extinguish burning splint with “pop sound”

(ii) With Zinc powder:

The Zinc powder turns red hot on strong heating

Yellow solid formed that turn white on cooling

White solid formed on cooling does not dissolve in water.

(iii)With Iron fillings:

The Iron fillings turn red hot on strong heating

Dark blue solid formed

Dark blue solid formed does not dissolve in water.

Procedure method 2

Put some water in a round bottomed flask

Polish magnesium ribbon using sand paper.

Coil it at the centre of a hard glass tube

Set up the apparatus as below.

Heat water strongly to boil so as to:

(i) drive away air in the glass tube.

(ii) generate steam

Heat the coiled ribbon strongly using another burner. Repeat the experiment using Zinc powder and fresh Iron filings.

Observations

(i)With Magnesium ribbon:

The Magnesium glows with a bright flame (and continues to burn even if heating is stopped)

White solid /ash formed

White solid /ash formed dissolve in water to form a colourless solution

Colourless gas produced/collected that extinguish burning splint with “pop sound”

(ii) With Zinc powder:

The Zinc powder turns red hot on strong heating

Yellow solid formed that turn white on cooling

White solid formed on cooling does not dissolve in water.

(iii)With Iron fillings:

The Iron fillings turn red hot on strong heating

Dark blue solid formed

Dark blue solid formed does not dissolve in water.

Explanations

(a)Hot magnesium burn vigorously in steam. The reaction is highly exothermic generating enough heat/energy to proceed without further heating.

White Magnesium oxide solid/ash is left as residue.

Hydrogen gas is produced .It extinguishes a burning splint with a “pop sound”.

Chemical Equation

Magnesium + Steam -> Magnesium oxide + Hydrogen

Mg(s) + H₂O(g) -> MgO(s) + H₂(g)

Magnesium oxide reacts /dissolves in water to form an alkaline solution of Magnesium oxide

Chemical Equation

Magnesium oxide + Water -> Magnesium hydroxide

MgO(s) + H₂O(l) -> Mg(OH)₂ (aq)

(b)Hot Zinc react vigorously in steam forming yellow Zinc oxide solid/ash as residue which cools to white.

Hydrogen gas is produced .It extinguishes a burning splint with a “pop sound”.

Chemical Equation

Zinc + Steam -> Zinc oxide + Hydrogen

Zn(s) + H₂O(g) -> ZnO(s) + H₂(g)

Zinc oxide does not dissolve in water.

(c)Hot Iron reacts with steam forming dark blue tri iron tetra oxide solid/ash as residue.

Hydrogen gas is produced .It extinguishes a burning splint with a “pop sound”.

Chemical Equation

Iron + Steam -> Tri iron tetra oxide + Hydrogen

2Fe(s) + 4H₂O(g -> Fe₂O₄(s) + 4H₂(g)

Tri iron tetra oxide does not dissolve in water.

(d)Aluminum reacts with steam forming an insoluble coat/cover of impervious layer of aluminum oxide on the surface preventing further reaction.

(e) Lead, Copper, Mercury, Silver, Gold and Platinum do not react with either water or steam.

HYDROGEN

Occurrence

Hydrogen does not occur free in nature. It occurs as Water and in Petroleum.

School laboratory Preparation

Procedure

Put Zinc granules in a round/flat/conical flask. Add dilute sulphuric (VI) /Hydrochloric acid.

Add about 3cm3 of copper (II) sulphate (VI) solution.

Collect the gas produced over water as in the set up below.

Discard the first gas jar. Collect several gas jars.

Observation/Explanation

Zinc reacts with dilute sulphuric (VI)/hydrochloric acid to form a salt and produce hydrogen gas.

When the acid comes into contact with the metal, there is rapid effervescence/ bubbles /fizzing are produced and a colourless gas is produced that is collected:

(i) Over water because it is insoluble in water

(ii) Through downward displacement of air/upward delivery because it is less dense than air.

The first gas jar is impure. It contains air that was present in the apparatus.

Copper (II) sulphate (VI) solution act as catalyst.

Chemical equation

(a) Zinc + Hydrochloric acid -> Zinc chloride + Hydrogen

Zn(s) + 2HCl (aq) -> ZnCl₂ (aq) + H₂ (g)

Ionic equation

Zn (s) + 2H⁺(aq) -> Zn²⁺(aq) + H₂(g)

Zinc + Sulphuric (VI) acid -> Zinc Sulphate (VI) + Hydrogen

Zn(s) + H₂SO₄ (aq) -> ZnSO₄ (aq) + H₂ (g)

Ionic equation

Zn (s) + 2H⁺(aq) -> Zn²⁺(aq) + H₂(g)

(b) Chemical equation

Magnesium + Hydrochloric acid -> Magnesium chloride + Hydrogen

Mg(s) + 2HCl (aq) -> MgCl₂ (aq) + H₂(g)

Ionic equation

Mg (s) + 2H⁺(aq) -> Mg²⁺(aq) + H₂(g)

Magnesium + Sulphuric (VI) acid -> Magnesium Sulphate(VI) + Hydrogen

Mg(s) + H₂SO₄ (aq) -> MgSO₄ (aq) + H₂(g)

Ionic equation

Mg (s) + 2H⁺(aq) -> Mg²⁺(aq) + H₂(g)

(c) Chemical equation

Iron + Hydrochloric acid -> Iron (II) chloride + Hydrogen

Fe(s) + 2HCl (aq) -> FeCl₂ (aq) + H₂ (g)

Ionic equation

Fe (s) + 2H⁺(aq) -> Fe²⁺(aq) + H₂(g)

Iron + Sulphuric (VI) acid -> Iron (II) Sulphate (VI) + Hydrogen

Fe(s) + H₂SO₄ (aq) -> FeSO₄ (aq) + H₂ (g)

Ionic equation

Fe (s) + 2H⁺(aq) -> Fe²⁺(aq) + H₂(g)

Note

Hydrogen cannot be prepared from reaction of:

(i)Nitric (V) acid and a metal. Nitric (V) acid is a strong oxidizing agent. It oxidizes hydrogen gas to water.

(ii) Dilute sulphuric (VI) acid with calcium/Barium/Lead because Calcium sulphate (VI), Barium sulphate (VI) and Lead (II) sulphate (VI) salts formed are insoluble. Once formed, they cover/coat the unreacted calcium/Barium/Lead stopping further reaction and producing very small amount/volume of hydrogen gas.

(iii) Dilute acid with sodium/potassium. The reaction is explosive.

Properties of Hydrogen gas

(a)Physical properties

Hydrogen is a neutral, colourlessand odorless gas. When mixed with air it has a characteristic pungent choking smell
It is insoluble in water thus can be collected over water.
It is the lightest known gas. It can be transferred by inverting one gas jar over another.

(b)Chemical properties

(i)Burning

Hydrogen does not support burning/combustion. When a burning splint is inserted into a gas jar containing Hydrogen, the flame is extinguished /put off.
Pure dry hydrogen burn with a blue quiet flame to form water. When a stream of pure dry hydrogen is ignited, it catches fire and continues to burn with a blue flame.

III. Impure (air mixed with) hydrogen burns with an explosion. Small amount/ volume of air mixed with hydrogen in a test tube produce a small explosion as a “pop” sound. This is the confirmatory test for the presence of Hydrogen gas. A gas that burns with a “pop” sound is confirmed to be Hydrogen.

(ii)Redox in terms of Hydrogen transfer

Redox can also be defined in terms of Hydrogen transfer.

(i)Oxidation is removal of Hydrogen

(ii)Reduction is addition of Hydrogen

(iii)Redox is simultaneous addition and removal of Hydrogen

Example

When a stream of dry hydrogen gas is passed through black copper (II) oxide, hydrogen gas gains the oxygen from copper (II) oxide.

Black copper (II) oxide is reduced to brown copper metal.

Black copper (II) oxide thus the Oxidizing agent.

Hydrogen gas is oxidized to Water. Hydrogen is the Reducing agent.

Set up of apparatus

(a)Chemical equation

(i) In glass tube

Copper (II) Oxide + Hydrogen -> Copper + Hydrogen gas

(oxidizing agent) (reducing agent)

(black) (brown)

CuO (s) + H₂(g) -> Cu(s) + H₂O(l)

(ii) when excess Hydrogen is burning.

Oxygen + Hydrogen -> Water

O₂ (g) + 2H₂ (g) -> 2H₂O (l)

(b)Chemical equation

(i) In glass tube

Lead (II) Oxide + Hydrogen -> Lead + Hydrogen gas

(oxidizing agent) (reducing agent)

(brown when hot/ (grey)

yellow when cool)

PbO (s) + H₂ (g) -> Pb(s) + H₂O (l)

(ii) when excess Hydrogen is burning.

Oxygen + Hydrogen -> Water

O₂ (g) + 2H₂ (g) -> 2H₂O(l)

(c)Chemical equation

(i) In glass tube

Iron (III) Oxide + Hydrogen -> Iron + Hydrogen gas

(oxidizing agent) (reducing agent)

(Dark grey) (grey)

Fe₂O₃ (s) + 3H₂ (g) -> Fe(s) + 3H₂O (l)

(ii) when excess Hydrogen is burning.

Oxygen + Hydrogen -> Water

O₂ (g) + 2H₂ (g) -> 2H₂O (l)

(iii) Water as an Oxide as Hydrogen

Burning is a reaction of an element with Oxygen. The substance formed when an element burn in air is the oxide of the element. When hydrogen burns, it reacts/ combines with Oxygen to form the oxide of Hydrogen. Theoxide of Hydrogen is called water. Hydrogen is first dried because a mixture of Hydrogen and air explode. The gas is then ignited .The products condense on a cold surface/flask containing a freezing mixture. A freezing mixture is a mixture of water and ice.

The condensed products are collected in a receiver as a colourless liquid.

Tests

(a) When about 1g of white anhydrous copper (II) sulphate (VI) is added to a sample of the liquid, it turns to blue. This confirms the liquid formed is water.

(b) When blue anhydrous cobalt (II) chloride paper is dipped in a sample of the liquid, it turns to pink. This confirms the liquid formed is water.

(c)When the liquid is heated to boil, its boiling point is 100^oC at sea level/one atmosphere pressure. This confirms the liquid is pure water.

Uses of Hydrogen gas

Hydrogenation/Hardening of unsaturated vegetable oils to saturated fats/margarine.

When Hydrogen is passed through unsaturated compounds in presence of Nickel catalyst and about 150^oC, they become saturated. Most vegetable oil is unsaturated liquids at room temperature. They become saturated and hard through hydrogenation.

In weather forecast balloons.

Hydrogen is the lightest known gas. Meteorological data is collected for analysis by sending hydrogen filled weather balloons to the atmosphere. The data collected is then used to forecast weather conditions.

In the Haber process for the manufacture of Ammonia

Hydrogen is mixed with Nitrogen in presence of Iron catalyst to form Ammonia gas. Ammonia gas is a very important raw material for manufacture of agricultural fertilizers.

In the manufacture of Hydrochloric acid.

Limited volume/amount of Hydrogen is burnt in excess chlorine gas to form Hydrogen chloride gas. Hydrogen chloride gas is dissolved in water to form Hydrochloric acid. Hydrochloric acid is used in pickling/washing metal surfaces.

As rocket fuel.

Fixed proportions of Hydrogen and Oxygen when ignited explode violently producing a lot of energy/heat. This energy is used to power/propel a rocket to space.

In oxy-hydrogen flame for welding.

A cylinder containing Hydrogen when ignited in pure Oxygen from a second cylinder produces a flame that is very hot. It is used to cut metals and welding.

Sample revision questions

A colourless liquid was added anhydrous copper (II) sulphate (VI) which turned blue.

(a)Why is it wrong to conclude the liquid was pure water?

Anhydrous copper (II) sulphate (VI) test for presence of water. Purity of water is determined from freezing/melting/boiling point.

(b)Write an equation for the reaction that takes place with anhydrous copper (II) sulphate (VI)

Anhydrous copper (II) sulphate (VI) + Water -> hydrated copper (II) sulphate (VI)

CuSO₄(s) + 5H2O (l) -> CuSO₄.5H₂O(s)

(c)(i)Which other compound would achieve the same results asanhydrous copper (II) sulphate (VI)

Anhydrous cobalt (II) chloride/CoCl₂.6H₂O

(ii)Write the equation for the reaction

Anhydrous cobalt (II) chloride + Water -> hydrated cobalt (II) chloride

CoCl₂ (s) + 6H₂O(l) -> CoCl₂.6H₂O (s)

(d)Complete the equation

(i) Sulphur (VI) oxide + Water -> Sulphuric (VI) acid

(ii) Sulphur (IV) oxide + Water -> Sulphuric (IV) acid

(iii) Carbon (IV) oxide + Water -> Carbonic (IV) acid

(iv) Nitrogen (IV) oxide + Water -> Nitric (V) acid

(v) Phosphorus (V) oxide + Water -> Phosphoric (V) acid

(vi) Sodium oxide + Water -> Sodium hydroxide

(vi) Sodium peroxide + Water -> Sodium hydroxide

Metal B reacts with steam. Metal C reacts with cold water. Metal A does not react with water.

(a)Arrange the metals as they should appear in the reactivity series.

(b)A product residue in D which was brown when hot but turned yellow on cooling during the reaction of metal B was formed. Gas E was also evolved. Identify

(i)Metal B Lead/Pb

(ii)Residue D Lead (II) oxide/PbO

(iii)Gas E Hydrogen/H₂

(c)A portion of product residue in D was added dilute nitric (V) acid. Another portion of product residue in D was added dilute sulphuric (VI) acid. State and explain the observations made.

When added dilute nitric (V) acid, D dissolves to form a colourless solution.

Lead (II) Oxide + dilute nitric (V) acid -> Lead (II) nitrate (V) + Water

PbO (s) + 2HNO₃ (aq) -> Pb (NO₃)₂ (aq) + H₂O (l)

When added dilute sulphuric (VI) acid, D does not dissolve. A white suspension/precipitate was formed. Lead(II)Oxide reacts with sulphuric(VI)acid to form insoluble Lead(II)sulphate(VI) that cover/coat unreacted Lead(II)Oxide, stopping further reaction.

Lead (II) Oxide + dilute sulphuric (VI) acid -> Lead (II) sulphate (VI) + Water

PbO (s) + H₂SO₄ (aq) -> PbSO₄ (s) + H₂O (l)

(a) Hydrogen can reduce copper (II) Oxide but not alluminium oxide. Explain

(b) When water reacts with potassium metal the hydrogen produced ignites explosively on the surface of water.

(i) What causes this ignition? (ii) Write an equation to show how this ignition occurs

In an experiment, dry hydrogen gas was passed over hot copper (II) oxide in a combustion tube as shown in the diagram below:

(a) Complete the diagram to show how the other product, substance R could be collected in the laboratory.

(b) Describe how copper could be obtained from the mixture containing copper (II) oxide

The setup below was used to investigate the reaction between metals and water.

(a) Identify solid X and state its purpose

Solid X .…………………………………………………………………..

Purpose …………………………………………………………………..

(b) Write a chemical equation for the reaction that produces the flame. 4. Gas P was passed over heated magnesium ribbon and hydrogen gas was collected as shown in the diagram below:

(i) Name gas P …………………………………………………………………………………………………

(ii) Write an equation of the reaction that takes place in the combustion tube (iii) State one precaution necessary at the end of this experiment

When hydrogen is burnt and the product cooled, the following results are obtained as shown in the diagram below:

(a) Write the equation for the formation of liquid Y

(b) Give a chemical test for liquid Y

Jane set-up the experiment as shown below to collect a gas. The wet sand was heated before

heating Zinc granules

Wet sand

(a) Complete the diagram for the laboratory preparation of the gas (b) Why was it necessary to heat wet sand before heating Zinc granules?

(a) Between N and M which part should be heated first? Explain

(b) Write a chemical equation for the reaction occurring in the combustion tube.

The set-up below was used to investigate electrolysis of a certain molten compound;-

(a) Complete the circuit by drawing the cell in the gap left in the diagram

(b) Write half-cell equation to show what happens at the cathode

Hydrogen can be prepared by reacting zinc with dilute hydrochloric acid.
a) Write an equation for the reaction.
b) Name an appropriate drying agent for hydrogen gas.
c) Explain why copper metal cannot be used to prepare hydrogen gas.
d) Hydrogen burns in oxygen to form an oxide.

(i) Write an equation for the reaction.

(ii) State two precautions that must be taken before the combustion begins and at the end of the combustion.

e) Give two uses of hydrogen gas.
f) When zinc is heated to redness in a current of steam, hydrogen gas is obtained. Write an equation for the reaction.
g) Element Q reacts with dilute acids but not with cold water. Element R does not react with dilute acids. Elements S displaces element P from its oxide. P reacts with cold water. Arrange the four elements in order of their reactivity, starting with the most reactive.
h) Explain how hydrogen is used in the manufacture of margarine.

a) The set-up below is used to investigate the properties of hydrogen.

On the diagram, indicate what should be done for the reaction to occur
Hydrogen gas is allowed to pass through the tube for some time before it is lit. Explain

iii) Write an equation for the reaction that occurs in the combustion tube iv) When the reaction is complete, hydrogen gas is passed through the apparatus until they cool down. Explain

v) What property of hydrogen is being investigated?
vi) What observation confirms the property stated in (v) above?

vii) Why is zinc oxide not used to investigate this property of hydrogen gas?

The set up below was used to collect gas K, produced by the reaction between water and

calcium metal.

(a) Name gas K ……………………………………………………………..

(b) At the end of the experiment, the solution in the beaker was found to be a weak base. Explain why the solution is a weak base

ACIDS, BASES AND INDICATORS

INTRODUCTION TO ACIDS, BASES AND INDICATORS

In a school laboratory:

(i)An acid may be defined as a substance that turns litmus red.

(ii)A base may be defined as a substance that turns litmus blue.

Litmus is lichen found mainly in West Africa. It changes its colour depending on whether the solution it is in, is basic/alkaline or acidic. It is thus able to identify/show whether another substance is an acid, base or neutral.

(iii)An indicator is a substance that shows whether another substance is a base/alkaline,acid or neutral.

Common naturally occurring acids include:

Name of acid	Occurrence
1.Citric acid	Found in ripe citrus fruits like passion fruit/oranges/lemon
2.Tartaric acid	Found in grapes/baking powder/health salts
3.Lactic acid	Found in sour milk
4.Ethanoic acid	Found in vinegar
5.Methanoic acid	Present in ants, bees stings
6.Carbonic acid	Used in preservation of fizzy drinks like coke, Lemonade, Fanta
7.Butanoic acid	Present in cheese
8.Tannic acid	Present in tea

Most commonly used acids found in a school laboratory are not naturally occurring. They are manufactured. They are called mineral acids.

Common mineral acids include:

Name of mineral acid	Common use
Hydrochloric acid (HCl)	Used to clean/pickling surface of metals Is found in the stomach of mammals/human beings
Sulphuric(VI) acid (H₂SO₄)	Used as acid in car battery, making battery, making fertilizers
Nitric(V)acid (HNO₃)	Used in making fertilizers and explosives

Mineral acids are manufactured to very high concentration. They are corrosive (causes painful wounds on contact with the skin) and attack/reacts with garments/clothes/metals.

In a school laboratory, they are mainly used when added a lot of water. This is called diluting. Diluting ensures the concentration of the acid is safely low.

Bases are opposite of acids. Most bases do not dissolve in water.

Bases which dissolve in water are called alkalis.

Common alkalis include:

Name of alkali	Common uses
Sodium hydroxide (NaOH)	Making soaps and detergents
Potassium hydroxide(KOH)	Making soaps and detergents
Ammonia solution(NH₄OH)	Making fertilizers, softening hard water

Common bases (which are not alkali) include:

Name of base	Common name
Magnesium oxide/hydroxide	Anti acid to treat indigestion
Calcium oxide	Making cement and neutralizing soil acidity

Indicators are useful in identifying substances which look-alike.

An acid-base indicator is a substance used to identify whether another substance is alkaline or acidic.

An acid-base indicator works by changing to different colors in neutral, acidic and alkaline solutions/dissolved in water.

Experiment: To prepare simple acid-base indicator

Procedure

(a)Place some flowers petals in a mortar. Crush them using a pestle. Add a little sand to assist in crushing.

Add about 5cm3 of propanone/ethanol and carefully continue grinding.

Add more 5cm3 of propanone/ethanol and continue until there is enough extract in the mortar.

Filter the extract into a clean 100cm3 beaker.

(b)Place 5cm3 of filtered wood ash, soap solution, ammonia solution, sodium hydroxide, hydrochloric acid, distilled water, sulphuric (VI) acid, sour milk, sodium chloride, toothpaste and calcium hydroxide into separate test tubes.

(c)Put about three drops of the extract in (a)to each test tube in (b). Record the observations made in each case.

Sample observations

Solution mixture	Colour on adding indicator extract	Nature of solution
wood ash	green	Base/alkaline
soap solution	green	Basic/alkaline
ammonia solution	green	Basic/alkaline
sodium hydroxide	green	Basic/alkaline
hydrochloric acid	Red	Acidic
distilled water	orange	Neutral
sulphuric(VI)acid	Red	Acidic
sour milk	green	Basic/alkaline
sodium chloride	orange	Neutral
Toothpaste	green	Basic/alkaline
calcium hydroxide	green	Basic/alkaline
Lemon juice	Red	Acidic

The plant extract is able to differentiate between solutions by their nature. It is changing to a similar colour for similar solutions.

(i)Since lemon juice is a known acid, then sulphuric (VI) and hydrochloric acids are similar in nature with lemon juice because the indicator shows similar colors. They are acidic in nature.

(ii)Since sodium hydroxide is a known base/alkali, then the green colour of indicator shows an alkaline/basic solution.

(iii) Since pure water is neutral, then the orange colour of indicator shows neutral solutions.

In a school laboratory, commercial indicators are used. A commercial indicator is cheap, readily available and easy to store. Common indicators include: Litmus, phenolphthalein, methyl orange, screened methyl orange, bromothymol blue.

Experiment:

Using commercial indicators to determine acidic, basic/alkaline and neutral solutions

Procedure

Place 5cm3 of the solutions in the table below. Add three drops of litmus solution to each solution.

Repeat with phenolphthalein indicator, methyl orange, screened methyl orange and bromothymol blue.

Sample results

Substance/ Solution	Indicator used
Substance/ Solution	Litmus	Phenolphthalein	Methyl orange	Screened methyl orange	Bromothymol blue
wood ash	Blue	Pink	Yellow	Orange	Blue
soap solution	Blue	Pink	Yellow	Orange	Blue
ammonia solution	Blue	Pink	Yellow	Orange	Blue
sodium hydroxide	Blue	Pink	Yellow	Orange	Blue
hydrochloric acid	Red	Colourless	Red	Purple	Orange
distilled water	Colourless	Colourless	Red	Orange	Orange
sulphuric(VI)acid	Red	Colourless	Red	Purple	Orange
sour milk	Blue	Pink	Yellow	Orange	Blue
sodium chloride	Colourless	Colourless	Red	Orange	Orange
Toothpaste	Blue	Pink	Yellow	Orange	Blue
calcium hydroxide	Blue	Pink	Yellow	Orange	Blue
Lemon juice	Red	Colourless	Red	Purple	Orange

From the table above, then the colour of indicators in different solution can be summarized.

Indicator	Colour of indicator in
Indicator	Acid	Base/alkali	Neutral
Litmus paper/solution	Red	Blue	Colourless
Methyl orange	Red	Yellow	Red
Screened methyl orange	Purple	Orange	Orange
Phenolphthalein	Colourless	Purple	Colourless
Bromothymol blue	Orange	Blue	Orange

The universal indicator

The universal indicator is a mixture of other indicator dyes. The indicator uses the pH scale. The pH scale shows the strength of bases and acids. The pH scale ranges from 1-14.These numbers are called pH values:

(i) pH values 1, 2, 3 shows a substance is strongly acid

(ii) pH values 4, 5, 6 shows a substance is a weakly acid

(iii) pH value 7 shows a substance is a neutral

(iv) pH values 8, 9, 10, 11 shows a substance is a weak base/alkali.

(v) pH values 12, 13, 14 shows a substance is a strong base/alkali

The pH values are determined from a pH chart. The pH chart is a multicolored paper with each colour corresponding to a pH value.i.e

(i) red correspond to pH 1, 2, 3 showing strongly acidic solutions.

(ii)Orange/ yellow correspond to pH 4, 5, 6 showing weakly acidic solutions.

(iii)Green correspond to pH 7 showing neutral solutions.

(iv)Blue correspond to pH 8, 9, 10, 11 showing weakly alkaline solutions.

(v)Purple/dark blue correspond to pH 12,13,14 showing strong alkalis.

The universal indicator is available as:

(i) Universal indicator paper/pH paper

(ii) Universal indicator solution.

When determining the pH of a unknown solution using

(i) pH paper then the pH paper is dipped into the unknown solution. It changes/turn to a certain colour. The new colour is marched/compared to its corresponding one on the pH chart to get the pH value.

(ii) universal indicator solution then about 3 drops of the universal indicator solution is added into about 5cm3 of the unknown solution in a test tube. It changes/turn to a certain colour. The new colour is marched/compared to its corresponding one on the pH chart to get the pH value.

Experiment: To determine the pH value of some solutions

(a)Place 5cm3 of filtered wood ash, soap solution, ammonia solution, sodium hydroxide, hydrochloric acid, distilled water, sulphuric (VI) acid, sour milk, sodium chloride, toothpaste and calcium hydroxide into separate test tubes.

(b)Put about three drops of universal indicator solution or dip a portion of a piece of pH paper into each. Record the observations made in each case.

(c)Compare the colour in each solution with the colors on the pH chart provided. Determine the pH value of each solution.

Sample observations

Solution mixture	Colour on the pH paper/adding universal indicator	pH value	Nature of solution
wood ash	Blue	8	Weakly alkaline
soap solution	Blue	8	Weakly alkaline
ammonia solution	green	8	Weakly alkaline
sodium hydroxide	Purple	14	Strongly alkaline
hydrochloric acid	red	1	Strongly acidic
distilled water	green	7	Neutral
sulphuric(VI)acid	red	1	Strongly acidic
sour milk	blue	9	Weakly alkaline
sodium chloride	green	7	Neutral
toothpaste	Blue	10	Weakly alkaline
calcium hydroxide	Blue	11	Weakly alkaline
Lemon juice	Orange	5	Weakly acidic

Note

All the mineral acids Hydrochloric, sulphuric (VI) and nitric (V) acids are strong acids
Two alkalis/soluble bases, sodium hydroxide and potassium hydroxide are strong bases/alkali. Ammonia solution is a weak base/alkali. All other bases are weakly alkaline.
Pure/deionized water is a neutral solution.
Common salt/sodium chloride is a neutral salt.
When an acid and an alkali/base are mixed, the final product has pH 7 and is neutral.

Properties of acids

(a)Physical properties of acids

Acids have a characteristic sour taste
Most acids are colourless liquids
Mineral acids are odorless. Organic acids have characteristic smell
All acids have pH less than 7
All acids turn blue litmus paper red, methyl orange red and phenolphthalein colourless.
All acids dissolve in water to form an acidic solution. Most do not dissolve in organic solvents like propanone, kerosene, tetrachloromethane, petrol.

(b)Chemical properties of acids

Reaction with metals

All acids react with reactive metals to form a salt and produce /evolve hydrogen gas.

Metal + Acid -> Salt + Hydrogen gas

Experiment: reaction of metals with mineral acids.

(a)Place 5cm3 of dilute hydrochloric acid in a small test tube. Add 1cm length of polished magnesium ribbon. Stopper the test tube using a thump. Light a wooden splint. Place the burning splint on top of the stoppered test tube. Release the thump stopper. Record the observations made.

(b)Repeat the procedure in (a) above using Zinc granules, iron filings, copper turnings, aluminum foil in place of Magnesium ribbon

(c)Repeat the procedure in (a) then (b) using dilute sulphuric (VI) acid in place of dilute hydrochloric acid.

Sample observations

(i) effervescence/bubbles produced/fizzing in all cases except when using copper

(ii) Colourless gas produced in all cases except when using copper

(iii) Gas produced extinguishes a burning wooden splint with an explosion/pop sound.

Explanation

Some metals react with dilute acids, while others do not. Metals which react with acids produce bubbles of hydrogen gas. Hydrogen gas is a colourless gas that extinguishes a burning splint with a pop sound. This shows acids contain hydrogen gas.

This hydrogen is displaced/removed from the acids by some metals like Magnesium, Zinc, aluminium, iron and sodium.

Some other metals like copper, silver, gold; platinum and mercury are not reactive enough to displace/remove the hydrogen from dilute acids.

Chemical equations

Magnesium + Hydrochloric acid -> Magnesium chloride + Hydrogen

Mg(s) + 2HCl (aq) -> MgCl₂ (aq) + H₂(g)

Zinc + Hydrochloric acid -> Zinc chloride + Hydrogen

Zn(s) + 2HCl (aq) -> ZnCl₂ (aq) + H₂(g)

Iron + Hydrochloric acid -> Iron (II) chloride + Hydrogen

Fe(s) + 2HCl (aq) -> FeCl₂ (aq) + H₂(g)

Aluminium + Hydrochloric acid -> Aluminium chloride + Hydrogen

2Al(s) + 3HCl (aq) -> AlCl₃ (aq) + 3H₂(g)

Magnesium + Sulphuric (VI) acid -> Magnesium sulphate (VI) + Hydrogen

Mg(s) + H₂SO₄ (aq) -> MgSO₄ (aq) + H₂(g)

Zinc + Sulphuric (VI) acid -> Zinc sulphate (VI) + Hydrogen

Zn(s) + H₂SO₄ (aq) -> ZnSO₄ (aq) + H₂ (g)

Iron + Sulphuric (VI) acid -> Iron (II) sulphate (VI) + Hydrogen Fe(s) + H₂SO₄ (aq) -> FeSO₄ (aq) + H₂(g)
Aluminium + Sulphuric (VI) acid -> Aluminium sulphate (VI) + Hydrogen

2Al(s) + 3H₂SO₄ (aq -> Al₂ (SO₄)₃ (aq + 3H₂ (g)

Reaction of metal carbonates and hydrogen carbonates with mineral acids.

All acids react with carbonates and hydrogen carbonates to form salt, water and produce /evolve carbon (IV) oxide gas.

Metal carbonate + Acid -> Salt + Water+ Carbon(IV)oxide gas

Metal hydrogen carbonate + Acid -> Salt + Water + Carbon (IV) oxide gas

Experiment: reaction of metal carbonates and hydrogen carbonates with mineral acids.

(a)Place 5cm3 of dilute hydrochloric acid in a small test tube. Add half spatula full of sodium carbonate. Stopper the test tube using a cork with delivery tube directed into lime water. Record the observations made. Test the gas also with burning splint.

(b)Repeat the procedure in (a) above using Zinc carbonate, Calcium carbonate, copper carbonate, sodium hydrogen carbonate, Potassium hydrogen carbonate in place of Sodium carbonate.

(c)Repeat the procedure in (a) then (b) using dilute sulphuric (VI) acid in place of dilute hydrochloric acid.

Set up of apparatus

Sample observations

(i) effervescence/bubbles produced/fizzing in all cases.

(ii) Colourless gas produced in all cases.

(iii) Gas produced forms a white precipitate with lime water.

Explanation

All metal carbonate/hydrogen carbonate reacts with dilute acids to produce bubbles of carbon (IV) oxide gas. Carbon (IV) oxide gas is a colourless gas that extinguishes a burning splint. When carbon (IV) oxide gas is bubbled in lime water, a white precipitate is formed.

Chemical equations

Sodium carbonate +Hydrochloric acid -> Sodium chloride + Carbon (IV) Oxide+ Water

Na₂CO₃(s) + 2HCl (aq) -> 2NaCl (aq) + H₂O(g) + CO₂ (g)

Calcium carbonate +Hydrochloric acid -> Calcium chloride + Carbon (IV) Oxide+ Water

CaCO₃(s) + 2HCl (aq) -> CaCl₂ (aq) + H₂O(g) + CO₂ (g)

Magnesium carbonate +Hydrochloric acid ->Magnesium chloride + Carbon (IV) Oxide+ Water

MgCO₃(s) + 2HCl (aq) -> MgCl₂ (aq) + H₂O (g) + CO₂ (g)

Copper carbonate +Hydrochloric acid ->Copper (II) chloride + Carbon (IV) Oxide+ Water

CuCO₃(s) + 2HCl (aq) -> CuCl₂ (aq) + H₂O (g) + CO₂ (g)

Copper carbonate +Sulphuric (VI) acid ->Copper (II) sulphate (VI) + Carbon (IV) Oxide+ Water

CuCO₃(s) + H₂SO₄ (aq) -> CuSO₄ (aq) + H₂O (g) + CO₂ (g)

Zinc carbonate +Sulphuric (VI) acid ->Zinc sulphate (VI) + Carbon (IV) Oxide+ Water

ZnCO₃(s) + H₂SO₄ (aq) -> ZnSO₄ (aq) + H₂O (g) + CO₂ (g)

Sodium hydrogen carbonate +Sulphuric (VI) acid ->Sodium sulphate (VI) + Carbon (IV) Oxide+ Water

NaHCO₃(s) + H₂SO₄ (aq) -> Na₂SO₄ (aq) + H₂O (g) + CO₂ (g)

Potassium hydrogen carbonate +Sulphuric (VI) acid ->Potassium sulphate (VI) + Carbon (IV) Oxide+ Water

KHCO₃(s) + H₂SO₄ (aq) -> K₂SO₄ (aq) + H₂O (g) + CO₂ (g)

Potassium hydrogen carbonate +Hydrochloric acid ->Potassium chloride + Carbon (IV) Oxide+ Water

KHCO₃(s) + HCl (aq) -> KCl (aq) + H₂O (g) + CO₂ (g)

Sodium hydrogen carbonate +Hydrochloric acid ->Sodium chloride + Carbon (IV) Oxide+ Water

NaHCO₃(s) + HCl (aq) -> NaCl (aq) + H₂O (g) + CO₂ (g)

Neutralization by bases/alkalis

All acids react with bases to form a salt and water only. The reaction of an acid with metal oxides/hydroxides (bases) to salt and water only is called neutralization reaction.

Since no effervescence/bubbling/fizzing take place during neutralization:

(i) The reaction with alkalis requires a suitable indicator. The colour of the indicator changes when all the acid has reacted with the soluble solution of the alkali (metal oxides/ hydroxides).

(ii) Excess of the base is added to ensure all the acid reacts. The excess acid is then filtered off.

Experiment 1: reaction of alkali with mineral acids.

(i)Place about 5cm3 of dilute hydrochloric acid in a boiling tube. Add one drop of phenolphthalein indicator. Using a dropper/teat pipette, add dilute sodium hydroxide dropwise until there is a colour change.

(ii)Repeat the procedure with dilute sulphuric (VI) acid instead of hydrochloric acid.

(iii)Repeat the procedure with potassium hydroxide instead of sodium hydroxide.

Sample observation:

Colour of phenolphthalein change from colourless to pink in all cases.

Explanation

Bases/alkalis neutralize acids. Acids and bases/alkalis are colourless. A suitable indicator like phenolphthalein change colour topink, when all the acid has been neutralized by the bases/alkalis. Phenolphthalein change colour frompink, to colourless when all the bases/alkalis has been neutralized by the acid.

Chemical equation

Sodium oxide + Hydrochloric acid -> Sodium chloride + Water

Na₂O(s) + HCl -> NaCl(aq) + H₂O(l)

Potassium oxide + Hydrochloric acid -> Potassium chloride + Water

K₂O(s) + HCl -> KCl(aq) + H₂O(l)

Sodium hydroxide + Hydrochloric acid -> Sodium chloride + Water

NaOH(s) + HCl -> NaCl(aq) + H₂O(l)

Ammonia solution + Hydrochloric acid -> Ammonium chloride + Water

NH₄OH(s) + HCl -> NH₄Cl (aq) + H₂O (l)

Potassium hydroxide + Hydrochloric acid -> Potassium chloride + Water

KOH(s) + HCl -> KCl(aq) + H₂O(l)

Sodium hydroxide + sulphuric (VI)acid -> Sodium sulphate(VI) + Water

2NaOH(s) + H₂SO₄ -> Na₂SO₄ (aq) + 2H₂O (l)

Potassium hydroxide + sulphuric (VI) acid -> Potassium sulphate (VI) + Water

2KOH(s) + H₂SO₄ -> K₂SO₄ (aq) + 2H₂O (l)

Ammonia solution + sulphuric (VI) acid -> Ammonium sulphate (VI) + Water

2NH₄OH(s) + H₂SO₄ -> (NH4)₂SO₄ (aq) + 2H₂O (l)

Magnesium hydroxide + sulphuric (VI) acid -> Magnesium sulphate (VI) + Water

Mg (OH)₂(s) + H₂SO₄ -> MgSO₄ (aq) + 2H₂O(l)

Magnesium hydroxide + Hydrochloric acid -> Magnesium chloride + Water

Mg (OH)₂(s) + HCl(aq) -> MgCl₂ (aq) + 2H₂O(l)

Teachers' Resources

FORM 4 CHEMISTRY NOTES

October 5, 2025 Hillary Kangwana Leave a comment

ORGANIC CHEMISTRY II: ALCOHOLS AND ALKANOIC ACIDS:

1. Alkanols

Nomenclature of alkanols
Primary secondary and tertiary alcohols
Preparation and properties of alcohols
Hydrolysis of halogenoalkanes
Hydration of alkenes
Fermentation of sugars and starches (Ethanol)
Physical properties of alcohols
Chemical properties of alcohols
Combustion
Reaction with metals (sodium)
Esterification
Oxidation
Reaction with acidified potassium dichromate
Reaction with acidified potassium permanganate
Reaction with copper metal
Dehydration reactions of alcohols
Uses of alkanols

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CHEMISTRY FORM FOUR NOTES: NEW

Alkanoic acids

Nomenclature
Preparation and properties of alkanoic acids
Physical properties
Chemical properties
Reaction with sodium carbonate
Reaction with sodium hydroxide
Reaction with magnesium metal
Esterification
Uses of ethanoic acid

Fats and oils

Soaps and soapless detergents

Soaps
Preparation of soaps
Role of soap in cleaning
Effect of hard water on soap
Removal of hardness
Temporary hardness
Permanent hardness

Soapless detergents
Preparation
Advantages

Polymers

Natural and synthetic polymers
Addition polymerization
Condensation polymerization

ALCOHOLS (ALKANOLS):

– They derivatives of alkanes in which a hydrogen has been replaced by a hydroxyl group (OH); which is the functional group.

– They form a homologous series of the general formula C_nH_2n+1OH, in which the OH can also be denoted as ROH, where R is an alkyl group.

Note: alkanol is the IUPAC name while alcohol is the common name;

Nomenclature of alcohols

The e of the corresponding alkane molecule is replaced with the suffix ol;
The parent molecule is the longest chain containing the – OH group;
The numbering of carbon atoms is done such that the carbon atom with the hydroxyl group OH- attains the lowest possible number
The constituent branch is named accordingly;

Examples:

CH₃(ii). CH₃CH₂CH₂CH₂OH (iii). H OH H

| Butan-1- ol;

CH₃-C-OH H C C C H

CH₃ H H H

2,2 dimethylpropanol-1-ol; Propan-2-ol;

(iv). CH₃ (v). H H OH H H (vi). H H

CH₃ CH₂ CCH₃ H C C C C C H H C C OH

OH H CH2 H H H H H

2 methyl butan-2-ol; Ethanol;

CH₃

3,4 ethyl methylhexan-3-ol;

Isomerism in alkanols.

– Alcohols exhibit positional isomerism due to the fact that the position of attachment of the functional group varies within the carbon chain;

Examples of isomeric alcohols

(a). Isomers of propanol

(i). H OH H (ii). H H H

H C C C H H C C O C H

H H H H H H

Propan-2-ol Ethyl methyl ether or methoxyethane;

(iii). H H H (iv).

H C C C O H

H H H

Propan-1-ol;

Note: Ethyl methyl ether is not actually an alcohol as it lacks the OH group;

– All have the molecular formula: C₃H₇OH

(b). Isomers of butanol

(i). CH₃CH₂CH₂OH (ii). CH₃CH₂CHCH₃ (iii). CH₃

Butan-1-ol;

OH CH₃ C OH

Butan-2-ol;

CH₃

2,2-dimethyl propan-2-ol

Note: All have the molecular formula: C₄H₉OH

(c). Draw the Isomers of pentanol, C₅H₁₁OH;

(i). H H H H H (ii). (iii) (iv)

H C C C O C C H

H H H H H

Ethylpropylether

Primary, secondary and tertiary alcohols.

Primary alcohols:

– Are alcohols in which the OH group is attached to a carbon atom to which 2 hydrogen atoms are attached;

– Thus they contain a CH₂OH group;

Examples:

H H H H

H C C C C – OH

H H H H

Butan-1-ol;

Secondary alcohols:

– Are alcohols in which the hydroxyl group is attached to the carbon atom to which only one other hydrogen atom is attached;

– The carbon atom with the OH group is thus bonded to two carbon atoms;

– They contain a CHOH group;

Example:

H H H H

H C C C C – H

H H OH H

Butan-2-ol;

Tertiary alcohols:

– The hydroxyl group is attached to a carbon atom with no hydrogen atoms attached

– The carbon atom with the OH group is bonded to 3 other carbon atoms; hence sorrounded by the methyl groups;

– Tertiary alcohols thus a contain a COH group;

Examples:

H OH H

H C C C – H

H CH₃ H

2-methyl propan-2-ol;

Preparation and properties of alkanols.

– Alkanols are prepared from three main methods.

Hydrolysis of halogenoalkanes;
Hydration of alkenes
Fermentation of starches and sugars (mainly for ethanol)

(a). Hydrolysis of halogenoalkanes;

– Halogenoalkanes are compounds in which one or more hydrogen atoms in an alkane are replaced by halogens;

– Addition of aqueous KOH or NaOH to a halogenoalkane and heating results to corresponding alcohol;

– Reaction involves replacement of the halogen atoms with the -OH from the alkali;

Examples:

(i). Preparation of methanol

CH₃Cl + NaOH ^heat CH₃OH + NaCl

Chloromethane sodium hydroxide; methanol sodium chloride

(ii). Preparation of propanol;

CH₃CH₂CH₂Br + KOH ^heat CH₃CH₂CH₂OH + KBr

1-bromopropane potassium hydroxide Propan-1-ol Potassium bromide

Note: –the conversion of a halogenoalkane to an alcohol is known as hydrolysis;

– Reagent in this case is an alkali and condition for reaction is heat;

(b). Hydration of alkenes.

– Conversion of an alkene to an alcohol is known as hydration;

– Main reagent for the reaction is water;

– Conditions for the reaction are:

An acid catalyst, mainly conc. H₂SO₄ or phosphoric acid (H₃PO₄);
High temperatures of about 80^oC;
High pressures of about 25-30 atmospheres;

Examples:

(i). Preparation of ethanol from ethene.

C₂H₄ + H₂O ^{Conc. H2SO4} C₂H₅OH;

^{80oC; 25-30 atm;}

H H H H

H C = C H + H OH H C C H

Ethane Water

H OH

Ethanol

(ii). Preparation of butanol from butene.

CH₃CHCHCH₃ + H₂O ^{Conc H3PO4} CH₃CH₂CH₂CH₂OH

Butene 80^oC, 25-30 atm Butanol

(c). Preparation (of ethanol) by fermentation.

– It is prepared from the fermentation of starches or sugars in the presence of yeast;

– Fermentation: Is a chemical decomposition brought by bacteria or yeast (anaerobically) usually accompanied by evolution of carbon (IV) oxide and heat.

The chemical process

– Starch is broken into sugars by the action of the enzyme amylase or diastase;

_{Break up into}

Starch molecule + water sucrose molecules

_Amylase

(C₆H₁₀O₅) n + nH₂O nC₆H₁₂O₆

Starch Water many sucrose molecules

– When yeast is added to dilute sucrose solution (ordinary sugar); the enzyme sucrase in yeast catalytically breaks down sugar (sucrose) into the simplest sugars, glucose and fructose i.e.

Equation:

C₁₂H₂₂O_11(aq) + H₂O_(l) Sucrase C₆H₁₂O_6(aq) + C₆H₁₂O₆

Sugar Water glucose fructose

– Finally the enzyme zymase, also produced by yeast converts glucose and fructose into ethanol and carbon (IV) oxide.

Equation:

C₆H₁₂O₆ _(aq) ^Zymase2C₂H₅OH_(aq) + CO₂

Glucose/fructose ethanol carbon (IV) oxide

Optimum conditions for fermentation:

Temperatures of 25-30^oC;
Yeast catalyst;
Absence of oxygen (airtight);

Note:

– When the reaction mixture contains about 12% by volume of ethanol, the activity of yeast ceases.

– This is because higher ethanol concentrations kill the yeast cells;

– Fermentation provides about 10% alcohol by volume;

– The concentration of resultant ethanol can be increased by fractional distillation.

– During the process, ethanol distills over fast due to its lower boiling point (78°C)

– The distillate at below 95°C is first collected (leaving water behind).

– The resultant fraction will have 95% alcohol by volume; and is called rectified spirit;

– Absolute ethanol; which is 99.5% by volume can be obtained by re-distillation of rectified ethanol between 78-82^oC to remove all the water in the mixture;

– This can be done in two main ways:

Addition of a small amount of benzene to the rectified spirit and then distilling; (benzene dissolves in the water in the alcohol)
Distillation of rectified spirit over a suitable drying agent like calcium oxide and then over calcium; (calcium reacts with steam, calcium oxide takes in condensed water)

Properties of alcohols (Ethanol)

(a) Physical properties

(i). It is a colourless, volatile liquid soluble in water in all proportions forming a neutral solution;

(ii). Has a characteristic smell and boils at78.5°C

Variation in physical properties of alkanols.

Name	Molecular formula	Molecular mass	Boiling point (^oC)	Melting point (^oC)	Solubility in 100g of water
Methanol	CH₃OH	32	64.5	-94	Soluble
Ethanol	C₂H₅OH	46	78.5	-117	Soluble
Propanol	C₃H₇OH	60	97	-127	Soluble
Butanol	C₄H₉OH	74	117	-90	Slightly soluble
Pentanol	C₅H₁₁OH	88	138	-79	Slightly soluble
Hexanol	C₆H₁₃OH	102	158	-52	Slightly soluble
Heptanol	C₇H₁₄OH	116	175	-34.6	Very slightly soluble
Octanol	C₈H₁₆OH	130	194	-16	Very slightly soluble;

Note:

– Solubility of alkanols decreases with increase in molecular mass;

– Both melting and boiling points increases with increase in the relative molecular mass; due to progressive increase in number of van der waals forces;

– Alkanols have higher melting and boiling points than their corresponding alkanes with the same molecular formula;

Reason:

– Alkanols have hydrogen bonding between their molecules, caused by the presence of the OH group; alkanes have van der waals between its molecules; Hydrogen bonds are stronger than weak van der waals;

Diagram: Hydrogen bonding between four ethanol molecules.

(b) Chemical properties /main reactions of ethanol.

Note; the main reactions of ethanol are those of its functional group OH;

(iii). Combustion

Procedure;

– A few drops of ethanol are placed in a watch glass and lit.

– A dry gas jar is held over the flame and the gas collected tested with limewater.

Observation:

– It burns with a blue flame, which is almost colourless.

– The resultant gas turns limewater into a white precipitate, indicating it is carbon (IV) oxide.

Explanations:

– Ethanol (alcohols) burns in air (oxygen) producing carbon (IV) oxide, water and heat energy;

– The lower members of the homologous series burn with a blue or non-luminous flame leaving no residue;

– As the hydrocarbon chain increases the flame becomes more luminous and smoky and a black residue remains;

Equation

C₂H₅OH_(l) + 3O_2(g) 2CO_2(g) + 3H₂O_(l)

Note:

– If an alkanol is burnt in a limited supply of oxygen, then the combustion is incomplete and the products include carbon (II) oxide or carbon and water

Equations

(i). C₂H₅OH_(l) + 2O_2(g) 2CO_(g) + 3H₂O_(l)

(ii). 2C₂H₅OH_(l) + 2O_2(g) 4C_(s) + 6H₂O_(l)

(iv). Reaction with metals (sodium);

Procedure:

– Tiny pieces of sodium one at a time; are added 1cm³ of pure ethanol in a boiling tube.

Observation;

– Sodium metal darts on the surface of the ethanol and then dissolves /disappears;

– The beaker becomes warmer indicating an exothermic reaction.

– Effervescence occurs and bubbles of a colourless gas are observed; gas burns with a pop sound.

Explanation:

– Sodium reacts with alcohol much as it does with water but the reaction is more gentle.

– Sodium reacts with ethanol to produce hydrogen gas, which on testing burns with a pop sound.

– The reaction is exothermic producing heat hence warmer beaker.

– A clear solution of sodium ethoxide is left formed in the boiling tube.

Equation;

2C₂H₅OH + Na_(l) 2CH₃CH₂ONa_(l) + H_2(g)

Note:

Alkanols react with electropositive metals such as sodium, potassium and aluminium to liberate hydrogen gas and form a solution of the metal salt, the metal alkoxide;

Examples

(i). 2CH₃CH₂CH₂OH + Na 2CH₃CH₂CH₂ONa + H₂

Propanol Sodium metal Sodium propoxide hydrogen

(ii). CH₃CH₂CH₂CH₂CHOH + K 2CH₃CH₂CH₂CH₂CHOK + H₂

Pentanol Potassium Potassium pentoxide Hydrogen

Note: The reactivity of alkanols with metals decreases as the hydrocarbon chain increases;

(v). Esterification;

– It is the production of esters (alkyl alkanoates) from the reaction between alcohols and carboxylic acids;

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

Observation

A fruity sweet smell (of ethyl ethanoate);

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

Equation Conc. H₂SO₄; warm

C₂H₅OH_(l) + CH₃COOH (aq) CH₃COOC₂H_5(l) + H₂O_(l)

Ethanol ethanoic acid ethyl ethanoate (ester)

Structurally;

H H H O H O H H

Conc. H₂SO₄

H C C OH + H C C H C C O C C H + H₂O

Water

H H H OH H H H

Ethanol Ethanoic acid Ethylethanoate

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the alkanoic acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric acid which act as a catalyst and warm (heat) conditions, the reaction is enhanced;

Further examples:

Write balanced equations for each of the following esterification reactions:

(i). Ethanol and propanoic acid;

(ii). Propanol and ethanoic acid;

(iii). Ethanol and methanoic acid;

Note:

– Generally a reaction between alcohol and a carboxylic acid (-COOH-) produces an ester and water in a process called Esterification;

General equation:

Conc. H₂SO₄, warm

Organic acid + alcohol ester + water.(Esterification), while;

Inorganic acid + alkali salt + water.(neutralization)

Differences between neutralization and Esterification

Esterification is slower than neutralization as the reaction is between molecules and not ions as in neutralization
Esterification is reversible; the forward reaction is esterification and the backward reaction is hydrolysis.
Esterification results to esters which are covalent compound; neutralization forms salts which are electrovalent.

Name and formulae of some common esters.

Alkanol	Alkanoic acid	Ester
Methanol CH₃OH	Propanoic acid O CH₃CH₂ C OH	Methylpropanoate O CH₃CH₂ C OCH₃
Ethanol C₂H₅OH	Methanoic acid O H C OH	Ethylmethanoate O H C OCH₂CH₃
Propan-1-ol	Ethanoic acid
Butan-1-ol	Ethanoic acid

(vi). Oxidation of primary alkanols;

– On heating in presence of oxidising agents, primary alkanols are oxidised to alkanoic acids;

Note:

– During oxidation of alcohols, they first lose hydrogen to form compounds called aldehydes (compounds ending in al)

– The resultant aldehydes (alkanal) then gain oxygen to form alkanoic acids.

Equations:

(i). Alkanol hydrogen alkanal + water

Then;

(ii). Alkanal + oxygen alkanoic acid;

General equation:

H O

R C OH + 2 [O] R C OH + H₂O

From oxidizing agent

Example: Oxidation of ethanol

– Ethanol like all other alcohols is oxidized by strong oxidizing agents such as potassium dichromate (VI) and potassium manganate (VII) to form ethanoic acid.

(a). Reaction with acidified potassium dichromate (VI):

Procedure

– A little solution of acidified potassium dichromate (VI) is added to a little solution of ethanol in a test tube and then warmed gently;

Observation:

– The solution (dichromate) changes from yellow to green;

Explanation

– The acidified potassium dichromate (VI) oxidizes the ethanol to ethanal then to ethanoic acid, while the dichromate undergoes reduction (chromate (VI) to Cr³⁺ changing colour from yellow to green;

Equations

H⁺ from Conc. H₂SO₄

C₂H₅OH_(l) + [ O ] CH₃CH_(aq) + H₂O_(l)

Ethanol From Slow reaction Ethanal water

K₂Cr₂O₇

Structurally:

H H H H

H C C OH + [ O ] H C C O + H₂O

H H H

Then;

O O

CH₃CH_(aq) + [ O ] CH₃C OH

Ethanal H⁺reaction Ethanoic acid

General equation

H H O

H C C OH_(l) + 2 [ O ] CH₃C OH + H₂O

H H

(b). Reaction with acidified potassium manganate (VII).

Procedure

Acidified potassium permanganate solution is added to ethanol in a test tube end the mixture is warmed gently.

Observation

The permanganate solution turns from purple to colourless.

The characteristic smell of ethanoic acid is felt.

Explanation

The ethanol is oxidized to ethanal then to ethanoic acid. The reduced permanganate decolorizes (turns from purple to colourless);

-The H⁺/KMnO₄ is decolourised due to reduction of manganate (VII) ions to Mn²⁺

General equation:

H H O

H C C OH_(l) + 2 [ O ] CH₃C OH + H₂O

H⁺/ KMnO₄

H H

(c). Catalytic oxidation of alkanols.

– Catalytic oxidation of alkanols results to a dehydrogenation (removal of hydrogen) resulting to the formation of an alkanal (aldehyde);

– These are compounds with the functional group COH;

Example: Catalytic oxidation of ethanol with hot copper metal

When ethanol is passed over heated copper at about 300 °C, it is dehydrogenated i.e. hydrogen is removed.

This results into formation of an ethanal; and hydrogen gas is liberated.

Equation _Cu(s)

CH₃CH₂OH(g) CH₃CHO_(l) + H_2(g)

Ethanol 250^oC Ethanal

(vii). Dehydration reactions of alcohols.

– Dehydration is the removal of water molecules from a compound;

– Excess concentrated sulphuric (VI) acid dehydrates alkanols and forms corresponding alkenes.

Conditions required: –

– High temperatures of 140-180°C.

– Catalysts such as conc. sulphuric acid, phosphoric acid and aluminium oxide;

Example: Dehydration of ethanol

(i). Apparatus

(ii). Procedure

– 15cm³ of absolute ethanol are put in around bottomed flask and 5cm³ of con H₂SO₄ added.

– The contents are mixed thoroughly by swirling the flask.

– The flask is then heated (warmed) gently with little shaking for about 1 minute.

– The gas collected is tested with acidified potassium manganate (VII) and bromine water.

(iii). Observations

– Evolution of a colourless gas that decolourises the purple acidified potassium manganate (VII)

– The resultant gas also decolourises the red brown bromine water.

(iv). Explanations

– On heating the alkanol (ethanol) undergoes an elimination reaction.

– It loses both a hydrogen (H) and a hydroxyl (OH) from two adjacent carbon atoms.

– The H and OH combine to form water; and the remnants form ethene.

– The sulphuric acid acts as a catalyst.

Note: Disadvantage of concentrated sulphuric acid over phosphoric acid.

– Being a strong oxidising agent the concentrated sulphuric acid oxidizes some of the alkanol formed to CO₂ and it is itself reduced to SO₂;

– This reduces the volume and purity of resultant alkanol; and is also a potential source of pollution;

Equation

H H H H

Conc. H₂SO₄

H C C H H C = C H + H₂O

170^oC

H OH

Note:

With cold concentrated sulphuric (VI) acid, alkanols react to form alkyl hydrogen sulphates.

Example: Ethanol and cold conc. sulphuric acid.

C₂H₅OH + H₂SO_4(l) C₂H₅HSO_4(l) + H₂O_(l)

Ethanol Ethyl hydrogen sulphate

(Ethoxyethane)

Uses of ethanol

Used as solvents; in the preparation of drugs, perfumes, liquors, vanish and paints.
Source of fuel e.g ethanol when blended with gasoline to form gasohol.

Note:

– Addition of ethanol to petrol improves the antiknock propertied of petrol, due to its low ignition point;

– The alkanol also absorbs any traces of moisture that may enter and damage the petrol system;

– Alkanols also ignite on their own to liberate heat;

Example:

CH₃CH₂OH_(l) + 3O_2(g) 2CO_2(g) + 3H₂O_(l) ∆H = -1368KjMol^–

Starting material for the manufacture of polyvinyl chloride (P.V.C)
Ethanol is used as a disinfectant (antiseptic) at special concentrations e.g for cleaning tissues and surgical equipments during operations and in dressing wounds;
Manufacture of alkanoic acids e.g ethanoic acid
Ethanol is used as thermometer liquid for measuring low temperatures;
Manufacture of alcoholic drinks
Large amounts of methanol are used in the manufacture of formaldehyde a chemical used in preservation of corpses;
They are used as antifreeze mixtures in car radiators e.g a mixture of ethanol and water freezes at lower temperatures that pure water;
Manufacture of esters to giver fruity flavourings for confectionery and drinks;

Tests for primary alkanols.

Reaction with sodium metal

– Alkanols liberate hydrogen gas, a colourless gas that burns with a pop sound;

Equation:

2CH₃CH₂OH_(l) + 2Na_(s) 2CH₃CH₂ONa + H_2(g)

Reaction with phosphorus (V) chloride:

– Alkanols liberate misty fumes of hydrogen chloride gas;

Equation:

PCl₅(s) + CH₃CH₂OH 2CH₃CH₂OP + HCl₂

Note:

– Both alkenes and alkanols decolourise the purple acidified potassium manganate (VII);

– However, alkenes decolourise the red bromine water; while alkanols do not.

Summary on preparation of alkanols (ethanol).

Ester RCOOC₂H₅

Heat with NaOH_(aq)

Ethyl iodide Boil with KOH_(aq) Ethanol Fermentation Starch, sugars

(Iodoethane) cellulose

CH₃CH₂I

H₂O + H₂SO₄

Ethene

CH₂CH₂

Summary on reactions of alkanols.

CO_2(g) + H₂O_(l)

Combustion

Ester ^{Esterification} Ethanol ^{Sodium metal} Sodium ethoxide + Hydrogen

RCOOC₂H₅ H⁺/ RCOOH CH₃CH₂OH CH₃CH₂ONa + H₂

Conc. H₂SO₄ at 170^oC Warm with acidified K₂Cr₂O₇or KMnO₄

Ethene PCl₅ Ethanoic acid

CH₂CH₂ CHCOOH

Chloroethane

CH₃CH₂Cl

ALKANOIC ACIDS (CARRBOXYLIC ACIDS)

– Also called organic acids and form a homologous series with a general formula of C_nH_2n-1OOH

– The formula can also be written as C_nH_2n _{+ 1}COOH; in which case n = (no. of carbon atoms – 1)

– Members differ from each other by an additional CH₂ group.

– Their functional group is the carboxylic group (-COOH) which is attached to the alkyl group.

– Graphical representation of carboxyl group; O

C OH

Note: All carboxylic acids have the COOH as the functional group but alkanoic acids are strictly alkanoic acids derived from alkanes

Nomenclature of alkanoic acids

– The ending of the corresponding alkane is replaced by ¢oic acid.

Examples;

– Methane to methanoic acid.

– Ethane to ethanoic acid.

– They are named as if they are derived from alkanes through replacement of one of the hydrogen atoms by the -COOH group.

Note: –Unlike alkanols the functional group (COOH) in alkanoic acids can only be at the end of the carbon chain.

– The C in the COOH is always given the first position, while the substituents are given locants (numbers in reference to the first position).

Examples:

IUPAC name	Old (traditional ) name	Structural formula
Methanoic acid	Formic acid	OH H C = O
Ethanoic acid	Acetic acid	OH CH₃ – C = O
Propanoic acid	Propionic acid	OH CH₃CH₂ C = O
Ethanedioic acid	Oxalic acid	O O C – C HO OH
Butanedioic acid	Succinic acid	O CH₂C OH O CH₂C OH

Branched alkanoic acids

– The naming of branched alkanoic acids follow the same general rules like that of alkanes; as long as the carbon atom with the COOH group is given the first position.

– The branch can either be an alkyl group or a halogen other than hydrogen.

Examples:

Compound	IUPAC name
O Cl – CH₂ – C OH	2-chloroethanopic acid;
O CH3 – CH – C OH CH₃	3-methylpropanoic acid;
O CH₃ – CH – C OH OH	2-hydroxypropanoic acid;
OH O CH₃ – C – C OH CH₃	2-hydroxy, 2-methylpropanoic acid;
O CH₂ – CH – CH₂ – C OH Br Cl	4-bromo, 3-chlorobutanoic acid;

Isomerism in alkanoic acids.

– Due to the existence of branched alkanoic acids, it is possible to obtain various isomers for a given alkanoic acid;

Example:

Draw all the isomers of pentanoic acid. (3 marks)

Preparation of alkanoic acids

(a). Industrial manufacture

– Is done by the oxidation of primary alkanols using air (oxygen) as the oxidising agent.

Conditions:

– Moderate temperatures

– 5 atm pressure

– Hot copper catalyst;

Laboratory preparation.

– Is done by the oxidation of primary alkanols using acidified potassium dichromate (VI).

Example: Laboratory preparation of ethanoic acid.

(i). Apparatus

(ii). Procedure

– Acidified potassium dichromate (VI) is heated in a water bath and ethanol added slowly from a water bath.

-The mixture is heated further and then distilled.

– The distillate is collected at about 105^oC.

(iii). Observations and explanations.

Colour of potassium dichromate (VI) changes from orange to green.

Reason:

The dichromate ions are reduced to chromium (III) ions.

Equation:

Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6e- 2Cr³⁺_(aq) + 7H₂O_(l)

– The ethanol is oxidised to ethanal (acetaldehyde); which is further oxidised to alkanoic acid.

Equations

H⁺ from Conc. H₂SO₄

C₂H₅OH_(l) + [ O ] CH₃CH_(aq) + H₂O_(l)

Ethanol From Slow reaction Ethanal water

K₂Cr₂O₇

Structurally:

H H H H

H C C OH + [ O ] H C C O + H₂O

H H H

Then;

O O

CH₃CH_(aq) + [ O ] CH₃C OH

Ethanal H⁺reaction Ethanoic acid

General equation

H H O

H C C OH_(l) + 2 [ O ] CH₃C OH + H₂O

H H

Properties of alkanoic acids

Gradation in physic al properties of alkanoic acids

Name of acid

Formula (structural)

Molecular formula

M.P °C

B.P °C

Solubility

Methanoic acid

Ethanoic acid

Propanoic acid

Butanoic acid

Pentanoic acid

Hexanoic acid

HC OOH

CH₃COOH

CH₃CH₂COOH

CH₃CH₂CH₂COOH

CH₃CH₂CH₂CH₂COOH

CH₃(CH₂)₄COOH

CH₂O₂

C₂H₄O₃

C₃H₆O₂

C₄H₈O₂

C₅H₁₀O₂

C₆H₁₂O₂

8.4

16.6

-20.8

-6.5

-34.5

-1.5

101

118

141

164

186

205

Most Soluble

increasing

solubility

Least soluble

Physical properties of ethanoic acid

– A colourless liquid with a sharp pungent smell.

– B.P is 118^oC and freezes at 17°c forming ice like crystals termed as glacial ethanoic acid.

– It is soluble in water and is weakly acidic with a P.H of approximately4.8.

Note

– Concentrated ethanoic acid is only slightly ionized and is a poor conductor of electricity

– On dilution its conductance steadily improves as the extent of ionization increases

Chemical properties (reactions) of alkanols.

(i). Reaction with carbonates.

– Alkanoic acids react with metal carbonates to form a salt (metal alkanoate), carbon (IV) oxide and water.

Examples:

Ethanoic acid and sodium carbonate

Ethanoic acid reacts with sodium carbonate to form sodium ethanoate and water with the liberation of carbon (IV) oxide gas.

Equation

Na₂CO₃ + 2CH₃COOH 2CH₃COONa + CO_2(g) + H₂O_(l)

Sodium carbonate Sodium ethanoate

Zinc carbonate and ethanoic acid.

ZnCO₃ + 2CH₃COOH (CH₃COO)₂Zn + CO_2(g) + H₂O_(l)

Zinc carbonate Zinc ethanoate

(ii). Reaction with metal hydroxides (Neutralization)

Alkanols neutralize alkalis like sodium hydroxide forming a salt (metal alkanoate) and water only.

Examples:

Sodium hydroxide and ethanoic acid.

CH₃COOH + NaOH CH₃COONa + H₂O(l)

Ethanoic acid Sodium hydroxide Sodium ethanoate

Potassium hydroxide and methanoic acid

HCOOH + KOH HCOOK + H₂O(l)

Methanoic acid Potassium hydroxide Potassium methanoate

(iii). Reaction with metal oxides (neutralization).

– Alkanoic acids react with metal oxides to produce salt (metal alkanoate and water only.

Examples:

Ethanoic acid and copper (II) oxide.

2CH₃COOH + CuO (CH₃COO)₂Cu + H₂O(l)

Ethanoic acid Copper (II) oxide Copper ethanoate

(iv). Reaction with metals

Alkanoic acids react with reactive metals to form a salt (metal alkanoate) and hydrogen gas;

Examples:

Ethanoic acid and sodium metal

2CH₃COOH + Na_(s) 2CH₃COONa + H_2(g)

Ethanoic acid Sodium ethanoate

Propanoic acid and magnesium metal

2CH₃CH₂COOH + Mg (CH₃CH₂COO)₂Mg + H_2(g)

Propanoic acid Magnesium propanoate

(v). Reaction with alkanols (Esterification)

– Alkanoic acids react with alkanols to form esters;

Conditions:

– Drops of concentrated sulphuric acid.

– Gentle warming.

Reaction of ethanoic

– Ethanoic acid reacts with ethanol in the presence of a few drops of concentrated sulphuric acid forming a sweet fruity smelling compound called ester.

– The process is called esterification.

Procedure

– 2-3 drops of concentrated sulphuric acid are added to a mixture of equal proportions of ethanol and pure ethanoic acid in an evaporating dish.

– The mixture is warmed gently in a water bath for sometime.

– The mixture is poured into a beaker and smelt;

Observation

A fruity sweet smell (of ethyl ethanoate);

Explanation

– Ethanol reacts with ethanoic acid in the presence of a few drops of concentrated sulphuric acid to form ethyl ethanoate and water.

– The reaction is very slow and so catalyzed by the hydrogen ions from the sulphuric acid.

Equation Conc. H₂SO₄; warm

C₂H₅OH_(l) + CH₃COOH (aq) CH₃COOC₂H_5(l) + H₂O_(l)

Ethanol ethanoic acid ethyl ethanoate (ester)

Structurally;

H H H O H O H H

Conc. H₂SO₄

H C C OH + H C C H C C O C C H + H2O

H H H OH H H H

– The alkyl part of the ester is derived from the alkanol, while the alkanoate part is derived from the acid;

– The alkanol attaches itself at the group in the carboxylic acid thereby displacing hydrogen atom;

– Under ordinary conditions the reaction takes place slowly; but in presence of concentrated sulphuric catalyst and warm (heat) conditions, the reaction is enhanced;

If propanol were used in place of ethanol, the reaction would yield the ester propyl ethanoate, according to the following equation;

CH₃COOH_(aq) + CH₃CH₂CH₂OH_(aq) CH₃COOCH₂CH_3(aq) + H₂O_(l)

Ethanoic acid propanol Propylethanoate

Note: –

Esters react with water to form the respective alkanoic acid and alkanol.

This reaction is termed hydrolysis and occurs in presence of concentrated sulphuric acid and heat as conditions.

Example: Conc. H₂SO₄

CH₃COOCH₃ + H₂O CH₃COOH + CH₃OH

Methyl ethanoate Heat Ethanoic acid Methanol

(vi). Reaction with ammonia.

Alkanoic acids react with ammonia to produce the ammonium salt of the acid.

General formula of the ammonium alkanoate salt is RCOONH₄

Example:

Ethanoic acid and ammonia gas.

CH₃COOH + NH₃ CH₃COONH₄

Uses of alkanoic acids

In pharmaceuticals, for making medicines e.g ethanoic acid is used in the manufacture of aspirin.
Manufacture of dyes and insecticides
Seasoning food as vinegar
Coagulation of rubber latex
Preparation of polyethenyl ethanoate and cellulose ethanoate which are used are used to make artificial fibres such as rayon.
Manufacture of soaps.
Preparation of perfumes and artificial favours used in food manufacture.

Tests for alkanoic acids.

The following tests can be used to test for alkanoic acids.

Reaction with carbonates and hydrogen carbonates

Esterification

Summary: Draw a summary flow chart to show all the reactions of a named alkanoic acid.

FATS AND OILS.

– Are esters of long chain carboxylic acids and glycerol

(a). Oils

– Oils occur naturally in plants and animals.

Examples: Whale oil, groundnut oil, corn oil and castor oil.

– Oils are liquids at room temperature.

Reason:

– They have high proportion of esters derived from the unsaturated oleic acid in them.

(b). Fats.

– These occur naturally in animals only.

Examples: Tallow, butter from milk, lard from pigs etc.

– Fats are solids at room temperature

– Oils can be converted to fats /hardened into fats by hydrogenation.

– This is the conversion of oils into fats by use of hydrogen; and forms the basis of margarine manufacture.

– During hydrogenation:

Hydrogen is bubbled into oils under high pressure and temperatures of about 400^oC in the presence off a nickel catalyst

Note

– Fats and oils are important raw materials in the manufacture of soaps.

Soaps and soapless detergents

Soaps

– Are a variety of compounds produced when oils or fats are reacted with sodium hydroxide.

– They are similar in that they contain a long hydrocarbon chain ending in a carboxylate anion to which is attracted a sodium cation.

– A typical soap is sodium stearate; C₁₇H₃₅COO^–Na⁺

Structurally

CH₃CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂CH₂C

O^–Na⁺

Preparation of soaps

– 2cm³of castor oil and 10 cm³ of 4M sodium hydroxide are poured into a 100cm³ beaker.

– The mixture is then for about 10 minutes, stirring continuously and adding distilled water to make up for evaporation.

Explanation:

– On boiling an alkali with fat or oil, a hydrolysis reaction occurs.

Equation:

CH₂ COO C₁₇H₃₅ CH₂OH

CH – COO C₁₇H₃₅ + 3NaOH 3C₁₇H₃₅COO^– Na⁺ + CHOH

Sodium stearate (soap)

CH₂ COO C₁₇H₃₅ CH₂OH

Fat Glycerol

– When hydrolysis reaction occurs in the presence of an alkali (sodium hydroxide), the process is known as saponification (the chemical reaction between a fat and an alkali)

– In the fat hydrolysis NaOH neutralizes the acid formed (i.e. stearic acid) to form the sodium salt of the acid removing it from the aqueous mixture.

– Thus in excess alkali, all the fat is utilized.

– The sodium salt (sodium stearate) of the acid is termed soap if the number of carbon atoms per molecule is more than eight.

Note:

– KOH may be used in place of NaOH as the alkali.

To the boiled mixture 3 spatula measures of sodium chloride are added, stirred well and allowed to cool.

Reason:

– The NaCl helps in separating the soap from the glycerol.

– This step is called salting and it reduces the solubility of the soap in the aqueous layer.

– The lower layer consists of glycerol, salts and unused alkali solution.

General formula of ordinary soap

C_nH_2n+1COO^–Na⁺where n >8 (n is greater than eight)

The solid is filtered off and washed with cold distilled water, to remove impurities like NaCl

The solid sample is placed in a test tube with distilled water, then with tap water.

Note:

The resultant soap may not have lathered easily with tap water.

Reason:

Some tap water contains a high proportion of calcium or magnesium ions that make water hard.

Summary on soap preparation

Fat (oil) Step I: saponification Solution of soap and alcohol

Add NaOH and boil Step II: salting

Add NaCl, stir well

And allow to cool

Ordinary soap

The role of soap in cleaning

Note: functions of soap in water

It makes the water able to wet material more effectively by lowering the surface tension.
Emulsification of oil and grease.

– Soap molecules have two dissimilar ends:

A hydrocarbon chain which is non-polar and has no attraction for water- hence oil soluble
A Carboxylate end, which is polar and is attracted to water; hence is water-soluble.

Note: the Carboxylate end is in fact negatively charged in water because after dissolution, the sodium ion and the carboxylate ion exist as separate entities.

Illustration:

CH₃(CH₂)₁₆– C O^–Na⁺

Non polar Carboxylate (Polar)

Effects of soap on oil water mixture (removal of oils and grease during washing)

Note: schematic representation of a soap molecule

On adding soap into oil water mixture the following build up occurs:

– A molecule of soap has a polar (hydrophilic) and non-polar (hydrophobic) parts;

– The non-polar end dissolves in oil and the polar end dissolves in water.

– When the mixture is agitated (thoroughly shaken) the hydrocarbon chain (tail) dissolves in grease while the carboxylate sodium end of the soap molecule (the head) remains dissolved in water.

Diagrams: role of soap in cleaning.

– Each oil drop ends up with a large cloud of negative charge around it as the polar heads are negatively charged.

– Consequently the oil drops repel each other, hence preventing them from coalescing.

– The water soluble sodium heads on the surface of the droplets keep the droplets emulsified (suspended) in the water.

– During rinsing the water carries away the oil droplets.

Effect of hard water on soap

– The Calcium and magnesium ions in hard water react with soap (sodium stearate) and remove it as an insoluble grey scum of magnesium or calcium stearate.

Equations:

(i). 2C₁₇H₃₅COO^–Na⁺_(aq) + Ca²⁺_(aq) (C₁₇H₃₅COO^–)₂Ca²⁺_(s) + 2Na⁺_(aq)

Sodium stearate (soap) Magnesium stearate

(ii). 2C₁₇H₃₅COO^–Na⁺_(aq) + Mg²⁺_(aq) (C₁₇H₃₅COO^–)2Mg²⁺_(s) + 2Na⁺_(aq)

Sodium stearate (soap) Magnesium stearate

– Soap is wasted in this way until all the calcium (II) and magnesium (II) stearate has been removed.

– The resultant scum is deposited on fabrics, giving them an unsightly dull appearance.

– Thus in hard water districts it is obviously advantageous to remove hardness before washing.

Removal of hardness:

Depends on whether the hardness is temporary or permanent

(a). Temporary hardness

Cause:

– The presence of calcium hydrogen carbonate or magnesium hydrogen carbonate dissolved in water.

Removal

– By boiling the water

– During the process the soluble calcium or magnesium hydrogen carbonate is precipitated out as insoluble calcium or magnesium carbonate.

Equation:

Ca(HCO₃)_2(aq) ^Heat CaCO_3(s) + CO_2(g) + H₂O_(l)

(b). Permanent hardness

Cause:

– Presence of calcium or magnesium chlorides and sulphates

Removal

Can be removed using the following methods:

(i). Distillation

– The water is distilled and the dissolved substances are left behind as water is evaporated and condensed

(i). Addition of washing soda (Na₂CO₃)

The washing soda reacts with the Mg²⁺ and Ca²⁺_(aq) ions precipitating them as the insoluble carbonates

Equations:

2Na⁺_(aq) + CO₃^2-_(aq) + 2Cl- + Ca²⁺_aq) CaCO_3(s) + 2Na⁺_(aq)+ 2Cl^– _(aq)

CO₃^2- _(aq) + Mg²⁺ MgCO_3(s)

Disadvantage of washing soda as a water softener

– It is alkaline and can cause damage to wool and silk.

(iii). Ion exchange process (e.g. in the permutit water softener)

– It involves use of resins and compounds which will exchange their own Na⁺ for Ca²⁺or Mg²⁺ dissolved in hard water

– Thus as the Na²⁺go into the water are left in the resin.

Equation:

2Na⁺ (resin^–)_(aq) + Ca²⁺_(aq) Ca²⁺ (Resin)₂ _(aq) + Na⁺_(aq)

Advantages and disadvantages of hard water

Advantages of hard water

It is good for drinking purposes as calcium contained in it helps to form strong bones and teeth.
When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO₃ are formed, coating the inside of the lead pipes preventing any further reaction
It is good for brewing and the tanning industries;

Disadvantages of hard water

Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

Note: For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg²⁺ and Ca²⁺

– Examples of soapless detergents: brand names such as omo, perfix, persil, and fab e.t.c.

Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

3. Formation of kettle fur which make electrical appliances inefficient and increases running costs.

Soap and pollution effects.

The wash water with soap and the dirt (grease) ends up in rivers and lakes thus affecting aquatic life; since plants do not grow well in soapy water.

Note: soaps are however biodegradable and so do not persist long in the environment.

Soapless detergents (synthetic detergents)

– Are cleansing agents lacking the carboxylate ions; but also act like soap in the cleaning process

– Instead they have sulphates (-OSO₃^–Na⁺) or sulphonate groups (-SO₃^–Na⁺) groups.

Are thus of two main types:

Sodium alkyl sulphates
Sodium alkylbenzene sulphonate.

(i). Sodium alkyl sulphates.

– Are detergents with the formula R OSO₃Na

Illustration.

– Consider the structure of sulphuric acid

HO S OH

– Replacing one of the hydrogen atoms with an alkyl group, R results into the compound

R OSO₃H;

– This is known as alkyl hydrogen sulphate;

– If the alkyl group (R) is a long chain such as dodecyl; CH₃(CH₂)₁₀CH₂– ; then the formula of the compound becomes CH₃(CH₂)₁₀ OSO₃H (alkylhydrogen sulphate)

– Reacting the alkylhydrogen sulphate with an alkali (NaOH) results into a compound with the formula CH₃(CH₂)₁₀ OSO₃Na.

– This is the soapless detergent, and is known as sodium dodecyl sulphate.

(ii). Sodium alkyl benzene sulphonates.

– Are formed when one of the OH^– groups in sulphuric acid is replaced with an alkyl benzene group.

– Have the general formula R – – SO₃Na

Illustration.

– If the alkyl benzene is of formula CH₃(CH₂)₁₀CH₂ – ; then the resultant compound is of formula CH₃(CH₂)₁₀CH₂ – – SO₃H (hydrogen dodecyl benzene sulphonate)

– The sodium alkylbenzene sulphonate is neutralized using sodium hydroxide to obtain the detergent, CH₃(CH₂)₁₀CH₂ – – SO₃Na

The compound CH₃(CH₂)₁₀CH₂ – – SO₃Na is called sodium dodecyllbenzene sulphonate.

Conclusion:

The two main types of soapless detergents are:

Sodium alkyl sulphates; CH₃(CH₂)₁₀ OSO₃Na
Sodium alkylbenzene sulphonate; CH₃(CH₂)₁₀CH₂ – – SO₃Na

Note:

– Due to lack of carboxylate ions, soapless detergents do not form scum with hard water. Advantage over soap

– They are not affected by hard water, as they do not form scum in hard water as follows:

Preparation of soapless detergents

– Most are made from residues from crude oil distillation

– The hydrocarbons are treated with concentrated sulphuric acid (instead of alkalis in cases of soap)

Procedure

– About 10cm³of olive oil in a small beaker, which is then stood in a larger one with ice-cold water.

– While stirring with a glass rod, concentrated sulphuric acid is carefully added to the olive oil using a dropping pipette.

– The acid is added until the yellow oil turns uniformly brown.

– 20cm³of 6M NaOH is then added; to neutralize the acid solution, resulting into a slightly basic product.

The soap is then tested with tap water and distilled water.

Explanation

– Some compounds like olive oil contain double bonds which can react concentrated sulphuric acid to form compounds of alkylhydrogen sulphate

Example:

R CH = CH CH CH₂ Est + H₂SO_{4 (l)} R CH CH CH – CH₂ Est

H OSO₃H

– In this case; R=alkyl group, either branched or straight; while Est = ester group.

– On adding NaOH, the alkyl hydrogen sulphate is neutralized where the hydrogen of the hydrogen sulphate is replaced by a sodium atom.

– The resultant compound is sodium alkylsulphate; R-CH – CH-CH₂-Est;

| |

H OSO₃Na

Equation:

R CH CH CH – CH₂ Est + NaOH_(aq) R CH – CH CH CH₂ Est + H₂O_(l)

H OSO₃H H OSO₃Na

– The R – CH – CH CH – CH₂– Est is the detergent and lathers easily with both tap and distilled water.

| |

H OSO₃Na

Note:

– Alkylhydrogen sulphates can also be from alcohols.

Example:

R CH CH₃ + H₂O_(l) R CH CH₃ + H₂O_(l)

OH OSO₃H

– Most soapless detergents are sodium alkyl sulphates with a general formula ROSO₃Na

– Sodium alkylbenzene soapless detergents are industrially manufactured from alkylbenzene;

– In this process alkylbenzene (a petroleum product) reacts with SO₃ to form sulphuric acid

– Upon neutralization with NaOH, sulphuric acid forms sodium alkylbenzene sulphonate,

R – – SO₃Na, a detergent.

Summary: manufacture of soapless detergents

Dodecene benzene

CH₃(CH₂)₉CH = CH₂ +

Chemical reaction

CH₃(CH₂)₁₁ Dodecylbenzene

Conc. sulphuric acid

CH₃(CH₂)₁₁ – SO₃H

Hydrogen dodecylbenzene sulphonate

Sodium hydroxide

CH₃(CH₂)₁₁ SO₃^– Na⁺

Sodium dodecylbenzene sulphonate

Mode of action of soapless detergent.

Soapless detergents have two ends; a long hydrocarbon part, the tail and a short ionic part the head.

Simple representation of a detergent molecule.

SO₃^– Na⁺

Non-polar tail (water hating) Polar head (water loving)

Examples:

(i). Sodium lauryl sulphate

CH₃(CH₂)₁₀CH₂ – O SO₃^–Na⁺

Tail head

(ii). Sodium alkylbenzene sulphonate

CH₃(CH₂)₁₀CH₂ – SO₃^–Na⁺

Tail head

– The tail is non-polar and dissolves in oil or grease (waterphobic) while the head is polar and dissolves in water (waterphilic).

– Each oil or grease gets sorrounded by the detergent molecules and hence a cloud of charged heads hence repel each other and do not coalesce.

– The dirt (grease loses its direct contact with the fabric being washed.

– Any agitation at this point then removes the dirt from the object.

Advantages of soapless detergents over soap;

They lather easily with hard water since the corresponding calcium and magnesium salts are soluble in water due to lack of the carboxylate ions.
Are mainly prepared from non-food raw materials.
They do not react with acidic water.

Note:

– Soapless detergents with branched chain alkyl groups are not easily broken down by bacteria and are therefore the cause of frothing in sewerage plants, rivers etc.

– Consequently modern industry is overcoming this disadvantage by making the detergents from alkylbenzene with straight chain alkyl groups.

Pollution effects of soapless detergents.

The active ingredients such as alkylbenzene sulphonates are non-biodegradable, hence accumulate in water sources and end up in human bodies.
Some of the additives such as phosphates cause eutrophication hence excessive build up of algae (algal blooms) which change the water taste and odour, and also reduce oxygen supply hence poor growth of aquatic organisms.
Because of their high lathering tendency, they cause excessive frothing and foaming in water sources especially after heavy rains.

Comparisons between soaps and soapless detergents.

Detergents	Soaps
Have strong cleansing action	Have weaker cleansing action
Are highly soluble in water; hence can be used in acidic or hard water.	Are not very soluble in water, and tend to be wasted when used in hard water; They cannot be used in acidic water.
Are made from byproducts of petroleum industry; which helps to conserve edible fats and oils	Are made from edible fats and oils;
They cuse water pollution;	Are biodegradable and have minimum pollution effects.
They are expensive.	Are cheaper than detergents.

POLYMERS

– A polymer is a macromolecule formed when two or more molecules link together to form a larger unit.

– Polymers have different properties different from those of the monomers.

– The process of polymer formation is called polymerization.

– Two types of polymers exist; natural and synthetic /artificial polymers.

(a). Natural polymers and fabrics

– Natural polymers occur naturally in living systems.

Examples:

– Rubber (latex), starch, cellulose, wool and proteins.

Natural rubber.

– Rubber trees give out a liquid called latex, which is collected from cuts in the trunks of rubber trees.

– Natural rubber is made out of latex from rubber trees.

– Latex is a hydrocarbon C₅H₈, called isoprene (2,methylbut 1,3 diene).

Formula of isoprene.

CH₂ = C CH = CH₂

CH₃

– Coagulation of latex leads to formation of a hydrocarbon polymer consisting of isoprene

(2, methylbut-1, 3-diene) units.

– This polymer is called polyisoprene with the formula; [ CH₂ – C = CH – CH₂ ]

CH₃

Characteristics of natural rubber.

– Soft and sticky;

– Low resistance and low tensile strength; thus breaks easily upon stretching;

– Loses its rubber like properties at temperatures above 60oC;

Note:

– These are not good qualities and for industrial purposes this quality must be improved.

– This is done through vulcanisation of rubber.

Vulcanization of rubber.

– Is a chemical reaction in which raw rubber is heated with sulphur and is done purposely to improve the wear quality of rubber.

Process of vulcanization;

– Rubber is heated with sulphur

– The sulphur atoms form links between chains of rubber molecules.

– This reduces the number of double bonds in the polymer; making the material tougher, less flexible and less softer.

– During the process the sulphur atoms attach themselves to the rubber molecule in such a way that the molecules become locked in place and are prevented from slipping.

Note: –

– Soft rubber has about 2% sulphur, while toughened rubber about 10 % sulphur.

– Rubber can also be made artificially in industries and this gives a form of synthetic rubber.

– Other than undergoing vulcanisation it has chemicals that give it desired properties.

– It is made from byproducts of petroleum industry.

– Examples of synthetic rubber include neoprene and thiokol.

Note: Neoprene is made by polymerization of chloroprene (2, chlorobut 1, 3 diene), C₄H₅Cl

Equation:

CH₂ = C CH = CH₂ CH₂ – C = CH – CH₂

Cl Cl

Chloroprene Neoprene or polychloroprene

Properties of synthetic rubber.

Are unreactive, hence they dont react with industrial chemicals like oils, grease, petrol etc.
Are non-inflammable, as they dont catch fire easily.
Capable of withstanding wide range of temperatures without changing shape.
Have high mechanical strength.

Uses of synthetic rubber.

Manufacture of insulating materials for electrical connections.
Making conveyor and seat belts.
Manufacture of car tyres and tubes.
Making gaskets, flexible pipes.

Advantages and disadvantages of natural polymers.

– Are biodegradable hence not a likely cause of environmental pollution.

– Are made from renewable resources such as wool and trees; hence not easily exhaustible.

– Most are not easily flammable hence good materials for items like clothing.

Disadvantages

– Are often very expensive compared to synthetic polymers.

– Some do not last for very long.

– Are easily affected by acids, alkalis, air etc.

(b). Synthetic polymers

– They are man-made e.g. polythene, Perspex

– They are of two main types; thermoplastic and thermosetting polymers.

Thermoplastic;

– Softens on heating and becomes rigid on cooling.

Examples: nylon, polythene, polystyrene, etc.

Thermosetting:

– Are those that become hard on heating and cannot be softened by heating.

Advantages of synthetic polymers and fibres over natural polymers

They can be made into different shapes easily.
They are cheaper.
Are often unaffected by ac ids, alkalis, water and air.
Are usually less denser and yet stronger

Disadvantages

Plastics are non-biodegradable hence causes a lot if problems in disposal
Plastics burn more readily than natural material
Some synthetic polymers give off poisonous gases when they burn. E. g. polyurethane gives off cyanide and carbon monoxide

Methods of polymerization:

Addition polymerization;

– Occurs when unsaturated molecules (monomers) join to form a long chain molecule (polymer) without the formation of any other product.

– Usually the monomers must have at least a double or triple bond.

– One of the bonds in the double or triple bonds in the monomer opens up, and the unbonded electrons form bonds with neighbouring molecules.

Conditions for polymerization.

– High pressures.

Examples.

Polymerization of ethene to Polythene

CH₂ = CH₂+ CH₂ = CH₂ [-CH₂ CH₂ CH₂ CH₂ -]

Calculations involving polymers.

A polyisoprene molecule is represented as: -[ CH₂ C = CH CH -]n

CH₃

– Given that the relative formula mass of polyisoprene is 748 000, calculate the number of isoprene units in the polymer.

Summary on common addition polymers.

Monomer	Polymer	properties	Uses
H H C = C H H Ethene	H H C – C H H Polythene	Light, tough, and durable	– Polythene bags, bowls packaging, electrical insulation; plastic pipes etc
H H H H C C = C H H Propene	H H C C CH₃ H Polypropylene	Light, tough and durable	– Making crates, boxes; and plastic ropes
H H C = C H Cl Chloroethene	H H C – C H Cl Polychloroethene(polyvinyl chloride)	Strong and hard (not as flexible as polythene)	– Making plastic pipes, -electrical insulators, floor tiles, credit cards,
CH = CH₂ Phenyl ethene (styrene)	H H C – C Cl Polyphenylethene (polystyrene)	Light, poor heat conductor, brittle;	-Insulation, -packaging materials and food containers;
F F C = C F F Tetrafluoroethene	F F C – C F F Polytetrafluoroethene (Teflon)	Non-stick surface; Withstands high temperatures	– Non-stick coatings on pans; Insulation;
O CH₂ = C C O CH₃ CH₃ Methylmethacrylate	H CH₃ C – C H C O CH₃ O Polymethylmethacrylate (Perspex)		-Optical components – transparent doors and windows; – Display signs; – dental fillings;

Condensation polymerization

– Occurs when monomers (similar or different) combine to form a long chain molecule; with the loss of small molecules like ammonia or water.

– The monomers should have at least two functional groups.

Reason:

– For molecules to join at both ends permitting chain formation;

Illustration

– Consider two molecules A and B; each with 2 functional groups:

Molecule A; HO A OH, with two OH functional groups;
Molecule B; HOOC B COOH, with two -COOH functional groups

On condensation

HO A O H + H O O C B COOH + HO A OH +..

Lost to form water

Hence;

HO A OH + HOOC B COOH [HO-A-OOC-B-CO]_n + 2H₂O_(l)

Types of condensation polymers.

– Are of two types:

Polyesters

– Are polymerized formed by an ester linkage; usually with the liberation of a water molecule.

Examples

O O O O

HO C C OH + H OCH₂CH₂O H – C C OH + H OCH₂CH₂O + H₂O

A diol n

A dioic acid Polyester

– Are condensation polymer involving monomers that have at least an amine group (in at least one of them); and thus usually result to the evolution of ammonia gas or water.

Example:

O O H H O O H H

H C (CH₂)₄ C OH + H N (CH₂)₆ N H C (CH₂)₄ C OH + H N (CH₂)₆ N + H₂O

Examples of condensation polymers and their uses.

Polymer	Monomer	Uses
Polyvinyl chloride	Vinyl chloride	Making rain coats, plastic discs, plastic water pipes, and electric insulation;
Starch	Glucose	Laundry
Cellulose.	Glucose	Paper and clothing manufacture
Silk and wool	Proteins (amino acids)	Making clothes
Polyester (terylene)	Terephthalic acid and Ethan 1,2 diol	Clothes and seat belts
Bakelite	Urea and methanol	Electrical fittings
Perspex	Methylmethacrylate	Safety belts, windscreen, and plastic lenses;
Nylon	Hexane-1,6-dioyl dichloride; and hexane- 1,6- diamine;	Used in making safety glass; reflectors; contact lenses; false teeth

Further examples of polymerization reactions.

Formation of nylon.

H H O O

NCH₂CH₂CH₂CH₂CH₂CH₂N + CCH₂CH₂CH₂CH₂C – N (CH2)6 N C (CH2)4 C –

H H Cl Cl n

Hexane 1,6 diamine hexane 1,6 dioyl dichloride Nylon

Formation of Terylene
UNIT 2: ACIDS, BASES AND SALTS

Unit Checklist:

Acids;

Meaning
Strong and weak acids
Concentrated and weak acids
Comparing strength of acids
Using evolution of hydrogen and carbon (IV) oxides
Using electrical conductivity
Using PH
Role of solvents in acidic properties of solvents
Hydrogen chloride in water
Hydrogen chloride in methylbenzene

Bases:

Meaning
Strong and weak bases (alkalis)
Measuring strength of alkalis
Using electrical conductivity
Using PH values
Effect of solvent type on properties of ammonia solution
Ammonia n water
Ammonia in methylbenzene

Uses of acids and bases

Oxides and hydroxides

Basic oxides
Acidic oxides
Neutral oxides
Amphoteric oxides
Meaning of amphoteric oxides
To verify amphoteric oxides
Reactions of amphoteric oxides

Salts

Meaning
Preparation methods (summary)
Solubility of salts
Qualitative tests for cations using NaOH and NH₄OH;
Effect of heat on metal oxides and hydroxides.
Effect of sodium carbonate on salt solutions
Properties of cations with sodium chloride, sodium sulphate and sodium sulphite

Solubility and solubility curves

Definition of solubility
Factors affecting solubility
Solubility curves and related calculations.
Fractional crystallization

Water

Hardness of water
Temporary hardness (Meaning; causes; and removal)
Permanent hardness (meaning; cause; and removal)

Acids

– Are substances whose molecules yield hydrogen ions in water; or

– Are substances, which contain replaceable hydrogen, which can be wholly or partially replaced by a metal.

HCl _(aq) H⁺_(aq) + Cl^–_(aq)

OR: – Acids are proton donors i.e. a substance which provides protons or hydrogen ions.

Strength of Acids

– Acids can be categorized as either strong or weak acids;

Strong acids

– Are those which dissociate or ionize completely to a large extent in water, to yield many hydrogen ions.

– They yield to the solution as many protons as they possibly can.

Examples

Hydrochloric acid; HCl_(aq) H⁺ _(aq) + Cl^–_(aq)

Sulphuric acid; H₂SO_4(aq) 2H⁺_(aq) + SO₄^2-_(aq)

Nitric acid; HNO_3(aq) H+ (aq) NO₃^– _aq)

Weak acids.

– Are acids, which undergo partial dissociation to yield fewer hydrogen ions.

– They do not ionize in water completely or to a large extent i.e. some of their molecules remained unionized in solution.

Examples:

Carbonic acid:

H₂CO_3(aq) ^water H⁺_(aq) + HCO₃^–_(aq)

Ethanoic acid

CH₃COOH (aq) ^water H⁺_(aq) + CH3COO(aq)

Note: – concentrated acids and dilute acids

Concentrated acids

– Is an acid with a high number of acid molecules per given volume.

Dilute acids:

Are acids with a low number of acid molecules per given volume.

– Thus there are concentrated strong acids or dilute strong acids; as well as concentrated weak acids and dilute weak acids.

Comparing the strength of acids

(i). Using rate of evolution of hydrogen

Apparatus:

– Boiling tubes; 1M HCl/ H₂SO₄/ HNO₃; Methanoic acid/ tartaric acid; magnesium ribbon.

Procedure:

– One boiling tube is half filled with 1M HCl; while another is half filled with 1M Ethanoic acid.

– 2 pieces of magnesium ribbons are cleaned to remove a layer of oxide on the surface.

– One of the two pieces is put in each tube of the acid.

Observations:

– Hydrochloric acid evolves hydrogen much more quickly than Ethanoic acid yet they were of equal concentration.

Conclusion

– Hydrochloric acid is a strong acid;

– Ethanoic acid is a weak acid.

Note: –

– The same experiment can be repeated with marble chips (CaCO₃) in acids of same concentration.

-The marble chips dissolve more quickly in HCl, which is a strong acid.

(ii). Using electrical conductivity

Procedure:

– 50cm³ of 2M-hydrochloric acid solution is placed into a beaker and set up apparatus as shown below.

– The switch is closed and the brightness of the bulb noted.

Diagram: Electrolytic circuit

Observations

– Strong acids like HCl, HNO₃ and sulphuric acid gave a brighter bulb light than weak acids like ethanoic, carbonic acids e.t.c

Explanations

– Strong acids are completely dissociated and have more H⁺ in solution and hence have got a higher electrical conductivity; than solutions of weak acids which are only partially ionized thus have fewer hydrogen ions in solution

(iii). Using PH

Procedure

– 2cm³ solutions of different acids of equal concentrations are paired into different test tubes.

– To each test tube 2 drops of universal indicator are added.

– Acids tested: HCl, H₂SO₄, HNO₃; ethanoic acid, carbonic acid, and tartaric acid.

– All acids are of 2M solutions

– The indicator colour and hence the PH number of each is noted; by comparing against the indicator chart.

Observations

Substance (1M)

Colour of universal indicator

Sulphuric acid

Hydrochloric acid

Nitric acid

Ethanoic

Carbonic acid

Tartaric acid

Red

Orange

Yellow

orange

Explanations

– Solutions of strong acids contain a higher concentration of hydrogen ions than those of weak acids

– Strong acids have low PH usually less than 3.

– Weak acids have higher PH values usually between 5 – 6.

Role of solvents on acidic properties of a solute.

Experiment: – to find out if solutions of HCl in different solvents display acidic properties

Apparatus and reagents

– Hydrogen chloride gas, water and methylbenzene

– Beakers and a funnel

– Blue and red litmus papers.

Procedure

– Solutions of hydrogen chloride gas are made by bubbling the dry gas from a generator into water and into methylbenzene contained in separate beakers.

– The hydrochloric gas is passed into the solution using an inverted funnel to prevent sucking back.

Apparatus

The resultant solutions are each separately subjected to various tests as shown below and observations recorded

Tests and observations

Test.

Aqueous HCl solution

Solution of HCl in methylbenzene

1. A piece of dry blue litmus paper is dropped into solution

2. Dry universal indicator paper

3. Add magnesium ribbon

4. Add small marble chips

5. Electrical conductivity

Blue litmus turns red

– Turns red (strong acid)

– Evolution of hydrogen

– CO₂ evolved

– Good conductor

No effect on litmus

– Turns green (neutral)

– No reaction

– Does not conduct

Explanation

– The results show that the aqueous solution of hydrogen chloride behaves as an acid; but the solution in methylbenzene lacks acidic properties

– When HCl gas dissolves in water it changes from molecules to ions;

Equation:

HCl _(aq) _water H⁺_(aq) + Cl^–_(aq)

– It is the hydrogen ions which give the acidic properties and these can only be formed in the presence of water

– HCl in water conducts electric current due to presence of free ions in solution

– HCl gas in methylbenzene does not conduct electric current because the HCl exists as molecules hence lack free ions

Note: – hydrogen chloride gas dissolves in water because both HCl and water polar molecules;

– This causes mutual attraction of both ends of HCl molecule by different water molecules causing the dissociation of HCl molecules into ions.

Illustration:

Hence:

HCl _(g) + water HCl _(aq)

HCl _(aq) H⁺_(aq) + Cl^–_(aq)

– The presence of hydrogen ions in aqueous solution of hydrogen chloride explains the electrical conductivity and acidic properties of hydrogen chloride

Acidic properties: –
turns blue litmus paper red;
evolves hydrogen gas when reacted with magnesium;
evolves carbon dioxide on reaction with CaCO₃;

2H⁺_(aq) + CaCO₃ _(s) Ca²⁺_(aq) + CO_2(g) + H₂O_(l)

2H⁺_(aq) + Mg_(s) Mg ²⁺_(aq) + H₂ _(g)

– Methylbenzene has a weak attraction for hydrogen chloride and hence hydrogen chloride remains as molecules in methylbenzene

Bases

– Are substances which accept the protons donated by acids and are hence proton acceptors

NH₃ _(aq) + H⁺_(aq) NH₄⁺_(aq)

CuO_(s) + 2H⁺ _(aq) Cu ²⁺_(aq) + H₂O_(l)

Alkalis

– An alkali is a soluble base i.e. a base that is soluble in water.

– They are compounds, which produce hydroxyl ions in aqueous solutions.

NaOH_(aq) Na⁺_(aq) + OH^– _(aq)

Note: –

When an acid proton reacts with a base (hydroxyl ions) in aqueous solution, a neutralization reaction occurs.

Strength of an Alkali

– Alkalis can be grouped as either strong or weak alkalis.

(a). Strong alkalis

– Are alkalis that undergo complete dissociation in aqueous solution; yielding a large number of hydroxyl (OH-) ions

Examples:

– Sodium hydroxide.

– Potassium hydroxide.

(b). Weak alkalis

– Are alkalis that undergo only partial dissociation in aqueous solution (water) yielding fewer numbers of hydroxyl ions.

Examples

– Calcium hydroxide

– Ammonium hydroxide

Measuring the strength of alkalis

(i). Using electrical conductivity

Procedure

50 cm³ of 2 M sodium hydroxide solution is put into a beaker and the apparatus set as shown below

Apparatus

Procedure

– The same procedure is repeated using other alkalis like NH₄OH; Ca (OH)₂ e.t.c.

Observation

– The bulb lights brightly with KOH and NaOH as electrolyte than with NH₄OH and Ca (OH)₂

Explanation

– NaOH and KOH are strong alkalis and are completely dissociated and have more ions in solution and hence have got a higher electrical conductivity than the weak alkalis of NH₄OH and Ca(OH)_2(aq)

(ii). Using PH values

Procedure

– 2 cm³ of NaOH and 2 cm³ of NH₄OH are each poured into 2 different test tubes separately

– Into each test tube 2 drops of universal indicator are added.

– The colour change is noted and the corresponding PH scale recorded

Observations

Alkali

Colour of universal indicator

– Ammonium hydroxide (1M)

– Calcium hydroxide (1M)

– Sodium hydroxide (0.1M)

– Sodium hydroxide (1M)

– Potassium hydroxide

Blue.

Purple.

Note: the PH scale

– Is a scale which gives a measure of the acidity of alkalinity of a substance.

Illustration: a PH scale.

1 2 3 4 5 6 7 8 9 10 11 12 13 14

Strong acid Weak acids Neutral Weak alkali Strong alkali

Increasing acidity Increasing alkalinity

(High H⁺ ion concentration) (Low H⁺ ion concentration)

Indicator colours:

PH	1	2	3	4	5	6	7	8	9	10	11	12	13	14
Colour	Red			Orange/ red	Yellow/ Green		Green	Green/ Blue		Blue/ Purple		Purple

Effects of type of solvent on the properties of ammonium solution

Procedure

– Ammonium solution is prepared by bubbling the gas from a generator into methylbenzene (toluene) and into water contained in separate beakers

– The solutions are each divided into 3 portions and tested with litmus paper; universal indicator and for electrical conductivity

Apparatus

Observations

Test	Solution of NH₃ in water	Solution of NH₃ in methylbenzene (toluene)
Dry litmus paper	Red litmus paper turns blue;	No effect
Dry universal indicator paper	Colour turns purple (alkaline PH)	Turns green (Neutral PH)
Electrical conductivity	Poor conductor	Non-conductor

Explanations

– When NH_3(g) dissolves in water it changes from molecules to ions.

Equation:

NH_3(g) + H₂O_(l) NH₄⁺_(aq) + OH^–_(aq)

– It is the hydroxide ions that cause alkaline properties.

– Since ammonium hydroxide is a weak alkali, it dissociates partially releasing fewer hydroxide ions hence the poor electric conductivity.

– Ammonium gas in methylbenzene or trichloromethane exists as molecules without free ions hence no alkaline properties and the electrical conductivity.

Uses of acids and bases

Acids

– Refer to the various acids for uses of sulphuric, nitric and hydrochloric acids.

Bases/ alkalis

– Some weak bases e.g milk of magnesia, are used to relieve stomach disorders.

Amphoteric oxides and hydroxides.

Oxides

– An oxide is a binary compound of oxygen and another element.

– Are of four categories:

Basic oxides
Acidic oxides
Neutral oxides
Amphoteric oxides

(i). Basic oxides

– Are usually oxides of metals (electronegative elements)

– They react with acids to form salt and water only.

Examples

CaO, MgO, CuO etc.

(ii). Acidic oxides

– Are usually oxides of non metals (electronegative elements).

– Many of them react with water to form (give) acids and are known as acid anhydrides

Examples

CO₂; SO₂; SO₃; P₂O₅ and NO₂

(iii). Amphoteric oxides

– Are oxides, which behave as both bases and acids.

– Are mainly oxides of certain metals in the middle group of the periodic table.

Examples

Oxides of Zn, Al, Pb

Experiment: – To verify amphoteric oxides

Procedure

– A small sample of aluminium oxide is placed in a test tube and 5 cm³ of 2M nitric acid added to it and the mixture shaken.

– The procedure is repeated in different test tubes with ZnO, PbO, CuO and CaO.

– The experiments are repeated using excess 2M sodium hydroxide in place of nitric acid

Observations

Name of solid

Observations when

acid is added

hydroxide is added

Aluminium oxide

Zinc oxide

Lead II oxide

Zinc hydroxide

Lead hydroxide

Aluminium hydroxide

Oxide dissolves

Hydroxide dissolves

“

Oxide dissolves

Hydroxide dissolves

“

Explanations

– These oxides are soluble in acids as well as in the alkalis (NaOH)

Reaction with acids

– Oxides react with acids to form a salt and water only in a reaction called neutralization reaction.

Equations

Oxides

(i).PbO_(s) + 2H⁺_(aq) Pb²⁺_(aq) + H₂O_(l)

(ii). Al₂O_3(s) + 6H⁺_(aq) 2Al³⁺_(aq) + 3H₂O_(l)

(iii). ZnO_(s) + 2HCl_(aq) ZnCl_2(aq) + H₂O_(l)

Hydroxides:

(i). Pb(OH)_2(s) + 2HCl_(aq) PbCl_2(aq) + 2H₂O_(l)

(ii). Zn(OH)_2(s) + 2HCl_(aq) ZnCl_2(aq) + 2H₂O_(l)

(iii). Al(OH)_3(s) + 3HCl_(aq) AlCl_3(aq) + H₂O_(l)

Note: in these reactions the metal oxides are reacting as bases

Reaction with alkalis

– These oxides and hydroxides also react with alkalis e.g sodium hydroxide in which case they are reacting as acids.

– Their reactions with alkalis involve the formation of complex ions; M(OH)^2-₄

Equations

Oxides

(i). PbO_(s) + 2NaOH_(aq) + H₂O_(l) Na₂Pb(OH)_4(aq)

Ionically: PbO_(s) + 2OH^–_(aq) + H₂O_(l) [Pb(OH)₄]^2-_(aq)

(ii). Al₂O_3(s) + 2OH^–_(aq)+3H₂O_(l) 2[Al(OH)₄]^–_(aq) + 3H₂O_(l)

Ionically: Al₂O_3(s) + 2OH^–_(aq) + 3H₂O_(l) 2[Al(OH)₄]^–_(aq)

(iii). ZnO_(s) + 2NaOH_(aq) + H₂O_(l) Na₂Zn(OH)_4(aq)

Ionically: ZnO_(s) + 2OH^–_(aq) + H₂O_(l) [Zn(OH)₄]^2-_(aq)

Hydroxides:

(i). Al(OH)_3(s) + NaOH_(aq) NaAl(OH)_4(aq)

Ionically: Al(OH)_3(s) + OH^–_(aq) [Al(OH)₄]^–_(aq)

(ii). Zn(OH)_2(s) + 2NaOH_(aq) Na₂Zn(OH)_4(aq)

Ionically: Zn(OH)_2(s) + 2OH^–_(aq) [Zn(OH)₄]^2-_(aq)

(iii). Pb(OH)_2(s) + 2NaOH_(aq) Na₂Pb(OH)_4(aq)

Ionically: Pb(OH)_2(s) + 2OH^–_(aq) [Pb(OH)₄]^2-_(aq)

Salts

– Is a compound formed when cations derived from a base combine with anions derived from an acid.

– Salts are usually formed when an acid reacts with a base i.e. when the hydrogen ions in an acid re wholly or partially by a metal ion or ammonium (NH₄⁺) radical.

Laboratory preparations of salts

– Salts are prepared in the laboratory using various depending on property of the salt especially solubility

Examples

(a). Preparations by direct synthesis

Equation:

Fe_(s) + Cl_2(g) 2FeCl_3(s)

(b). Reactions of acids with metals, metal oxides, metal hydroxides and metal carbonate

Equations:

Zn_(s) + H₂SO_4(aq) ZnSO_4(aq) + H_2(g)

CuO_(s) + H₂SO_4(aq) CuSO_4(aq) + H₂O_(l)

NaOH_(aq) + HCl_(aq) NaCl_(aq) + H₂O_(l)

PbCO_3(s) + 2NHO_3(aq) Pb(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

Note:

– Acid + metal method will not be suitable if:

The metal is too reactive e.g. sodium or potassium.
The salt formed is insoluble; as it will form an insoluble layer on the metal surface preventing further reaction.
The metal is below hydrogen in the reactivity series.

(c). Double decomposition/ precipitation

– Mainly for preparations of insoluble salts

– Involves formation (precipitation) of insoluble salts by the reaction between two solutions of soluble salts.

Equations:

Pb(NO₃)_2(aq) + 2NaCl_(aq) PbCl_2(s) + 2NaNO_3(aq)

AgNO_3(aq)+ HCl_(aq) AgCl_(s) + HNO_3(aq)

Types of salts:

– Are categorized into three main categories:

Normal salts
Acid salts
Double salts

Solubility of salts: – a summary

– All common salts of sodium, potassium and ammonium are soluble.

– All common nitrates are soluble.

– All chlorides are soluble except silver, mercury and lead chlorides.

– All sulphates are soluble except calcium, barium, lead and stomium sulphates.

– All carbonates are insoluble except sodium, potassium and ammonium carbonates.

– All hydroxides are insoluble except sodium, potassium ammonium and calcium hydroxides is sparingly soluble.

Note:

– Lead(II) chloride is soluble in hot water.

– Calcium hydroxide is sparingly soluble in water.

Reactions of some cations with NaOH_(aq) and NH₄OH_(aq) and solubilities of some salts in water

Cation	Soluble compounds in water	Insoluble compounds in water	Reaction with NaOH_(aq)	Reaction with NH₄OH solution
K⁺	all	None	No reaction	No reaction
Na⁺	all	None	No reaction	No reaction
Ca²⁺	Cl^–; NO₃^–	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	No precipitate
Al³⁺	Cl^–; NO₃^–; SO₄^2-;	CO₃^2-; O^2-; OH^–;	White precipitate soluble in excess	White precipitate insoluble in excess
Pb²⁺	NO₃^–; ethanoate;	All others	White precipitate soluble in excess	White precipitate insoluble in excess
Zn²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Mg²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; OH^–;	White precipitate insoluble in excess	No precipitate
Fe²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; OH^–;	(dark) green precipitate insoluble in excess	Green precipitate insoluble in water
Fe³⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; OH^–;	(red) brown precipitate insoluble in excess	Brown precipitate insoluble in excess
Cu²⁺	Cl^–; SO₄^2-; NO₃^–;	CO₃^2-; O^2-; OH^–;	Pale blue precipitate insoluble in excess	Pale blue precipitate soluble in excess forming a deep blue solution
NH₄⁺	all	none	Ammonium gas on warming	Not applicable

Explanations

– In these experiments NaOH forms insoluble hydroxides with ions of Zn²⁺, Al³⁺, Cu²⁺, Fe²⁺, Ca²⁺, Mg²⁺, Fe³⁺, and Pb²⁺.

– These hydroxides have a characteristic appearance, which form the basis of their identification

Examples

Equations:

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

(White ppt).

Cu²⁺_(aq) + 2OH^–_(aq) Cu(OH)_2(s)

(Pale blue ppt).

Fe²⁺_(aq) + 2OH^–_(aq) Fe(OH)_2(s)

(Dirty green ppt).

Fe³⁺_(aq) + 2OH^–_(aq) Fe(OH)_3(s)

(Red-brown ppt).

Pb²⁺_(aq) + 2OH^–_(aq) Pb(OH)_2(s)

(White ppt).

– The hydroxides of aluminum, zinc and lead dissolves in excess sodium hydroxide solution because of complexes are formed

Equations:

Al(OH)_3(s) + OH^–_(aq) [Al(OH)₄]^–_(aq)

(Tetra-hydroxyl-aluminium (III) ion)

Pb(OH)_2(s) + 2OH^–_(aq) [Pb(OH)₄]^2-_(aq)

(Tetra-hydroxyl-lead (II) ion)

Zn(OH)_2(s) + 2OH^–_(aq) [Zn(OH)₄]^2-_(aq)

(Tetra-hydroxyl-zinc (II) ion)

Note: – in these reactions KOH_(aq) may be used instead of sodium hydroxide

With ammonia solution

– Insoluble metals hydroxides are similarly formed.

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

(White ppt).

Cu²⁺_(aq) + 2OH^–_(aq) Cu(OH)_2(s)

(Pale blue ppt).

Fe²⁺_(aq) + 2OH^–_(aq) Fe(OH)_2(s)

(Dirty green ppt).

Fe³⁺_(aq) + 2OH^–_(aq) Fe(OH)_3(s)

(Red-brown ppt).

Pb²⁺_(aq) + 2OH^–_(aq) Pb(OH)_2(s)

(White ppt).

– However hydroxides of copper and zinc dissolve in excess ammonia solution due to formation of complex ions/ salts

Equations:

Zn(OH)_{2 (s)} + 4NH_3(aq) [Zn(NH₃)₄]²⁺_(aq)+ 2OH^–_(aq)

(White ppt) (Tetra-amine zinc (II) ion; colourless solution)

Cu(OH)_2(s) + 4NH_3(aq) [Cu(NH₃)₄]²⁺_(aq)+ 2OH^–_(aq)

(Pale blue ppt) (Tetra-amine copper (II) ion; deep blue solution)

Effects of heat on metal hydroxides

Procedure

– Hydroxides of Zn, Ca, Pb, Cu e.t.c are strongly heated in a test tube each separately

Observation

– Most metal hydroxides are decomposed by heat to form metal oxides and water

– Sodium and potassium hydroxides only decompose at very high temperatures.

– Hydroxides of metals lower in the reactivity series are readily decomposed by heat than those metals higher in the series.

Examples

Cu(OH)_2(s) ^Heat CuO_(s) + H₂O_(l)

(Blue) (Black)

Pb(OH)_2(s) ^Heat PbO_(s) + H₂O_(l)

(White) (Red brown when hot; yellow when cold)

Zn(OH)_2(s) ^Heat ZnO_(s) + H₂O_(l)

(White) (Yellow when hot; white when cold)

Ca(OH)_2(s) ^Heat CuO_(s) + H₂O_(l)

(White) (White)

Note:

– Both iron (II) and iron (III) hydroxides give iron (III) oxide when heated.

Equations:

2Fe(OH)_2(s) + ½ O_2(g) ^Heat Fe₂O_3(s) + 2H₂O_(l)

(Green) (Red-brown)

Fe(OH)_3(s) ^Heat Fe₂O_3(s) + 3H₂O_(l)

(Brown) (Red-brown)

These oxides do not decompose on further heating

Effects of sodium carbonate on various salt solutions

Procedure

– 3 drops of NaOH_(aq) are added to 2cm³ of 1M solution containing magnesium ions in a test tube

the procedure is repeated with salt solutions containing

Solution containing	Observations after adding sodium carbonate
Mg²⁺	A white precipitate is formed
Ca²⁺	A white precipitate
Zn²⁺	A white precipitate
Cu²⁺	A green precipitate
Pb²⁺	A white precipitate
Fe²⁺	A green precipitate
Fe³⁺	A brown precipitate and a colourless gas that forms a white ppt. in lime water;
Al³⁺	A white precipitate and a colourless gas that forms a white ppt. in lime water;

Explanations:

– Sodium carbonate, potassium and ammonium carbonate are soluble in water; all other metal carbonates are insoluble

– Hence their solutions may be used to precipitate the insoluble metal carbonates.

Ionic equations:

Ca²⁺(aq) + CO₃^2-_(aq) CaCO_3(s)

Note:

Iron (III) and Aluminium salts hydrolyse in water giving acidic solutions which react with carbonates to liberate carbon dioxide gas; hence effervescence.

Reaction of metal ions in salt solutions with sodium chloride, sodium sulphate and sodium sulphate

(i). Procedure

– 2cm³ of a 0.1M solution containing lead ions is placed in a test tube.

– 2-3 drops of 2M sodium chloride solution are added and the mixture warmed;

– The procedure is repeated using salt solutions containing Ba²⁺; Mg²⁺; Ca²⁺; Zn²⁺; Cu²⁺; Fe²⁺and Fe³⁺

– Each experiment (for each salt) is repeated using Na₂SO₄ and Na₂SO₃ respectively, in place of sodium chloride.

(ii). Observations

Solution containing	Sodium sulphate	Sodium chloride	Sodium sulphate
Zn²⁺	– Colourless solution	– Colourless solution	Colourless solution
Mg²⁺	– Colourless solution	– Colourless solution	Colourless solution
Cu²⁺	– Blue solution	– Blue solution	Blue solution
Fe²⁺	– Greenish solution	– Green solution	–
Fe³⁺	– Yellow solution	– Yellow/ dark brown solution	–
Pb²⁺	– White precipitate	– White precipitate which dissolve on warming	White precipitate
Ba²⁺	– White precipitate	– White precipitate	White precipitate

Explanations

– All the listed cations soluble salts except Ba²⁺ and Pb²⁺

– Lead sulphate and barium sulphate are insoluble in water;

– Lead chloride and barium sulphite are insoluble; however PbCl_2(s) dissolves on warming

Equations:

Pb²⁺_(aq) + 2Cl^–_(aq) PbCl_2(s)

Pb²⁺_(aq) + SO₄^2-_(aq) PbSO_4(s)

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

Ba²⁺_(aq) + SO₃^2-_(aq) BaSO_3(s)

Note:

– To distinguish the precipitate of barium sulphate from barium sulphite; dilute HNO_{3 (aq)} or HCl_(aq) is added to both;

– BaSO_3(s) will dissolve in the dilute acid but barium sulphate will not.

Uses (importance) of precipitation reactions.

– Precipitation of metal carbonate from aqueous solutions is useful in softening hard water; usually by removing calcium and magnesium ions from water as insoluble carbonate

Useful information on salts (qualitative analysis)

Colours of substances in solids and solutions in water.

COLOUR
SOLID	AQUESOUS SOLUTION (IF SOLUBLE)
1. White	Colourless	Compound of K⁺; Na⁺, Ca²⁺; Mg²⁺; Al³⁺; Zn²⁺; Pb²⁺; NH₄⁺
2. Yellow	Insoluble	Zinc oxide, ZnO (turns white on cooling); Lead oxide, PbO (remains yellow on cooling, red when hot)
2. Yellow	Yellow	Potassium or sodium chromate;
3. Blue	Blue	Copper (II) compound, Cu²⁺
4. Pale green Green	Pale green (almost colourless) Green	Iron (II) compounds,Fe²⁺ Nickel (II) compound, Ni²⁺; Chromium (II) compounds, Cr³⁺; (Sometimes copper (II) compound, Cu²⁺)
5. Brown	Brown (sometimes yellow) Insoluble	Iron (III) compounds, Fe³⁺; Lead (IV) oxide, PbO₂
6. Pink	Pink (almost colourless) Insoluble	Manganese (II) compounds, Mn²⁺; Copper metal as element (sometimes brown but will turn black on heating in air)
7. Orange	Insoluble	Red lead, Pb₃O₄ (could also be mercury (II) oxide, HgO)
8. Black	Purple Brown Insoluble	Manganate (VII) ions (MnO^–) as in KMnO₄; Iodine (element)-purple vapour Manganese (IV) oxide, MnO₂ Copper (II) oxide, CuO Carbon powder (element) Various metal powders (elements)

Reactions of cations with common laboratory reagents and solubilities of some salts in water

CATION	SOLUBLE COMPOUNDS (IN WATER)	INSUOLUBLE COMPOUNDS (IN WATER)	REACTION WITH AQUEOUS SODIUM HYDROXIDE	REACTION WITH AQUEOUS AMMONIA SOLUTION
Na⁺	All	None	No reaction	No reaction
K⁺	All	None	No reaction	No reaction
Ca²⁺	Cl^–; NO₃^–;	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	White precipitate insoluble in excess, on standing;
Al³⁺	Cl^–; NO₃^–; SO₄^2-	O^2-; OH^–;	White precipitate soluble in excess	White precipitate insoluble in excess
Pb²⁺	NO₃^–; ethanoate;	All others;	White precipitate soluble in excess	White precipitate insoluble in excess
Zn²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Fe²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Dark) pale green precipitate insoluble in excess	(Dark) pale green precipitate insoluble in excess
Fe³⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Red) brown precipitate insoluble in excess	(Red) brown precipitate insoluble in excess
Cu²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	Pale blue precipitate insoluble in excess	Pale blue precipitate soluble in excess forming a deep blue solution
NH₄⁺	All	None;	Ammonias gas on warming	Not applicable.

Qualitative analysis for common anions.

	SO₄^2-_(aq)	Cl^–_(aq)	NO₃^–_(aq)	CO₃^2-_(aq)
TEST	Add Ba²⁺_(aq) ions from Ba(NO₃)_2(aq); acidify with dilute HNO_3(aq)	Add Ag⁺_(aq) from AgNO_3(aq). Acidify with dilute HNO₃ Alternatively; Add Pb²⁺ from Pb(NO₃)₂ and warm	Add FeSO_4(aq); Tilt the tube and carefully add 1-2 cm³ of concentrated H₂SO_4(aq)	Add dilute HNO_3(aq); bubble gas through lime water;
OBSERVATION	The formation of a white precipitate shows presence of SO₄^2- ion;	The formation of a white precipitate shows presence of Cl^– ion; Formation of a white precipitate that dissolves on warming shown presence of Cl^–_(aq) ions	The formation of a brown ring shows the presence of NO₃^– ions	Evolution of a colourless gas that forma a white precipitate with lime water, turns moist blue litmus paper red; and extinguishes a glowing splint shows presence of CO₃^2- ions
EXPLANATION	Only BaSO₄ and BaCO₃ can be formed as white precipitates. BaCO₃ is soluble in dilute acids and so BaSO₄ will remain on adding dilute nitric acid	Only AgCl and AgCO₃ can be formed as white precipitates. AgCO₃ is soluble in dilute acids but AgCl is not; – PbCl₂ is the only white precipitate that dissolves on warming	Concentrated H₂SO₄ forms nitrogen (II) oxide with NO₃^–_(aq) and this forms brown ring complex (FeSO₄.NO) with FeSO₄;	All CO₃^2- or HCO₃^– will liberate carbon (IV) oxide with dilute acids

Checklist:

Why is it not possible to use dilute sulphuric acid in the test for SO₄^2- ions;
Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO₃^2-ions?
How would you distinguish two white solids, Na₂CO₃ and NaHCO₃?

What to look for when a substance is heated.

1. Sublimation	White solids on cool, parts of a test tube indicates NH₄⁺ compounds; Purple vapour condensing to black solid indicates iodine crystals;
2. Water vapour (condensed)	Colourless droplets on cool parts of the test tube indicate water of crystallization or HCO₃^– (see below)
3. Carbon (IV) oxide	CO₃^2- of Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺; Cu²⁺;
4. Carbon (IV) oxide and water vapour (condensed)	HCO₃^–
5. Nitrogen (IV) oxide	NO₃^–of Cu²⁺; Al³⁺; Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺
6. Oxygen	NO₃^– or BaO₂; MnO₂; PbO₂;

Reduction-oxidation (Redox reactions)

(a). Displacement reactions.

(i) More reactive halogens metals will displace less reactive metals from solutions of their salts in the series:

Zn Fe Pb Cu

More reactive Less reactive

Example:

– Zinc powder placed in a solution of copper (II) sulphate, which contains Cu²⁺_(aq) ions, will become Zn²⁺_(aq) ions and brown copper solid (metal) will be deposited.

– The Cu²⁺_(aq) is reduced to copper by addition of electrons and the zinc is oxidized to Zn²⁺_(aq) by removal of electrons.

(ii). More reactive halogens will displace less reactive halogens from solutions of their salts in series:

Cl₂ Br₂ I₂

More reactive Less reactive.

Example:

– Chlorine bubbled into a solution of potassium iodide (colourless), which contains I^–_(aq) ions will turn grey (black) as iodine is liberated.

– The chlorine is reduced to Cl^–_(aq) ions by addition of electrons and the I^–_(aq) ions are oxidized to iodine by removal of electrons.

(b). Decolourisation of purple potassium manganate (VII) ions.

When a few drops of purple KMnO₄ solution are added to a compound and the purple colour disappears, then this shows that the MnO₄^–_(aq) ions have been reduced to almost colourlessMn²⁺_(aq) ions

The substance in the solution has been oxidized.

Example:

– KMnO₄ will oxidize Fe²⁺_(aq) ions to Fe³⁺_(aq) ions; pale green solution turns red-brown.

– KMnO₄ will oxidize Cl^–_(aq) ions to Cl_2(g); colourless solution results to a green gas with a bleaching action;

(c). Orange potassium chromate (VI) turning to a green solution.

– Orange solution of dichromate ions, Cr₂O₇^2-_(aq), changes to green Cr³⁺_(aq) ions when the dichromate is reduced.

– The substance causing this change is oxidized.

Example:

– K₂Cr₂O₇ will oxidize Fe²⁺_(aq) to Fe³⁺_(aq)

– K₂Cr₂O₇ will oxidize SO_2(g) to SO₄^2-_(aq)

– Formation of sulphate ions in solution from sulphur (IV) oxide gas is often used in the test for sulphur (IV) oxide gas.

(d). Oxidation of Fe²⁺_(aq) to Fe³⁺_(aq) ions by concentrated nitric acid.

Solubility and solubility curves

Solubility

– Is the maximum number of grams of a solid which will dissolve in 100g of solvent (usually water) at a particular temperature

– A solution is made up of two parts: – a solute and a solvent.

Solute

– The solid part of a solution usually dispersed in the solvent e.g. a salt.

Solvent

– The liquid part of the solution into which the solute is dissolved.

Experiment: to determine the solubility of potassium nitrate at 20^oC.

(i). Materials

– Beakers, evaporating dish, measuring cylinder, burner, scales, thermometer, distilled water and potassium nitrate

(ii). Apparatus

(iii). Procedure

– About 50cm³ of distilled water is placed in a beaker

– Potassium nitrate is added to it a little at a time stirring continuously.

– The nitrate is added until no more will dissolve and there is an excess undissolved salt present. This is the saturated solution of KNO₃ at the temperature.

Note:

Saturated solution: solution that cannot dissolve any more of the solid/ solute at a particular temperature

– The solution is allowed to settle and it is temperature recorded.

– About 25 cm³ of clear solution is poured in a previously weight evaporating dish.

– The mass of the dish and solution is recorded.

– The dish is then heated in a water bath (to avoid spurting) till the solution is concentrated.

– The concentrated solution is allowed to cool and the dish weighted with its contents.

Results and calculations

Temperature	20.0^oC
Mass of evaporating dish + solution	100.7g
Mass of evaporating dish	65.3g
Mass of solution	35.4g
Mass of evaporating dish + dry salt	73.8g

Calculating:

Mass of salt dissolved = (73.8 65.3)g = 8.5g;

Mass of water (solvent) = (35.4 8.5)g = 26.9g

Thus:

If 26.9g of water dissolves 8.5g of KNO3 at 20oC;

Then 100g of water will have ? = 100 x 8.5 = 31.6g of salt;

26.9

Therefore the solubility of KNO₃ at 20^oC = 31.6g per 100g of water

Factors affecting solubility.

(i). Temperature

– For most salts solubility increases with rise/ increase in temperature.

Reason

– Increased temperature increases the kinetic energy, and hence the momentum and velocity of the solvent molecules so that they can disintergrate the solute molecules more effectively.

– However solubilities of certain salts remain almost constant with temperature change

– Solubility of gases however decreases with increase in temperature;

Reason:

Increase in temperature causes the gas molecules to expand and hence escape from the solvent.

Experiment: To investigate the effect the effect of temperature on solubility.

Requirements: potassium nitrate, distilled water, test tube, thermometer, stirrer, bunsen burner, 250 cm³ glass beaker, 4.5g of potassium nitrate.

(ii). Procedure.

Using a 10ml measuring cylinder, measure 5 cm³ of distilled water and add it to the boiling tube containing solid potassium nitrate. Insert a thermometer into the boiling tube and heat the mixture gently in a water bath or while shaking to avoid spillage. Continue heating until all the solid has dissolved. Stop heating and allow the solution to cool while gently stirring with a thermometer. Record the temperature at which the crystals of potassium nitrate first appear. Note this in the table below.

Retain the boiling tube and its contents for further experiments.

Measure 2 cm³ of distilled water and add to the mixture in the boiling tube. Heat until the crystals dissolve, then cool while stirring with a thermometer. Record the temperature at which the crystals again first stat to reappear. Repeat this procedure, each time adding more 2 cm³ of distilled water, heating, cooling and recording the crystallization temperature until the table is completely filled.

Table 2:

Experiment number	I	II	III	IV	V
Volume of water added	5	7	9	11	13
Temperature at which crystals appear (^oC)
Solubility of K in g/100g of water

Questions:

(a). Complete the table and calculate the solubility of solid X in g/100g of water at different temperatures. (2 marks)

(b). Using the table above, plot a graph of solubility of solid X in g/100g of water against temperature. (5 marks)

(c). From the graph:

(i). calculate the mass of K that would be obtained if the saturated solution is cooled from 60^oC to 40^oC. (2 marks)

(ii). determine the solubility at 70^oC. (1mark)

(iii). at what temperature would solubility of K be 100g/100g of water? (1mark)

(ii). Stirring

– Stirring increases the solubility of a solid

Reason

– Stirring causes the molecules of solvent and solute to move faster causing the solute particles to disintergrate more effectively

Solubility curves

– Are curves showing the variation of solubility with temperature.

Uses / importance of solubility curves

– Can be used to determine the mass of crystals that would be obtained by cooling a volume of hot saturated solution from one known temperature to another.

– Solubility differences can be used to separate substances i.e. recrystallization or fractional crystallization (refer to separation of mixtures)

– Separation of salts from a mixture of salts with differing solubilities e.g. extraction of sodium carbonate from Trona (refer to carbon and its compounds)

– Manufacture of certain salts e.g. sodium carbonate by the Solvay process (refer to carbon and its compounds)

Worked examples

An experiment was carried out to determine the solubility of potassium nitrate and the following results were obtained.

Temperature	10	15	30	40	50	60
Mass of KNO₃ per 100g of water	20	25	45	63	85	106

(a). What is meant by solubility? (1 mark)

(b). Plot a graph of mass of potassium nitrate against temperature. (3 marks)

(c). From the graph work out the mass of KNO₃ that would crystallize if a solution containing 70g of KNO₃ per 100g of water was cooled from 45^oC to 25^oC. (2 marks)

(d). Explain what would happen if 100g of KNO₃ was put in cold water and heated to 50^oC. (2 marks)

The table below shows the solubility of sulphur (IV) oxide at various temperatures.

Temperature (^oC)	0	5	10	15	20	25	35	40	45	50	55	60
Mass of SO₂per 100g of water	22	18.4	15.4	13.0	10.8	9.05	7.80	6.80	5.57	4.80	4.20	3.60

(a). On the grid provided plot a graph of solubility against temperature. (3 marks)

(b). From the graph determine:

(i). The lowest temperature at which 100cm³ of water would contain 11.6g of sulphur dioxide.(1 mark)

(ii). The maximum mass of sulphur (IV) oxide that would dissolve in 2 litres of solution at 10^oC. (Assume that the density of the solution is 1gcm^-3) (3 marks)

(ii). Write the equation for this reaction. (1 mark)

(iii). Using the information from the graph, determine the volume of the saturated sulphur (IV) oxide solution that can neutralize 153 cm³ of 2M sodium hydroxide solution at 25^oC. (3 marks)

Water

– Can be pure or impure

Pure water

– Is a pure substance which is a compound of hydrogen and oxygen; that boils at 100^oC; melts at 0^oC and has a density of 1gcm^-3 at sea level.

Impure water

– Are the natural waters constituted of dissolved solutes in pure water.

Hardness of water

– Water without dissolved substances (salts) hence lathers easily with soap is referred to as soft water while water with dissolved substances that does not lather easily with soap is termed as hard water

Experiment: effect of water containing dissolved salts on soap solution

Procedure

– 2 cm³ of distilled water is put in a conical flask.

– Soap solution from a burette is added into the water and shaken until formation of lather is noted.

– If the soap fails to lather more soap solution is added from the burette till it lathers and the volume of the soap required for lathering recorded.

– The procedure is repeated with each of the following: tap water, rain water, dilute solutions of MgCl₂, NaCl, CaCl_2,a(NO₃)₂, CaSO₄, MgSO₄, Mg(HCO₃)₂, Ca(HCO₃)₂, ZnSO₄ , NaHCO₃, and KNO₃.

– The procedure is repeated with each of the solutions when boiled.

Observations

Explanations

– Distilled water requires very little soap to produce lather because it lacks dissolved salts and hence termed soft water.

– Solutions containing NaCl, ZnSO₄, KNO₃ and NaHCO₃ do not require a lot of soap to form lather

Water containing Ca²⁺ and Mg²⁺ ions do not lather easily (readily) with soap

Reason:

– These ions react with soap (sodium stearate) to form an insoluble salt (metal stearate) called (Mg and Ca stearate respectively); which is generally termed scum.

Equations:

With Ca²⁺

2C₁₇H₃₅COO^–Na⁺_(aq) + Ca²⁺_(aq) (C₁₇H₃₅COO^–)₂Ca_(s) + 2Na⁺_(aq)

Sodium stearate calcium stearate

With Mg²⁺

2C₁₇H₃₅COO^–Na⁺_(aq) + Mg²⁺_(aq) (C₁₇H₃₅COO^–)₂Mg_(s) + 2Na⁺_(aq)

Sodium stearate Magnesium stearate

– Thus water with Mg and Ca is termed hard water and can only be made soft by removing these ions upon which the water will lather easily with water

– When Ca(HCO₃)_2(aq) and Mg(HCO₃)_2(aq) are boiled the amount of soap required for lathering decreases than before boiling

Reason

– Boiled decomposes the 2 salts into their respective carbonate s which precipitates from the solution leaving soft water which leathers easily with water

– The amount of soap solution used with solutions containing sulphates and chlorides of calcium and magnesium did not change significantly even after boiling

Reason

– The soluble sulphates and chlorides of Mg and Ca do not decompose upon boiling hence can not be precipitated out.

Types of water hard ness

Temporary hardness

– Is hardness due to the presence of CaHCO₃ or Mg(HCO₃)₂ in water; and can usually be removed by boiling.

Removal of temporary hardness in water:

(i). Boiling:

– Boiling decomposes and an insoluble chalk of CaCO₃ and MgCO₃ respectively is deposited in the sides of the vessel.

– This forms an encrustation commonly known as furr the process being furring.

Equations:

Ca(HCO₃)_2(s) ^Heat CaCO_3(s) + 2CO_2(g) + H₂O_(l)

Mg(HCO₃)_2(s) ^Heat MgCO_3(s) + 2CO_2(g) + H₂O_(l)

(ii). Distillation:

– Water containing dissolved salts is heated in a distillation apparatus;

– Pure water distils over first leaving dissolved salts in the distillation flask (refer to separation of mixtures)

– Is of less economic value as it is too expensive hence disadvantageous.

(iii). Addition of calcium hydroxide:

– Involves adding correct amount of lime water where CaCO₃ is precipitated out.

– This method is cheap and can be used on large scale at water treatment plants.

– However if excess lime (Ca²⁺) ions is added this will make water hard again.

Equation:

Ca(HCO₃)_2(aq) + Ca(OH)_2(aq) 2CaCO_3(s) + 2H₂O_(l).

(iv). Addition of ammonia solution:

– Addition of aqueous ammonia to water containing calcium and magnesium hydrogen carbonates (temporary hard) precipitates calcium and magnesium ions as corresponding carbonates.

Equations:

Ca(HCO₃)_2(aq) + 2NH₄OH_(aq) CaCO3_(s) +2H₂O_(l) + (NH₄)₂CO_3(aq)

(ii). By permutit softener (ion exchange).

– Uses a complex sodium salt (NaX), such as sodium aluminium silicate commonly known as sodium permutit.

– Permutit is a manufactured ion exchange resin.

Iron exchange resin: materials that will take ions of one element out of its compounds and replace it with ions another element

Working principle

– The permutit is contained in a metal cylinder

– The hard water is passed through the column of permutit in the cylinder and it emerges softened at the other end

– As hard water passes through the column ion exchange takes place.

– The Ca²⁺ and Mg²⁺ remain in the column while sodium ions from the permutit pass into water thus softening it.

Diagram: permutit water softener.

Equations:

NaX_(aq) + Ca²⁺_(aq) CaX_(s) + 2Na⁺_(aq)

NaX_(aq) + Mg²⁺_(aq) MgX_(s) + 2Na⁺_(aq)

– When all the Na⁺ ions in the permutit have been replaced by Ca²⁺ and Mg²⁺ ions the permutit can not go on softening water.

– It is then regenerated by washing the column with brine (a strong NaCl solution); during which calcium and magnesium chlorides are washed away.

Equation:

CaX_(s) + 2NaCl_(aq) CaCl_2(aq) + Na₂X_(s)

MgX_(s) + 2NaCl_(aq) MgCl_2(aq) + Na₂X_(s)

Permanent hardness

– Is that due to soluble sulphates and or chlorides of calcium and or magnesium and cannot be removed by boiling

Removal of permanent hardness

(i). By the addition of washing soda (sodium carbonate)

– Washing soda softens hard water by causing the formation of insoluble CaCO₃ or MgCO₃

– The soluble sodium salts left in water do not react with soap.

Equations:

Na₂CO_3(aq) + CaCl_2(aq) 2NaCl_(s) + CaCO_3(S)

Ionically:

Ca²⁺_(aq) + CO₃^2-_(aq) CaCO_3(s)

Na₂CO_3(aq) + MgSO_4(aq) Na₂SO_4(s) + CaCO_3(S)

Ionically:

Mg²⁺_(aq) + CO₃^2-_(aq) MgCO_3(s)

– This method is very convenient and economical on large scale. It softens both temporary and permanent hardness

(ii). By permutit softener (ion exchange); explanations as before

(iii). Distillation.

Advantages of hard water

(i). It is good for drinking purposes as calcium ions contained in it helps to form strong bones and teeth.

(ii). When soft water flows in lead pipes some lead is dissolved hence lead poisoning. However when lead dissolves in hard water insoluble PbCO₃ are formed, coating the inside of the lead pipes preventing any further reaction; this reduces any chances of lad poisoning.

(iii). It is good for brewing and the tanning industries; it improves wine or beer flavour in brewing industries.

Disadvantages of hard water

(i). Soap forms insoluble salts with magnesium and calcium ions; scum (calcium or magnesium stearate) thereby wasting soap.

– For these reason soapless detergents are preferred to ordinary soaps because they do not form scum; but rather form soluble salts with Mg²⁺ and Ca²⁺

Examples of soapless detergents: omo, perfix, persil, fab e.t.c.

(ii). Deposition of insoluble magnesium and calcium carbonates and sulphates formed from hard water result into blockage of water pips due to the formation of boiler scales

(iii). Formation of kettle fur which makes electrical appliances inefficient hence increasing running costs.

(iv). Formation of scum on clothing reduces their durability and aesthetic appearance

UNIT 3: ENERGY CHNAGES IN CHEMICAL AND PHYSICAL PROCESSES (THERMOCHEMISTRY)

Unit Checklist.

1. Introduction

2. Specific heat capacity and enthalpy of a system

3. Endothermic and exothermic reactions.

4. Activation energy

5. Determination of heat changes

Ø Heat of combustion

Meaning
Determination
Calculations

Ø Fuels

Meaning
Heating value of a fuel
Fuel pollution
Choice of a fuel

Ø Heat of neutralization

Meaning
Determination
Calculations

Ø Enthalpy of solution

Meaning
Determination
Calculations

Enthalpy of displacement
- Meaning
- Determination
- Calculations
Enthalpy of precipitation.
- Meaning
- Determination
- Calculations

Energy level diagrams
Thermochemical cycles
Hesss law
Enthalpy of formation.
Heat of formation and bond energies.
Latent heat

Latent heat of fusion
Latent heat of vapourisation
Relationship between vapourisation, fusion and structure.

Introduction:

– Most chemical and physical process are accompanied by energy changes which occur in the form of heat measured in joules (J) and kilojoules (KJ).

– The heat results from the motion of atoms and molecules.

Specific heat capacity

– Specific heat capacity of a substance is the number of joules required to raise the temperature of one gram of the substance by one degree Kelvin e.g. SHC for water is 4.18Jg^-1K^-1

The heat content (Enthalpy) of a system.

– Heat content is denoted by H; while heat change is denoted as ∆H1;

– And; ∆H = H_Products– H_Reactants

= H₂-H₁

– Heat changes in chemical reactions can either be exothermic and endothermic.

Endothermic and exothermic reactions.

(a). Endothermic reactions

– Is a reaction accompanied by a fall in temperature and energy is absorbed from the surroundings.

– The enthalpy change is normally positive since heat of products (H_Products) is higher than heat of reactants (H_reactants).

– Thus H₂-H₁ is a positive value; since H_I is less than H₂.

Graphically this can be denoted as follows;-

∆H = ⁺Ve; reaction is endothermic

Reactants; H₁

Reaction path

Examples:

(i). When NH₄NO₃ dissolves in water, the temperature of the solution drops.

Procedure

– A spatula end-full of ammonium nitrate is dropped into a test tube of water.

– The bottom of the test tube is felt with the hand.

Observations:

– The hand feels cold.

Reason:

– Energy is absorbed by the products, cooling the test tube.

Thus H₁ is less than H₂ giving a positive ∆ H1 showing an endothermic reaction.

(ii) N_2(g) + O_{2 (g)} 2NO_(g); ∆H=+91KJMol^-1

(b). Exothermic reactions.

– Are reactions accompanied by a rise in temperature and energy is liberated to the surroundings.

– The enthalpy change is normally negative, since heat of reactants (H₁) is normally higher than heat of products (H₂).

– Thus, H₂-H₁ gives a negative value since H₁ is higher than H₂.

Exothermic reactions can be represented graphically as follows;-

∆H = ^–Ve; reaction is exothermic

Products; H₂

Reaction path

– In exothermic reactions H₂ is lower than H₁.

Examples:

(i). Manufacture of ammonia in the Haber process;

i.e. N_2(g) + 3H_{2 (g)} 2NH_3(g); ∆H=-46Kjmol^–

(ii) Dissolving sodium hydroxide pellets in water,

-The temperature of the resulting NaOH_(aq) is higher than the temperature of water at room temperature.

-This implies the internal temperature of products is lower than the reactants’ original temperature.

-Thus (H products H reactants) = -ve value, an exothermic reaction.

Activation energy.

– Is the energy required to activate the reactants before a reaction can take place.

– Thus activation energy is the energy required to initiate a reaction.

– The size of activation energy will differ from one reaction to another and so will be the gap between the energy of reactants and the energy is products.

Examples:

(i). Exothermic reactions

E₁

A B Activation energy

Reactants

C_(s) + O_2(g)

∆H = -ve.

E₂

Products

CO_2(g)

Reaction progress (pathway)

At A; bonds are being broken and the energy is absorbed.
At B; bonds are now being formed, and so energy is evolved.

Endothermic reactions

A B

E₂

Products

CaO_(s) + CO_2(g)

∆H = +ve

E₁

Reactants

CaCO_3(s)

Reaction progress (pathway)

At A; bonds are broken and energy is absorbed.
At B; bonds are formed and energy is evolved.

Determination\ measurement of Enthalpy (heat) changes.

Calorimeters are sued and have to be insulated to reduce heat loss to the surrounding.

Source of errors:

– Heat loss to the surrounding.

– Absorption of heat by the calorimeter (vessels).

Main heat changes under consideration:

Enthalpy of combustion (∆H_c)
Enthalpy of neutralization (∆H_neut)
Enthalpy of solution (∆H_soln)
Enthalpy of precipitation (∆H_precip)
Enthalpy of displacement (∆H_Disp)
Enthalpy of formation (∆H_f)

Heat of combustion ∆H_c

– Is the heat changes when the mole of a substance is completely burned in oxygen, at one atmospheric pressure.

– Since heat is usually evolved and hence ∆H is usually negative.

Example;

C_(s) + O_2(g) CO_2(g) ; ∆ H= -394 Kjmol^-1

Apparatus for finding ∆H_c of a fuel.

Heat evolved = Specific heat capacity of water x Mass of water x Temperature rise

= CM∆T; joules, where; C = specific heat capacity.

M = mass of water.

∆T= temperature change.

Examples

Assume

Volume of water in calorimeter = 100cm³

Initial temperature of water = 225^oC

Final temperature of water = 50.5^oC

Change in temperature of water = 28.0^oC

Mass of water = 100g

Mass of lamp before burning = 30.46g

Mass of lamp after burning = 30.06g

Mass of ethanol burnt = 30.46 -30.06

= 0.40g

(a). Determine the heat evolved;

= ∆ H= CM∆ T

=4.2 x 100 x 28

=11760 Joules

=11.760 KJ

(b). Hence calculate the molar heat of combustion of ethanol;

Moles of ethanol burnt = Mass of ethanol burnt; 0.4 = 0.008695 moles.

RFM of ethanol 46

Thus if 0.008695 moles = 11.76 kilojoules,

Then 1mole = ?

1 mole = 11.76 x 1 = -1352.50 KJMol^–

0.008695

= -1352.5 KJMol^-1 (negative value as the reaction is exothermic since heat is evolved).

When ethanol was burnt in the apparatus shown (in example 1), the results were: mass of fuel burnt, M₁=1.50g; mass of water, M₂=500g; ∆T= 13.0^oC. (C=12; H=1; O=16; SHC of water=4.18KJKg^–K^–). Find the molar enthalpy of combustion of ethanol.

Compare the experimental value with the listed value of –1368 KJMol^– and explain any difference.

Solution:

Heat evolved = CM∆T

= 4.18 x 500 x 13

1000

= 27.17KJ

Molar mass of ethanol, C₂H₅OH = [(12 x 2) + (1 x 6) + (1 x 16)] = 46

Thus if 1.5g = 27.17KJ

Then 46 g = ?

= 46 x 27.17

1.5

= 833.213KJMol^–

In an experiment to determine the heat of combustion of methanol, CH₃OH a student used a set up like the one shown in the diagram below. Study the set up and the data and answer the questions that follow.

Data:

Volume of water =500 cm³

Final temperature of water =27.0^oC

Initial temperature of water =20.0^oC

Final mass of lamp + menthol = 22.11g

Initial mass of lamp + methanol =22.98g

Density of water- 1gcm^-3

(Heat change = mass x temperature change x 4.2Jg^-1oC^–1

Questions:

(a). Write an equation for the combustion of methanol. {1mark}

(b). calculate:

(i). The number of moles of methanol used in the experiment {2marks}

(ii). The heat change for this experiment. {1mark}

(iii). The heat of combustion per mole of methanol. {2marks}

(c) Explain why the molar heat of combustion for methanol obtained in this experiment is different from the theoretical value. {2marks}

When 0.6g of element J were completely burned in oxygen, all the heat evolved was used to heat 500 cm³ of water, the temperature of the water rose from 23.0^oC to 32.0^oC. Calculate the relative atomic mass of element J given that the specific heat capacity of water = 4.2jg^-1k^-1, density of water=1.0gcm^-3 and molar heat of combustion of J is 380JMol^– {3marks}

Hydrogen peroxide contained in 100cm³ of solution with water was completely decomposed. The heat evolved was 1380J. Determine the rise in temperature due to the reaction.

(SHC of water = 4.2Jg^-1k^-1; density of water = 1gcm^-3; O =16; H = 1) {3 marks}

Fuels:

– Are compounds which produce a high heat of combustion.

– Fuels can be:-

Solids; such as charcoal, wood, coal.
Liquids; such as ethanol, gasoline
Gaseous; such as methane, water, gas etc.

Basic concepts:

Calorific value

– Is the energy content of a fuel;

– Is the heat evolved when a given mass of fuel is completely burnt in oxygen;

Note:

– Sometimes fuels may undergo incomplete combustion.

– Incomplete combustion of fuel is disadvantageous in that:-

It reduces the energy content.
It leads to pollution.

Heating value:

– Is the amount of heat energy given out when a unit mass or unit volume of a fuel is completely burnt in oxygen.

Fuel pollution.

– Is commonly caused by internal combustion engine.

– Fuel in engine (e.g. petrol) burns completely to water and carbon (IV) oxide, only under ideal conditions.

Causes of fuel pollution:

(i). Incomplete combustion which causes production of (produces) CO and unburnt carbon (soot).

(ii). Some fuels contain sulphur and nitrogen and on combustion release SO₂ and NO₂. These gases are acidic, resulting to acidic rain which corrodes buildings and affects trees and animals in various ways.

(iii). Fuel additives; e.g. tetraethyl lead, Pb (C₂H₅)₄ added to petrol to enhance burning efficiency produces volatile lead compounds in the exhaust fumes.

– Lead is very poisonous and affects the nervous system and the brain in children.

Factors to consider when choosing a fuel.

(i) Heating value,

(ii) Ease and rate of combustion.

(iii) Availability

(iv) Ease of transportation

(v) Ease of storage

(vi) Environment effects.

Heat of neutralization;

– Is the heat evolved when acid and a base react to form and mole of water.

– Alternatively;

– It is the heat evolved when one mole of hydrogen ions from an acid reacts with one mole of hydroxide ions from an alkali to form\ give one mole of water.

Equation:

OH^–_(aq) + H⁺_(aq) H₂O_(l)

– Neutralization reactions are exothermic.

– Are determined by:

– measuring the temperature rise produced when a known volume of acid is neutralized by a known volume of alkali.

Examples.

(a). Strong acids reacting with strong alkalis.

The ∆H is always about –57KJMol^–and higher than that for weak acids – weak alkalis.

Reason:-

– Strong acids and alkalis are already fully ionized in aqueous solution and no heat is lost in ionizing the acid or the alkali.

Consider:

– Reaction between sodium hydroxide and hydrochloric acid

NaOH_(aq) Na⁺_(aq) + OH^–_(aq)

HCl_(aq) H⁺_(aq) + Cl^–_(aq)

On reacting; OH^–_(aq) + H⁺_(aq) H₂O_(l)

Diagrammatically (energy level diagram)

∆H = -57KJMol^-1

H₂O_(l)

Reaction progress (pathway)

(b). Neutralization reactions involving weak acids or weak alkalis.

– The ∆H is lower than expected, e.g. only –52Kjmol-; and hence lower than that for strong acids-strong alkalis.

Reason:

– Weak acids and weak alkalis are NOT fully ionized in aqueous solutions and some heat is used in ionizing them.

Consider; reaction between ethanoic acid and ammonia solution.

NH₄⁺_(aq) + OH^–_(aq) + CH₃COO^–_(aq) + H⁺_(aq)

∆H = +5KJ

∆H = -57KJ

NH_3(aq) + H₂O_(l)+ CH₃COOH_(aq)

CH₃COO^–_(aq) + NH₄⁺_(aq) + H₂O_(l)

Reaction pathway

Therefore;

∆H is given by;

∆H = ∆H_I + ∆H_II

= 5 + (-57)

= -52KJMol^-1

Note: –

– Dibasic acids e.g. H₂SO₄ contains two replaceable hydrogen atoms hence on incomplete neutralization, they form two moles of water.

– Therefore, H neutralization = ½ x ∆H reaction.

Experiment: – To determine the heat of neutralization of hydrochloric acid by sodium hydroxide.

(i) Procedure:-

– A clean dry 250 cm³ glass or plastic beaker is wrapped with a newspaper leaf.

– Exactly 50 cm³ of 2.0M hydrochloric acid solution is transferred into the beaker.

– The temperature T₁ of the acid solution is noted.

– Using another clean dry measuring cylinder, exactly 50 cm³ of 2.0M NaOH solution is measured and its steady temperature T₂ is noted.

– The contents of the beaker (acid), are carefully stirred with a thermometer while adding NaOH.

– The highest temperature T₄ attained by the resulting mixture is noted.

(ii) Apparatus

(iii). Results:

– Temperature of the acid, T₁= ^oC;

– Temperature of the hydroxide, T2 = ^oC;

– Average temperature of the two solutions; T₁ + T₂ = T₃= ^oC

-The highest temperature of the mixture; T₄ = ^oC;

-The temperature change, ∆T= (T₄ –T₃) = ^oC;

Sample calculations:

Given: Temperature of hydrochloric acid solution T₁= 22.75^oC;

Temperature of sodium hydroxide solution T₂= 22.80^oC;

Average temperature of acid and alkali T₁ + T₂= T₃= 22.78^oC;

Highest temperature of alkali and acid mixture, T₄= 36.40^oC;

Temperature change, ∆T= T₄ – T₃ = (36.40 – 22.78)^oC;

=13.62^oC;

Assumption: the specific heat capacity of the solution= 4.2KJKg^-1K^-1.

– In the experiment, 50 cm³ of 2M HCl are neutralized by 50 cm³ of NaOH, thus; volume of the mixture= (50 + 50) cm³ = 100 cm³.

– Taking density of the resultant solution to be 1gcm-3, then;-

Mass of solution M;

=density x volume

=1g/cm³ x 100cm³= 100g.

=0.1 kg i.e. (100/1000)

Thus heat evolved,

Mass of solution x specific heat capacity x temperature change

= MC∆T,

= 0.1Kg x 4.2kj/kg.k x 13.62^oC

=- 5.7 KJ.

– But 50 cm³ of 2M HCl contains 2 x 50 moles, = 0.1 moles of H⁺ ions.

1000

– Similarity 50 cm³ of 2M NaOH contains 2 x 50 = 0.1 moles of OH^– ions.

1000

– This implies that the two solutions neutralize each other completely.

Therefore;

Molar heat of neutralization (heat liberated when one mole of each reagent is used;-

Mole = 5.72KJ

1mole = 1 x 5.72 = -57.2 KJMol^-1

0.1

– Since heat is evolved, the reaction is exothermic, the molar heat of neutralization= -57.2 KJMol^-1.

Thus, enthalpy change;

H⁺_(aq) + OH^–_(aq) H₂O_(l); ∆H_(neut) = -57.2 KJMol^–. (Thermochemical equation)

1mole 1mole 1mole

Note:

Thermochemical equation: refers to a chemical equation which shows the enthalpy change.

Worked examples:

Given, T₁= 21.0^oC; T2=22.0^oC; T₄ = 34.5^oC; volume of hydrochloric acid= 100 cm³, volume of NaOH_(aq)=100 cm³, molarity of the solutions are each 2M. Calculate the heat of neutralization for the two reagents. (Assumptions, density of the mixture=1gcm^-3 and SHC=4.2KJKg^-1K^-1)

2 (a). When 100g of water at 94.0^oC were added to a calorimeter at 17.5^oC, the temperature rose to 80.5^oC. Determine the heat capacity of the calorimeter. What assumption did you make in your calculations?

Solution:

Heat given out by water = heat received by calorimeter of heat capacity C.

Heat = MC∆T,

0.100 x 4180 x (94.0 – 80.5) = C (80.5 – 17.5)

C = 0.100 x 4180 x 13.5

62.5

= 90.288Jg^-1k^-1

Assumption; specific heat capacity of water is 4180jg^-1k^-1

(b). 250 cm³ of sodium hydroxide were added to 250 cm³ of hydrochloric acid in the calorimeter. The temperature of the two solutions was 17.5^oC initially and rose to 20.1^oC. Calculate the standard enthalpy of neutralization.

Solution:

– Assuming the specific heat capacities of the solutions is the same as that of water, 4180jg^-1k^-1.

Heat from neutralization = heat by calorimeter + solutions.

= (C∆T) + MC∆T,

Mass of solutions = 250+250) =500 cm³

= density x volume; = 1gcm^-3 x 500 cm³=500g

Hat from neutralization = 90(20.1-17.5) + (0.500 x 4180 x (20.1-17.5)

= 5670J.

The following results were obtained in an experiment to determine the heat of neutralization of 50 cm³ 2M HCl and 50 cm³ 2M sodium hydroxide.

Mass of plastic cup = 45.1g

Initial temperature of acid = 27.0^oC

Initial temperature of alkali = 23.0^oC

Mass of plastic cup + HCl + NaOH = 145.1g

Temperature of the mixture of acid and alkali = 38.5^oC.

(a) Define heat of neutralization; {1mark}

Solution:-

Enthalpy change when one mole of water is formed from a reaction between an acid and a base.

(b) Write an ionic equation for the neutralization of hydrochloric acid and sodium hydroxide. {1mark} Solution:

H⁺_(aq) + OH^–_(aq) H₂O_(l)

(c) Calculate;

(i) The amount of heat produced during the experiment. {3marks}

(Specific heat capacity of solution=4.2kjkg^-1k^-1, density of solution= 1gcm^-3).

Solution:

Amount of heat = MC∆T

Mass of solution = (145.1 – 45.1) = 100g

Temperature change; ∆T = (38.5)-(27.0 +23.0) = 38.5 – 25 =13.5^oC;

Heat produced ∆H = 100g x 4.2kjg^-1k^-1 x 13.5^oC

=5670 joules

=5.67KJ

(ii). Molar heat of neutralization for the reaction. {3marks}

Solution:

Number of moles involved; taking only NaOH or HCl

1000 cm³ = 2moles

50 cm³= 2 x 50 = 0.1moles

1000

If 0.1 mole produced 5.67 kg; 1 mol = ?

0.1=5.67kg

1mol= (5.67 x 1) = 56.7 KJMol^-1

0.1

(d). Explain why the molar heat of neutralization of NaOH and ethanoic acid of equal volume and molarity would be less than the value obtained in c (ii) above. (1mark)

Solution:

– Some of the heat produced during neutralization is used up by the weak acid to dissociate fully hence the lower value.

(e). Write down the Thermochemical equation for the reaction between NaOH and dilute hydrochloric acid above.

Solution

NaOH_(aq)+ HCl_(aq) NaCl_(aq) + H₂O_(l); ∆H= -56.7KJMol^-1

(f). Draw an energy level diagram for the neutralization reaction in 4 (c ) above.

Na⁺_(aq) + Cl^–_(aq)

∆ H=-56.7KJMOL-1

H₂O_(l)

Reaction pathway

Heat of solution/ enthalpy of solution.

– Is the heat change when a given mass (moles) of a substance is dissolved in a stated amount of solvent (water).

Molar enthalpy of solution:

– Is the heat change when one mole of a substance is dissolved in a stated amount of solvent (water).

– Alternatively;

– It is the heat change when one mole of a substance dissolves in water to give an infinitely dilute solution i.e. (a solution which shows no change in its properties when more water is added).

– Is determined as;

∆H_solution= – ∆H_lattice + ∆H_hydration ; when ∆H_lattice is given as a negative value;

Alternatively;

∆H_solution = ∆H_lattice + ∆H_hydration; where ∆H_lattice is given a s appositive value;

Where,

Lattice energy

– Is the leaf evolved when one mole of a compound is formed from its separate gaseous ions.

Alternatively;

– It is the energy required to break the ionic bonds within a crystal (solid) lattice.

Hydration (solvation) energy,

– Is the heat evolved when one mole of ions are hydrated by water molecules.

Note:

– Water is a good solvent because it has a high negative(-ve) enthalpy of solvation resulting from powerful interaction between polar molecules and solute ions due to large dipoles on water molecules.

– Therefore,

– Heat of solution can either be exothermic or endothermic depending on the magnitude of hydration and lattice energies.

– Solution process occurs into two stages.

Example:

– Consider the dissolution of sodium chloride solid.

(i). Energy is taken in to break the crystalline lattice.

NaCl_(s) Na⁺_(g) + Cl^–_(g)

∆H = +∆H_lattice

=+776KJ.

(ii). Heat is evolved when one mole of icons are hydrated by water molecules.

Na⁺_(g) + Cl^–_(g) H₂O_(l) Na⁺_(aq) + Cl^–_(aq)

∆H hydration=-771KJ

Illustrations:

(a) Energy level diagram for an endothermic dissolving process for a solid MX_(s)

M⁺_(g) + X^–_(g) + H₂O_(l)

Step II

Step I ∆H_hydration

Final state

∆H_lattice M⁺_(aq) + X^–_(aq)

∆H_solution

MX_(s) + H₂O_(l)

-The lattice energy is larger than the hydration energy hence;-

∆H_solution = +ve

(b) Energy level for the diagram for an exothermic dissolving process for solid MX_(s)

M⁺_(g) + X^–_(g) + H₂O_(l)

Step I Step II

∆H_lattice ∆H_hydration

Initial state

MX_(s) + H₂O_(l)

∆H_solution

Final state

M⁺_(aq) + X^–_(aq)

The lattice energy is smaller than hydration energy hence;

∆H solution = -ve.

Note: All gases dissolve with evolution of heat;

Reason:

– There are no intermolecular forces or bonds to break before hydration occurs.

Worked examples:

The equation below represents changes in physical states for iron metal.

Fe_(s) Fe_(l); ∆H = 15.4KJMol^–

Fe_(l) Fe_(g); ∆H = 354KJMol^–

Calculate the amount of heat required to change 11.2g of solid into iron to gaseous iron. (Fe=56.0)

Solution:

Total heat needed to convert 1 mole of iron from solid to gas = (15.4Kj + 354Kj); = 369.4Kj;

1 mole = 56g;

Thus if 56g requires 369.4Kj;

Then 11.2g =?

= 11.2 x 369.4; = 73.88Kj;

The lattice and hydration enthalpies for lithium chloride and potassium chloride are given below.

Salt	∆H_lattice	∆H_hydration.
LiCl	-861	-884
KCl	-719	-695

(a). Calculate the enthalpy of solution, ∆H_solution; Kj mol^– for;

Lithium chloride. (2 marks)

Solution:

H_solution = -∆H_lattice + ∆H_hydration

= – (-861) + (-884)

= 861- 884;

= -23Kjmol-

Potassium chloride: (2 marks)

Solution:

H_solution = -∆H_lattice + ∆H_hydration

= – (-719) + (-695)

= 719- 695;

= -24Kjmol-

(b) Which of these two salts above has a higher solubility in water? Give reasons for your answer.

Solution:

Potassium chloride; the difference in ∆H_solution is greater;

Given that the lattice energy of NaCl_(s) is +77kjmol^-1 and hydration energies of Na⁺_(g) and Cl^–_(g) are

-406kjmol^-1 and -364kjmol^-1 respectively. Calculate the heat of solution, ∆H_soln for one mole of NaCl_(s)

Experiment: To measure the molar enthalpy change for the dissolution (enthalpy of solution) for various compounds.

(i) Procedure;

– A clean 250 cm³ glass or plastic beaker is wrapped with a newspaper leaf.

– About 50 cm³ of tap water is measured into the beaker and the steady temperature noted.

– The beaker is held in a tilted position and 2 cm³ of and sulphuric acid added into the water.

(ii) Apparatus.

Caution:

– Concentrated sulphuric acid should always be added to water and never vise versa.

Reason:

– The mixture is then carefully but vigorously stirred using a thermometer and the highest temperature of the solution recorded.

– The experiment is repeated with other compounds:

Ammonium nitrate solid
Potassium nitrate solid
Sodium hydroxide pellets.
Sodium nitrate solid

Results and observations.

	Concentrated sulphuric acid	Ammonium nitrate solid	Potassium nitrate solid	Sodium hydroxide pellets
Temperature of water(˚c)
Highest temperature of solution (˚c)
Change in temperature(˚c)

Calculate the enthalpy change (∆H_soln) in each experiment done.

Additional information;

– Density of concentrated Sulphuric acid =1.84gcm^-3

– Density of water =1gcm^-3

– No change in volume when any of the solids dissolved in water

– The specific heat capacities of the solutions=4.2kjkg^-1k^-1

(a). Sulphuric acid:

– Total volume of solution = (50 +2) =52 cm³

– Temperature change =A rise of 1ºC

– Enthalpy change; ∆H_soln; =MC∆T

M = 52 cm³ x 1gcm^-3= 52g = 0.52kg.

C = 4.2kjkg^-1k^-1

∆T = 5K.

Thus enthalpy change = 0.52 x 4.2 x 1

= – 2.184kj

Calculating molar enthalpy of solution of concentrated sulphuric acid:

Mass of acid used = density x volume

= 1.84 x 2 = 3.68g;

Formula mass of Sulphuric acid = 98g

3.6g of acid liberated 2.184kj

Thus 98g of acid liberated; 2.198 x 98 =56Kj;

3.68

Because heat was liberated (+ve ∆T), the ∆H_soln = -56Kjmol^-1

(b). Sodium nitrate:

Data:

– Volume of water used = 100 cm³

– Mass of sodium nitrate = 2.5g

– Initial temperature of water = 21.0ºC

– Final temperature of water =19.5ºC

– Temperature change; = A fall of 1.5ºC i.e. (19.5 – 21.0) ºC;

– Mass of 100 cm³ of water = 100g = 0.1kg;

Enthalpy change i.e. heat absorbed = MC∆T

= 0.1 x 4.2 x 1.5

= 0.63Kj.

Molar heat of dissolution;

– Molar mass of NaNO3 =23 + 14 + 48 = 85g;

– 2.5g of NaNO₃ absorbed 0.63Kj;

– 85g of NaNO₃ will absorb =?

= 0.63 x 85

2.5

= +21.42Kjmol^-1

Since heat is absorbed (-ve heat change), the reaction is endothermic thus ∆H_soln = +21.42Kjmol^-1

Note: energy level diagram for the dissolution of sodium nitrate:

NaNO_3(aq)

∆H_soln

Reaction path

Potassium hydroxide;

– Calculate the molar enthalpy of dissolution of potassium hydroxide (based on values obtained).

Heat of displacement.

– This refers to the enthalpy change that occurs when one mole of a substance is displaced from a solution of its ions.

Experiment: To determine the molar enthalpy change in the reaction between Cu²⁺ ions and zinc or iron.

(i) Procedure:

– A plastic cup or glass beaker is wrapped with a newspaper leaf.

– 25 cm³ of 0.2M copper (II) sulphate solution is transferred into the beaker.

– The steady temperature of the solution is noted.

– 0.5g of zinc powder are carefully transferred into the plastic cup and stirred with a thermometer.

– The highest temperature attained by the solution is recorded.

Spatula

(ii). Procedure:

Thermometer

Observations:

– The blue colors of copper (II) sulphate fades.

– Brown deposits of copper metal are formed in the plastic cup.

Explanation:

– Zinc is higher in the electrochemical series than copper;

– Zinc therefore displaces copper ions from its solution.

Equations:

Zn_(s) + CuSO_4(aq) ZnSO_4(aq) + Cu_(s)

Ionically:

Zn_(s) + Cu²⁺_(aq) Zn²⁺ + Cu_(s)

– During the reaction the blue Cu²⁺ in the solution are replaced by the colorless Zn²⁺.

– Consequently the blue colour of the solution fades, as brown deposits of copper metal are formed in the plastic cup.

– Excess solid (zinc powder) was used to ensure complete displacement of Cu²⁺.

Results:

– Initial temperature of copper sulphate solution, T₁= 23˚C

– Highest temperature of the mixture T₂ = 33ºC

– Temperature change, ∆T; = T₂-T₁=10ºC

– Volume of copper sulphate solution used = 25.0 cm³.

– Mass of zinc powder taken = 0.6g

– Density of the solution = 1gcm^-3.

– Specific heat capacity of the solution = 4.2Kjkg^-1k^-1.

Assumptions:

– The volume of the solution remains unchanged after the reaction.

(Cu = 63.5, S = 32, O =16, Zn = 65, Fe =56)

Question:

Using the above data, calculate, calculate the;

(i). Heat change for the above reaction.

(ii). The molar heat of displacement of copper (II) ions.

Solutions:

(i) Heat change = MC∆T

Mass = density x volume

1gcm^-3 x 25cm³ =25g.

∆H = 25kg x 4.2 kjkg^-1k^-1x 10k

1000

= 1.050Kj;

(ii). Molar heat of displacement of Cu²⁺ displacement.

1000 cm³ = 0.2 moles.

25 cm³ = 0.2 x 25 = 0.005moles

1000

Thus,

0.005moles =1.050Kj

1mole =1 x 1.050 = 1050 = 210Kj;

0.005 5

– Since the final temperature of the mixture is higher than initial temperature; it means the H_products is lower than H_reactants.

– The reaction is thus exothermic.

– Thus molar heat of displacement of Cu²⁺ = -210Kjmol^–

Thermochemical equation:

Zn_(s) + Cu²⁺_(aq) Cu_(s) + Zn²⁺_(aq); ∆H=-210kJmol^-1

Note:

– The experimental value for the heat liberated in this reaction is lower than the theoretical value of 216Kjmol^-1.

Reasons:

– The heat lost to the surroundings and the heat absorbed by the apparatus is not accounted for in the calculations.

– This reaction is typically a Redox reaction.

i.e. Oxidation

Cu²⁺_(aq) + Zn_(s) Cu_(s) + Zn²⁺_(aq)

Reduction.

Heat of precipitation:

– Is the heat change which occurs when one mole of a substance is precipitated from the solution.

Experiment: To determine the heat of precipitation of silver chloride.

Procedure:

– 50 cm³ each of 2M AgNO₃ and 2M NaCl are left in separate beakers and their constant temperatures noted and recorded.

– The NaCl solution is added to the silver nitrate solutions.

– The beaker is covered with a card board and shaken gently to allow mixing of the solution.

– The highest temperature of the mixture is noted and recorded.

(ii). Apparatus: 50cm³ 2M NaCl_(aq)

Thermometer

Beakers

50cm³ 2M AgNO_3(aq50cm³ 2M HCl_(aq)50cm³ 2M AgNO_3(aq)Mixture (1ooml)

(ii) Data (results)

– Temperature of AgNO₃ solution = 21.0ºC

– Temperature of NaCl solution =19.0ºC

– Mean temperature of the solutions = {19+21} = 20.0ºC

– Final temperature after mixing = 34.0ºC;

– Temperature change = a rise of 14^oC;

– Volume of solution mixed =100 cm³

– Mass of solution mixed =100g (assuming density=1gcm^-3)

Questions:

– From the above data, calculate;-

(i) Energy change/ heat change for the reaction

(ii) The molar ∆H_{precipitation} for AgCl.

Solution:

(i) Heat change, = MC∆T

= 100 x 4.2 x 14

1000

= – 5.88kj

(ii) Equation,

AgNO_3(aq) + NaCl_(aq) AgCl_(s) + NaNO_3(aq)

Mole ratio 1 : 1 : 1 : 1

Moles of AgNO₃;

1000 cm³ = 2 moles

50 cm³= 2 x 50 = 0.1mole;

Since reaction ratio AgNO₃: AgCl = 1:1, then moles of AgCl_(s) precipitated = 0.1mole

If 0.1mole = 5.88kj

1 mole = (1 x 5.88) = 58.8kJmol^–

0.1

Reaction being exothermic, ∆H_precip = -58.8kJmol^–

Thermochemical equation:

Ag⁺_(aq) + Cl^–_(aq) AgCl_(s); ∆H= -58.8kjmol-

Suitable energy level diagram,

Ag⁺_(aq) + Cl^–_(aq)

∆H_prep= -58.8kJmol^–

AgCl_(s)

Reaction pathway

Enthalpy of formation.

– Is the heat change when one mole of a substance is formed from its constituent elements under standard conditions.

Note: this will be dealt with under thermochemical cycles and Hesss law.

Standard conditions for measuring enthalpy changes.

– Experimental evidence shows that enthalpy changes are affected by physical stats of the reactants, temperature, concentration and pressure.

– Consequently, for comparison purposes certain conditions have been chosen as standard for measurement and determination of enthalpy changes.

– These conditions are:

A temperature of 25^oC (298K)
Pressure of 1 atmosphere (101.325K)

Standard enthalpy changes:

– Are values of enthalpy changes that are measured under standard conditions.

– They are denoted by he symbol ∆H⁰

– A subscript is also added to the symbol to indicate the type of enthalpy change involved.

Common standard enthalpy changes.

(i). ∆H⁰_f; refers to the standard molar enthalpy change of formation.

Example:

H_2(g) + ½ O_2(g) H₂O_(l); ∆H⁰_{f (H2O)} = -286KJmol^–

(ii). ∆H⁰_c; refers to the standard molar enthalpy change of combustion;

Example:

CH_4(g) + 2O_2(g) CO_2(aq)+ 2H₂O_(l); ∆H⁰_c(CH4)= -890KJmol^–

(iii). ∆H⁰_neut; refers to the standard molar enthalpy change of neutralization.

Example:

HCl_(aq) + NaOH_(aq) NaCl_{(aq) +}H₂O_(l); ∆H⁰_neut = -58KJmol^–

(iv). ∆H⁰_soln; refers to the standard molar enthalpy change of solution (dissolution).

Example:

NaNO_3(s) + H₂O_(l) NaNO_3(aq); ∆H⁰_{soln (NaNO3)} = +21KJmol^–

(v). ∆H⁰_hyd; refers to the standard molar enthalpy change of hydration.

Example:

Na⁺_(s) + Cl^–_(s) + H₂O_(l) Na⁺_(aq) + Cl^–_(aq); ∆H⁰_{hyd (NaCl)} = -774KJmol^–

(vi). ∆H⁰_latt; refers to the molar enthalpy change of lattice formation.

Example:

Na⁺_(g) + Cl^–_(g) NaCl_(aq); ∆H⁰_{latt (NaCl)} = -774KJmol^–

(vi). ∆H⁰_at; refers to the standard molar enthalpy change of atomization.

Example:

Na_(s) Na_(g); ∆H⁰_{at (Na)} = 108.4 KJmol^–

Energy level diagrams, thermochemical cycles and Hess’s law.

Hess’s law (the law of constant heat summation);

States that:

– If a reaction can occur by more than one route the overall change in enthalpy is the same, whichever route is followed.

Route 1; ∆H₁

Illustration:

Route 2

Route 2 ∆H₂∆H₃

Therefore,∆H₁= ∆H₂ + ∆H₃

Worked examples:

Use the information below to determine the enthalpy of combustion of carbon (formation of carbon (IV) oxide).

C_(s) + ½O_2(g) CO_(g); ∆H = -110.45KJ

CO_(g) + ½O_(g) CO_2(g); ∆H = -282.0KJ

Solution:

Route 2

Route 2 ∆H₂∆H₃

CO_(g)

Therefore; ∆H₁ = ∆H₂ + ∆H₃

= -110-45 + (-282.0) = – 392.45kJ

∆H₁= ∆H_combustion

= -392.45KJ;

Note: – Enthalpy of formation, ∆H_f;

– Is the heat absorbed or evolved when one mole of a substance is formed from its elements under standard conditions.

– This can not be determined experimentally because reactions cannot take place under standard conditions.

– In such cases the enthalpy change is determined theoretically using other measurable enthalpies.

– In the case of enthalpy of formation, the enthalpies of combustion of the compound and that of its constituent elements are linked using an energy level or energy cycle diagram.

– The enthalpy of formation is then calculated using the law of constant heat summation (Hesss law).

Worked examples:

One mole of butane(C₄H₁₀) burns completely in oxygen and liberates 2877kj.

(a) Write an equation for the combustion of butane.

2C₄H_10(g)+ 13O_2(g) 8CO_2(g) + 10H₂O_(l)

(b) Draw an energy cycle, hence an energy level diagram for the reactions concerned.

Solution:

Equations for combustion:

C_(s) + O_2(g) CO_2(g); then 8C_(s) + 😯_2(g) 8CO_2(g)(based on balancing for butane)

H_2(g) + ½ O_2(g) H₂O_(g); then 10H_2(g) + 5O_2(g) 10H₂O_(g); (based on balancing for butane)

2C₄H_10(g)+ 13O_2(g) 8CO_2(g) + 10H₂O_(l)

Then, energy cycle for the changes,

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c

8CO_2(g) + 10H₂O_(l)

Thus energy level diagram for the reactions.

8C_(s) + 10H₂O_(l)

∆H₂ = ∆H_f

2C₄H_10(l)

+½13O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_c

+13O_2(g)

8CO_2(g) + 10H₂O_(l)

Reaction pathway (progress)

(c) If the following heats of combustion are given

∆Hº_c (graphite) = -393kjmol^–

∆Hº_c (H_2(g)) = -286KJmol^–

∆Hº_c (C₄H₁₀) = -2877kjmol^–

Calculate the heat of formation of butane, C₄H₁₀ from its elements in their normal states at S.T.P.

Solution:

∆H₁=∆H₂ + ∆H₃ where:

∆H₂= ∆H_formation of C₄H₁₀;

∆H₁=∆Hº_c(graphite) + ∆Hº_c(hydrogen);

While ∆H₃= ∆Hº_c(butane)

From the equation,

∆H₁= ∆H₂ + ∆H₃,

∆H₂ = ∆H₁– ∆H₃

= {8(-393) +10(-286)]-[2(-2877)]

Dividing all through by 2;

= {4(-393) +5(-286)]-2877]; i.e. in order to work with 1 mole of butane;

= (-1572) + (-1430) – (-2877)

= -3002 + 2877

= -125KJmol^–

2 (a). Define standard heat of combustion of a substance.

Solution:

– The heat change when one mole of a substance is completely burnt in oxygen under standard conditions;

(b). Ethanol, CH₃CH₂OH; cannot be prepared directly from its elements and so its standard heat of formation must be obtained indirectly.

(i) Write an equation for the formation of ethanol from its elements in their normal physical states at standard conditions of temperature and pressure.

Solution:

2C_(s) + 3H_2(g) + ½O_2(g) C₂H₅OH_(l)

(ii). Draw an energy level diagram linking the heat of formation with its heat of combustion and the heats of combustion of its constituent elements.

Solution:

Equations

(i). C₂H₅OH_(l) + 3O_2(g) 2CO_2(g) + 3H₂O_(l)

Balancing based on equation for combustion of ethanol

(ii). 2C_(s) + 2O_2(g) 2CO_2(g)

(iii). 3H₂O_(g) + 1½O_2(g) 3H₂O_(l)

Then, energy cycle for the changes,

+3O_2(g)

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c

+3O_2(g)

2CO_2(g) + 3H₂O_(l)

Thus energy level diagram for the reactions.

2C_(s) + 3H_2(l)+ ½ O_2(g)

∆H₂ = ∆H_{f (ethanol)}

C₂H₅OH_(l)

+3O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_{c (ethanol)}

+3O_2(g)

2CO_2(g) + 3H₂O_(l)

Reaction pathway (progress)

(iii). If the following heats of combustion are given:

∆H_c(graphite)=-393kJmol^–

∆H_c(hydrogen)=-286kJmol^–

∆H_c(ethanol)= -1368kJmol^–

Calculate the heat of formation of ethanol, ∆H⁰_{f (ethanol)}

Solution:

∆H₁ = ∆H₂+ ∆H₃

∆H₂= ∆H₁– ∆H₃

= [2(-393 + 3(-286)]-[-1368]

= [(-786) + (-856)] – [-1368]

= -164 + 1368

= – 276kJmol^–

Given the standard enthalpies of combustion as

C_(s) + O_2(g) CO_2(g); ∆H₁= -394kJmol^–;

H_2(g) + ½O_2(g) H₂O_(l); ∆H₂= -286kJmol^–

C₂H_2(g) + 1½O_2(g) 2CO_2(g) + H₂O_(l); ∆H = -1300kJmol^–.

Find the standard enthalpy of formation of ethyne, C₂H_2(g).

Solution:

– Considering the H_c(ethyne), 2 moles of CO₂ and 1 mole of H₂O_(l) are produced;

– Rebalancing the equations for H_c(carbon) and H_c(hydrogen) likewise:

2C_(s) + O_2(g) 2CO_2(g);

H_2(g) + ½O_2(g) H₂O_(l);

Then, energy cycle for the changes,

+2½O_2(g)

∆H_c; ∆H₁ = ∆H₂ + ∆H₃ ∆H₃ = H_c_(ethyne)

+2½O_2(g)

2CO_2(g) + H₂O_(l)

Thus energy level diagram for the reactions.

2C_(s) + H_2(l)

∆H₂ = ∆H_{f (ethanol)}

C₂H_2(l)

+2½O_2(g)

∆H₁= ∆H₂ + ∆H₃ ∆H₃ = ∆H_{c (ethanol)}

+2½O_2(g)

2CO_2(g) + H₂O_(l)

Reaction pathway (progress)

Thus ∆H₁ = ∆H₂ + ∆H₃

∆H₂ = ∆H₁– ∆H₃

= [2(-394) + (-286)]-[-1300]

= (-788-286) + 1300

= -1074 + 1300

= +226kJmol^–

Since ∆H_f is +ve, ethyne is described as an endothermic compound.

Note:

– In the determination of ∆H_f of ethyne, Hess law is used because attempts to make ethyne from carbon and hydrogen {2C_(s) + H_2(g) C₂H_2(g)}will result to the formation of a mixture of hydrocarbons.

Heat of formation and bond energies

– Chemical reactions involve:

Bond breaking; this requires energy.
Bond formation; releases energy.

Example:

– Formation of methane from carbon and hydrogen.

C_(s) + 2H_2(g) CH_4(g); ∆H = -74.9kJmol^–

Energy level diagonal for the formation of CH₄.

C_(s) + 4H_(g)

II ∆H_at;

C_(s) + 2H_2(g)

I ∆H_at = +715KJmol^–III ∆H = -4(C-H); bond energy

C_(s) + 2H_2(g)

∆H_IV = I + II + III

Final enthalpy = -74.9Kj

CH_4(g)

Reaction pathway (progress)

Energy changes involved are:

∆H_I; Enthalpy of atomization – this is the energy absorbed when a substance decomposes to form one mole of gaseous atoms.
∆H_II; enthalpy of atomization of H₂; i.e. dissociation of hydrogen molecules to free hydrogen atoms.
∆H_III; enthalpy of formation of four C – H bonds, during which heat is liberated;

Therefore, enthalpy of reaction ∆H_R = ∆H_IV

∆H_R= ∆H₁+ ∆H_II + ∆H_III

= +715 + 4(218) + -4(C-H)

-74.9 =+715 + 872 +-4 (bond energy)

4 (C-H) =1587 + 74.9

=1661.9

C –H =1661.9 = 415.4kjmol-

– This is the amount of energy released when one C-H bond is formed;

Worked examples.

Use the bond energies given below to calculate the heat of reaction for;-

H_2(g) + Cl_2(g) 2HCl_(g)

Bond	Bond energy
H-H	435kjmol^–
Cl-Cl	243kjmol^–
H-Cl	431kjmol^–

Solution:

Bond breaking,

Total energy absorbed = +678 KJ

H_2(g) = H-H 2H_(g); ∆H = +435Kj;

Cl_2(g) = Cl-Cl 2Cl_(g); ∆H= +243Kj

Total energy released = -862 KJ

Bond formation,

2H_(g) + 2Cl_(g) 2(H-Cl)_(g); ∆H=2(-431) = -862Kj;

Heat of reaction:

Energy absorbed in bond breakage + energy released in bond formation.

= [+435 + 243] + [-862]

= [+678-862] kJ

= -184kj

Thus, H – H + Cl – Cl 2[H – Cl]; ∆H=-184Kj

Note: energy level diagram.

2H_(g) + 2Cl_(g)

∆H_II ∆H_{at (chlorine)}

+243KJ

2H_(g) + Cl_2(g)

∆H_I ∆H_at = +435KJmol^– ∆H_III= -2 (H-Cl); bond energy

H_2(g) + Cl_2(g)

∆H_IV = I + II + III

Final enthalpy = -74.9Kj

2HCl_(g)

Reaction pathway (progress)

Study the information given in the table below and answer the questions that follow;-

*Bond*	*Bond energy(kJmol^–)***
C H	414
Cl Cl	244
C Cl	326
H Cl	431

(a). Calculate the enthalpy change for the reaction

CH_4(g) + Cl_2(g) CH₃Cl_(g) + HCl_(g)

Solution:

Bond breaking

(i). 4(C-H) 4CH_(g); ∆H =4(+414) = +1656kj;

(ii). Cl – Cl 2Cl_(g); ∆H1 = +244kj

Bond formation

(i). 3(CH) 3(C-H); ∆H= 3(-414) =-1242kj

(ii). CCl (C-Cl); ∆H= -326kj

(iii). HCl (H-Cl); ∆H= -431kj

Heat change = ∆H_{bond breakage} + ∆H_{bond formation}

= [(+1656) + (+244)] + [(-1242) + (-326) + (-431)]

= -1900 + (-1999)

∆H = -99kJ

(b). Draw an energy level diagram for the reaction

C4H_(g) + 2Cl_(g)

∆H_II ∆H_{at (chlorine)}

+244KJ

C4H_(g) + Cl_2(g)

∆H_I ∆H_at = +1656Kj ∆H_III= -2 (H-Cl); bond energy

CH_4(g) + Cl_2(g) = -99Kj;

∆H_IV = I + II + III

Final enthalpy

CH₃Cl_(g) + HCl_(g)

Reaction pathway (progress)

Study the information given in the table below and answer the questions that follow.

*Bond*	*Bond energy*
C C	346
C = C	610
C – Br	280
Br Br	193
C H	413

(i) Calculate the enthalpy change for the reaction.

CH₂=CH₂ + Br₂ CH₂-CH₂

Br Br

Solution:

Bond breaking,

4(C-H) 4CH_(g); ∆H = 4 (+413) = +1652kj

Br₂ 2Br ∆H = +193kj

C = C C = C; ∆H= + 610kj;

Bond formation:

4(C-H), 4CH; = 4(-413) = -1662kj

2(C-Br), 2CBr; = 2(-280) = -560kj

(C-C ) 2C; = -346kj

Heat change:

∆H bond breakage + ∆H bond formation

= [+1652 + 193 + 610] + [-1652 + -560 + -346kJ]

= [+2455] + (-2558)

∆H change = -103kJ;

Molar heat of fusion, ∆H_f;

– Is the amount of heat required to change one mole of a solid into a liquid at its melting point.

– It is a measure of intermolecular forces between solid particles since it is the heat energy used to separate the solid particles.

Molar heat of vaporization, ∆H_vap;

– Is the amount of heat required to change one mole of liquid into vapor at its boiling point.

– It is a measure of the intermolecular forces between the liquid particles since it is the heat energy used to separate liquid particles.

Relationship between heats of fusion and vaporization to structure.

They are;-

(a). Higher for substances with giant structures e.g.

Ionic solids
Metals
Giant covalent structures, e.g. diamond, graphite and silicon (IV) oxide

(b). Lower in simple molecular structures whose particles are held together by weak intermolecular forces.

Examples:

Iodine crystals
Sulphur

Examples:

	NaCl	C_(graphite)	Cu	H₂O	I_(s)	S
Structure type	Giant ionic	Giant atomic	Giant Metallic	Molecular	Molecular	Molecular
∆H_fusionkjmol^–	28	–	13	6	8	–
∆H_vap kjmol^–	171	717	305	41	21	10

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

UNIT OUTLINE:

Reactions rates:

Ø Measurements of reaction rates

Rate of product yield pre unit time
Rate of disappearance of reactants

Ø Apparatus for measuring reaction rates

Collision theory and activation energy
Factors affecting reaction rates

§ Temperature

Particles size (surface area)
Catalysts
Pressure

Reversible reactants

Meaning
Chemical equilibrium

Ø Factors affecting position of equilibrium

§ Temperature

Pressure
Concentration (of reactants or products)
Catalysts
LeChartaliers principle.

Reaction rate:

– The rate of a chemical reaction is taken as the rate at which products are formed or the rate at which reactants disappear.

Measurements of reaction rates

Reaction rates are measured in terms of

How much product appears in a given time.
How much reactant disappears in a given time

– Rates curves are then plotted

– These are plots of concentration or properties, which are functions of concentration against time

Graph: concentration of products against time.

Time

Graph: concentration of reactants against time.

Time

Apparatus used to determine rates of reaction

Examples:

– Reaction between dilute HCl and marble chips.

– This can be by:

(a). Measuring the decrease in mass with time

Apparatus:

Graph:

Time

(b). Measuring the volume of the gas produced with time and then plotting a graph.

Apparatus:

Graph:

Time

Similar experiments maybe done using zinc granules and zinc powder with an acid

Worked example:

Marble chips and dilute hydrochloric acid were mixed. The mass of the reaction mixture was measured and recorded with time. The results are shown in the table below.

Time (sec)	0	30	60	90	12	150	180	21	240	270	300
Mass of mixture (g)	42.0	41.5	41.0	40.7	40.4	4.02	40.1	40.0	40.0	40.0	40.0
Loss in mass (g)

(a). Complete the table. (5 marks)

(b). Draw a graph of loss of mass against time. (3 marks)

(c). On the same axes sketch the curves that would be obtained if:

(i). the acid was more concentrated. (1mark)

(ii). larger marble chips were used. (1mark)

Collision theory and activation energy.

The collision theory of reacting particles postulates that:

Ø Reacting particles must collide before a chemical reaction occurs

Not all collisions are effective/ result in chemical reaction
Only those particles with sufficient energy result in effective collisions i.e. energy equal to or greater than the activation energy.
Any factor, which increases the rate of a chemical reaction, does so by increasing the number of effective collisions;

Activation energy (E_A):

– Is the minimum amount of energy required by reacting particles to cause a successful collision to form products;

– It refers to the energy an energy barrier that must be overcome by the reactants to be converted to products;

– This energy barrier determines the magnitude of the activation energy of the reactants;

Factors determining the value of activation energy:

– Strength of the bonds in the particles of the reactants;

– Whether the reaction is exothermic or exothermic;

– Presence of catalysts;

Graph: activation energy:

Factors affecting the rate of reaction

– Factors that increase the rate of a reaction does so by:

Increasing number of effective collisions;
Lowering the activation energy; since a reaction with high activation energy is slow at ordinary conditions;

These factors are:

Concentration of reactants
Temperature
Surface area (size of the particles
Catalyst Light
Pressure

(a). Concentration of reactants.

– Increase in concentration of one of the reactants increases the rate of the reaction;

Reason:

– Higher concentration results in a greater probability of collision hence a higher rate of reaction

Example: reaction between Mg _s) and 2M and 4M hydrochloric acid.

Graphical representation

Time (seconds)

– The maximum volume of a gas collected in both experiments is the same because the number of moles of HCl is the same

– Curve for 4M HCl is steeper indicating a faster rate because the acid is more concentration

– The 4M HCl contains more H⁺ions per unit volume than 2M HCl. This reaction takes a shorter time to go to completion

– The start of the flat part of the curve indicates the end of the reaction;

Diagram:

Magnesium and hydrochloric acid mixture

Note: Rate of reaction can also be verified by measuring the rate of disappearance of reactants per unit time.

– Reactants with 4M HCl will disappear faster;

– Curve for a faster reaction rate is always to the left of the reference curve;

Worked example:

In an experiment to determine the effect of concentration on the rate of a reaction, various concentrations of sodium thiosulphate were reacted with equal volume of hydrochloric acid and the time taken for the precipitate to obscure a cross on paper put under the reaction beaker was determined.

The results are shown in the table below.

Volume of S₂O₃^2- (cm³)	Volume of H₂O, cm³	Volume of HCl, cm³	Time for cross to disappear, (s)
50	0	10	20
40	10	10	25
30	20	10	34
20	30	10	52
10	40	10	70

(a). Plot a graph of time (vertical axis) against volume of aqueous sodium thiosulphate. (3marks)

(b). From the graph estimate the time for the cross to disappear when 10 cm³ of hydrochloric acid is added to a mixture of 35cm3 of aqueous thiosulphate and 15cm³ of water. (2 marks)

(c). What is the effect on the rate of the reaction of adding more water to the aqueous sodium thiosulphate? Explain your answer. (2 marks)

(d). On the same graph, plot the curves you would obtain if the experiment were repeated at:

(i). at 45^oC; (1mark)

(ii). using 10 cm³ of less concentrated hydrochloric acid. (1mark)

(b). Effect of temperature.

– An increase in temperature increases the rate of a chemical reaction.

Reasons:

– Increase in temperature increases the kinetic energy of the reacting particles;

– The particles therefore move faster leading to more frequent and more effective collisions.

– The rate of reaction therefore increases;

Example: reaction between sodium thiosulphate and dilute HCl_(aq)

Stoichiometric equation:

Na₂S₂O_3(aq) + 2HCl_(aq) 2NaCl_(aq) + H₂O_(l) + S_(s) + SO_2(g)

Ionic equation:

S₂O₃^2- _(aq) + 2H⁺_(aq) S_(s) + SO_2(g) + H₂O_(l);

– The time taken to precipitate the same amount of sulphur by the same volume of acid at different temperatures is measured.

– Concentration of thiosulphate solution and the acid are kept constant.

Graph: relationship between time and temperature in a reaction.

Temperature (^oC)

– The rate is proportionate to 1/ t;

Hence;

– The higher the temperature the shorter the time for completion of the reaction and the higher the rate of the reaction;

Alternatively:

You can plot a graph of 1/ t against temperature;

Graphically:

Temperature (^oC)

(c). Effect of particle size (surface area).

– A decrease in particle size (surface area) increases the rate of a chemical reaction; and an increase in particle size decreases the rate of a chemical reaction.

Reason:

– A smaller particle size means a larger total surface area and this offers a large surface on which the reacting particles can collide;

– This means more collisions, hence more chances of effective collisions leading to a higher rate of reaction;

Example: Reaction between dilute HCl and marble chips

The larger CaCO₃ granules undergo a slower rate of reaction than the finely powdered CaCO₃

Equation:

CaCO_{3 (S)}+ 2H⁺ _aq) Ca²⁺_(aq) + CO_2(g) + H₂O_(l)

Graph 1: volume of gas produced against time (for granules and powder)

Time (seconds)

Graph 2: loss in mass of calcium carbonate against time

Time (seconds)

Worked example:
1. exactly 3.0g of powdered carbonate of metal M of formula MCO₃ were mixed with excess dilute hydrochloric acid. The mass of the reaction vessel and its contents were recorded at various times. From these readings, the total loss in mass of the reaction vessel and its contents was calculated and recorded as shown in the table below. The experiment was carried out at room temperature.

Time (secs)	0	30	60	90	120	150	180	210
Total loss in mass (g)	0	0.08	0.37	0.90	1.19	1.28	1.32	1.32

(a) (i). Plot a graph of total loss in mass against time. (1mark)

(ii). Determine the total loss in mass after 100seconds. (2 marks)

(b) (i). Write an equation for the reaction that occurs. (1mark)

(ii). calculate the number of moles of:

Carbon (IV) oxide gas produced. (2 marks)

MCO₃ (2 marks)

(iii). Calculate the relative atomic mass of metal M. (3 marks)

(c). On the same axis sketch a curve that you are likely to obtain if the experiment was repeated:

(i)9. at 50^oC. Explain. (2 marks)

(ii). Using MCO₃ granules instead of the powder. Explain. (2 marks)

(d). Effect of catalysts

– Catalysts are substances which alter the rate of a chemical reaction without being consumed, e.g. finely divided iron in the harber process;

– They alter the rate of a chemical reaction but remain unchanged at the end of the reaction;

– They can therefore be reused at the end of the reaction;

Modes of action of catalysts:

Provision of a surface over which particles can react.

Note: Due to this solid catalysts are more effective when used in powder form;

Formation of short-lived intermediate compounds with products which then break up to give the products and the catalyst;
By adsorption;

Note: – Generally, a catalyst increases the rate of a reaction by providing a different pathway of lower activation energy.

– This means more collisions can overcome this energy barrier and result in a reaction.

– It lowers the activation energy of the reaction;

– Catalysts are also reaction-specific;

Some common catalysts:

Catalyst	Reaction catalyzed.
Manganese (IV) oxide	Dissociation of hydrogen peroxide to give oxygen gas and water;
Vanadium (V) oxide or platinum	Contact process for the conversion of sulphur (IV) oxide to sulphur (VI) oxide;
Platinum	Manufacture of nitric acid in the Oswalds process;
Finely divided iron or vanadium	Haber process in the manufacture of ammonia;
Nickel	Hydrogenation of oils to make fats; Hydrogenation of unsaturated hydrocarbons
Copper (II) oxide	Oxidation of ethanol to produce ethanal;
Titanium (IV) chloride	Ziegler method of polymerizing alkenes

Graph: reaction rates with and without a catalyst.

Time (seconds)

Worked example:

Hydrogen peroxide decomposes slowly to water and oxygen under normal conditions.

When a little manganese (IV) oxide is added to the solution the rate of decomposition is enhanced. The results of an experiment on the decomposition of hydrogen peroxide in presence of manganese (IV) oxide are shown below. Study the results and answer the questions that follow.

Volume of oxygen (cm³)	0	19	27	33	36	38	39	40	40
Time (seconds)	0	30	60	90	120	150	180	210	240

(a). Plot a graph of volume of oxygen produced (vertical axis) against time. (3 marks)

(b). What is the rate of reaction:

(i). during the first 30 seconds. (2 marks)

(ii). between 30 seconds and 60 seconds. (2 marks)

(c). Explain why: (i). the slope of the curve is steeper at the start of the reaction. (2 marks)

(ii). the curve goes flat at the end of the reaction. (2 marks)

(d) (i). Suggest the role of manganese (IV) oxide I the decomposition of hydrogen peroxide. (1mark)

(ii). Comment on change in mass of manganese (IV) oxide at the end of the reaction. Explain.(2 marks)

(e). Effect of light on the rate of reaction

-The effect of heating and illuminating substances is the same;

– In both cases the constituent particles absorb radiant energy leading to an increase in the number of particles with activation energy resulting in increased rate of reaction;

– Light energizes the particles involved in a reaction;

– This increases the chances of effective collisions per unit time thus increasing the rate of reaction.

– Light of higher frequencies give higher reaction rates e.g. UV Light;

– Examples of reactions affected by light are those involving halogens:

Examples of reactions affected by light.

Reaction between Cl_2(g) and H_2(g) does not take place in the dark but is explosive in bright light;

Cl_2(g) + H_2(g) ^{bright light} 2HCl _(g)

For the reaction between methane and bromine; decolourisation of bromine only occurs in presence of light;

CH_4(g) + Br_2(l) ^Light CH₃Br_(g) + HBr_(g)

Experiment: Effect of light on the decomposition of silver bromide.

Procedure:

– About 20cm³ of 0.1M potassium bromide is put in a glass beaker.

– 5cm³ of 0.05M silver nitrate solution is added.

– The resulting pale yellow precipitate is divided into three portions in 3 separate test tubes.

– One of the test tubes in immediately placed in a dark cupboard; the second on a bench and the third is placed in a direct source of light e.g. sunlight.

– Formation of a pale yellow precipitate of silver bromide when silver nitrate reacts with potassium bromide.

Equation:

KBr _(aq) + AgNO_3(aq) AgBr_(s) + KNO_3(aq)

Pale yellow

Test tube in light: precipitate changes colour from pale yellow to grey.
Test tube on the bench: slight change in colour from pale yellow to slight grey.
Test tube in dark cupboard: no observable (noticeable) colour change in precipitate.

Explanation:

– Light decomposes silver bromide to metallic silver (hence the grey colour) and bromine.

Equation:

_Light

2AgBr_(aq) 2Ag_(s) + Br_2(g)

grey

– No observable change in test tube placed in darkness due to lack of light;

– The degree of decomposition and hence change depends on the light intensity falling on the test tubes;

– The rate of decomposition of silver bromide increases with increase in light intensity.

Conclusion:

– Light affects the rate of some chemical reactions by energizing the particles involved in a reaction hence increasing the chances of effective collisions per unit time thus increasing the rate of reaction.

Application of effect of light on reaction rate.

– Processing of black and white photographic films is done in dark to prevent the decomposition of the silver bromide that is usually used to coat photographic plates;

Further examples of reactions affected by light:

(i). Cl_2(g) + H_2(g) ^UV ^Light 2HCl_(g)

CH_4(g) + Br_2(g) ^{UV Light} CH₃Br_(g) + HBr_(g)

(iii). 6CO_2(g) + 6H₂O_(l) ^Light C₆H₁₂O_6(aq) + 6O_2(g)

(f). Pressure

– Increase in pressure increases the rate of reaction involving the formation of small volume of product because of increase in concentration and slight increase in temperature

– In a given volume the higher the number of molecules of a given gas in a container, the greater the pressure;

– Increase in pressure causes the same effect as increase in concentration;

– Thus the rate of reactions involving gases can be increased by increasing the pressure of the gases

Note:

From gas laws, decreases in volume results in increase in pressure. Thus, reactions accompanied by decrease in volume move faster to completion.

Reversible reactions

– Are reactions which can be made to go to either direction (forward or backward) by changing conditions such as temperatures, pressure etc.

– In such cases none of the reactants is completely used up and so the reaction does not go into completion.

Note:

Reversible reactions are of two types:

Reversible physical changes e.g. heating ice, iodine etc.
Reversible chemical changes e.g. heating hydrated copper (II) sulphate, Haber process, decomposition of limestone, heating blue cobalt chloride, ammonium chloride etc.

Examples:

– Reaction of nitrogen and hydrogen to give ammonia in the Haber process.

N_2(g) + 3H_2(g) 2NH₃ _(g)

– The reversible sign means the reaction can reach a state of equilibrium if left undisturbed.

Explanations of reversible reactions:

Consider the curves below:

Time

Point (a): forward reaction.

– At the start of the reaction, the rate of the forward reaction is faster.

– The concentration of the reactants is greatest at the beginning.

– The rate of forward reaction and the concentration of reactants decreases with time as the reaction proceeds.

Point (b): backward reaction.

– At zero time the rate of the backward reaction is zero.

– The concentration of the products is lowest at this time.

– The rate of the backward reaction and concentration of products increases as the forward reaction proceeds.

Point (c): equilibrium.

– It comes to a time (t) when the rate of forward reaction is equal to the rate of backward reaction.

– This balance is called equilibrium.

Equilibrium:

– Is the state at which the rate of forward reaction is equal to the rate of backward reaction.

Types of equilibrium:

Static equilibrium:

– refers to a situation when two opposing forces balance each other and whatever was happening before comes to a standstill.

Dynamic equilibrium:

– Refers to a situation when two opposing processes, the forward and reverse, continue taking place but at the same rate.

– The equilibrium is said to be dynamic and this state of balance can be reached from either direction.

Examples:

3Fe _(s) + 4H₂O _(g) Fe₃O₄ _(s) + 4H_{2 (g)}

– The Equilibrium state here is established when:

Either steam is passed over heated iron in a closed container; or
When hydrogen is passed over heated iron oxide;

Characteristics of equilibrium systems.

(i) The concentrations of reactants and products do not change after the equilibrium has been reached unless the system is disturbed.

Note:

– Observable properties of reactants or products such as colour, mass, volume, PH, temperature etc can be used to detect whether a system is in equilibrium or not.

Examples:

Property	How it can be used to detect attainment of equilibrium
Colour	The colour intensity remains constant at equilibrium
Volume.	Total volume of solution remains constant at equilibrium;
Precipitate	The height of the precipitate remains constant;

(ii). The equilibrium can be reached from either direction.

(iii). All the reactants and products are present in the system.

Factors affecting the position of equilibrium

– Such factors act as a strain on the state of equilibrium and the system reacts in a way to oppose the change.

– The factors are:

Temperature
Pressure
Concentration

(a). Temperature

Endothermic reactions

– Increase in temperature favours the endothermic reactions;

Example:

N₂O₄ _(g) 2NO₂ _(g) ; ∆H = + ve

Pale yellow dark brown

– Increase in temperature shifts the equilibrium to the right; since the reaction is endothermic and hence more yield of nitrogen (IV) oxide; hence dark brown fumes will be observed;

– Decrease in temperature shifts the equilibrium to the left; since the reaction is endothermic favoured by low temperatures; hence formation of N₂O₄; and the pale yellow colour is observed;

Exothermic reactions:

– Decrease in temperature favours exothermic reactions.

Example:

2 SO_2(g) + O_2(g) 2SO_3(g); ∆H = -ve

Thus:

– Decrease in temperature causes more yield of sulphur (VI) oxide; because the equilibrium shifts to the right; since the reaction is exothermic favoured by low temperatures;

– Increase in temperature causes SO₃ to decompose decreasing its yields. Increase in temperature shifts the equilibrium to the left; because the reaction is exothermic which is favoured by low temperatures;

(b). Pressure

-Increase in pressure favours the side with fewer numbers of gaseous molecules since this is the side where pressure is reduced (low).

Example:

N₂ _(g) + 3H₂ _(g) 2NH₃ _(g)

– Increase in pressure favours the production of ammonia; by shifting the equilibrium position to the right; because the volume of gaseous reactants is higher than the volume of gaseous products.

– Decrease in pressure leads to the production of less ammonia (ammonia decomposes); by shifting the equilibrium position to the left; because the volume of gaseous reactants is higher than the volume of gaseous products.

N₂O₄ _(g) 2NO₂ _(g)

Pale yellow dark brown

Increase in pressure:

– Shifts the equilibrium to the left; leading to formation of more N₂O₄ hence mixture turns pale yellow; because the volume of gaseous products (NO₂) is higher than the volume of gaseous reactants (N₂O₄).

Decrease in pressure:

– Shifts the equilibrium position to the right; leading to formation of more NO₂ hence colour turns dark brown; because the volume of gaseous reactants (N₂O₄) is lower than the volume of gaseous products (NO₂);

Note: –

Change in pressure has no effect on the equilibrium mixture where gaseous molecules on the two sides are equal

Example:

N₂ _(g) + O₂ _(g) 2NO_(g)

– Increase or decrease in pressure has no effect on position of equilibrium; because the volume of gaseous reactants is equal to volume of gaseous products;

Note:

– Generally reactions involving only solids and, or liquids are not affected by pressure change because they are not compressible;

Summary: effects of pressure on equilibrium:

Reaction	Effects of pressure change on position of equilibrium
Reaction	Increase	Decrease
N₂O_4(g) 2NO_2(g)	Equilibrium shifts to the left; more N₂O₄ is formed;	Equilibrium shifts to the right; more NO₂is formed;
N_2(g) + 3H_2(g) 2NH_3(g)	– More NH₃ is formed; equilibrium shifts to the right; since volume of gaseous reactants is higher than volume of gaseous products	– More N₂ and H₂ are formed; equilibrium shifts to the left; since volume of gaseous reactants is higher than volume of gaseous products
2SO_2(g) + O_2(g) 2SO_3(g)	– More SO₃ is formed; equilibrium shifts to the right (forward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products	– Less SO₃ is formed (more SO₂ and O₂ are formed); equilibrium shifts to the left (backward reaction is favoured); since volume of gaseous reactants is higher than volume of gaseous products
4NH_3(g) + 5O_2(g) 4NO_(g)+ 6H₂O_(g)	– More NH₃ and O₂ is formed; equilibrium shifts to the left; since volume of gaseous reactants is lower than volume of gaseous products	– Less NH₃ and O₂ is formed (more NO and H₂O are formed); equilibrium shifts to the right; since volume of gaseous reactants is lower than volume of gaseous products
H_2(g) + Cl_2(g) 2HCl_(g)	– No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same;	– No effect on the equilibrium; because volume of gaseous reactants and volume of gaseous products is the same;

(c). Concentration

– Consider the equilibrium in bromine water system:

Br_2(aq) + H₂O_(l) OBr^–_(aq)+ Br^–_(aq) + 2H⁺_(aq)

Yellow orange Colourless

Addition of sodium hydroxide to the equilibrium:

– When NaOH_(aq) is added to the equilibrium, the concentration of H⁺ decrease and the rate of forward reaction are favoured (equilibrium shifts to the right); the reaction of bromine with water increases and there is loss of colour of bromine water.

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the right of equilibrium) to form water.

Equation:

OH^–_(aq) + H⁺_(aq) H₂O_(l)

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the right hence formation of more products;

– This leads to a change in colour from yellow orange to colourless;

Addition of hydrochloric acid:

– Addition of HCl_(aq) is added concentration of H⁺ increases; more bromine is formed and the orange-yellow colour of bromine water becomes more intense;

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the colourless bromide and hypobromite ions to form yellow-orange aqueous bromine;

– This shifts the equilibrium to the left hence the increase in the intensity of the yellow-orange colour of bromine water;

Further examples:

(i). Given the equilibrium:

SO_2(g) + O_2(g) 2SO_3(g)

Removal of sulphur (VI) oxide:

– Reducing the concentration of SO₃ by removing it causes more sulphur (IV) oxide to be converted to sulphur (IV) oxide.

Addition of either oxygen or sulphur (IV) oxide.

– Addition of either sulphur (IV) oxide or oxygen to the equilibrium shifts the equilibrium to the right; due to increase of concentration of products hence more yield of sulphur (VI oxide;

Addition of pyrogallic acid:

– Addition of pyrogallic acid into the equilibrium shifts the equilibrium to the left; since pyrogallic acid dissolves oxygen gas reducing concentration of reactants on the left hence less yield of sulphur (VI) oxide;

(ii). Given the equilibrium:

2CrO₄^2-_(aq) + 2H⁺_(aq) 2Cr₂O₇^2-_(aq) + H₂O_(l)

Yellow Orange

Addition of sodium hydroxide to the equilibrium:

– When NaOH_(aq) is added to the equilibrium, the concentration of H⁺ decrease and the rate of backward reaction are favoured (equilibrium shifts to the left); the reaction of dichromate ions with water to form chromate ions and hydrogen ions increases and there is change in colour to yellow.

Explanation:

– Addition of sodium hydroxide provides hydroxyl ions into the equilibrium.

– The hydroxyl ions react with the hydrogen ions (on the left of equilibrium) to form water.

Equation:

OH^–_(aq) + H⁺_(aq) H₂O_(l)

– This process removes hydrogen ions from the equilibrium mixture.

– This shifts the equilibrium to the left hence formation of more reactants (chromate and hydrogen ions);

– This leads to a change in colour from orange to yellow;

Addition of hydrochloric acid:

– Addition of HCl_(aq) is added concentration of H⁺ increases; more dichromate solution is formed and the orange colour of dichromate ions become more intense;

Explanation:

– Addition of hydrochloric acid introduces more hydrogen ions into the equilibrium;

– The hydrogen ions react with the yellow chromate solution to form orange dichromate solution (and water);

– This shifts the equilibrium to the right hence the increase in the intensity of the orange colour of dichromate solution;

Generally:

– A change in concentration disturbs the already established equilibrium by making the reaction rate in one direction faster

– The reaction then proceeds predominantly in that direction until equilibrium is re established

(d). Effect of catalyst

– Presence of a catalyst has no effect on the position of the equilibrium but alters the rate at which the equilibrium is attained.

– Catalysts usually allow the equilibrium to be reached in a shorter period of time by increasing the rates of the reactions

Le Chateliers Principle

– States that:

When stress is applied to a system in equilibrium, the system reacts so as to oppose the stress.

– This implies that when a change in condition is applied to a system in equilibrium, the system moves so as to oppose the change.

Note:

– The effect of different factors on equilibrium was first investigated in 1888 by a French Chemist; Henri Louis Le Chatelier.

– All explanations so far described are based on Le Chateliers principle.

Industrial applications of chemical equilibrium.

– The ability to change the position of an equilibrium by varying the conditions has been important in industrial processes as industrialists aim at obtaining maximum products at minimum cost and shortest time possible.

– Conditions required to obtain greatest yield of products at minimum costs and shortest time possible are called optimum conditions.

– Such optimum conditions are obtained by continuous removal of products hence reduction in its concentration or varying external factors like temperature and pressure.

Summary:

Optimum conditions for common industrial processes.

Condition	Optimum condition for:
Condition	Ø Haber process N_2(g) + 3H_2(g) 2NH_3(g); ∆H=- 92KJMol-	Ø The contact process 2SO_3(g) + O_2(g) 2SO_3(g); ∆H = -197Kjmol-
Temperature	– The reaction is exothermic hence favoured by low temperatures; which shift the equilibrium to the right hence more yield of ammonia gas; – However rate at which the NH_3(g) will be produced would be too slow and thus uneconomical; an optimum temperature of about 450^oC is thus normally used;	– Forward reaction is exothermic thus increase in temperature favours backward reaction; shifts the equilibrium to the left hence less yield of SO_2(g); – Low temperature will favour forward reaction thus shifts the equilibrium to the right; hence more production of SO_2(g); However rate of yield will be too slow and hence optimum temperature for maximum yield are set at about 450^oC;
Pressure	Increase in pressure favours the forward reaction; shifting the equilibrium to the right hence more yield of ammonia; because the volume of gaseous reactants is higher than the volume of gaseous products; – However the cost of producing and maintaining high pressures in a system is very high; – Thus for maximum yield the optimum pressure is 200 atmospheres;	– An increase in pressure will shift the equilibrium to the left leading to low yield of sulphur (IV) oxide; since the volume of gaseous reactants is lower than the volume of gaseous products; – Thus the optimum pressure used is atmospheric pressure, which gives a percentage conversion of SO_2(g) to SO_3(g) of about 96%.
Concentration	– Removal of ammonia gas shifts the equilibrium to the right hence more yield of ammonia; since this removal lowers concentration of NH_3(g) – thus for maximum yield ammonia is removed as soon as it is produced; so that more H_2(g) and N_2(g) can continue reacting;	– Removal of sulphur (VI) oxide gas shifts the equilibrium to the right hence more yield of SO_3(g); since this removal lowers its concentration; – thus for maximum yield SO_3(g) is removed as soon as it is produced; so that more SO_2(g) and O_2(g) can continue reacting;
Catalyst	– Main catalyst used is finely divided iron; platinum is a better catalyst but is very expensive and easily poisoned by impurities, hence may increase cost of production;	– A catalyst of vanadium (V) oxide is used to increase the rate of reaction;

UNIT 4: ELECTROCHEMISTRY

Checklist:

Meaning of electrochemistry
Displacement and Redox reactions.
Oxidation and Redox reactions in terms of electron gain and electron loss
Oxidation numbers (states)

Rules of assigning oxidation numbers
Calculating oxidation numbers

Redox reactions involving halide ions (halogens)

Further examples of Redox reactions.

Tendency of metals to from ions.
Measurements of tendency of metals to ionize

Electrochemical cells
Salt bridge
Cell diagrams
Cell equations

Standard electrode potential

Definition
Standard hydrogen electrode
The hydrogen half cell
Electrode potentials (negative and positive values)
Calculating E⁰ values from Redox reactions.

Voltaic cells

Primary cells
- Structure
- Reactions
- Zinc (negative) terminal
- Brass (positive) terminal
- Functions of various components
Secondary cells (lead acid accumulators)
- Structure
- Reactions
- During discharge
- During recharging;

Electrolysis

Definition
Terminologies used in electrolysis (basic concepts)
Preferential discharge of ions
Electrolysis of various substances
- Dilute sulphuric acid
- Dilute sodium chloride
- Concentrated sodium chloride (brine)
- The mercury cathode cell
- Copper (II) sulphate
  - Using inert electrodes
  - Using copper electrodes
- Factors affecting electrolysis and electrolysis products
- Applications of electrolysis
  - Extraction of reactive metals
  - Purification of metals
  - Electroplating
  - Anodizing aluminium utensils
  - Manufacture of sodium hydroxide, chlorine and hydrogen;
    - The mercury cathode cell
    - Diaphragm cell;
    - Membrane cell;
  - Quantitative aspects of electrolysis
    - Basic terminologies
      - The ampere
      - The coulomb
      - The faraday
    - Faradays laws of electrolysis
      - Statements

Definition:

– Electrochemistry is the chemistry of electrochemical reactions; which deal with the relationship between electrical energy and chemical reactions.

– Electrochemical reactions involve transfer of electrons and are essentially REDOX reactions.

Displacement and REDOX reactions

Experiment 1:- Displacement reactions among metals

(i). Procedure

(a). 5 cm³ of 1M CuSO₄ _(aq) is put in a test-tube and its temperature recorded.

– In the solution, a spatula end-full of iron fillings is added.

– Any observations and temperature change are determined and recorded.

– The procedure is repeated with fresh samples of CuSO₄ with Zn, Mg, and Cu powders.

(b). The procedure is repeated with 1M magnesium sulphate solution instead of CuSO_{4 (aq}₎.

(ii). Observations:

Metal solid	Copper (II) Sulphate	Magnesium Sulphate
Iron fillings	– A red brown solid (Cu) is formed – The blue colour of the solution (Cu²⁺) fades then changes to green (Fe²⁺)	– No reaction
Zinc powder	– A red brown solid, copper metal is deposited. – The blue colour of the solution (Cu²⁺) fades then turns colourless;	– No observable reaction (change)
Copper powder	– No reaction	– No reaction

(iii). Explanations

– Reactions between metals and ions of another metal involve transfer of electrons from the metal to the other metal ion in solution.

Examples:

Fe_(s) and CuSO₄ _(aq)

– Copper being lower in the electrochemical series accepts electrons easier (than Fe) to form copper atoms (brown solid);

Half equations

Fe _(s) Fe²⁺_(aq) + 2e^–

Cu²⁺_(aq) + 2e^– Cu _(s)

Overall reaction

Cu²⁺_(aq) + 2e- + Fe_(s) Fe²⁺_(aq) + 2e^– + Cu_(s)

Then;

Cu²⁺_(aq) + Fe_(s) Fe ²⁺ _(aq) + Cu _(s)

(Blue) (Green) (Brown solid)

Oxidation and reduction in terms of electron loss and gain

– The loss of electrons is oxidation and the species that gains electrons (causes electron loss); Cu²⁺ in this case is called oxidizing agent; and is itself reduced.

– Reduction: refers to gain of electrons, and the species that donates electrons (iron solid in this case) is called a reducing agent and is itself oxidized.

– Displacement reactions generally involve reduction and oxidation simultaneously and are thus termed Redox Reactions.

Further examples

(i). Zinc solid and CuSO_4(aq)

(i). Zn_(s) Zn²⁺_(aq) + 2e^–(Oxidation)

(ii). Cu²⁺_(aq) + 2e^– Cu_(s) (Reduction)

Oxidation

Then: Cu²⁺_(aq) + Zn_(s) Zn²⁺_(aq) + Cu_(s) (Redox)

Blue Colourless; Red brown

Reduction

(ii). Magnesium solid and copper (II) sulphate

(i). Mg_(s) Mg²⁺_(aq) + 2e^–(Oxidation)

(ii). Cu²⁺_(aq) + 2e^– Cu_(s) (Reduction)

Oxidation

Then: Mg_(s)+ Cu²⁺_(aq) Mg²⁺_(aq) + Cu_(s) (Redox)

Blue Colourless; Red brown

Reduction

(iii). Silver nitrate and copper solid.

(i). Cu_(s) Cu²⁺_(aq) + 2e^–(Oxidation)

(ii). 2Ag⁺_(aq) + 2e^– 2Ag_(s) (Reduction)

Oxidation

Then: Cu_(s)+ 2Ag⁺_(aq) Cu²⁺_(aq) + 2Ag_(s) (Redox)

Brown Colourless Blue Grey

Reduction

Note:

– Amount of heat evolved in these redox reactions depends on the position of the metal in the activity series relative to the metal ion in solution.

– The closer the metals are in the activity series; the less readily displacement occurs and the lower the heat evolution during the displacement.

E.g.: Heat evolved Mg//Cu²⁺is higher than that evolved between Fe//Cu²⁺.

Conclusion:

– Metals displace from solutions, those metals lower than themselves in the activity series.

Note:

– The more the reactive a metal is; the stronger a reducing agent it is and the weaker an oxidizing agent it is.

Example:

– Potassium is stronger reducing agent; but weaker oxidizing agent than silver, gold etc.

Summary:

Strength of reducing/ oxidizing agent

Weakest oxidizing agent Potassium, K Strongest reducing agent

Sodium, Na

Magnesium, Mg

Aluminium, Al

Zinc; Zn

Iron, Fe

Lead, Pb

Copper, Cu

Silver, Ag

Strongest oxidizing agent Mercury, Hg Weakest reducing agent

Summary on displacement reactions

Metal ion (in solution)	Mg	Al	Zn	Fe	Pb	Cu
K⁺	C	C	C	C	C	C
Na⁺	C	C	C	C	C	C
Ca²⁺	C	C	C	C	C	C
Mg²⁺	C	C	C	C	C	C
Al³⁺		C	C	C	C	C
Zn²⁺			C	C	C	C
Fe²⁺				C	C	C
Pb²⁺					C	C
Cu²⁺						C
Ag²⁺

Key:

A cross (x) indicates no reaction hence no redox reaction occurs.

A tick indicates redox reaction occurs.

Oxidation numbers/ oxidation state.

– Is the apparent charge that atoms have in molecules or ions.

– For monoatomic ions, the oxidation number (state) is the magnitude and sign of charge;

Example:

Oxidation no of Aluminium in Al³⁺ is +3;

Importance of oxidation numbers:

– Helps in keeping track of electron movement in redox reactions; hence determination of the reduced and oxidized species.

Oxidation and reduction in terms of oxidation numbers

Oxidation

– Is an increase in oxidation number.

Reduction

– Refers to a decrease in oxidation number

Rules in assigning oxidation numbers

Oxidation number of an uncombined element is zero (0)
The charge on a monoatomic ion is equivalent to the oxidation number of that element;
The oxidation number of hydrogen in all compounds is +1 except in metal hydrides where its 1;
The oxidation number of oxygen in all compounds is 2 except in peroxides where it is 1 and 0F₂ where it is +2.
In complex ions the overall charge is equal to the sum of the oxidation states of the constituent elements.
In compounds, the sum of oxidation numbers of all constituent atoms is equal to zero.

Worked examples

Calculate the oxidation number of nitrogen in:

(i). NO₃^–

Solution:

N + (-2 x 3) = -1

N = -1 + 6

= +5

Note: thus nitric acid with a nitrate ion (NO₃^–) is called nitric (V) acid since the oxidation number of nitrogen in it is +5;

(ii). NO₂;

Solution:

N + (-2 x 2) = 0

N = 0 + 4

= +4

Note: Thus the gas NO₂ is referred to as nitrogen (IV) oxide because the oxidation umber of nitrogen in it is +4

(iii). NO₂^–;

Solution:

N + (-2 x 2) = -1

N = -1 + 4

= +3

Note: thus nitrous acid containing nitrite ion is called nitrous (III) acid since the oxidation number of nitrogen in it is +3.

(iii). AgNO₃;

Solution:

1 + N + (-2 x 3) = 0

1 + N + (-6) = 0

N = 0 1 + 6;

N = +5

Determine the oxidation number of manganese in each of the following, and hence give the systematic names of the compounds.

(i). MnSO₄

Solution:

Mn + 6 + (-2 x 4) = 0

Mn = 0 6 + 8;

Mn = +2

Systematic name: Manganese (II) sulphate;

(ii). Mn₂O₃;

Solution:

2Mn + (3 x -2) = 0;

2Mn = 0 + 6

Mn = ½ x 6;

Mn = +3;

Systematic name: Manganese (III) oxide;

(iii). KMnO₄

Solution:

1 + Mn + (-2 x 4) = 0

Mn = 0 1 + 8;

Mn = +7;

Systematic name: Potassium manganate (VII) oxide

(iv). MnO₃^–;

Solution:

Mn + (-2 x 3) = -1

Mn = -1 + 6;

Mn = +5

Systematic name: Manganese (V) ion;

Determination of redox reactions using oxidation numbers.

Worked examples:

Redox reactions involving Halide ions and halogens

Experiment

(i). Procedure:

– 2 cm³ of chlorine gas are bubbled into each of the following solutions: – KI, KCl, KBr, and KF.

– The observations are made and recorded.

– The procedure is repeated using fluorine, bromine and iodine in place of chlorine.

Precaution:

– Chlorine and bromine are poisonous.

(ii). Observations

Halogen	Potassium fluoride	Potassium Chloride	Potassium Bromide	Potassium Iodide
Fluorine (F₂)	No visible, change	Green-yellow gas is evolved	Colourless solution changes to red brown	Colourless solution turns black;
Chlorine (Cl₂)	No visible colour change;	No visible colour change;	Colourless solution turns red brown	Colourless solution turns black
Bromine (Br₂)	No visible colour change;	No visible colour change;	No visible colour change;	Colourless solution turns black;
Iodine (I₂)	No visible colour change;	No visible colour change;	No visible colour change;	Colourless solution turns black;

Note: Colours of halogens in tetra chloromethane:-

Halogen	Colour in tetrachloromethane
Fluorine
Chlorine	Yellow
Bromine	Red-brown
Iodine	Purple

(iii). Explanations

– Fluorine displaces all the other halogens; Cl₂, Br₂ and I₂ because it has a greater tendency to accept electrons than all the rest.

– Chlorine displaces both Bromine and Iodine from their halide solutions

– Cl₂ takes electrons from the bromide and iodide ions i.e. oxidizes them, to form bromine and iodine respectively.

Equations:

(i). Chlorine and potassium bromide:

Cl_{2 (g)} + 2KBr_(aq) 2KCl_(aq) + Br_{2 (l)}

Ionically:

Cl_{2 (g)} + 2Br^–_(aq) 2Cl^–_(aq) + Br_{2 (l)}

Green-yellow Red brown

Redox equation: Reduction

Cl_{2 (g)} + 2Br^–_(aq) 2Cl^–_(aq) + Br_{2 (l)}

Oxidation

(ii). Chlorine and Potasium iodide:

Cl_{2 (g)} + 2KI_(aq) 2KCl_(aq) + I_{2 (l)}

Ionically:

Cl_{2 (g)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Green-yellow Black

Half cell reactions:

Oxidation: 2I^–_(aq) I_2(aq) + 2e-
Reduction: Cl_{2 (g)} + 2e- 2Cl^–_(aq)

Redox equation: Reduction

Cl_{2 (aq)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Oxidation

– Bromine takes electrons form iodide ions but not from fluorine and chlorine.

– Iodine is formed i.e. due to oxidation of iodide ions by the Bromine.

Equations

(ii). Bromine and Potasium iodide:

Br_{2 (g)} + 2KI_(aq) 2KBr_(aq) + I_{2 (l)}

Ionically:

Br_{2 (g)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Red brown Black

Half cell reactions:

Oxidation: 2I^–_(aq) I_2(aq) + 2e-
Reduction: Br_{2 (g)} + 2e- 2Br^–_(aq)

Redox equation: Reduction

Cl_{2 (aq)} + 2I^–_(aq) 2Cl^–_(aq) + I_{2 (l)}

Oxidation

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iodine increases from -1 to 0; hence oxidation;

(iv). Conclusion:

– The stronger the tendency of an element to accept electrons, the stronger is its oxidizing power.

– Fluorine is the strongest oxidizing agent of the 4 halogens considered.

Order of oxidizing power for halogens.

Fluorine; F₂;

Chlorine; Cl₂ Increasing oxidizing power.

Bromine; Br₂

Iodine; I₂

Further examples of redox reactions

(a). Action of acid on metals

Mg_(s) + 2HCl_(aq) 2MgCl_(aq) + H_{2 (g)}

Ionically:

Mg_(s) + 2H⁺_(aq) Mg²⁺_(aq) + H_{2 (g)}

Half cell reactions:

Oxidation: Mg_(s) Mg²⁺_(aq) + 2e-
Reduction: 2H⁺_(g) + 2e- H_2(g)

Redox equation: Reduction

2H⁺_(aq) + Mg_(s) H_{2 (aq)} + Mg²⁺_(aq)

Oxidation

Note: oxidation number of hydrogen decreases from 1 to 0 hence reduction; while oxidation number of magnesium increases from 0 to 2; hence oxidation;

(b). Reaction of active metals with water

Example

2Na_(s) + 2H₂O_(l) 2NaOH_(aq) + H_{2 (g)}

Ionically:

2Na_(s) + 2H₂O_(l) 2Na⁺_(aq) + 2OH^–_(aq) + H_{2 (g)}

Half cell reactions:

Oxidation: 2Na_(s) 2Na⁺_(aq) + 2e-
Reduction: 2H₂O_(g) + 2e- 2OH^–_(aq) + H_2(g)

Redox equation: Oxidation

2Na_(s) + 2H₂O_(aq) 2Na⁺_(aq) + 2Cl^–_(aq) + I_{2 (l)}

Reduction

Note: oxidation number of water decreases from 0 to -1(total) hence reduction; while oxidation number of sodium increases from 0 to 1; hence oxidation;

(c). Reaction of heated Iron with dry chlorine

2Fe_(s) + 3Cl_2(g) 2FeCl_3(s)

Ionically (assumed):

2Fe_(s) + 3Cl_2(aq) 2Fe³⁺_(aq) + 6Cl^–_(g)

Half cell reactions:

Oxidation: 2Fe_(s) 2Fe³⁺_(aq) + 6e^–
Reduction: 3Cl_2(g) + 6e^– H_2(g)

Redox equation: Oxidation

2Fe_(s) + 3Cl_2(g) 2Fe³⁺_(aq) + 6Cl^–_(aq)

Reduction

Note: oxidation number of chlorine decreases from 0 to -1 hence reduction; while oxidation number of iron increases from 0 to 3; hence oxidation;

(d). Reaction between Bromine and Iron (II) ions

Ionically:

Br_2(l) + 2Fe²⁺_(aq) 2Fe³⁺_(aq) + 2Br^–_(aq)

Half cell reactions:

Oxidation: Br_2(l) 2Br^–_(aq) + 2e-
Reduction: 2Fe²⁺_(g) + 2e- 2Fe³⁺_(aq)

Redox equation: Reduction

Br_{2 (aq)} + 2Fe²⁺_(s) 2Br^–_(aq) + 2Fe³⁺_(aq)

Oxidation

Note: oxidation number of bromine decreases from 0 to -1 hence reduction; while oxidation number of Fe²⁺increases from 2 to 3 (in Fe³⁺); hence oxidation;

(e). Oxidation by potassium Manganate (VII) (KMnO₄)

Procedure:

– Purple Potassium manganate (VII) is added into a solution containing iron (II) ions in a test tube.

– A few drops of concentrated sulphuric (VI) acid are added.

Observations:

– The purple solution (containing Manganate (VII) ions) turns to colourless (manganate (II) ions) i.e. the purple solution is decolourised;

Explanation:

– The Manganate (VII) ions which give the solution a purple colour are reduced to Manganese (II) ions which appear colourless. This is a redox reaction.

Equations

Ionically:

MnO₄^–_(aq) + 8H⁺_(aq) + 5Fe²⁺_(aq) Mn²⁺_(aq) + 5Fe³⁺_(aq) + 4H₂O_(l)

Purple Colourless

Half cell reactions:

Oxidation: 5Fe²⁺_(s) 5Fe³⁺_(aq) + 5e^–; (since number of electrons is 5)
Reduction: MnO₄^–_(aq) + 8H⁺_(aq)+ 5e^– Mn²⁺_(aq) + 4H₂O_(l)

Redox equation: Reduction

MnO₄^–_(aq) + 5Fe²⁺_(aq) Mn²⁺_(aq) + 5Fe³⁺_(aq)

Oxidation

Note:

oxidation number of Manganate ions in KMnO₄ decreases from 7 to 2 (in Mn²⁺) hence reduction; while oxidation number of iron increases from 2 (in Fe²⁺) to 3 (in Fe³⁺); hence oxidation;

The presence of Fe³⁺ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)₃;

(f). Action of potassium dichromate (VI) on iron (II) ions (Fe²⁺):

(i). Procedure:

– A solution containing iron (II) ions is added into a solution of oxidized potassium dichromate (VI).

(ii). Observations:

– The orange solution of potassium dichromate turns green.

(iii). Explanations:

– The iron (II) ions are oxidized to iron (III) ions

– The chromium (VI) ions (orange) are reduced to chromium (III) ions

– This is thus a REDOX reaction.

Equations

Ionically:

Cr₂O₇^2-_(aq) + 14H⁺_(aq) + 6Fe²⁺_(aq) 2Cr³⁺_(aq) + 6Fe³⁺_(aq) + 7H₂O_(l)

Orange Green

Half cell reactions:

Oxidation: 6Fe²⁺_(s) 6Fe³⁺_(aq) + 6e^–; (since number of electrons is 6)
Reduction: Cr₂O₇^2-_(aq) + 14H⁺_(aq)+ 5e^– 2Cr³⁺_(aq) + 7H₂O_(l) + 6Fe³⁺_(aq)

Redox equation: Reduction

Cr₂O₇^2-_(aq) + 6Fe²⁺_(aq) 2Cr³⁺_(aq) + 6Fe³⁺_(aq)

Oxidation

Note:

The oxidation number of dichromate ions in K₂Cr₂O₇ decreases from 6 to 3 (in Cr³⁺) hence reduction; while oxidation number of iron increases from 2 (in Fe²⁺) to 3 (in Fe³⁺); hence oxidation;

The presence of Fe³⁺ at the end of the reaction can be detected by adding sodium hydroxide solution to form a red brown precipitate of Fe(OH)₃;

(g). Action of acidified potassium permanganate on Hydrogen Peroxide.

– The overall reaction is a Redox reaction

Redox equation

2MnO₄^–_(aq) + 5H₂O_(l) + 6H⁺_(aq) 2Mn²⁺_(aq) + 5H₂O_(l) + 5O_{2 (g)}

Purple Colourless

Half cell reactions:

Oxidation: 5H₂O_2(aq) 10H⁺_(aq) + 5O_2(g) + 10e^–;
Reduction: 2MnO₄^–_(aq) + 16H⁺_(aq)+ 10e^– 2Mn²⁺_(aq) + 8H₂O_(l)

Redox equation: Reduction

2MnO₄^–_(aq) + 5H₂O_{2 (aq)} 2Mn²⁺_(aq) + 10H⁺_(aq) + 5O_2(g)

Oxidation

Note:

The oxidation number of manganate ions in KMnO₄ decreases from 7 to 2 (in Mn²⁺) hence reduction; while oxidation number of hydrogen increases from -1 (in H₂O₂) to 1 (in H⁺); hence oxidation;
Thus the acidified potassium manganate (VII) oxidizes hydrogen peroxide to water and hydrogen;

(h). H₂O₂oxidizes Iron (II) salts to Iron (III) salts in acidic medium

Oxidation:

2Fe²⁺_(aq) 2Fe³⁺_(aq) + 2e^–

Reduction:

2H⁺_(aq) + H₂O_2(aq) + 2e^– 2H₂O_(l)

Overall redox:

2Fe²⁺_(aq) + H₂O_{2 (aq)} + 2H⁺_(aq) 2Fe³⁺_(aq) + 2H₂O_(l)

Terms used in describing oxidation reduction

Term

Electron change

Oxidation number change

Oxidation

Reduction

Oxidizing agent

Reducing agent

Substances oxidized

Substance reduced

loss of electrons

gain of electrons

receives electrons

loses electrons

gains electrons

increases

decreases

increases

decreases

The tendency of metals to form ions

– When a metal is placed in an aqueous solution of its ions, some of the metal dissolves;

Equation:

M_(s) Mⁿ⁺_(aq) + ne^–

– Dissolution of the metal causes electron build up on its surface; making it negatively charged, while the surrounding solution becomes positively charged.

Diagram: dissolution of metal and electron build up.

Metal rod

Solution containing metal ions

– The positive charge of the solution increases and some of the cations start recombining with the electrons on the metal surface to form atoms.

Equation:

Mⁿ⁺ _(aq) + ne^– M(s)

– Consequently, an electric potential difference is created between the metal rod and the positively charged ions in solution.

– This arrangement of a metal rod (electrodes) dipped in a solution of its ions constitutes a half cell.

Note:

– The tendency of a metal to ionize when in contact with the ions differs form one metal to another.

– This difference can be measured by connecting two different Half cells to make a full cell.

– The electrodes of the 2 half cells are connected by a metallic conductor; while the electrolytes (solutions) of the half cells are connected through a salt bridge.

Diagram: Connection of two half cells to a full cell.

Experiment: – to measure the relative tendency of metals to ionize

(i). Procedure:

– A Zinc rod is placed into 50 cm³ of 1M zinc sulphate in a beaker;

– Into another beaker containing 50 cm³ of 1M CuSO_{4 (aq)}, a copper rod is dipped;

– The two solutions are connected using a salt bridge.

– The two metal rods are connected through a connecting wire connected to a voltmeter

– The experiment is repeated using the following half-cells instead of the Zinc-half cells:-

Mg rod dipped in 1M MgSO₄ (aq)
Lead dipped in 1M lead Nitrate
Copper dipped in 1M CuSO₄ (aq)

(ii). Apparatus

Diagram:

(iii). Observation

– The zinc rod in the zinc-zinc ions half cell dissolves;

– The blue colour of the copper (II) Sulphate solution fades/ decrease;

– Red-brown deposits of copper appear on the copper rod in the copper-copper ions half-cell.

– A voltage of 1.10 V is registered in the voltmeter.

(iv). Equations/ reactions, at each half cell

In zinc-zinc ions half cell

Zn_(s) Zn²⁺_(aq) + 2e^–

In the copper-copper ion half cell

Cu²⁺_(aq) + 2e- Cu_(s);

(v). Explanations:

– Zinc rod has a higher tendency to ionize than the copper rod, when the metal rods are placed in solutions of their ions.

– Thus the zinc rod has a higher accumulation of electrons than the copper rod.

– This makes it more negative compared to the relatively more positive copper rod, which has a lower accumulation of electrons.

– On connecting the 2 half cells; electrons will flow form the zinc rod to the copper rod through the external wire.

– The copper rod gains the electrons lost by the Zinc rod.

Roles of the slat bridge:

Note: It forms a link between the 2 half cells, thereby completing the circuit

– It compete the circuit by:

Allowing its ions to carry charge from one half – cell to the other.
Maintaining the balance of charge in the two half-cells; by providing the ions which replace those used up at the electrodes.

– The overall reaction in the cells is a Redox reaction

Half cell reactions:

Zn_(s) Zn²⁺_(aq) + 2e^– (oxidation)

Cu²⁺_(aq) + 2e- Cu_(s); (reduction)

Overall redox equation

Oxidation

Zn_(s) + Cu²⁺_(aq) Zn²⁺_(aq) + Cu_(s)

Reduction

– The voltage of 1.10 V registered in the voltmeter is a measure of the difference between the electrode potential (E^θ) of Zinc and Copper electrodes, i.e. the potential difference/ the Electromotive force.

– Thus: An electrochemical/ voltaic cell;

Is the combination of two half –cells to give a full cell capable of generating an electric current from a redox reaction.

Cell diagram for a voltaic cell

Rules/conventions for cell representation

A vertical continuous line (/); represents the metal-metal ion or metal ion-metal interphase.
Vertical broken line ( ); between the 2 half-cells; represents the salt bridge.

Note: The salt bridge may also be represented by two unbroken parallel lines (//).

Example: – cell diagram for the above cell;

Zn_(s) / Zn²⁺_(aq) Cu²⁺_(aq) / Cu_(s);

Alternatively:

Zn_(s) / Zn²⁺_(aq) // Cu²⁺_(aq) / Cu_(s);

Electrode potential E⁰, values of other metal – metal ions relative tot he Cu/Cu ²⁺ half cell

Metal / metal ion half cell	Electrode potential E^θ relative to Cu²⁺ / Cu half cell
Mg_(s) / Mg²⁺_(aq)	+2.04
Zn_(s) / Zn²⁺_(aq)	+1.10
Pb_(s) / Pb²⁺_(aq)	+0.78
Cu_(s) / Cu²⁺_(aq)	+0.00
Ag_(s) / Ag⁺_(aq)	-0.46

Positive and negative E values

(i). Positive E values –

– If the E^θ value for a metal / metal ion is positive then the metal undergoes oxidation (loses electrons) while the reference electrode undergoes reduction (accepts electrons)

Example:

– The E^θ value for Zn_(s) / Zn²⁺_(aq) relative Cu²⁺_(aq) / Cu_(s)is positive because the zinc metal is oxidized to zinc ions while the copper ions are reduced to copper metal.

(ii). Negative E values:

– Implies that the reference half cell undergoes oxidation (donates electrons) while the other metal ions in the other half cell undergoes reduction (accepts electrons)

Example:

– The E value for Ag _(s) / Ag⁺_(aq) is negative because Cu is more reactive than silver and gives out electrons (oxidation); while the less reactive Ag has its ions accepting electrons (reduction) to form Ag solid.

(iii). The 0 (Zero) E value:

– Always indicate the reference electrode / half-cell; in which case there would be no potential difference (with itself)

Example:

– The 2 half cells of Cu_(s) / Cu²⁺_(aq) or Cu²⁺_(aq) / Cu_(s) have no potential difference in between them hence a zero (0) E value.

Note:

– Any other element could be chosen as the reference electrode in place of copper Cu and difference electrode potentials values would be obtained for the same elements.

– The electrode potential of a single element is usually determined by measuring the difference between the electrode potential of the element and a chosen standard electrode.

– This gives the standard electrode potential (E⁰) of the element.

The standard electrode potential (E⁰)

Definition

– Is the potential difference for a cell comprising of a particular element in contact with 1 molar solution of its ions and the standard hydrogen electrode.

– It is denoted with the symbol E⁰.

Importance

– It is useful in comparing the oxidizing and reducing powers of various substances.

The standard Hydrogen Electrode

– Is the hydrogen half-cell, which has been conventionally chosen as the standard reference electrode.

– It has an electrode potential of zero at:-

A temperature of 25^oC
A hydrogen pressure of 1 atmosphere
A concentration of 1M hydrogen ions

Note: The ions in the other half-cell must also be at a concentration of 1 molar.

Components of the Hydrogen half cell;

– Consist of an inert platinum electrode immersed in a 1M solution of Hydrogen ions

– Hydrogen gas at 1 atmosphere is bubbled into the platinum electrode.

– The hydrogen is adsorbed into the platinum surface and an equilibrium (state of balance) is established between the adsorbed layer of molecular hydrogen and hydrogen ions in the solution.

Equation:

½ H_{2 (g)} H⁺ _(aq)+ e ^–

Platinised platinum

– Is platinum loosely coated with finely-divided platinum.

– This enables it to retain comparatively large quantity of hydrogen due to its porous state.

– Platinised platinum also serves as a route by which electrons leave or enter the electrode.

– The hydrogen electrode is represented as: H_{2 (g)} / H⁺_(aq); 1M

Diagram: The standard hydrogen electrode:

– The electrode potential of any metal is taken as the difference in potential between the metal electrode and the standard hydrogen electrode.

Negative and positive electrode potentials

(a). Negative electrode potential

– If the metal electrode has a higher/ greater tendency to loose electrons than the hydrogen electrode; then the electrode is negative with respect to hydrogen electrode; and its electrode potential is negative.

Examples: Zinc, Magnesium etc.

(b). Positive electrode potential

– If the tendency of an electrode to loose electrode is lower than the hydrogen electrode, then the electrode is positive with respect to the hydrogen electrode; and its potential is positive.

Examples: copper, silver etc

Reduction potentials

– Is a standard electrode potential measured when the electrode in question is gaining electrons.

– The lower the tendency of an electrode to accept/ gain electrons; the lower (more negative) the reduction potential and vise versa.

Examples:

K⁺ _(aq) + e _(aq) K_(s); E⁰ = -2.92V

F_{2 (g)}+ e^– _(aq) 2F^–_(aq); E⁰ = +12.87V

Mg _(s) + 2e^– Mg²⁺_(aq); E^θ = -2.71 V

– Thus potassium ions with E⁰= -2.92V have a lesser tendency to gain electrons than magnesium ions.

– Thus Potassium is the weakest oxidizing agent; but the strongest reducing agent, since it has the greatest tendency to donate electrons.

Note:

– Oxidation potentials will be the potentials of electrodes measured when they are losing electrons hence undergoing oxidation.

Standard electrode potentials reduction potentials) of some elements

Reduction equation

E^θvolts

Least readily reduced

Most readily reduced

– Increasing strength of oxidizing agent

– Decreasing strength of oxidizing agent

F₂ _(g) + 2e^– 2F^– _(aq)

+2.87

Cl₂ _(g) + 2e^– 2Cl^– (_aq)

+ 2.87

Br_2(g)+ 2e 2Br^– _(aq)

+ 1.36

Ag⁺_(aq)+ 2e^– Ag _(s)

+ 0.80

I₂ _(g) + 2e^– 2I^–_(aq)

+ 0.54

Cu²⁺_(aq)+ 2e^– Cu _(s)

+ 0.34

2H⁺ _(aq)+ 2e^– H₂ _(g)

+ 0.00

Pb²⁺_(aq) +2e^– Pb _(s)

– 0.13

Fe²⁺⁽_aq) + 2e^– Fe _(s)

– 0.44

Zn²⁺ _(g) + 2e^– Zn _(s)

– 0.76

Al ³⁺ _(aq)+ 3e^– Al _(s)

– 1.66

Mg²⁺ _(aq) + 2e^– Mg _(s)

– 2.71

K⁺ _(aq) + e^– K _(s)

– 2.92

Note:

– The standard electrode potentials in the above table are reduction potentials.

– The greater the tendency to undergo reduction, the higher (more positive) the E^θ value.

– The reverse reaction (oxidation) would have a potential value equal in magnitude but opposite in sign to the reduction potential.

Example:

Zinc

Reduction potential

Zn²⁺ _(aq) + 2e- Zn (s); E⁰ = -0.76V;

Oxidation potential

Zn _(s) Zn²⁺ _(aq) + 2e^–; E^θ = + 0.76V;

Uses of E⁰ values

Comparing the reducing powers and oxidizing powers of various substances;
Predicting whether or NOT a stated REDOX reaction will take place.

The E⁰ value for a REDOX reaction

– Is usually calculated as the sum of the E⁰ value for the half cells involved.

Note:

– If the sum is positive then the reaction can occur simultaneously;

– If the value of the sum is negative the reaction cannot occur;

Sample calculations

Cu ²⁺ can oxidize Zinc but Zn ²⁺ cannot oxidize Cu;

(a). Cu ²⁺/ Cu // Zn / Zn²⁺

Cu ²⁺ _(aq) + 2e^– Cu _(s); E⁰ = + 0.34 V

Zn _(s) 2e^– + Zn ²⁺ _(aq); E⁰ = + 0.76 V

Cu ²⁺ _(aq) + Zn _(s) Cu_(s) + Zn _(aq); E⁰ = – 1.10 V

– The overall reaction is positive, hence zinc can be oxidized by copper (II) ions, and hence reaction occurs;

(b). Zn ²⁺ _(aq) / Zn _(s) // Cu _(s) / Cu ²⁺ _(aq)

Zn²⁺ _(aq) + 2e- Zn_(s); E⁰ = – 0.76 V

Cu_(s) 2e^– + Cu²⁺_(aq); E⁰ = – 0.34 V

Zn ²⁺_{(aq) +} Cu_(s) Zn _(s) + Cu ²⁺_(aq); E⁰ = – 1.10 V

– The overall E⁰ is negative; thus Zn ²⁺ cannot oxidize Cu to Cu²⁺ (Cu cannot reduce Zn ²⁺to Zn); and hence the reaction cannot occur.

Can chlorine displace bromine form bromide solution?

Cell diagram: Cl_{2 (g)}/ Cl^–_(aq) // 2Br^–_(aq) / Br_{2 (g)}

Cl₂ _(g) + 2e^– 2Cl^– _(aq); E⁰ = + 1.36 V

2Br^– _(aq) 2e^– + Br₂ _(g); E⁰ = – 1.09 V

Cl₂ _(g) + 2Br^– _(aq) 2Cl^–_(aq) + Br₂ _(g); E⁰ = +0.27 V

– The overall E⁰ for the reaction is positive, so chlorine can displace bromine from a bromide solution.

Worked examples

The diagram below represents part of the apparatus to be used for the determination of the standard electrode potential of Aluminum, E^θ Al ³⁺_(aq)/Al _(s)

(a). Name the solutions which could be placed in beakers A and B; specifying their concentrations.

Answer:

– In beaker A; 1 M HCl _(aq) i.e. any solution with 1 M hydrogen ions

– In beaker B; 1 M Al (NO₃) _(aq);i.e. any aqueous solution with 1M Al ^3+.

(b). One essential part of the cell has been omitted. Name the missing part and give its functions.

Answer:-

Missing part: salt bridge

Function: completes the circuit by;

Allowing its ions to carry charge form one half cell to another.
Providing ions which repose those used up of the electrodes, hence maintaining a balance of charge in the 2 half cells.

(c). The voltmeter reading was found to be 1.66 V.

(i). Give the standard electrode potential for the aluminum electrode.

Solution: it is -1.66 since the Hydrogen-hydrogen ions half-cell = 0.00V.

(ii). Show the direction of flow of electrons in the circuit;

Solution: From Al³⁺ / Al half cell to the H₂/ 2H⁺ _(aq) half cell

(d). Give the half-cell equations and the overall cell equation

Solution:

3H⁺ _(aq) + 3e^– 1½ H_{2 (g)}; E⁰ = + 0.00V;

Al _(s) 3e^– + Al³⁺ _(aq); E⁰ = + 1.66V;

Overall:

Al _(s) + 3H⁺ _(aq) Al³⁺_(aq) + 1½H₂ _(g); E⁰ = +1.66 V

You are given the following half-equations;

Mg²⁺_(aq) + 2e- Mg _(s); E^θ = – 2.71 V

Zn²⁺_(g) + 2e- Zn _(s); E^θ = – 0.76 V

(i) (a). Obtain an equation for the cell reaction.

Mg _(s) 2e^– + Mg²⁺ _(aq); E⁰ = +2.71V

Zn²⁺_(aq) + 2e^– Zn _(s); E⁰ = – 0.76V

Mg _(s) + Zn²⁺ _(aq) Mg²⁺_(aq) + Zn_(s); E⁰ = +1.95V

Thus equation:

Mg _(s) + Zn²⁺ _(aq) Mg²⁺_(aq) + Zn_(s); E⁰ = +1.95V

(b). Calculate the E⁰ value for the cell.

Mg _(s) 2e^– + Mg²⁺ _(aq); E⁰ = +2.71V

Zn²⁺_(aq) + 2e^– Zn _(s); E⁰ = – 0.76V

Mg _(s) + Zn²⁺ _(aq) Mg²⁺_(aq) + Zn_(s); E⁰ = +1.95V

(c). Give the oxidizing species.

– Reducing species

Magnesium i.e. The species that undergoes oxidation since its oxidation number increases (from 0 to 2); as it reduces the other;

– Oxidizing species

Zinc/Zinc ions; – the species that undergoes reduction; since its oxidation number decreases (from 2 to 0) as it oxidizes the other species (Mg).

Given the following half-equations

I_{2 (g)} + 2e- 2I^– _(aq); E^θ= + 0.54V

Br_{2 (g)} + 2e- 2Br^– _(s); E^θ = +1.09 V

(a). Obtain an equation for the all reaction

2I^–_(aq) 2e^– + I² _(g); E⁰ = – 0.54V

Br₂_(g) + 2e^– 2Br^– _(s); E⁰ = +1.09V

Br₂ _(g) + 2I^– _(aq) 2Br^–_(aq) + I_{2 (g)}; E⁰ = +0.55V

(b). Calculate the E⁰ value for the cell

2I^–_(aq) 2e^– + I² _(g); E⁰ = – 0.54V

Br₂_(g) + 2e^– 2Br^– _(s); E⁰ = +1.09V

Br₂ _(g) + 2I^– _(aq) 2Br^–_(aq) + I_{2 (g)}; E⁰ = +0.55V

(c). Give the oxidizing species;

Oxidation

Br₂ _(g) + 2I^– _(aq) 2Br^–_(aq) + I_{2 (g)}; E⁰ = +0.55V

Reduction

0 -1 -1 0

Oxidizing species: Bromine; Br_{2 (aq)}

Consider the following list of electrodes and electrode potential values.

Electrode reaction

E^θ volts

A²⁺/ A

B²⁺/ B

C²⁺/ C

D⁺/ D

E²⁺/ E

F²⁺/ F

+0.34

-0.71

-0.76

+0.80

-2.87

-2.92

(a). Which of the ions is the strongest oxidizer?

D⁺; because it is most readily reduced/ have the highest tendency to accept electrons as evidenced by its highest positive E^θvalue when the ions change to element (D⁺/ D^–)

(b). Which of the ions is the strongest reducer?

– Is least readily reduced hence lowest E⁰ value (-2.92 V); accepts electrons least readily i.e. shows the lowest E⁰ when its ions gain electrons/ are reduced (F⁺/ F, = -2.92 V)

The following is a list of reduction standard electrode potentials.

Metal

E^θ volts(standard electrode potential)

Magnesium

Zinc

Iron

Hydrogen

Copper

Silver

-2.36

-0.76

-0.44

0.00

+0.34

+0.79

(a). Which two metals, if used together in a cell would produce the largest e.m.f?

Magnesium-silver cell;

Mg²⁺_(aq) + 2e^– Mg _(s); E^θ = – 2.36 V

2Ag⁺_(g) + 2e^– 2Ag _(s); E^θ = – 0.79 V

Thus;

Mg _(s) 2e^– + Mg²⁺ _(aq); E⁰ = +2.36V

2Ag⁺_(aq) + 2e^– 2Ag _(s); E⁰ = – 0.79V

Mg _(s) + 2Ag⁺ _(aq) Mg²⁺_(aq) + 2Ag_(s); E⁰ = +3.15V

(b). What would be the voltage produced by:-

(i). Zinc-copper cell

Cu²⁺_(aq) + 2e- Cu _(s); E^θ = + 0.34 V

Zn²⁺_(g) + 2e- Zn _(s); E^θ = – 0.76 V

Thus; Zn_(s) / Zn²⁺_(aq) // Cu²⁺_(aq) / Cu_(s);

Zn _(s) 2e^– + Zn²⁺ _(aq); E⁰ = + 0.76V

Cu²⁺_(aq) + 2e^– Cu _(s); E⁰ = + 0.34V

Zn _(s) + Cu²⁺ _(aq) Zn²⁺_(aq) + Cu _(s); E⁰ = +1.10V

(ii). Copper-silver cell

Cu²⁺_(aq) + 2e- Cu _(s); E^θ = + 0.34 V

Ag⁺_(g) + 2e- Ag _(s); E^θ = + 0.79 V

Thus; Cu_(s) / Cu²⁺_(aq) // 2Ag⁺_(aq) / 2Ag_(s);

Cu _(s) 2e^– + Cu²⁺ _(aq); E⁰ = – 0.34V;

2Ag⁺_(aq) + 2e^– 2Ag _(s); E⁰ = + 0.79V;

Cu _(s) + 2Ag⁺ _(aq) Cu²⁺_(aq) + 2Ag_(s); E⁰ = + 0.45V

(c). Explain the meaning of the positive and negative signs;

Positive signs

– The metal in question has a lower tendency to loose electrons than hydrogen hence more relatively positive to hydrogen;

– They are stronger oxidizing agents but weaker reducing agents but weaker reducing agents than hydrogen;

Negative signs

– The particular metal has a higher tendency to loose electrons than hydrogen; hence relatively more negative than hydrogen.

– They are weaker oxidizing agents but stronger reducing agents than Hydrogen.

The following are some half-cell electrode potentials of some elements.

Reaction

(a). Select two half cells which when oxidized give the burst E value; and fill the cell representation.

Solution: The silver copper cell; i.e.

Cell representation

(b). Calculate the E⁰ value

(c). Give the strongest reducing agent and strongest oxidizing agent.

Strongest reducing agent

Strongest oxidizer – silver – has highest reduction potential

Study the table below and answer the questions that follow.

(a). Which two metals would form a metallic couple with the highest EMS.

(b). Calculate the e.m.f. of the cell that would be produced by (i) above.

(c). Write down the cell representation for the cell above.

(d). Which metal is the strongest reducing agent in the above list

Metal A – have the lowest reduction potential.

8 (a). The table below gives reduction potentials obtained when the half-cells for each of the metals represented by letters J, K, L, M and N where connected to a copper half- cell as the reference electrode.

(i). What is metal L likely to be? Give a reason

Copper It has an E⁰ value/ reduction potential of 0.00, with copper as the reference electrode

(ii). Which of the metals cannot be displaced from the solution of its salt by any other metal in the table? Give a reason.

Metal J: Has the lowest reduction potential; meaning it least readily accepts electrons (most readily donates electrons) than any other metal.

(iii). Metal K and M were connected to form a cell as shown in the diagram below;

Indicate on the diagram, the direction of flow of electrons. Explain.

from K to M.

K is a stronger reducing agent than M, as evidenced by its lower reduction potential.

It thus loses electrons faster becoming more Negative than M; hence electrons move from K through external wires to M.

Write the equations for the half-cell reactions that occur at:-

Metal K electrode:-

Metal M electrode

III. If the slat bridge is filled with saturated sodium Nitrate solution, explain how it helps to complete the circuit.

Answer

It allows its ions (Na⁺ _(aq) , and No^{– 3} _(aq) ) to carry charge from one half cell to another. Providing ions which replace those used up at the electrodes.

VOLTAIC CELLS

– Are also called electrochemical cells.

Are cells in which electrical energy is generated from chemical reactions.

Types of electrochemical cells

(i). Primary cells –

Electrochemical cells which are not rechargeable

(ii). secondary cells-

– Voltaic/electrochemical cells which are rechargeable.

Primary cells/dry cells:

– Are of various types and an example is the Le clanche dry cell.

The Leclanche dry cell

Structure

Consist of a Zinc can with carbon rod at the centre.
The central graphite/carbon rod is surrounded by powdered Manganese (IV) oxide and carbon; which are inturn surrounded by a paste of NH4Cl (s) and Zinc Chloride.
The protruding portion of the carbon rod is covered with a brass cap; and the zinc can covered with a sealing material.

Chemical reaction

The Zinc can is the negative terminal; while the carbon/graphite rod is the positive terminal;

a the Zinc can/negative terminal
positive terminal /brass cap

Hydrogen gas

(NH₃ (g) + 2e^– 2NH₃ (g) + H₂(g)

Note: These gases (NH₃ _(aq) and H₂ _(g) ) are NOT used immediately but are used in more complex reactions.

The ammonia gas forms a complex with the zinc chloride in the paste.
The hydrogen gas is oxidised to water by the Manganese (IV) oxide.

Functions of the various components.

a) Brass cap – Functions as the positive terminal where the reduction reaction

b) Zinc can – Is the Negative terminal; where the oxidation reaction occurs.

c) Carbon rod – It serves as the positive electrode.

Acts as the connecting wire between the positive and negative terminal through it electron flow form the Zinc can to the brass cap.

d) Manganese (IV) Oxide – To oxidise the Hydrogen gas produced at the anode/positive terminal/brass cap to water.

A single dry cell can produce a potential of 1.5 V

Note: Dry ammonium chloride does not conduct electric current. This explains why a paste, which is a conductor, is used.

The dry cells cannot provide a continuous supply of electricity for an undulate period of time.

Reason – The reactants (electrolytes) are used up and cannot be replaced.

Secondary cells

Are voltaic/electrochemical cells that are rechargeable.

A common example is the lead acid accumulator.

The lead acid accumulator.

Structure: – The positive plate is a lead grill filled with lead (IV) oxide; while the negative plate consists of a similar lead grill filled with spongy lead.

– The grills are immersed in sulphuric acid; which serves as the electrolyte.

Reactions

i) During discharge/when in use

– At the negative terminal/lead – The lead dissolves forming lead (II) ions

equation

At the positive terminal (lead (IV)

Lead (IV) oxide reacts with the Hydrogen ions (H⁺) in sulphuric acid; also forming lead (II) ions.

Then; the lead (II) ions formed at both electrodes react instantly with the Sulphate ions to form lead (II) Sulphate.

– The insoluble lead Sulphate adheres to the electrodes.

Overall reaction

Note: The Lead Sulphate should NOT be left for too long to accumulate on the electrodes

Reason: The fine PbSO_{4 (s)}will charge to coarse non reversible and inactive form and the accumulator will become less efficient.

During use/discharge; the lead and the lead (IV) oxide are depleted, and the concentration of sulphric acid declines.

ii) During recharging – Is usually done by applying a suitable voltage to the terminals of the

– At the negative terminal – the lead ions accept electrons to form lead solid.

Overall reaction

This process restores the original reactants.

ELECTROLYSIS

Defination

Is the decomposition of molten or aqueous solutions by passage of electric current through it.

Terminologies used in electrolysis

i) Electrolyte

– Is a solution which allows electric current to pass through while it gets decomposed.

Electric current transfer in electrolyte occur through ions.
The electrolyte can be aqueous solutions or molten solutions
Electrolyte with may ions are called strong electrolyte; while those with few ions are called weak electrolytes.

Examples

ii) Electrodes – Are the solid conductors, usually roots, which usually complete the circuit between electrolytes and cell/battery

– Are of two types

a) Anode – The electrode connected to the positive terminal f a battery/cell
b) Cathode– Electrode connected to the negative terminal of the battery/cell.

Note: Graphite rods are commonly preferred as electrodes in most access.

Reasons: They are inert/unreactive

Are cheap

Platinum is also relatively inert; but not less preferred to Graphite because its expensive.

Preferential discharge of ions

The products of electrolysis of any given electrolyte depend on the ions present in an electrolyte.
Commonly most molten electrolytes have only two ions; a cation and an anion and are termed Binary electrolytes.
As the electrolyte decomposes, ions collect/move to the opposite poles.
Negatively charged ions move to the Anode; the positive electrode, while the positively charged ions move to the cathode, the negative electrode;
Regardless of how many ions move to an electrode; only one can be discharged ot give a product.
Both cations and anions have a preferential discharge series.

Discharge for cations

Cations are discharged by reduction (accepting electrons) to form their respective products.

The ease of reduction of cations depends on their position of electrochemical series.

Thus Ag⁺ is most readily discharged as its the weakest reducing agent.

Discharge for anions

Anions are discharged by oxidation (electron loss) to form their respective products.

Discharge of anions is viz.

ELECTROLYSIS OF VARIOUS SUBSTANCES

Electrolysis of various substances
i)

ii) Procedure – An electric current is passed through the dilute sulphuric acid.

iii) Observation

At the Anode

– A colourless gas; collects

The gas collected relights a glowing splint; and its volume is half the volume of the gas at cathode. The gas is oxygen.

At the cathode

A colourless gas collects
The collected gas burns with a pop-sound; and its volume is double the volume of gas at the anode.
The gas is Hydrogen gas.

Explanations – Ions present in the electrolyte
i) Hydrogen ions and Sulphate ions form sulphuric acid.
ii) Hydrogen ions and hydroxide ions from water.

At the Anode (Positive electrode)

– The negatively charged Sulphate ions and hydroxide ions migrate to the anode.

Reason: OH^– _(aq) ions have a greater tendency to loose electrons than the SO₂^{– 4} _(aq) ions

Anode equation

At the cathode (positive electrodes) – The positively charged hydrogen ions migrate to the cathode.

Equation

Note:

The volume of oxygen produced at the anode is half the volume of hydrogen produced at the cathode.

Reason: The 4 electrons lost by the hydroxide ions to form 1 mole (1 volume) of oxygen molecules are gained by the four hydrogen ions which form 2 molecules (2 volumes) of hydrogen molecules.

During the electrolysis, the concentration of the electrolyte (H₂SO₄ (aq), increases

Reasons: The volumes of hydrogen and oxygen gas liberated are in the same ratio as they are combined in water.

Thus the amount of water in the electrolyte progressively decrease; hence the increased electrolyte concentration.

Conclusion

Electrolysis of dilute sulphuric acid is thus the electrolysis of water.

Note: The Hoffmans voltmeter can be used instead of the circuit above. Viz.

Electrolysis of dilute sodium chloride

i) Apparatus

ii) Procedure

An electric current is passed through dilute sodium chloride solution; with carbon rods as the electrodes.
Gases evolved of each electrode are collected and tested.

iii) Observations

At the anode:
A colourless gas is collected
The gas relights a glowing splint, and its volume is half the volume of the gas collected at the cathode.
The gas is Oxygen, O₂

At the Anode
A colourless gas collects
The gas burns with a pop sound; and its volume is twice the volume of the gas collected at the anode.
The gas is hydrogen gas;

iv) Explanations

The Ions present in the electrolyte

At the Anode

Cl^– and OH^– migrate to the anode
OH- are preferentially discharged coz they have greater tendency to lose electrons than the chloride ions; – the OH^– _(aq) lose electrons to form water and O₂ _(g) at anode.

Anode equations

At the Cathode

– The positively charged Na⁺ _(aq) , and H⁺ _(g) migrate to the cathode

the H⁺_(aq) are preferentially discharged.

Reason: They H⁺ _(aq) have a greater tendency to gain electrons than Na+ (aq) ions

The H⁺ _(aq) gain elect5rons to form Hydrogen atoms (H) which ten form molecules of hydrogen which bubble off at the electrode.

Cathode equations

Conclusion

Ratio of the volumes of H₂ _(g) and O₂ _(g) evolved at cathode and anode is 2:1 respectively.

Electrolysis f dilute NaCl is thus the electrolyze of water since only water is decomposed.

Electrolysis of Brine/concentrate sodium chloride.

i) Apparatus

ii) Procedure

An electric current is passed though concentrated sodium chloride/brinc

iii) Observation

a) At the Anode

A greenish yellow gas is evolved.
The gas has a pungent irritating smell; and its volume is equal to the volume of the gas evolved at he cathode.
The gas is chlorine Cl₂ _(g)

b) At the Cathode

A colourless gas is liberated
The gas burns with a pop sound; and its volume is equal to volume of gas evolved at the anode.
The gas is Hydrogen gas; H₂ (g)

iv) Explanations

The ions present in the electrolyte are:-
Na⁺ _(aq) and Cl^– from sodium chloride
H+ _(aq) and OH- _(aq) from water.

At the Anode

Cl^– _(aq), and OH^– _(aq), migrate to the anode
The chloride ions are preferentially discharged.

Reason; – OH^– _(aq) have higher tendency to lose electrons than Cl^– ions.

However coz of the higher concentration Cl^– _(aq) , relative to OH^– _(aq), the Cl^– _(aq),are preferentially discharged hence the evolution of Chlorine gas.

A the Cathode

Na⁺ _(aq) and H⁺ _(aq) migrate to the cathode.
H+ with a higher tendency to gain electrons are preferentially discharged; hence the evolution of hydrogen gas at the cathode.

NB: 1. The pH of the electrolyte becomes alkaline/increases with time.

Reason: The removal of H⁺ _(aq) which come form water leaves excess hydroxide ions (OH^– _(aq),hence the alkalinity.

Evolution of chlorine gas at anode soon stops after sometime and is replaced by O₂ _(g)

Reason: Evolution of Cl₂ _(g) decrease/lowers the concentration of Cl^– _(aq) in the electrolyte.

As soon as the Cl^– _(aq) concentration becomes equal to that of OH^– _(aq)

The mercury cathode cell

Is an electrolytic arrangement commonly used for the large scale manufacture of chlorine and sodium hydroxide.

i) Apparatus

Electrolyte in the mercury cell is Brine (concentrated NaCl)
Anode is carbon or titanium
Cathode is a moving mercury film.

ii) Reactions
At the Anode

Both Chloride and Hydroxide ions are attracted
Due to their high concentrations the chloride ions are preferentially discharged.
The Cl^– _(aq) lose electrons to form Chlorine gas. (greenish yellow)

Cathode (moving mercury)

The Na⁺ _(aq) and H⁺ _(aq) are attracted
The discharge of H⁺ _(aq) is more difficult than expected
Hydrogen has a high over voltage at the moving mercury electrode and so sodium is discharged.

Equation

The discharged sodium atoms combine with mercury to form sodium amalgam

Equation

The sodium amalgam reacts with water to form sodium hydroxide, hydrogen and mercury.

Equation

Hydrogen is pumped out while the Mercury is recycled.
The resultant NaOH is of very high party.

Limitations/disadvantages of the Mercury Cathode cell.

Its expensive due to the high cost of mercury.
Pollution form Mercury; i.e. Mercury is poisonous and must be removed from the effluent.

Electrolysis of Copper (II) Sulphate solution.

NOTE: The products of electrolysis of copper (II) Sulphate solution depends on the nature of the elctrodes used.

i) Apparatus

ii) Ions present in the elctrolyte:

From copper (II) Sulphate

From water

During electrolysis

Using carbon/platinum electrodes

iii) Observations

At the Anode:

A colourless gas is liberated
The gas relights a glowing splint; hence its oxygen.

At the cathode

A reddish brown coating (of Cu solid) is deposited.

In the electrolyte

The blue colour of the solution (CuSO₄) _(aq) / becomes pale and finally colourless after a long time.

Reason: The blue colour is due to Cu2+. As the Cu²⁺ _(aq) are continuously being discharged at the cathode; the concentration of CU ²⁺ decreases i.e. decrease in the concentration of Cu 2⁺ _(aq) in the solution

The electrolyte become acidic/pH decreases (declines)

Reason: Accumulation of H⁺ _(aq) in the solution since only OH^– (from water) are being discharged (at the anode).

iv) Explanation

At the anode

The SO ^2- ₄ _(aq) and OH ^– _(aq) migrate to the anode.
The hydroxide ions have a higher tendency to lose electrons than the SO ^2- ₄ _(aq)
They (OH^–) easily loose electrons to form the neutral and unstable hydroxide radical (OH^–)
The hydroxide radical (OH) decomposes to form water and Oxygen.

At the cathode

Copper ions and H⁺ _(aq) migrate to the cathode
Cu²⁺ _(aq) have a greater tendency to accept electrons than H⁺ _(aq)
The Cu ²⁺ _(aq) are thus reduced to form copper metal which is deposited as a red-brown coating on the cathode.

Cathode equation.

Using copper rods electrodes

observations

At the anode – Mass of the anode (Copper anode) decreases

At the cathode – reddish brown deposit

cathode increases in mass

Electrolyte no apparent change

Note: The gain in mass of the cathode is equal to the loss in mass of the anode.

Explanations

At the anode

– The SO ^2- _4(aq) and OH^– _(aq) are attracted to the anode.

However, none of them is discharged;
Instead; the copper anode itself gradually dissolves; hence the loss in mass of the anode;

Reason: its easier to remove electrons form the copper anode itself than format the hydroxide ions

At the cathode

H⁺ _(aq) migrate to the cathode
The Cu ²⁺ _(aq) are preferentially discharged; because they have a greater tendency to accept electrons
The copper cathode is thus coated with a reddish brown deposit of copper metal hence increase in mass.

Cathode equation

Factors affecting electrolysis and electrolytic products.

Electrochemical series

Electrolytic products at the anode and cathode during electrolysis depends on its position in the Electrochemical series.

Cations: The higher the cation in the electrochemical series; the lower the tendency of discharge at the cathode.

Reason: Most electropositive cations require more energy in order to be reduced and therefore are more difficult to reduce.

Reduction order

Anions: Discharge is through oxidation ad is as follows.

Concentration of electrolytes

A cation or anion whose concentration is higher is preferentially discharge if the ions are close in the electrochemical series.

Example: dilute and concentrated NaCl

Product at the anode.

The electrodes used: Products obtained at electrodes depend on the types of electrodes used

Examples: in the electrolysis of CuSO₄ _(aq) using carbon and copper rods separately.

APPLICATIONS OF ELECTROLYSIS

Extraction of reactive metals

Reactive metals/elements like sodium, magnesium, aluminum are extracted form their compounds by electrolysis.

Example: Sodium is extracted from molten sodium chloride using carbon electrodes.

Purification of metals

It can be used in refining impure metals

Examples: Refining copper

The impure copper is made of the anode.
Their strips of copper are used as the cathode
Copper (II) Sulphate are used as the electrolyte.

During the electrolysis the anode dissolves and pure copper is deposited on the cathode.

The impurities (including valuable amounts of silver and gold) from the crude copper collect as a sludge become the anode.

Electroplating

Is the process of coating one metal with another, using electrolysis so as to reduce corrosion or to improve its appearance.

During electrolysis:

the item to be electroplated is made the cathode
the metal to be used in electroplating is used as the anode
the electrolyte is made from a solution containing the ions of the metal to be sued in electroplating.

Examples

Gold-plated watches; silver plated utensils
Steel utensils marked EPNS. I.e. Electroplated Nickel Silver.

Anodizing Aluminum

Is the reinforcement of the oxide coating on Aluminum utensils/articles

Is done by electrolysis of dilute sulphuric acid using Aluminum articles as anode.

Importance: Prevention arrosion of Aluminum articles

Manufacture of sodium hydroxide, chlorine and hydrogen

Sodium hydroxide is prepared by the electrolysis of brine, for which 3 methods are available
The method depends o the type of electrolytic cell.
These cells are
The mercury cell
The diaphragm cell
The membrane cell

The mercury cell

Components
The electrolyte is concentrated sodium chloride
The anodes are made of graphite or titanium, which are placed above the cathode.
The cathode consists of mercury, which flows along the bottom of the cell.

Chemical reactions

Anode

Both chloride and hydroxide ions are attracted.
Chloride ions are preferentially discharged due to their high concentration
The chloride ions undergo oxidation to form green yellow chlorine gas.

Equation

At the cathode (flowing mercury)

Na⁺ _(aq) migrate to the cathode
Sodium ions are preferentially discharged.
They undergo reduction to form sodium solid.

Equation

the discharged sodium atoms combine with mercury to form sodium diagram

Equation

The sodium amalgam is then passed into another reactor containing water.
The amalgam reacts with water forming hydrogen and sodium hydroxide.

The mercury is regenerated and it is recycled into the main cell.

– Main product: Sodium hydroxide

– By products: Sodium and chlorine.

Advantages of mercury cathode cell

The resultant sodium hydroxide is very pure; as it has no contamination from sodium chloride.
It is highly concentrated; e. about 50%.

Disadvantages

Some of the mercury said its way into the environment leading to mercury pollution; a common case of brain damage in humans.
At the operating temperatures (70⁰C 80⁰C), mercury vapours escape into the atmosphere and cause irritation and destruction of lungs tissues.
Its operation requires highly skilled man power.

Diaphragm cell

Components

An asbestos diaphragm; to separate the electrolytic cell into two compartments; thus preventing mixing of H₂ and Cl₂ molecules
The anode compartment contains a graphite rod.
The cathode compartment contains a stainless steel cathode.

Diagram

Chemical reactions

The asbestos diaphragm is permeable only to ions, but not to the hydrogen or chlorine molecules.
It thus prevents H_{2 (g)}and Cl _(g) form mixing and reacting to yield HCl _(g)
It also separates NaOH and Cl2 which would otherwise react.

Chemical reactions

At the anode

Chloride ions undergo oxidation to form chlorine gas.

Equation

– At the cathode

H= and Na⁺ (a) migrate to the cathode compartment.
The H⁺ are preferentially discharged.
They (H⁺ _(aq) undergo reduction to form hydrogen gas.

Equation

the discharge of H+ causes more water molecules to dissociate, thus increasing the concentration of OH- in the solution.
therefore, the Na+ and OH- ions also react in the cathode compartment to form sodium hydroxide.

Equation

Advantage

Does not result into pollution

Disadvantage

The resultant NaOh is dilute (12% NaOH)
It is also not pure due to contamination with NaCl (12% NaOH + 15% NaCl by mass.

Note: – The concentration of the NaOH can be increased by evaporating excess water, during which NaCl with a lower solubility crystallizes out first, leaving NaOH at a higher concentration.

– The solid NaCl (Crystals) are then filtered off.

This is a case of fractional crystallization.

The membrane cell

Is divided into 2 compartments by a membrane
Most commonly used type of Membrane is the cation exchange membrane.
This membrane type allows only cations to pass through it.

Components

A cation exchange that divides the cell into 2 compartments; an anode and a cathode compartments.
Both electrodes are made of graphite
The electrolyte in the anode compartment is purified brine
The electrolyte in the cathode compartment is pure water.

Diagram

Chemical reactions

The anode

Chloride ions undergo oxidation to form green yellow chlorine gas.

The cathode

As current passes through the cell H+ and Na+ pass across the membrane to the cathode

H+ are preferentially discharged.
They undergo reduction to liberate hydrogen gas.

Continuos discharge of H+ leaves the OH- at a higher concentration

The OH- react with Na+ to form sodium hydroxide

Advantages

Resultant sodium hydroxide is very pure, since it has no contamination from NaCl.
The sodium hydroxide has a relatively high concentration; at about 30 35% NaOH by mass.

Uses of sodium hydroxide, chlorine, and hydrogen

Sodium Hydroxide

React with chlorine to form sodium chlorate sodium hypochlorite or I, NaOCl. This is a powerful oxidising agent which is used for sterilization and bleaching in textiles, paper and textile industries.

i.e. 2NaOH _(aq) + Cl₂ _(g) NaOCl _(aq) + NaCl _(g) + H₂O _(l)

Manufacture of sodas, detergents and cosmetics.
Neutralization of acidic solutions in the laboratories.

Hydrogen

For hydrogenation in the manufacture of margarine
Manufacture of ammonia
Production of hydrochloric acid

Chlorine

Formation of sodium chlorate I; for bleaching in pulp, textile and paper industries.
Sewage and water treatment
Manufacture of polymers such as polyvinyl chloride.
Manufacture of pesticides.

QUANTITATIVE ASPECTS OF ELECTROLYSIS

Basic terminologies and concepts

Ampere

Is the standard unit used to measure an electric current; the flow of electrons

Is usually abbreviated as amps.

Coulomb

Is the quantity of electricity, when a current of 1 ampere flows for one second. I.e. 1 Coulomb = 1 Ampere x 1 second

Generally:

Quantity of Electricity = current x Time in seconds

A = It; Where

Q = Quantity of electricity in coulombs

I = Current in Amperes

T = time in seconds

Faraday

Is the quantity of electricity produced by one mole of electrons; and is usually a constant equivalent to 96487 (approx. 96500) coulombs

Faradays laws of Electrolysis.

First law;

The mass of substance liberated during electrolysis is directly proportional to the quantity of electricity passed.

Worked examples

A current of 2.0 Amperes was passed through dilute potassium sulphate solution for two minutes using inert electrodes.
Write the equation for the reaction at anode.

Work out the mass of the product formed at the cathode. (H = 1.0, Faraday = 96,000 C)

Solution; quantity of electricity = current x time

= 2 x 2x 60

= 240 coulombs

cathode reaction

1 mole of electrons = 96000 C

4 moles of e- = 4 x 96000 C

= 384,000 C

What mass of copper would be coated on the cathode from a solution of copper (II) Sulphate by a current of 1 amp flowing for 30 minutes.

(Cu = 63.5; Faraday constant = 96487 Cuo/-)

solution

Cathode reaction

Cu 2+ (aq) + 2e- Cu (s)

– 1 mole of Cu requires 2 moles of electrons

Quantity of electricity passed; = 1 x 30 x 60 coulombs; = 1800C
1 mole of electrons carriers a charge of 96487 coulombs = 192974 coulombs
- coulombs deposit 63.5 g of Cu.

Thus 1800 C will deposit 63.5 x 1800 = 0.592 grams

192974

An element x has relative atomic mass of 88g. when a current of 0.5 amperes was passed through a solution of x chloride for 32 minutes, 10 seconds; 44 g of x was deposited at the cathode. (1 faraday = 96500 c)

Calculate the charge on the ion of x.

In the electrolysis of dil CuSO4 solution, a steady current of 0.20 Amperes was passed for 20 minutes. (1 Faraday = 96, 500 C Mol-, Cu = 64)

Calculate

The number of Coulombs of electricity used

The mass of the substance formed at the cathode.

2 moles of electrons liberate 1 mole of Cu. i.e. Cu ²⁺ + 2e Cu (s)

An element p has a relative atomic mass of 44. When a current of 0.5 Amperes was possed through a fused chloride of p for 32 minutes and 10 seconds; 0.22g of p were deposit

UNIT 5: METALS: EXTRACTION PROPERTIES AND USES.

Introduction:

Introduction
Extraction methods
Concentration of the ores
Metal extraction

Sodium metal
Main ores
Extraction process
Properties of sodium
Uses of sodium

Aluminium metal
Main ores
Qualitative analysis
Extraction from bauxite.
Electrolysis of purified bauxite
Properties of aluminium
Uses of aluminium

Zinc metal
Main ores
Qualitative analysis
Extraction process
By oxidation
By electrolysis
Properties of zinc
Uses of zinc

Iron metal
Main ores
Qualitative analysis
Extraction from haematite
Properties of iron
Uses of iron

Copper metal

Main ores
Qualitative analysis
Extraction process from copper pyrites
Properties of copper
Uses of copper

Lead metal

Main ores
Extraction process
Properties of lead
Uses of lead

Introduction:

– Only most unreactive metals occur naturally in their elementary form.

Examples: – Gold, Silver, Platinum.

– Other elements occur as ores i.e. metal bearing rocks.

Examples:

– Oxides

– Sulphides

– Carbonates

– Chlorides.

Note:
An ore is a mineral deposit with reasonable composition of a desired metal.

Methods of Extraction

– Depend on position of the metal in the reactivity series.

– Main methods are:

Electrolytic Method:

– Used for metals high up in the reactivity series

E.g. – Sodium and Potassium

– Calcium and Magnesium

– Aluminium.

– These metals occur in very stable ores

Reduction method:

– For less reactive metals.

E.g. Iron, Zinc, and Copper.

– Is achieved using;

(i) Carbon in form of coke.

(ii) Carbon (II) oxide

(iii) Hydrogen

– Oxidation is also used followed by reduction.

Preliminary steps before extraction.

– Minerals (mineral) are usually mined with several impurities which lower the concentration of the metal per given mass or volume.

– Thus the ore is first concentrated before the actual extraction.

– Concentration is possible due to difference in properties between the mineral compound and the earthy materials.

Methods of ore concentration.

Physical methods.

(a). Optical sorting.

– Used to separate ore particles that have sufficiently different colours to be detected by the naked eye.

– It involves physical handpicking of the desired particles.

– Mainly used for minerals containing transition elements such as chromium.

(b). Hydraulic washing.

– Also called sink and float separation.

– Utilizes the difference in density between the minerals and the unwanted materials

– The ore is washed with streams of water.

– The denser ore particles will sink to the bottom of the washing container and can then be collected.

– Examples in ores of tin and lead.

(c). Magnetic separation.

– Is used when either the ore particles or the earthy materials (unwanted materials) are magnetic.

– A strong magnet is used to attract the magnetic components and leaving the non-magnetic materials behind.

– Examples: in ores like magnetite (Fe₃O₄) and chromite which are magnetic.

(d). Electrostatic separation.

– Used to separate particles which have different electric charges.

– The particles are subjected into an electric field.

– The oppositely charged particles follow different paths and can then be separated.

(e). Froth floatation

– Is mainly use for sulphide ores.

– Takes advantage of two facts.

Oil can wet the surfaces of ores.
Oil floats on water

The process:

– The ore is ground into a fine powder; to increase the surface area for upcoming reactions.

– It is then mixed with water and a suitable oil detergent e.g pine or eucalyptus;

– The mixture is then agitated by blowing compressed air through it;

– Small air bubbles attach to the oiled ore particles; which are thenn buoyed up and carried to the surface where they float.

– A froth rich in mineral is formed at the top while impurities sink at the bottom.

– The froth is skimmed off and dried.

– Froth floatation process is used for copper, lead and zinc metals;

Diagram: froth floatation apparatus.

Chemical concentration.

– Involves the use of chemical reactions to concentrate the ores.

Examples:

– Bauxite, the main aluminium ore is chemically concentrated by a process known as Bayers process.

– This takes advantage of the amphoteric nature of aluminium oxide, which can thus react with both acids and bases.

– Chemical concentration can also be done by leaching.

– This involves reacting the ore with a compound such as sodium cyanide;

– The cyanide ions form complex ions with the metal.

– The complex ions formed are water soluble, and can be separated by filtration, leaving the unwanted materials in the residue.

The metals

Sodium

Main ores;

Rock salt / sodium chloride; NaCl
Chile saltpetre / sodium Nitrate; NaNO₃

iii. Soda ash/sodium carbonate; Na₂CO₃.

Other ores include;

(i). Borax; Na₂B₄O₇.10H₂O

(ii). Sodium Sulphate, Na₂SO₄;

Extraction;

– Sodium is obtained by the electrolysis of fused sodium chloride in the electrolytic cell.

– Calcium chloride and calcium fluoride are added to the electrolyte.

Reasons;

– To lower the melting point of sodium chloride from 800^oC to 600^oC;

– Once molten, the electrical resistance within the cell is sufficient to maintain the temperature without external heating.

– Steel or iron is used as the cathode, while carbon/graphite is used as the anode.

– Thus steel is not used as the anode.

Reason;

– At high temperatures, steel would react with chloride formed at the anode, but graphite is inert even at high temperatures.

– Steel wire gauze separates the electrodes.

Reason;

– To prevent products of electrolysis (sodium and chlorine) from mixing and reacting to form sodium chloride.

– The electrolytic apparatus used in sodium extraction is called the Downs cell.

Diagram: The Downs cell.

– During electrolysis, fused sodium chloride dissociates according to the equation;

NaCl_(l) Na⁺_(l) +Cl^–_(l)

At the cathode:

Observation;

– Soft silvery metal

Explanation

– Na⁺ ions are attracted and undergo reduction (accept electrons) to form/ produce molten sodium metal.

Equation;

Na⁺_(l) + e^– Na_(l)

– Molten sodium is lighter than fused sodium chloride and floats on the surface where it overflows into a separate container / sodium reservoir.

Note;

– The resultant sodium is usually collected in liquid / molten state, floating on top of the electrolyte.

Reasons;

– Less dense than molten sodium chloride

– Has a low melting point.

At the anode;

Observations;

– Evolution of a green-yellow gas.

Explanation:

– Chlorine gas is evolved as a by product and collected separately.

– Negatively charged Cl^– ions migrate to the positive anode and undergo oxidation to form chlorine gas;

Equation:

2Cl^–_(l) Cl_2(g) + 2e-

Properties of Sodium;

– Is a soft silvery metal with low density; 0.979gcm^-3

– Has a low melting point, 97^oC, and a low boiling point of 883^oC.

Chemical reactions

(a) With air

– Na is very reactive and tarnishes in moist air to form an oxide layer.

4Na_(s)+ O_2(g) 2Na₂O_(s);

– The oxide layer reacts with more air moisture to form hydroxide

Na₂O_(s) + CO_{2 (g)} Na₂CO_3(s)+ H₂O_(l)

Note

– Due to those series of reactions sodium is stored under a liquid hydrocarbon e.g. petroleum, kerosene.

– Sodium burns in oxygen with a golden yellow flame to form sodium peroxide

Equation:

2Na_(s) + O_2(g) Na₂O_2(s)

(White)

(b). With water

– Na reacts vigorously with water to form NaOH and Hydrogen.

Equation:

2Na_(s) + 2H₂O_(l) 2NaOH_(aq) + H_2(g)

– The resulting solution is highly alkaline with a PH of 14.

– Sodium is stored under oil to prevent contact with moisture from the atmosphere.

Note:

– The reaction between Na and dilute acids would be explosive and not safe to investigate

(c). With chlorine

– Sodium burns in chlorine gas;

2Na_(s) + Cl_2(g) 2NaCl_(s)

(d) With ammonia gas;

– Sodium forms hydrogen and a solid;

2Na_(s) + NH_3(g) 2NaNH_2(s) + H_2(g)

Sodamide

And;

NaNH_2(s) + H₂O_(l) NaOH_(aq) + NH_3(g)

Uses of sodium;

Is alloyed with lead in the preparation tetraethyl (IV) lead, which is added to petrol as an anti-knock.
Provides the glow in sodium vapours lamps, for street lighting (orange-yellow street lights).
Is an excellent conductor of heat and electricity with low melting point hence used;

In nuclear reactors to conduct away heat.
Modern aeroplane engines.

Manufacture of sodium peroxide, and sodium cyanide used in the extraction of silver and gold.

Question;

– Although electrolysis is an expensive way of obtaining metals, it must be used for some metals. Explain.

Solution;

– Group 1 and 2 metals together with Al are themselves such powerful reducing agents that their oxides cannot be reduced by chemical reducing agents.

Worked example

Below is a simplified diagram of the Downs cell in which sodium metal is manufactured.

(a) (i) Identify; –

Electrolyte X: – Molten sodium chloride

Gas Y: -Chlorine gas

(ii) Write an equation for the reaction at the cathode.

Na⁺_(l) + e- Na_(l).

(iii). In what state is sodium collected?

– Molten state/liquid state.

(iv). Give two properties of Na that makes it possible to be collected as in (b) (iii) above.

– Its less dense than molten sodium chloride.

– Has a low melting point.

(v). The cathode is made of steel but the anode is made of graphite.

Why is this yet steel is a better conductor?

– At high temperature steel would react with chlorine formed but graphite is inert even at high temperatures.

(vi). In this process, the naturally occurring, raw material is usually mixed with another compound. Identify the compound and state its use.

Compound; – Calcium chloride

Use;-To lower melting point of Nacl2 from 800^oc to 600^oc

(vii). What is the function of the steel gauze cylinder?

– Prevents sodium reacting with chlorine forming NaCl

(viii). Give one industrial use of sodium

– A coolant in nuclear reactors;

– Alloy with lead in tetraethyl (IV) lead;

(ix). Explain why sodium metal is stored under paraffin;

– Keep it out of air; reacts very fast with air forming a dull surface.

– Can react with water.

(b). State an industry that can be built next to a sodium extracting plant.

(c). A current of 100 Amperes flows through an electrolyte of molten sodium chloride for 15 hours. Calculate the mass of sodium produced in kg (Na = 23; 1F = 96500C)

Solution:

Q = It

=100 x 15 x 60 x 60

=5400000C.

Cathode equation

Na⁺_(l) + e^– Na_(l)

96500C = 23g of Na

5400000 C =23 x 5400000 =1287.04g

96500

=1.287kg

(d) .For the same quantity of electricity as in (c) above ; calculate the volume of the gaseous product produced in the cell at 15⁰c and 800mmhg.(Molar gas volume at s.t.p = 22.4dm³)

Aluminium

– Forms 7% of the earths crust and is the most common metal.

Main ores;

– Bauxite; Al₂O₃.H₂O

– Mica; K₂Al2S₆0₁₆.

– China clay;Al₂S₂O₇2H₂O

– Corundum;Al₂O₃.

Chemical test;

– Crush the ore into a fine powder;

– Add dilute nitric (V) acid to the powder

– Filter to obtain a solution of the ore;

– To a solution of the ore add NaOH_(aq) dropwise till in excess, and then repeat the same procedure using Ammonia solution, NH₄OH.

Observations:

With NaOH_(aq):

– White precipitate in soluble in excess;

With NH₄OH_(aq):

– White precipitate insoluble in excess;

Extraction from Bauxite;

-Involves two main processes;-

Purification of Bauxite.
Electrolysis of purified bauxite (alumina)

Purification of bauxite

– Chief impurities are small quantities of silica and iron (III) oxide.

– The oxide ore is ground and treated under pressure/ dissolved in hot aqueous sodium hydroxide.

During the process;

– The amphoteric bauxite dissolves in NaOH forming sodium aluminate;

Equation:

2NaOH_(aq) + Al₂O₃.3H₂O_(s) 2NaAl(OH)_4(aq).

Ionically:

Al₂O_3(s) + 2OH^–_(aq) + 3H₂O_(l) 2[Al(OH)₄]^–_(aq);

– Silica impurities also dissolve forming sodium silicate

Equation:

SIO_2(s) + 2NaOH_(aq) Na₂SIO_3(aq) + H₂O_(l)

– The iron impurities (mainly iron (III) oxide) DO NOT dissolve.

– This mixture is then filtered, during which iron (III) oxide remain as residue of red mud while a filterate of sodium aluminate and sodium silicate is collected.

– Carbon (IV) oxide is bubbled through the filterate, followed by dilution then addition of a little aluminium hydroxide to cause precipitation (seeding) of Aluminium hydroxide.

Ionically:

2[Al(OH)₄]^–_(aq) + CO_2(g) 2Al(OH)_3(s) + H₂O_(l);

Seeding

Al(OH)₃

Alternatively:

Al(OH)₄^–_(aq) 2Al(OH)_3(s) + OH^–_(aq);

General equation:

NaAlO_2(aq) + 2H₂O_(l) ^hydrolysis NaOH_(aq) + Al(OH)_3(s).

– The precipitated Aluminium hydroxide is then filtered off, washed and ignited to give pure aluminium oxide (Alumina);

Equation:

2Al(OH)_3(s) Al₂O_3(s) + 3H₂O_(l)

Alumina

Electrolysis of purified bauxite (alumina)

– The Alumina (Al₂O₃), has a high melting point, 2015^oC and a lot of heat would be required to melt it.

– Additionally the molten compound is a very poor conductor of electricity.

– Consequently, cryolite (Na₃AlF₆) is mixed with the oxide.

Reason;

– To lower the melting temperature of Al₂O₃ from 2015^oC to around 800^oC;

– At this lower temperature the molten oxide also conducts well.

– The molten alumina mixed with bauxite is then electrolysed in a steel cell lined with carbon graphite as the cathode.

Note;

– Other than being an electrolyte the graphite cathode lining also prevents alloy formation, as it ensures no contact between the resultant aluminium and the steel tank;

-The anodes also made of Graphite dip into the steel tank at intervals.

Diagram: electrolytic steel cell for the extraction of Aluminium.

Electrolytic reactions;

– The Aluminium oxide dissociates to give constituent ions;

Equation:

Al₂O_3(l) 2Al³⁺_(l) +3O^2-_(l)

At the cathode;

Observation;

– A silvery white metal which quickly becomes dulled.

Explanation:

– Aluminium ions move to the cathode and are reduced to form aluminium metal.

Equation;

2Al³⁺_(l) + 6e- 2Al_(l)

At the anode;

Observation;

– Effervescence of a colourless gas.

Explanations:

– Oxygen ions migrate to the anode and get oxidized to form oxygen gas.

– The resultant oxygen gas reacts further with the graphite anode to form carbon (IV) oxide.

– This is due to the high temperatures involved during the process.

Note;

– Consequently the carbon anode should be replaced from time to time.

Equations;

3O^2-_(l) 3O_2(g) + 6e-

Then;

C_(s) (anode) + O_2(g) CO_2(g)

Note:

– Cryolite usually adds Na⁺; and F^– ions into the electrolyte.

– Thus the anions are O^2- and F^– ions into the electrolyte.

– However oxygen is discharged in preference to Fluorine.

Reason;

– Fluorine is a stronger oxidizing agent than oxygen. Thus oxygen easily gives electrons than fluorine, hence discharge.

– Aluminium is discharged in preference to sodium.

Summary: – Flow chart on the Extraction of Aluminium from bauxite.

Properties of Aluminium

Physical properties

– Is a silvery white metal which quickly becomes dulled with a thin oxide layer.

– Has very low density (2.7gcm^-3), with ability to be rolled into wires / foil.

– Is a good conductor of heat and electricity.

Chemical properties.

Reaction with air;

– In air it acquires a continuous very thin coating of oxide, which resists further reaction.

– Removal of this protective cover renders the metal reactive.

– Consequently steel wool or wood ash should NOT be used in aluminium utensils.

– Usually, salty water attacks the oxide film allowing the aluminium to corrode and for this reason, ordinary aluminium is not used for marine purposes.

– Aluminium will burn in air at 800^oC to form is oxide and nitrate.

Equations:

4Al_(s) + 3O_2(g) 2Al₂O_3(s);

2Al_(s) + N_2(g) 2AlN_(s)

Reaction with Acids.

Note:

– The protective Aluminium oxide (being covalent and insoluble) layer makes its reactivity with acids less than expected.

With nitric (V) acid;

– Has hardly any effect on the metal, at any concentration.

Reason:

– Being a powerful oxidizing agent, it simply thickens the oxide layer thereby preventing further reaction.

With sulphuric (VI) acid;

– Only hot concentrated sulphuric (VI) acid breaks down the oxide layer and reacts with the metal.

Equation:

2Al_(s)+ 6H₂SO_4(l) Al₂(SO₄)_3(aq)+ 6H₂O_(l)+ 3SO_2(g)

With Hydrochloric acid;

– Dilute HCl dissolves aluminium slowly; liberating hydrogen.

Equation:

2Al_(s) + 6HCl_(l) 2AlCl_3(aq)+ 3H_2(g)

With concentrated HCl the rate of reaction is increased.

Reaction with chlorine;

– Hot aluminium burns in chlorine gas with a white light, forming dense white fumes of Aluminium (III) Chloride.

-The white fumes cool and collect on the cooler parts of the apparatus as a white solid.

Equation:

2Al_(s)+ 3Cl_2(g) 2AlCl_3(g)

Note:

– The apparatus for the preparation of AlCl₃ is kept very dry.

Reason:

– Aluminium chloride is readily/easily hydrolysed by water/moisture, and so it fumes in damp air with the evolution of hydrogen chloride gas.

Equation:

AlCl_3(s)+ 3H₂O_(l) Al(OH)_3(s)+3HCl_(g)

Reaction with water.

– Aluminium does not react with cold water, due to the formation of an insoluble coating of Aluminium oxide.

Note;

– If the oxide film is removed, the metal reacts slowly with cold water.

Reaction with caustic soda.

– The metal, especially in powder form, reacts with caustic soda solution, liberating hydrogen and leaving sodium aluminate in solution.

– The reaction is exothermic and once started, it is very vigorous.

Equation:

2NaOH_(aq) + 2Al_(s) + 2H₂O_(s) 2NaAlO_2(aq)+ 3H_2(g)

Ionically:

2Al_(s)+ 2OH^–_(aq)+ 2H₂O_(l) 2AlO^-2_(g)+ 3H_2(g)

Note:

– Thus aluminium has an amphoteric nature as it reacts with both acids and alkalis.

Uses of aluminium

Making parts of airplanes, railway, trucks, buses, tankers, furniture, and car e.t.c.

Reason;

– It is Very light due to a very low density.

Making cooking vessels/ utensils such as sufurias.

Reason:

– It is a good conductor of heat and electricity.

– It is not easily corroded by cooking liquids due to the unreactive coating of aluminium oxide.

Making overhead cables

Reason:

– It is a good conductor of electricity.

– It is light hence can easily be supported by poles and ductile to be rolled into wires (cables).

Aluminium powder mixed with oil is used as a protective paint.

Making Aluminium foils due to its high malleability. The foil is used in cooking; packaging and for milk bottle tops.

Making alloys, which have high tensile strength and yet light.

Examples:

Alloy	Component
Duralumin	Aluminium, copper, manganese and magnesium
Magnalium	Aluminium (70%) and magnesium (30%)

As a reducing agent in the Thermite process in the production of some elements such as chromium, cobalt manganese and titanium.

Example:

Cr₂O_3(s) + 2Al_(s) 2Cr_(s) + Al₂O_3(s)

Note:

The thermite process

– Is a process of reducing oxides of metals which are ordinarily difficult to reduce using Aluminium powder.

Examples:

Iron (III) Oxide (Fe₂O₃)
Chromium (III) Oxide (Cr₂O₃)
Compound oxide of manganese (Mn₃O₄)

– The oxide and the Al powder are well mixed together, forming Thermite.

– The thermite is ignited using magnesium ribbon fuse, since the reaction will not start at low temperatures.

– The high heat of formation of Aluminium oxide , results into a vigorous exothermic reaction that leads to a molten metal.

Example: -In the reduction of chromium (III) Oxide.

Cr₂O_3(s) + 2Al_(s) 2Cr_(l) + Al₂O_3(s) + Heat.

Sample question:

Zinc

Main ores;

(i). Zinc blende; ZnS

(ii). Calamine; ZnCO₃.

Extraction:

– Is done by electrolysis or reduction of its oxide using carbon.

Preliminary steps:

– The ore is first concentrated by froth floatation.

– The ore is roasted in air to convert it to the oxide.

Equations:

From Zinc blende:

2ZnS_(s) + 3O_2(g) 2ZnO_(s) + 2SO_2(g)

From Calamine:

ZnCO_3(s) ZnO_(s) + CO_2(g)

– After obtaining the oxide the metal is extracted by either reduction or reduction:

(a). The reduction method.

– The oxide is mixed with coke and limestone and heated in a furnace.

Diagram: Furnace for zinc extraction by reduction:

– The limestone (CaCO₃) decomposes to liberate CO₂ which is then reduced by coke to form carbon (II) oxide.

Equations:

Heat

CaCO_3(s) CaO_(s) + CO_{2 (g)}

Then:

CO_{2 (g)} + C_(s) 2CO_(g)

– The resultant carbon (II) oxide and coke are the reducing agents in the furnace, at about 1400^oC.

– They reduce the oxide to the metal; which is liberated in vapour form.

Equations:

ZnO_(s)+ C_(s) Zn _(g)+ CO _(g)
ZnO_(s)+ CO_(s) Zn _(g)+ CO₂ _(g)

– At the furnace temperatures zinc exists in vapour form, and leaves at the top of the furnace.

– Liquid zinc being lighter settles above molten lead and is run off;

– The vapour is condensed in a spray of molten lead to prevent re-oxidation of zinc.

– The resultant zinc is 98-99% pure and can be further purified by distillation.

– SO₂ is a by-product and is the main source of pollution in the extraction of zinc.

– Usually it is channeled to a contact process plant for the manufacture of sulphuric acid.

– Alternatively it can be scrubbed off to prevent pollution of the environment.

– Less volatile impurities remain in the furnace.

– The silica impurities combine with the quicklime/ calcium oxide (CaO) from limestone to form calcium silicate.

Equation:

CaO_(s)+ SiO_(s) CaSiO_3(s)

– The silicates together with other less volatile impurities form slag, at the bottom of the furnace from where it is run off.

Summary: Flow chart and the extraction of zinc

Sulphur (IV) oxide Coke and limestone CO₂ and excess CO_(g)

ZnO_(s)

ZnS (ore)

Zinc + impurities

Separation chamber

Slag

Air

Zinc liquid (pure)

(b). Electrolytic extraction of zinc.

– Zinc metal is obtained from the oxide via a series of steps:

Step I: Preparation of electrolyte:

– The ZnO obtained from roasting the ore is converted to zinc sulphate by reacting it with dilute sulphuric (VI) acid.

Equation:

ZnO_(s) + H₂SO_4(aq) ZnSO_4(aq) + H₂O_(l)

– Any lead (II) oxide impurity present in the zinc oxide reacts with the acid to form lead (II) sulphate.

Equation:

PbO_(s) + H₂SO_4(aq) PbSO_4(s) + H₂O_(l)

– The insoluble lead (II) sulphate is then precipitated and separated by filtration;

– The zinc sulphate is then dissolved in water and the solution electrolysed.

Step II: The electrolytic process:

Electrolyte:

– Zinc (II) sulphate solution;

Ions present:

– Zn²⁺ and H⁺ as cations; and SO₄^2- and OH^– as anions;

Cathode:

Lead containing 1% silver.

Anode:

– Aluminium sheets;

Chemical reactions:

Cathode:

Observations:

– Deposits of a grey solid.

Explanations:

– Zn²⁺and H⁺ migrate to the cathode.

– The Zn²⁺ are discharged in preference to H⁺;

Reason:

– The cathode is relatively reactive. Thus since zinc is more reactive thn hydrogen, its ions undergo reduction faster;

Equation:

Zn²⁺_(aq) + 2e- Zn_(s);

Note:

If graphite electrodes were used, hydrogen gas would have been evolved instead;

Anode:

Observations:

– Evolution of a colourless gas that relights a glowing splint;

Explanations:

– OH^– and SO₄^2-migrate to the cathode.

– The OH^– are discharged in preference to SO₄^2-; giving off oxygen gas

Reason:

The OH^– ions have a higher oxidation potential than SO₄^2- and therefore easily giving electrons for reduction at the cathode

Equation:

4OH^–_(aq) 2H₂O_(l) + O_2(g) + 4e-

Note:

– Over 80% of zinc is extracted by the electrolytic method.

– Zinc extracted by the electrolytic method is much more pure.

Note: – Industrial plants that can be set up near the zinc extraction plant.

– Contact process plant, to make use of the SO₂ by-product.

– Lead accumulators factories, to utilize the zinc produced.

– Paper factory using, SO₃ and hence SO₂ in bleaching.

– Brass factory for alloying zinc and copper.

– Steel factory to use zinc in galvanization.

Properties of zinc

Physical properties

– Is a blue-grey lustrous metal with:

Density of 7.1gcm^-3
Melting point of 420^oC
Boiling point of 907^oC

Chemical reactions

(a). with air;-

– Zinc tarnishes slowly forming a protective layer which prevents further reaction i.e. oxide layer or basic carbonate.

Equations:

2Zn_(s) + O_2(g) 2ZnO_(s)

Then;

ZnO_(s) + CO_2(g) ZnCO_3(s)

– It burns with a blue-green flame when strongly heated in air to form an oxide which is yellow when hot and white when cold.

Equation:

2Zn_(s) + O_2(g) 2ZnO_(s)

(b). With water

– Zinc does not react with water

– Steam reacts with red-hot zinc, forming zinc oxide and liberating hydrogen gas.

Equation:

Zn_(s) + H₂O_(g) ZnO_(s) + H_2(g)

(c). with dilute acids

– Zinc is above hydrogen in the reactivity series hence displaces hydrogen from steam (water) and dilute acids like H₂SO₄and HCl.

Equation:

Zn_(s) + 2H⁺_(aq) Zn²⁺_(aq) + H_2(g)

– Pure zinc reacts slowly while impure zinc reacts faster/ more quickly.

– Copper (II) sulphate is used as a catalyst to speed up the reaction.

(d). Concentrated acids.

(i). Concentrated sulphuric (VI) acid.

Equation:

Zn_(s) + 2H₂SO_4(l) ZnSO_4(aq) + 2H₂O_(l) + SO_2(g)

(ii). 50% concentrated nitric (v) acid.

– It reacts with 50% concentrated nitric (V) acid to liberate nitrogen (II) oxide.

Equation:

3Zn_(s) + 8HNO_3(l) Zn(NO₃)_2(aq) + 2H₂O_(l) + 2NO_(g)

(iii). Concentrated nitric (V) acid.

– It reduces concentrated nitric (V) acid to nitrogen (IV) oxide.

Equation:

Zn_(s) + 4HNO_3(l) Zn(NO₃)_2(aq) + 4H₂O_(l) + 2NO_2(g)

(e). with alkalis.

– Zinc is amphoteric and dissolves in hot alkalis to give the zincate ion and hydrogen gas.

Equation:

Zn_(s) + 2OH^–_(aq) + 2H₂O_(l) H_2(g) + [Zn(OH)₄]^2-_(aq)

(Zincate ion)

(f). Other reactions.

(i). Zinc burns in chlorine to give zinc chloride

Zn_(s) + Cl_2(g) ZnCl_2(s)

(ii). Zinc combines with sulphur

Zn_(s) + S_(s) 2ZnS_(s)

Uses of zinc:

Galvanization of iron sheets to prevent corrosion and rusting.

Note:

Rusting does not occur even when galvanized iron sheets are scratched and exposed.

Reason:

– The rest of the zinc protects the iron from rusting. This is because zinc being more reactive gets oxidized in preference to iron, and is hence sacrificed in the protection of iron.

– This method is referred to as cathodic or sacrificial protection.

Making brass; an alloy of copper and zinc.
Making outer casings of dry batteries;
Die-castings contain zinc and aluminium, and are used for making radio and car parts;
Zinc cyanide is used for refining silver and gold;

Sample question:

Iron

– Is the second most abundant metal after aluminium, forming about 7% of the earths crust.

Main ores

– Haematite, Fe₂O₃;

– Magnetite, Fe₃O₄;

– Siderite, FeCO₃;

Qualitative analysis for presence copper in an ore sample.

– Crush the ore into fine powder;

– Add dilute nitric (V) acid to the ore, to dissolve the oxide filter to obtain the filtrate.

– To the filtrate add aqueous sodium hydroxide / ammonium hydroxide dropwise till in excess.

– Formation of a red brown / brown precipitate in both cases indicates presence of Fe³⁺

Extraction from haematite (Fe₂O₃).

Summary of the process

– The ore-haematite is crushed and mixed with coke and limestone.

– The mixture is called charge.

– The charge is loaded into the top of a tall furnace called blast furnace.

– Hot air-the blast is pumped into the lower part of the furnace.

– The ore is reduced to iron as the charge falls through the furnace.

– A waste material called slag is formed at the same time.

– The slag floats on the surface of the liquid iron produced.

– Each layer can be tapped off separately.

Details of the extraction process.

(i) The blast furnace

Is a tall, somewhat conical furnace usually made of silica and lined on the inside with firebrick.

Details of the extraction process

Raw materials.

– Iron ore; i.e. haematite

– Coke; C

– Limestone; CaCO₃;

– Hot air;

Conditions

– Temperature at the bottom of furnace, 1400-1600^oC

– Temperature at the top of the furnace, 400^oC

Reactions and processes

Step-1 crushing and loading

– The ore is crushed into powder form, to increase the surface area for the upcoming reduction/ redox reactions.

– It is then mixed with coke and limestone and then fed at the top of the furnace using the double bell (double-cone devise) changing system

Note:

– The double bell charging system ensures that the furnace can be fed continuously from the top with very little heat loss, by preventing any escape of hot gases.

– This in turn reduces production costs.

Step 2: -Pre heating of the blast furnace.

– Air that has been preheated to about 700^oC is blown/ fed into the base of the blast furnace through small pipes called tuyers.

– This provides the required temperatures for the reactions in the blast furnace.

– This results into highest temperatures, about 1600^oC at the hearth (bottom of the furnace) which then decreases upwards the furnace.

Step 3: -Generation of reducing agents.

– Two reducing agents are used in this process: Coke and carbon (II) oxide; with carbon (II) oxide being the main reducing agent.

(i). Oxidation of coke;

– Coke burns in the blast at the bottom of the furnace.

– The reaction temperatures is about 1600^oC and the product is Carbon (IV) oxide gas

– This reaction is exothermic, producing a lot of heat in the blast furnace.

Equation:

C_(s) + O_2(g) CO_2(g)

(ii). Decomposition of limestone;

– The limestone in the charge decomposes in the blast furnace to calcium oxide (Quicklime) and carbon dioxide.

Heat

Equation:

CaCO_3(s) CaO_(s) + CO_2(g)

– The calcium oxide will be used in the removal of the main ore impurity/ silicates/ silica in the form of silicon (IV) oxide.

– The CO₂ then moves up the blast furnace to regenerate carbon (II) oxide, the chief reducing agent.

(iii). Production of carbon monoxide

– The CO₂ from oxidation of coke and decomposition of limestone (calcium carbonate) react with (excess) coke, to form carbon (II) oxide

– The reaction occurs higher up in the blast furnace at about 700^oC;

Equation:

CO_2(g) + C_(s) 2CO_(g)

Step 4: The actual reduction process

– Reduction of the ore is by either CO or coke, depending on temperatures.

(i). Reduction by coke

– This occurs much lower down the furnace at higher temperatures of about 800^oC and above.

– This reaction is ordinarily slow and thus serves to only reduce the part of the ore reduced by CO at lower temperatures in the upper parts of the furnace.

Equation:

2Fe₂O_3(s) ₊ 3C_(s) 4Fe_(s) + CO_2(g)

Note:

– The resultant CO₂ is quickly reduced to CO by the white-hot coke to more carbon (II) oxide as per step 3(iii) above.

(ii). Reduction by carbon (II) oxide

– This is the main reducing agent.

– The reaction between CO and Fe₂O₃ is relatively faster and occurs at lower temperatures of 500^oC-700^oC, higher up the furnace.

Equation:

Fe₂O_3(s) + 3CO_(g) 2Fe_(s) + CO_{2 (g)}

– The resultant carbon (IV) oxide is also quickly recycled by being reduced to CO by coke to from more reducing agent

(iii). Melting

– The iron produced in both of the reduction processes is in solid state.

– As the iron drops / falls down the furnace, it melts as it passes through the melting zone/ molten zone (1500^oC-1800^oC)

– The molten iron runs to the bottom of the furnace.

– Temperatures at the hearth (bottom of the furnace) is maintained at approx. 1400^oC and yet pure iron melts at about 1525^oC.

– Consequently the molten iron would easily solidify at the base (Temp =1400^oC)

– However this is not usually the case;

Reason:

-Impurities absorbed by iron during melting (mainly carbon) reducing the melting point to below 1400^oC.

– The molten iron is then easily tapped off.

Step 5: -Removal of earthy impurities.

– The earthy impurities in the ore (mainly silica) react with calcium oxide from decomposition of limestone to form calcium silicate.

Equation:

CaO_(s) + SiO_2(s) CaSiO_3(s)

– These earthy impurities form molten slag whose main component is calcium silicate.

– The slag does not mix with iron but rather floats on top of it, at the base of the furnace.

Importance of the slag

– As it floats on top of molten iron it protects it from being re-oxidized by the incoming hot air.

Uses/application of the slag

Light-weight building material.
Manufacture of cement.
Road building material.

Step 6:- Removal of furnace (waste) gases.

– Hot unreacted/waste gases leave at the top of the furnace.

– Main components include Nitrogen, unreacted CO₂, unreacted CO, oxygen and Argon (Noble gases)

– Additionally they contain dust particles.

Note:

– Upon removal of dust particles, the furnace gases, being hot can be used to pre-heat the air blown in at the base.

Summary: flow chart for the extraction of iron.

Properties of iron:

Physical properties:

– It has a melting point of 420oC and a boiling point of 907^oC;

– Have a good thermal and electrical conductivity;

– It is ductile and malleable;

Chemical properties.

(i). Reaction with air.

– It readily rusts in presence of moist air hydrated brown iron (III) oxide; Fe₂O₃.H₂O_(s)

Equation:

4Fe_(s) + 2H₂O_(l) + 3O_{2 (g)} 2Fe₂O₃.H₂O_(l)

– When heated it reacts with oxygen to form tri-iron tetroxide; Fe₃O₄;

Equation:

3Fe_(s) + 2O_{2 (g)} Fe₃O_4(s)

(ii). Reaction with water.

– It does not readily react with cold water.

– It however reacts with steam liberating hydrogen gas and forming tri-iron tetroxide.

Equation:

3Fe_(s) + 4H₂O_(g) Fe₃O_4(s) + 4H_2(g)

(iii). Reaction with chlorine.

– Hot iron glows in chlorine without further heating, forming black crystals of iron (III) chloride;

– Iron (III) chloride sublimes on heating and will thus collect on the cooler parts of the apparatus;

Equation:

2Fe_(s) + 3Cl_2(g) 2FeCl_3(s)

Note:
– Iron (III) chloride fumes when it is exposed to damp (moist) air;

Reason:

– It is readily hydrolysed by water with evolution of hydrogen chloride gas;

Equation:
FeCl_3(s) + 3H₂O_(l) Fe(OH)_3(s) + 3HCl_(g)

(iv). Reaction with acids:

Hydrochloric acid:

– Iron reacts with hydrochloric acid to liberate hydrogen gas.

Equation:

2Fe_(s) + HCl_(aq) FeCl_2(aq) + H_2(g)

Sulphuric (VI) acid:

Fe_(s) + H₂SO_{4 (aq)} FeSO4 _(aq) + H_{2 (g)}

Note: With hot concentrated H₂SO₄;

– The iron reduces hot concentrated H₂SO₄ to sulphur (IV) oxide and it is itself oxidized to iron (III) sulphate.

Equation:

2Fe_(s) + 6H₂SO_{4 (l)} Fe₂ (SO₄)_3(aq) + 6H₂O_(l) + 3SO_{2 (g)}

Nitric (V) acid.

– Iron reacts with dilute nitric (V) acid to form nitrogen (IV) oxide and ammonia which then forms ammonium nitrate.

Equation:

10HNO_{3 (aq)} + 4Fe_(s) 4Fe(NO₃)_2(aq) + NH₄NO_3(aq) + 3H₂O_(l)

– Warm dilute nitric (V) acid gives iron (II) nitrate.

– Concentrated nitric (V) cid renders the iron unreactive.

Reason:

– Formation of iron oxide as a protective layer on the metal surface.

(vi). Reaction with sulphur.

– Iron when heated in sulphur forms iron (II) sulphide.

Equation:

Fe_(s) + S_(s) FeS_(s)

Uses of iron.

– Iron exists in different types and alloys, depending on percentage composition of iron, and other elements.

– Each type of alloy of iron has different uses depending on properties.

Iron alloy or type	Properties.	Uses
Cast iron	– Refers to iron just after it has been produced in the blast furnace; – Contains 3-5% carbon, 1% silicon, and 2% phosphorus; Disadvantage: very brittle hence easily breaks; Advantage: It is extremely very hard;	– Making: Ø Furnaces; Ø Grates; Ø Railings; Ø Drainage pipes; Ø Engine blocks; Ø Iron boxes; Note: This is due to its very hard nature; – Manufacture of wrought iron and steel;
Wrought iron	– Refers to cast iron with 0.1% carbon. – It is malleable hence can easily be moulded or welded;	– making iron nails; horse shoes; agricultural implements like pangas; Note: Its use is declining due to increased use of mild steel
Steel	– Are alloys whose main component is iron; – Other components may be carbon; vanadium; manganese; tungsten; nickel and chromium; Examples: Mild steel: has about 0.3% carbon, 99.75% iron; Special steel: has a small percentage of carbon together with other small substances; Stainless steel: -Contains 74% iron, 18% chromium, and 8% nickel; Cobalt steel: – Contains about 97.5% iron and 2.5% cobalt; – Very tough and hard; – Slightly magnetic;	– Mild steel is used for making: Ø Nails; Car bodies; Ø Railway lines; Ship bodies; Ø Rods for reinforced concrete, pipes; Note: Advantage of mild steel: – It is easy to work on; – That with 10-12% chromium and some nickel is used to make: cutlery; sinks; vats; – Steel containing 5-18% tungsten is used for: making high speed cutting and drilling tools; – For making electromagnets;

Copper

Description: – A red brown metal.

Distribution: – Canada, USA, Zambia, and Tanzania.

Main ores;

– Copper pyrites, CuFeS₂;

– Cuprite, CuO

– Chalcocite, Cu₂S

– Malachite, CuCO₃.Cu (OH)₂

Qualitative analysis / test for presence in an ore sample.

– Crush the ore and then add dilute nitric or hydrochloric or sulphuric acid to dissolve the ore.

– Filter to obtain Cu²⁺ filtrate.

– Divide filtrate into 2 different test tubes.

– To one sample add aqueous Ammonium hydroxide dropwise till in excess formation of a pale blue precipitate soluble in excess NaOH to form a deep blue solution.

Equations

With little NaOH:-

Cu²⁺_(aq) + 2OH^–_(aq) Cu(OH)_2(s)

Pale blue ppt.

In excess:

Cu(OH)_2(aq) + 4NH_3(aq) [Cu(NH₃)₄]²⁺_(aq) + 2OH^–_(aq)

Deep blue solution

-To the second portion add sodium hydroxide solution dropwise till in excess, formation of a pale blue precipitate insoluble in excess confirms presence of Cu²⁺.

Extraction- from copper pyrites.

Crushing the ore

– The ore is crushed to increase the surface area for the succeeding chemical reactions.

– The ore is then concentrated.

Concentration of the ore.

– The ore is concentrated by froth floatation.

– The fine ore powder is mixed with water and oil, after which air is blown into the mixture, usually from below.

– Bubbles of the air forms froth, resulting to concentration of the ore.

– The lighter oil floats on top of the water, with the ore floating on top of the oil.

– The denser water sediments the earthy impurities like soil particles.

– The concentrated ore is then tapped off.

– This process involves formation of an oil froth onto which the ore floats hence the name froth formation.

First Roasting

-The concentrated copper pyrite, CuFeS₂ is then roasted in air to remove some of the sulphur impurities as sulphur (IV) oxide.

Equation:

2CuFeS_2(s) + 4O_{2 (g)} 3SO_{2 (g)} + 2FeO_(S) + Cu₂S_(s);

Note:

– During 1^st roasting limestone and silica (SiO₂) are added to the roasted ore and the mixture heated in the absence of air.

Importance

– Removal of iron impurities.

– The iron (II) so formed during roasting is converted to iron (II) silicate, FeSiO₃.

– The iron (II) silicate constitutes the major portion/component of the slag.

Equation:

FeO_(s) + SiO_2(g) FeSiO_3(s)

– The slag separates itself from the copper (I) sulphide.

– The sulphur (IV) oxide escapes into the atmosphere and is the major pollutant in this process.

Pollution control mechanisms.

– Scrubbing the gas using calcium hydroxide;

Equation:

SO_2(g) + Ca (OH)_2(s) CaSO_3(s) + H₂O_(l)

– Construction of a contact process nearby.

Second Roasting.

– The CuS is heated in a regulated supply of air where some of it is converted to Cu₂O

Equation:

2Cu₂S_(s) + 3O_2(g) Cu₂O_(s) + 2SO_2(g)

Reduction of copper (II) oxide

Note: – Not all the Cu₂S was oxidized to copper (I) oxide Cu₂O.

– The unreacted (unoxidized) Cu₂S serves as the reducing agent in this step.

i.e. The copper (I) oxide formed in step 4 is reduced to copper metal by the (unreacted) copper (I) sulphide.

– This is called blister copper.

Equation:

Cu₂S_(s) + 2Cu₂O_(s) 6Cu_(s) + SO_{2 (g)};

Electrolysis

-The copper metal from reduction in step 6 is impure and is thus purified by electrolysis.

Main impurities

– Traces of gold

– Traces of silver

– Iron

– Sulphur

Electrolytic apparatus

Anode: Impure copper

Cathode: Pure copper plates/ sheets;

Electrolyte: Dilute copper (II) sulphate solution (containing Cu²⁺; H⁺; SO₄^2- and OH^–)

Diagram of electrolytic apparatus.

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a brown solid.

Explanations:

– The copper (II) ions, Cu²⁺ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Cu²⁺ and H⁺ but Cu²⁺ are preferentially discharged due to their easy tendency to undergo reduction.

Equation:

Cu²⁺_(aq) + 2e- Cu_(s);

At the anode

Observations:

– Dissolution of the anode, hence the impure copper rod decreases in size.

Explanation

– Since the metal rod is dipped into a solution of its ions, the copper solid undergoes oxidation, losing electrons to form copper ions, Cu²⁺

– Consequently as more copper ions, Cu²⁺ get reduced at the cathode; more are released by the dissolving anode.

Equation:

Cu_(s) Cu²⁺_(aq) + 2e-

Overall reaction

Cu_(s) + Cu²⁺_(aq) Cu²⁺_(aq) + Cu_(s)

– The electrolytic product is 99.98% copper.

– Traces of silver and gold collect as sludge at the bottom of the cell.

Note:-To improve purity of the product of electrolysis, the following steps are advisable;

(i). Increase the dilution of the electrolyte/ use a very dilute electrolyte.

(ii). Reduce the amount of current / use a low current.

Summary of extraction of copper from copper pyrites.

Uses of copper

Making copper wires and contacts in switches, plugs and sockets

Reason:-Copper is a good conductor of electricity

Note:-For this purpose, pure copper is necessary, since impurities increase electrical resistance.

Making soldering instruments.

Reason:-Copper has a high thermal conductivity.

Manufacture of alloys.

Examples

Alloy	Components
Brass	Copper and zinc
Bronze	Copper and tin
German silver	Copper, zinc and nickel;

Making coins and ornaments.

Reason:-it is durable and aesthetic.

Properties of copper.

Physical properties

– Soft red-brown metal.

– Melting point of 1083^oC and a boiling point of 2595^oC

– Density is 8.92gcm^-3 and electrical conductivity is about 5.93 x 10^-9Ώ^-1m^-1

Chemical properties

(i). It does not react with cold water or steam.

(ii). Heating in air

– When heated in air it forms a black layer of copper (II) oxide on its surface.

– Finely divided copper burns with a blue flame.

Equation:

2Cu_(s) + O_2(g) 2CuO_(s)

(Red brown) (Black)

(iii). Reaction with chlorine

– Cu reacts with chlorine in presence of heat to form green copper (II) chloride.

Equation:

Cu_(s)+ Cl_2(g) CuCl_2(s)

(iv). Reaction with Acids.

– Copper does not react with dilute hydrochloric, nitric and sulphuric acids.

– However it reacts with 50% Nitric acid, concentrated Nitric acid and concentrated sulphuric acid.

With 50% Nitric (V) acid

– Copper reduces the nitric acid to nitrogen monoxide.

Equation;

3Cu_(s) + 8HNO_3(l) 3Cu(NO₃)_{2 (aq)} + 4H₂O_(l) + 2NO_(g);

With concentrated Nitric (V) acid

– Copper reduces the acid to brown nitrogen (IV) oxide/Nitrogen dioxide gas.

Equation:

Cu_(s) + 4HNO_3(l) Cu (NO3)_2(aq) + 2H₂O_(l) + 2NO_2(g)

With concentrated sulphuric acid.

-The sulphuric acid is reduced to sulphur (IV) oxide.

Equation:

Cu_(s) + 2H₂SO_4(aq) CuSO₄ _(aq) + 2H₂O_(l) + SO₂ _(g)

Worked example.

The flow chart below outlines some of the processes involved in the extraction of copper from copper pyrites. Study it and answer the questions that follow.

Hot air Air

(a) (i). Write an equation for the reaction in the first roasting chamber. (1mark)

(ii). Name gas K. (1mark)

(iii). Give the name and formula of slag M. (1mark)

(iv). Give the name of the reaction in chamber N. (1mark)

(v). Name the impure copper X. (1mark)

(b). Pure copper is obtained from impure copper by electrolysis.

(i). Name the anode the cathode and the electrolyte. (3 marks)

(ii). Write equations for the reactions at the anode and cathode. (2 marks)

(iii). calculate the time taken for a current of 10 amperes to deposit 32kg of pure copper. (Cu = 64, 1F = 96000C) (3 marks)

(c). Draw a diagram to show how you would plate an aluminium spoon with copper

6. Lead.

– Is a transition element that combines with other elements to form compounds with 2 oxidation states.

– It is among the group 4 elements;

Main ores:

– Galena, PbS (lead sulphide); the main ore;

– Cerrusite, PbCO₃ (lead carbonate)

– Anglesite, PbSO₄(lead II) sulphate);

Qualitative analysis / test for presence of Zn²⁺ in an ore sample.

– The ore is crushed and then dilute nitric or hydrochloric or sulphuric acid added to dissolve the ore.

– It is then filtered to obtain Zn²⁺ filtrate.

– The filtrate is divided into 2 different test tubes.

– To the first portion sodium hydroxide solution is added dropwise till in excess, formation of a white precipitate soluble in excess confirms presence of either Zn²⁺; Al³⁺ or Pb²⁺.

– To the second sample aqueous ammonium hydroxide is added dropwise till in excess; formation of a white precipitate soluble in excess NH₄OH_(aq) confirms presence of Zn²⁺ only;

Equations

With little NH₄OH:-

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

White ppt.

In excess:

Zn(OH)_2(aq) + 4NH_3(aq) [Zn(NH₃)₄]²⁺_(aq) + 2OH^–_(aq)

Colourless solution

Extraction of lead:

– Occur in three main steps:

Ø Ore concentration;

Ø Extraction by reduction;

Ø Purification (refining) by electrolysis;

1. Ore concentration:

– Is done by selective froth floatation;

– The ore is ground into a fine powder, then water and a suitable oil added;

– Air is then blown into the mixture; facilitating formation of a low density froth that floats on top;

– Additionally, chemicals such as sodium cyanide and zinc sulphate are added to facilitate separation of zinc sulphide present in the ore.

– The separated PbS is then dried and broken into smaller pieces, then subjected to reduction;

2. Reduction:

Step I: Roasting the ore:

– The crushed and concentrated ore is roasted in a furnace to convert it to lead (II) oxide;

Equation:

2PbS_(s) + 3O_{2 (g)} 2PbO_(s) + 2SO_{2 (g)}

– During roasting some of the lead (II) sulphide is converted to lead (II) sulphate;

Equation:

PbS_(s) + 2O_2(g) PbSO_4(s);

– Any lead sulphate formed is converted to lead silicate by silicon (IV) oxide;

– The fate of lead (II) silicate;

Note:

– Additionally the lead (II) sulphate may further react with lead sulphide to form lead metal;

Step II: Ore reduction:

– The lead oxide obtained is mixed with coke, limestone and silica and some scrap iron;

– The mixture is fed into the top of the Imperial smelting furnace (ISF); where it is melted using hot air blasts introduced near the bottom of the furnace;

Diagram: The Imperial Smelting furnace for Lead extraction.

Main reactions:

(i). The lead (II) oxide is reduced to lead by the coke.

Equation:
PbO_(s) + C_(s) Pb_(s) + CO_(s);

(ii). The resultant carbon (IV) oxide produced in reaction (i) above further reduces any remaining lead (II) oxide;

Equation:
PbO_(s) + CO_(s)Pb_(s) + CO_2(g);

(iii). The scrap iron is added so as to react with any lead sulphide that may be present.

Equation:

Fe_(s) + PbS_(s) Pb_(s) + FeS_(s)

(iv). The limestone undergoes decomposition to give calcium oxide and liberate carbon (IV) oxide;

Equation:

CaCO_3(s) CaO_(s) + CO_{2 (g)};

– The carbon (IV) oxide gets reduced by coke to form more carbon (IV) oxide for reduction as in reaction (ii);

Equation:
CO_{2 (g)} + C_(s) 2CO_(g);

(v). The calcium oxide reacts with silica in form of SiO₂ to form calcium silicate;

Equation:

CaO_(s) + SiO_{2 (g)} CaSiO_{3 (l)};

Waste gases and residues.

– The iron sulphide and calcium silicate form a molten slag which is less dense and floats on top of molten lead at the bottom of the furnace;

– From here the slag is separately tapped off;

– Excess gases and air that did not react in the blast furnace escape through outlets at the top of the furnace;

– These waste gases can be trapped and recycled;

– These gases include: excess CO; excess CO₂; oxygen; nitrogen; some SO₂; and argon;

Pollution effects:
– Main pollutant is sulphur (IV) oxide from roasting of the ore.

Pollution control:
– It is directly fed into a contact process plant or scrubbed using calcium hydroxide forming calcium sulphite;

Purification (refining) of lead:
– The molten lead obtained in this process contains impurities such as gold, silver, copper, arsenic, tin and sulphur;

– The impure lead is refined by electrolysis.

Electrolysis of molten lead.

(i). Electrolyte:

Any aqueous solution containing lead ions;

(ii). The anode:

– Impure lead;

(iii). The cathode:

– Pure lead;

Electrolytic reactions;

At the cathode.

Observations:

– Deposition of a grey solid.

Explanations:

– The lead (II) ions, Pb²⁺ move to the cathode, where they accept electrons and undergo reduction.

– Cations in the electrolyte are Pb²⁺ and H⁺ but Pb²⁺ are preferentially discharged due to their easy tendency to undergo reduction.

Equation:

Pb²⁺_(aq) + 2e- Pb_(s);

At the anode

Observations:

– Dissolution of the anode, hence the impure lead rod decreases in size.

Explanation

– Since the metal rod is dipped into a solution of its ions, the impure lead solid undergoes oxidation, losing electrons to form lead (II) ions, Pb²⁺

– Consequently as more lead (II) ions, Pb²⁺ get reduced at the cathode; more are released by the dissolving anode.

Equation:

Pb_(s) Pb²⁺_(aq) + 2e-

Overall reaction

Oxidation at anode;

Pb_(s) + Pb²⁺_(aq) Pb²⁺_(aq) + Pb_(s)

Reduction at cathode;

Summary: flow chart on extraction of lead.

Properties of lead:

Physical properties;

– Has a low melting point but a high density;

– The unusually low melting point of lead is difficult to explain using simple metallic bonding theory;

– It is rather soft and pliable;

– Relatively malleable;

Chemical properties.

Note: – Lead is fairly unreactive to most other metals;

Uses of lead:

It is used in several alloys e.g. solder and also added to bronze alloys to make them stronger;
Lead ingots are used in the manufacture of accumulators;
Being so malleable and so chemically inert lead sheeting was used for roofing (look at roofs of old churches and cathedral)-cost and pollution effects have however brought this to a stop;
Making lead pipes for water supply; this is also discouraged particularly in soft water areas due to threat of lead poisoning;
Making tetraethyl lead (IV) which for many years was used as a fuel additive to increase octane rating of fuels;
Used in weights, clock pendulums, plumb bobs etc; due to its high density;
It absorbs X-rays and hence lead aprons and lead glass are used to shield hospital radiographers;
Used for safe disposal or storage of radioactive substances since no radioactive emission has been known to pass through thick lead blocks;

Sample question:

UNIT 6: RADIOACTIVITY

Checklist

Meaning of radioactivity
Natural and artificial radioactivity
Nuclear equations and chemical equations.
Types of radiations

Alpha particles
Beta particles
Gamma rays

Half life of a radioisotope
Radioactive decay curves and half life
Radioactive disintegration and nuclear equations.
Radioactive decay series, nuclear fission and nuclear fusion;
Uses of radioactivity;
Dangers and pollution effects of radioactivity;

Definitions:

Radioactivity is an automatic spontaneous disintegration of nuclei of some heavy elements emitting some kinds of radiant energy
Elements which exhibit this are said to be radioactive and emit different types of radiation

Types of radiation from radioactive elements

They can be identified by

Measuring their penetration power

They were directed to thin leaves of paper, aluminium, lead plates of varying thickness.
Observations:

One penetrated only thin foils of paper but not aluminium of 0.05mm thick and lead
One penetrated both paper and Al but not lead
One penetrated paper Al and lead
Directing them to air

Observations
Penetrated less through air but caused a lot of ionization to the air molecule
Penetrated more through air but caused less ionization
Penetrated most through air but caused less ionization

These penetrated just as far as X-rays

Directing through magnetic /electric fields

S N

– +

The radioactive element radiation emits all the 3types of radiation when they were directed to poss through electric magnetic fields to the fields

Were slightly deflected towards the negative

Were strongly deflected towards the positive

Were not deflected

Therefore

A and B are charged particles

-Have opposite charges

A is positive, B is negative

And

Deflections obeyed through L.H Rule

A were deflected less because they were moving with higher momentum (MV) because of high mass

Experimental evidence

Experiments carried out by Rattlerford revealed

Were positively charged and had a mass of 4 units and thus a charge of +2 these are Alpha- particles therefore double charged helium ions,
Were negatively charged and had similar properties to cathode rays. Measurement of charge/ mass confirmed they were electrons and carries a unit of 1

They are beta particles

Were carrying no charge. They are electromagnetic waves similar to light rays and X- rays with a wave length of only 10 metres

Their emission enables a nucleus to lose surplus energy

They are gamma rays

Evidence for the nature of gamma rays

Gamma rays
Are unaffected by an electric field
Are an affected by a magnetic field
Can penetrate several centimeters of lead
Can be diffracted by the lattice of a crystal
Have no change in atomic number mass of the atoms emitting them

NATURE AND PROPERTY OF α , β ,AND γ

Nature	α- particles helium nuclei ⁴He (He )	Β-particles electrons	Rays electromagnetic waves radiation
Relative penetrating power	Least 5cm in air stopped by paper and Al foils	Several metres of air thin Al foil (100)	Penetrate air, Al and many mm of Pb (10000)
Range in air	A few cm	A few m	A few km
Effect of electric and magnetic fields	Small deflection	Large deflection	No deflection
Ionization of gases	Cause much ionization Discharges electroscopes rapidly	Cause less ionization Discharges electroscopes slowly	Negligible ionization

Radioactive decay curve

Is always a symptotic to the x-axis

Number of atoms of the

Radioactive element

– The number of atoms disintegrating per unit of time /second is always proportional to the number of atoms, N at that time

This number N decreases slowly / exponentially with time

At any time to the number of atoms of a radioactive element is N/ No and at time T½, only ½ the total number of atoms of the original radioactive element will be present i.e.

T½ = ½N

This time is referred to as the half-life period of a radioactive element

Therefore

Half-life period T½ of a radioactive element is defined as the time taken for ½ the atoms to disintergrate

Thus

In T½ the radioactivity of the element diminishes to half its value

Example

Radioactive decay curve of radon

With half life of 4 days

No = 6 X 10³ atoms

T½ = 4days

No. Of atoms

10³ x 6

10³x 5

10³ x 4

10³ x 3

Radon emits α- particles

Initial no of radon atoms = 10³ X 6

After 4 days the number of atoms present=10³ X 3

“ “ “ “8 days “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ “ =10³ X 1.5

“ “ “ “12 days “” “ “ “ “ “ “ “ “ “ “ “ “ “ = 10³ X 0.75

It is impossible to predict the atoms that will disintergrate next

The half lives of some radioactive isotopes

Radioactive isotopes	Half life
Uranium-238	4.5 X 10^{9 years}
Radium-226	1.6 X 10^{3 years}
Carbon-14	5.7 X 10^{3 years}
Strontium-90	28 years
Iodine-131	8.1days
Radon-222	4days
Bismuth-214	19.7minutes
Polonium-218	3minutes
Polonium-214	1.5 X 10 ^-4sec

-The half-life of a radioactive isotope provides aquantitave measure of its stability

Thus

-The shorter the half-life the faster the isotope decays and the more unstable it is

-The longer the half-life the slower the decay process and the more stable the isotope

RADIOACTVE DISINTERGRATION EQUATIONS

-Different radioactive elements disintergrate to emit different radiations

Emission of α particles

-Occurs in isotopes with an atomic number greater than 83 (Z> 83) because they are unstable since their nuclei are so heavy and their atomic mass is too large

-They attain stability by ejecting an alpha particle containing two protons and two neutrons

Examples

²³⁸U-α partially-

(b) Emission of β particles

Occur in isotopes with more neutrons than stable isotopes of the same element, therefore heaviest isotopes of an element are likely to emit β particles to attain stability
During the β-decay process, a neutron splits up forming a proton &an electron

The proton remains in the nucleus while the electron /β particles is ejected.

Result

Number of neutrons in the isotope decreases by one while number of protons increase by one.

(2) Emissions of beta particles in two stage

Application of Radioactivity /Using radioactive isotopes.

Carbon dating

Radio active Carbon 14 nuclide produced when Nitrogen 14 is bombarded with radiocum from the sum is used.
When they die carbon-14 resent starts to reduce as decay takes place by emission of β particles.
Using the decay curve of carbon –14 it is possible to estimate age of the animals/plants since the carbon –14 is present in their tissues.

2.Medication/treatment of cancer

Gamma rays are used to treat /kill cancer cells when the tumour is subjected to the radiations

E.G Gamma rays /penetrating from CO are used in treating inaccessible growths.

Superficial /skin cancers can be treated by less penetrating radiation from or in plastic sheets strapped on the affected

3.Studying metabolic pathways

-Radioactive Isotopes can be used to trace the uptake of metabolism of various elements by animals /plants.

E.g Uptake of phosphate & metabolism of phosphorus by plants can be studied using a fertilizer containing

Radioactive tracer studies using have helped in the exudation of photosynthesis and protein synthesis

I have been used in the diagnosis & treatment of thyroid diseases &in research into thyroid gland functioning.

4.Thickness gauge and empty packet detectors

-Radiation passing through a material decreases as the material gets thicker

Hence:

Amount of penetrating Beta or-gamma- radiation can be used to estimate the thickness of various materials like paper, metal or plastic

-Radiation thickness gauges can be used to control the thickness of sheet steel emerging from a high-speed rolling mill.

-β –Rays measure thickness upto ~ 0.2cm of steel γ-rays can be used with steel upto 10cm thick.

-Level gauges are used to measure amount of liquid in fire extinguisher &gas cylinders

-Empty packet detectors can be set to reject empty/insufficiently filled packets filled packets of biscuits/cigarettes. \

5.decting pipe bursts

Can be underground pipes carrying water /oil
If the water / oil is mixed with radioactive substances form the mixture will leak at the point where there is a burst and the radiations can be detected if a detector is passed

Effect on static electricity

In textile industry the presence of static charges can attract dust and cause fires
When a radioactive element is placed in such industries the radiations emitted will ionize air and ions formed will attract the static charges : this minimises problems due to static charges

Hazards of radioactive

Arise from

Exposure of the body to external radiation
Ingestion/ inhalation of the radioactive matter

They damage body cells/ tissues
Cause mutation/deformities

Precautions

Protect the body with lad/ concrete shielding
Never pick/ hold radioactive elements with bare hands ; use forceps and well protected tongs
Use radiation absorbers

SAMPLE QUESTIONS

Particles from a radioactive source move through 7cm in air at ordinary pressure
The radioactive emission of radium are α-, β- and γ. Draw labelled diagram to show how the rays can be separated
the table below shows nuclides which are radioactive products of Their ½ lives and K.E during decay are shown

nucleide	Half life	energy
Th	1.39 X 10¹⁰	3.98
Th	1.9yrs	5.42
Ra	3.64days	5.66
Rn	54.5 sec	6.28
Po	0.16 sec	6.77
At	3 X 10^-4 sec	7.64
Bi	60.5 min	X
Po	2.9 X 10^-7 sec	8.78

Identify pairs of isotopes of same element
Identify 2 nuclides by in this table, which have been produced directly by α- decay of other nuclides in the table. Use equations

Identify 2 nuclides in this table, which have been produced by β- decay of other nuclides in the table. Use equations

Suggest a series of decays for formation of
Deduce how ½ life varies with energy of emitted rays

Suggest the value of X for

Teachers' Resources

CHEMISTRY FORM 1, 2, 3 & 4 SUMMARIZED NOTES

October 5, 2025 Hillary Kangwana Leave a comment

UNIT 1: ATOMIC STRUCTURE AND THE PERIODIC TABLE.

Checklist

Structure of the atom
The subatomic particles;

Protons
Neutrons
Electrons;

GET THE FREE CHEMISTRY NOTES IN PDF BELOW;

Chemistry High School notes for form 1-4 (Free updated pdf downloads)

CHEMISTRY FORM ONE NOTES

CHEMISTRY FORM FOUR NOTES: NEW

Atomic number and mass number
Isotopes;
Energy levels and electron arrangements.
The periodic table;

Groups
Periods;

Relative atomic mass and isotopes;
Ion formation;

Ion;
Cations;
Anions;
Ionization energy;
Electron affinity;

Valency and oxidation numbers;
Chemical formula;
Chemical equations
Balancing chemical equations;

The atom:

– Refers to the smallest particle of an element hat can take part in a chemical reaction;

– It has an average diameter of 10^-8 cm with a nucleus of about 10^-13 cm;

Parts of an atom

– The atom is made of two main parts:

The nucleus
The energy levels;

The nucleus:

– Is the positively charged part of an atom;

– The nucleus contains two subatomic particles; neutrons and protons;

– The positive charge is due to presence of protons;

– The nuclei of all atoms contain neutrons except the hydrogen atom;

– The protons and the neutrons are together referred to as the nucleons;

The energy levels.

– They contain the electrons;

– Electrons are so small and move so fast that their path cannot be traced directly;

– Thus the energy level simple represents the region where the electrons are most likely to be found;

Structure of the atom.

Note:

– The atom can still however be split into smaller particles termed the sub-atomic particles;

The sub-atomic particles.

– Are generally three:

Protons;
Neutrons;
Protons;

Protons.

– Are the positively charged sub-atomic particles;

– Are found in the nucleus and thus form part of the nucleons;

– The number of protons in the nucleus is equal to the number of electrons in the energy levels;

Neutrons.

– Are neutrally charged sub-atomic particles found in the nucleus of the atom;

– They are thought to probably prevent the positively charged protons from getting too close to each other;

Electrons.

– Are negatively charged sub-atomic particles found in the energy levels;

– The number of electrons in the energy levels is equal to the number of protons in the nucleus;

– This makes the atom to be electrically neutral;

Atomic number and mass number.

Atomic number.

– Refers to the number of protons in the nucleus of an atom;

Examples.

Sodium has 11 protons I the nucleus and thus said to have atomic number 11;
Chlorine has 17 protons in the nucleus and thus said to have atomic number 17;

Mass number;

– Refers to the sum of the number of protons and neutrons in an atom of an element;

Examples:

Sodium has 2 neutrons and 11 protons hence a mass number of 23;
Chlorine has 18 neutrons and 17 protons hence a mass number of 35.

Notation of atomic number and mass number;

– Both atomic number and mass number of an element can be written along with the symbol of an element;

Mass number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; ²³Na;

Magnesium ²⁴Mg;

Atomic number;

– Is conventionally represented as a superscript to the left of the symbol;

Examples:

Sodium; ₁₁Na;

Magnesium ₁₂Mg;

Thus the elements can be conventionally represented as:

Sodium ²³Na

Magnesium ²⁴Mg

Atomic properties of the first 20 elements.

Element	Symbol	Number of electrons	Number of protons	Number Of neutrons	Atomic number	Mass number
Hydrogen	H
Helium	He
Lithium	Li
Beryllium	Be
Boron	B
Carbon	C
Nitrogen	N
Oxygen	O
Fluorine	F
Neon	Ne
Sodium	Na
Magnesium	Mg
Aluminium	Al
Silicon	Si
Phosphorus	P
Sulphur	S
Chlorine	Cl
Argon	Ar
Potassium	K
Calcium	Ca

Isotopes.

– Are atoms of the same element with same atomic number but different mass number due to different number of neutrons.

Examples of isotopes.

Element	Isotope	Atomic No.	Number of protons	Number of neutrons	Mass number	Isotopic representation
Hydrogen


Carbon
Carbon
Oxygen


Chlorine
Chlorine

Energy levels and electron arrangements.

Energy levels:

– Are definite orbits in an atom that the electrons occupy.

– The energy levels are numbered 1, 2, 3starting with the one closest to the nucleus.

– Electrons occupying the same energy level have approximately the same amount of energy.

– Each energy level can only accommodate a given maximum number of electrons.

Maximum number of electrons per energy level

Energy level	Maximum number of electrons
1^st	2
2^nd	8
3^rd	8 (only for the first 20 elements)

Illustrations:

Hydrogen

– It has only one electron and thus this electron occupies the first energy level.

– Since the first energy level is not yet full, hydrogen does not have the second energy level;

– The electron arrangement of hydrogen is thus 1.

Helium:

– Helium is atomic number 2 and has only two electrons, which occupy the first energy level.

– The first energy level is thus completely full, but since there are no other r electrons lithium also has only one energy level;

– The electron arrangement is thus 2.

Chlorine:

– Chlorine has atomic number 17 and thus has 17 electrons;

– The first two electrons occupy the fist energy level which is thus completely filled up;

– The remaining 15 electrons occupy the second energy level, which can however accommodate only 8 to be completely filled up;

– Thus the remaining 7 electrons move to the third energy level; which needs 8 to be completely filled up;

– Since the third energy level is not yet full chlorine does not have a fourth energy level;

– The electron arrangement is thus 2.8.7.

Electron arrangement.

– Refers to the distribution of electrons in the energy levels of an atom.

Example: electron arrangement for the first 20 elements.

Element	Symbol	Atomic number	No. of electrons	Electron arrangement
Hydrogen
Helium
Lithium
Beryllium
Boron
Carbon
Nitrogen
Oxygen
Fluorine
Neon
Sodium
Magnesium
Aluminium
Silicon
Phosphorus
Sulphur
Chlorine
Argon
Potassium
Calcium

Dot and cross diagrams.

– Is a diagrammatic representation of the electron arrangements in an atom in which the energy levels are represented by concentric lines while electrons are represented by dots or crosses.

– However all electrons are the same regardless of whether they are represented as dots or crosses.

Examples:

Lithium	Magnesium	Aluminium	Carbon

No. of electrons =	No. of electrons	No. of electrons	No. of electrons
No. of protons =	No. of protons	No. of protons	No. of protons
No. of neutrons =	No. of neutrons	No. of neutrons	No. of neutrons

Beryllium	Nitrogen	Chlorine	Argon

No. of electrons =	No. of electrons =	No. of electrons =	No. of electrons =
No. of protons =	No. of protons =	No. of protons =	No. of protons =
No. of neutrons =	No. of neutrons =	No. of neutrons =	No. of neutrons =

Potassium	Boron	Silicon	Phosphorus

No. of electrons =	No. of electrons =	No. of electrons =	No. of electrons =
No. of protons =	No. of protons =	No. of protons =	No. of protons =
No. of neutrons =	No. of neutrons =	No. of neutrons =	No. of neutrons =

The Periodic table

– Is a table showing the arrangements of elements in order of their relative atomic masses.

– Is based on the ideas of Dmitri Ivanovich Medeleev.

– The modern periodic table is based on Meneleevs periodic law which states that:

The properties of elements are a periodic functions of their relative atomic masses

– The modern periodic law itself states that:

The properties of elements are a periodic functions of their atomic numbers

Design of the modern periodic table.

It has vertical columns called groups and horizontal rows called periods.

Groups:

– Are the vertical columns of a periodic table.

– Are eight in number; ad numbered in capital Roman numerals I all through to VIII.

Note: group VIII is also called group zero because the elements have little tendency to gain o lose electrons during chemical reactions

– Between group 2 and group 3 is a group of elements called the transition metals;

Periods:

– Are the horizontal rows in a periodic table.

– They are 8 in number in a modern periodic table.

Transition metals.

– Are elements that form a shallow rectangle between group II and group III.

– These elements are generally metallic and hence the name transition metals”

– They are not fitted in any group because they have variable valencies.

– Are sometimes called the d-block elements.

– They are much less reactive than the elements in groups I and II

– They have some unique characteristics that make them not fit in the 8 groups of the periodic table

– These include:

Have variable valencies; hence show different oxidation states in their compounds;
They form coloured compounds as solids and in aqueous solutions;
Have very high melting and boiling points (than metals in groups I and II).
They do not react with water;
Have very high densities (compared to metals in groups I ad II)

Lanthanides and Actinides.

– They form a block of elements within the transition metals.

– Are sometimes called the inner transition metals.

– The lanthanides consist of 14 elements from Cerium (Ce) to Lutetium (Lu).

– The Actinides are the 14 elements from Thorium (Tho) to Lawrencium (Lr).

Placing an elements in the periodic table

– The position of an elements in the periodic table is governed by the atomic number and hence the

electron arrangement.

The period:

– The period to which an element belongs is determined by the number of energy levels.

– The number of energy levels is equal to the period to which an elements belongs.

Examples:

Elements	Symbol	Atomic number	Electron arrangement	Number of energy levels	Period
Lithium	Li	3	2.1	2	2
Sodium	Na	11	2.8.1	3	3
Calcium	Ca	20	2.8.8.2	4	4
Nitrogen	N	7	2.5	2	2
Helium	He	2	2	1	1

Group:

– The group to which an element belongs to is governed by the number of electrons in the

outermost energy level.

– The number of electrons in the outermost energy level is equal to the group to which the element

belongs.

Examples:

Elements	Symbol	Atomic number	Electron arrangement	Outermost electrons	Group
Potassium	K	19	2.8.8.1	1	1
Aluminium	Al	13	2.8.3	3	3
Silicon	Si	14	2.8.4	4	4
Oxygen	O	8	2.6	6	6
Chlorine	Cl	17	2.8.7	7	7

Diagram of a full periodic table.

Note: In the modern periodic table atomic masses are used instead of mass numbers. The atomic

masses are preferable because they take care of elements with isotopes unlike mass numbers.

Diagram: part of the periodic table showing the first 20 elements

Relative Atomic Mass and Isotopes.

Introduction:

– The masses of individual atoms of elements are very negligible and thus quite difficult to weigh.

– On average the mass of an atom is approximately 10^-22g which cannot be determined by an

ordinary laboratory balance.

– For this reason the mass of atom has been expressed relative to that of a chosen standard element

hence the term relative atomic mass.

– The initial reference element was hydrogen which was later replaced with oxygen.

– Later the oxygen scale was found unsuitable;

Reason:

Oxygen exists I several isotopes and thus led to problems when deciding the mass of an oxygen atom.

– For this reason oxygen was replaced with carbon as the reference atom and to date relative atomic

masses of elements are based on an atom of carbon-12 (note that carbon is isotopic and exists as

Carbon -12 or carbon-14).

Definition:

Relative atomic mass (R.A.M) of an element refers to the average mass of an atom of the element

compared with a twelfth (1/12) of an atom of carbon-12.

RAM = Average mass of one atom of an element

¹/₁₂ mass of an atom of carbon-12

Measurement of Relative atomic mass

– RAM of elements is determined by an instrument called Mass Spectrometer.

– The instrument can also be used to determine the relative abundance of isotopes.

– The use of a mass spectrometer in determining the RAM of elements is called mass spectrometry.

How mass spectrometry works.

– In the mass spectrometer atoms and molecules are converted into ions.

– The ions are then separated as a result of the deflection which occurs in a magnetic field.

– Each ion (from an atom, isotope or molecule) gives a deflection which is amplified into a trace.

– The height of each peak measures the relative abundance of the ion which gives rise to that peak.

Note:

– Generally the relative atomic mass of an element is closest in value to the mass of the most

abundant isotope of the element.

Example: Diagram of a spectrometer trace for Lithium

Explanations:

– The trace has two peaks indicating that there are two isotopes for lithium.

– The fist peak occurs at a relative isotopic mass of 6 and the second at 7; these are the RAM of the two

isotopes respectively.

– The percentage abundance of the isotope with RAM of 6 (⁶Li) is 9 while the RAM of the isotope with

RAM 7 (⁷Li) is 91.

Calculating relative atomic masses of isotopic elements.

– Information form a spectrometer trace is usually extracted and used in calculation the relative atomic

mass of elements.

Worked examples.

The mass spectrum below shows the isotopes present in a sample of lithium.

(i). Use this mass spectrum to help you complete the table below for each lithium isotope in the sample. (3 marks)

Isotope	Percentage composition	Number of
Isotope	Percentage composition	Protons	Neutrons
⁶Li
⁷Li

(ii). Calculate the relative atomic mass of this lithium sample. Your answer should be given to three significant figures. (3 marks)

Element X with atomic number 16 has two isotopes. ⅔ of ³³X and ⅓ of ³⁰X. What is the relative atomic mass of element X? (2 marks)
Calculate the relative atomic mass of an element whose isotopic masses and relative abundances are shown below. (2 marks)

A neutral atom of silicon contains 14 electrons, 92% of silicon – 28, 5% silicon 29 and 3% silicon 30

(i). What is the atomic number of silicon? (1mark)

(ii). Calculate the relative atomic mass of silicon. (1mark)

Oxygen exists naturally as isotopes of mass numbers 16, 17 and 18 in the ratio 96:2:2 respectively. Calculate its R.A.M (2 marks)

Calculate the relative atomic mass of potassium from the isotopic composition given below.

Isotope Relative abundance

³⁹K 93.1

⁴⁰K 0.01

⁴¹K 6.89

Sulphur and sulphur compounds are common in the environment.

(a). A sample of sulphur form a volcano contained 88% by mass of ³²S and 12% by mass of ³⁴S.

(i). Complete the table below to show the atomic structure of each isotope of sulphur.

Isotope	Number of
Isotope	Protons	Neutrons	Electrons
³²S
³⁴S

(ii). Define relative atomic mass. (2 marks)

(iii). Calculate the relative atomic mass of the volcanic sulphur. (2 marks)

Iridium, atomic number 77, is a very dense metal. Scientists believe that meteorites have deposited virtually all the iridium present on earth. A fragment of a meteorite was analysed using a mass spectrometer and a section of the mass spectrum showing the isotopes present in iridium is shown below.

(a). Explain the term isotopes. (1mark)

(b). Use the mass spectrum to help you complete the table below for each iridium isotope in the meteorite.

Isotope	Percentage composition	Number of
Isotope	Percentage composition	Protons	Neutrons
⁶Ir
⁷Ir

(ii). Calculate the relative atomic mass of the iridium in this meteorite. (3 marks)

Ion formation.

Introduction:

– Atoms whose outermost energy levels contain the maximum possible number of electrons are said to be stable.

– Thus atoms with energy levels 2, 2.8 and 2.8.8 are said to be stable.

– Electron configuration 2 is said to have a stable duplet state while electron configuration 2.8 and 2.8.8 is said to have a stable octet state.

– These electron configurations resemble those of noble gases and as such they are stable and do not react.

– Atoms without this stability acquire it by either electron gain or electron loss.

– Whether an atom loses or gains electro(s) depend on the number of electrons in the outermost energy level.

– Take the case of sodium.

– Atomic number is 11 with an electron configuration of 2.8.1.

– Thus sodium has two options in to become stable:

to lose the single electron and acquire a stable electron configuration of 2.8.
to gain 7 electrons in its outermost energy level and acquire a stable electron configuration of 2.8.8

– Gaining a single electrons and losing a single electrons requires equal amounts of energy.

– Thus it is cheaper and faster in terms of energy for sodium to lose the single electron in the outermost energy level than to gain 7 electrons into its outermost energy level.

– Thus sodium acquires a stable electron configuration 2.8 by losing the single electron in its outermost energy level.

Diagram

Sodium atom Sodium ion.

Equation:

Na → Na⁺ + e^–

Further examples:

Element	Electron arrangement	Options for stability	Best (cheapest) option
Chlorine	2.8.7	2.8 or 2.8.8	2.8.8
Potassium	2.8.8.1	2.8.8 or 2.8.8.8	2.8.8
Aluminium	2.8.3	2.8 or 2.8.8	2.8
Magnesium	2.8.2	2.8 or 2.8.8	2.8
Carbon	2.4	2 or 2.8	2. or 2.8
Oxygen	2.6	2. or 2.8	2.8

Ion:

Definition: an ion is a charged particle of an element.

– Are formed when an atom of an element either loses or gains electrons.

Illustration:

– For a neutral atom the number of electrons in the energy levels (negative charges) is equal and thus

completely balances the number of protons in the nucleus (positive charges).

– Thus the net charge in a neutral atom is zero (0).

– When an atom gains electron(s), the number of electrons becomes higher than the number of protons

resulting to a net negative charge hence an ion.

– Oppositely when an atom loses electron(s) the number of protons becomes higher than the number of

electrons resulting into a net positive charge hence an ion.

– The charge on the ion is usually indicated as a superscript to the right of the chemical symbol.

– Thus ions are of two types:

Cations
Anions

Cations:

– Are positively charged ions.

– Are formed when atoms lose electrons resulting into the number of protons being higher than the number of electrons.

– Are mostly ions of metallic elements since most metals react by electron loss.

Examples:

(i). Magnesium:

– It has atomic number 12, with electron arrangement 2.8.2.

– It has 12 protons and 12 electrons hence a net charge of 0 hence the atom is written simply as Mg.

– It will form its ions by losing the two electrons from the outermost energy level.

– Thus the number of electrons decreases to 10 while the number of protons remains 12.

– This leads to a net charge of +2, giving the ion with the formula Mg²⁺.

Diagrammatic illustration:

(ii). Phosphorus

– It has atomic number 13, with electron arrangement 2.8.3.

– It has 13 protons and 13 electrons hence a net charge of 0 hence the atom is written simply as Al.

– It will form its ions by losing three (3) electrons out of the outermost energy level.

– Thus the number of electrons decreases by three to 10 while the number of protons remains 13.

– This leads to a net charge of +3, giving the ion with the formula Al⁺³.

Diagrammatic illustration:

Anions:

– Are negatively charged ions.

– Are formed when atoms gain electrons resulting into the number of electrons being higher than the number of protons.

– Are mostly ions of non-metallic elements since most non-metals ionize (react) by electron gain.

Examples.

(i). Chlorine:

– It has atomic number 17, with electron arrangement 2.8.7.

– It has 17 protons and 17 electrons hence a net charge of 0 hence the atom is written simply as Cl.

– It will form its ions by gaining a single electron into the outermost energy level.

– Thus the number of electrons increases to 18 while the number of protons remains 17.

– This leads to a net charge of -1, giving the ion with the formula Cl^–.

Diagrammatic illustration:

Note: the electros gained must be represented by a different notation from the initial electrons in the atom. E.g. if the initial electrons are represented with crosses (x) then the gained electrons should be represented by dots (.) and vise versa.

(ii). Phosphorus

– It has atomic number 15, with electron arrangement 2.8.5.

– It has 15 protons and 15 electrons hence a net charge of 0 hence the atom is written simply as P.

– It will form its ions by gaining three (3) electrons into the outermost energy level.

– Thus the umber of electrons increases by three to 18 while the number of protons remains 15.

– This leads to a net charge of -2, giving the ion with the formula P^-3.

Diagrammatic illustration:

Electron transfer during chemical reactions.

– Atoms react either by electron gain or electron loss.

– Generally metals react by electron gain while non-metals react by electron loss.

Illustration:

– Consider the reaction between sodium and chlorine.

– Sodium attains stability // reacts by losing the single electron form its outermost energy level.

– Chlorine attains stability // reacts by gaining a single electron into its outermost energy level.

– Thus during the reaction between the two elements the single electron lost by the sodium atom to

form the sodium ion is the same one gained by the chlorine atom to form the chloride ion.

Some basic concepts.

Valence electrons:

– Refers to the number electrons in the outermost energy level.

Examples:

– Calcium, with electron arrangement 2.8.8.2 has 2 valence electrons.

– Oxygen with electron arrangement 2.6 has 6 valence electrons.

– Phosphorus with electron arrangement 2.8.5 has 5 valence electrons.

Valency:

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– Valency is also known as the combining power of an element.

Examples:

– Calcium, with electron arrangement 2.8.8.2 loses 2 electrons during chemical reactions and hence has

a valency of 2.

– Oxygen with electron arrangement 2.6 gains 2 electrons during chemical reactions and thus has a

valency of 2.

– Phosphorus with electron arrangement 2.8.5 has gains 3 electrons during chemical reactions and

hence has a valency of 3.

– Aluminium, with electron arrangement 2.8.3 loses 3 electrons during chemical reactions and hence

has a valency of 2.

Note: Some elements have variable valencies and are usually termed the transitional elements (metals)

Examples:

– Iron can have valency 2 or 3;

– Copper can have valency 1 or 2

– Lead can have valency 2 or 4.

Summary on valencies of common elements.

Valency 1

Valency 2

Valency 3

Metals

Sodium

Potassium

Calcium

Barium

Magnesium

Zinc

Iron

Lead

Copper

Aluminium

Iron

Non-metals

Nitrogen

Chlorine

Fluorine

Hydrogen

Nitrogen

Oxygen

Sulphur

Nitrogen

Phosphorus

Radicals.

– Are groups of atoms with a net charge that exist and react as a unit during chemical reactions.

– Radicals also have a valency, which is equivalent to the value of its charge.

Summary on valencies of some common radicals.

Valency 1

Valency 2

Valency 3

Radicals

Ammonium (NH₄⁺)

Hydroxide (OH^–)

Nitrate (NO₃^–)

Hydrogen carbonate (HCO₃^–)

Hydrogen sulphate (HSO₄^–)

Carbonate (CO₃^2-)

Sulphate (SO₄^2-)

Sulphite (SO₃^2-)

Phosphate (PO₄^3-)

Oxidation number.

– Refers to the number of electrons an atom loses or gains during a chemical reaction.

– In writing the oxidation number the sign (+ or -) to show gain or loss is written followed by the

number of electrons lost or gained respectively.

Illustration.

– Atoms are electrically neutral and are thus assigned an oxidation state of 0 since the number of

protons in the nucleus is equal to the number of electrons in the energy levels.

– However when atoms react they either lose or gain electrons and thus acquire a new state.

– This new state is a new oxidation state and the atom thus acquires a new oxidation number

Examples:

Atom	E. arrangement	Ion formula	Valency	Oxidation number
Sodium	2.8.1	Na⁺	1	+1
Magnesium	2.8.2	Mg²⁺	2	+2
Aluminium	2.8.3	Al³⁺	3	+3
Nitrogen	2.5	N^3-	3	-3
Sulphur	2.8.6	S^2-	2	-2
Chlorine	2.8.7	Cl^–	1	-1

Further examples:

Particle	Oxidation number
Copper metal, Cu	0
Lead (II) ion, Pb²⁺	+2
Bromide ion, Br^–	-1
Aluminium ion, Al²⁺	+2
Sulphide ion, S^2-	-2
Magnesium metal, Mg	0

Note: oxidation number (state) and charge of an element.

– Oxidation state is written with the positive or the negative sign coming before the element.

Examples: -2, 3, +1, -1 etc.

– Charge on an element is write as a superscript of the element with the number coming before the

positive r the negative sign

Examples: Mg²⁺, Al³⁺, Na⁺, Cl^– etc.

Chemical formulae.

– Refers to a representation of a chemical substance using chemical symbols.

– In a single atom it is equivalent to the chemical symbol of the element.

– In a compound it shows the constituent elements and the proportions in which they are combined.

Deriving the chemical formula of compounds.

– In order to write the correct formula of a compound the following must be known:

The symbols of the constituent elements or radicals.
The valencies of the elements or radicals

– The chemical formula should start with the element which is more likely to lose electron (s) followed

by the element that is more likely to gain.

Worked examples.

Deriving the formula of sodium chloride.

Elements	Sodium	Chlorine
Formula	Na	Cl
valencies	1	1
balancing	x1	x1

Balancing ratios as subscripts: Na₁Cl₁

Formula: NaCl

Explanation:

– For sodium to combine with chlorine to form sodium chloride, sodium loses an electron while

chlorine gains an electron.

– Thus every sodium atom needs only a single chlorine atom for both t be fully stable

Note:

– When the balancing ratio // subscript is 1, it is usually not written since the symbol of the

element itself represents a single atom.

Deriving the formula of magnesium chloride.

Elements	Magnesium	Chlorine
Formula	Mg	Cl
valencies	2	1
balancing	X1	X2

Balancing ratios as subscripts: Mg₁Cl₂

Formula: MgCl₂

Deriving the formula of magnesium oxide.

Elements	Magnesium	Oxygen
Formula	Mg	O
valencies	2	2
balancing	X1	X1

Balancing ratios as subscripts: Mg₁O₁

Formula: MgO

Further worked examples.

Derive the chemical formula of each of the following compounds.

Calcium fluoride
Carbon (II) oxide
Carbon (IV) oxide
Aluminium nitrate
Calcium hydrogen carbonate

Complete the table below for elements A, B and C

Element	Valency	Chemical formula of various compounds
Element	Valency	hydroxides	Sulphates	carbonates	nitrates	phosphates	Hydrogen carbonates
A	1
B	2
B	3

Chemical equations.

– Refers to representations of a chemical reaction by means of chemical symbols and formula.

Key features of a chemical equation.

– The correct formulae of the reactants are on the left of the equation.

– The correct formulae of the products are on the right of the equation.

– The reactants and products are separated by an arrow pointing to the right.

– The state symbols of the reactants and products must be stated as subscripts to the right of the symbols

– The number of each atom on the reactants side must be equal to the number of the same atom on the products side.

Example.

Reaction between hot copper metals and oxygen gas.

Word equation: Copper + oxygen → Copper (II) oxide.

Chemical equation: 2Cu_(s) + O_2(g) → 2CuO_(s).

Balancing chemical equations.

– A chemical equation is only valid if it is balanced.

– A chemical equation is said to be balanced if the number of each atom on the reactants side is equal to that on the products side.

– This is because atoms are neither created nor destroyed during a chemical reaction.

Rules and guidelines in balancing chemical equations.

Step 1: Write the chemical equation in words.

Example: Copper metal + oxygen gas.

Step II: Write the correct formulae of both reactants and products

Example: Cu + O₂ → CuO

Step III: Check whether the number of atoms of each element on the reactants side is equal to that on the products side.

– If equal proceed to step (V);

– If not equal proceed to step (IV).

Example: Cu + O₂ → CuO

– In this case there are two oxygen atoms on the reactants side yet there is only one oxygen atom on the products side. Thus we proceed to step IV

Step IV: Multiply the chemical formula containing the unbalanced atoms with the lowest common multiple.

Example: Cu + O₂ → CuO

– In this case the chemical formula with the unbalanced atom is CuO on the products side. We thus multiply it by 2.

– The new equation now reads: Cu + O₂ → 2CuO

Step V: check again to ensure that all atoms are balanced.

– If all atoms are balanced proceed to step VI.

– If not then repeat step IV until all atoms are balanced.

Example: Cu + O₂ → 2CuO

– In this case multiplying CuO by 2 offsets the balancing of Cu; which is now unbalanced!

– We therefore repeat step IV in order to balance Cu.

– There is only 1 Cu atom on the reactants side yet there are 2 Cu atoms on the products side.

– We thus multiply the formula with the unbalanced atom (s) by the lowest common multiple, in this case 2.

– The new equation at this step thus becomes: 2Cu + O₂ → 2CuO
– We then repeat step V; in this case all atoms are now balanced.

Step VI: The physical states of the reactants and the products are then indicated.

– If this is not done the chemical equation is considered incorrect.

Types of state symbols.

– There four main state symbols.

Solid; denoted as (s)
Liquid; denoted as (l)
Aqueous (in solution in water); denoted as (aq)
Gaseous; denoted as (g)

– In a chemical equation the state symbols are written with their denotations as subscripts to the right of the chemical formulae.

Example: 2Cu_(s) + O_2(g) → 2CuO_(s)

Thus the balanced chemical equation for the reaction between copper metal ad oxygen is:

2Cu_(s) + O_2(g) → 2CuO_(s)

Worked examples:

Balance equations for each of the following reactions.

Sodium hydroxide and dilute hydrochloric acid

Zinc oxide and dilute sulphuric (VI) acid

Zinc metal and dilute nitric (V) acid

Calcium carbonate and dilute sulphuric (VI) acid

Sodium and water

Balance each of the following equations.

Mg_(s) + HCl_(aq) →MgCl_2(aq) + H_2(g)

Na_(s) + H₂O_(l) → NaOH_(aq) + H_2(g)

NaOH_(aq) + H₂SO_4(aq) →Na₂SO_4(aq) + H₂O_(l)

CuCO_3(s) + HNO_3(aq) → Cu(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

H₂S_(g) + O_2(g) → O_2(g) + H₂O_(l)

C₂H_6(g) + O_2(g) → CO_2(g) + H₂O_(l)

Pb(NO₃)_2(s) → PbO_(s) + NO_2(g) + O_2(g)

Fe_(s) + Cl_2(g) → FeCl_3(s)

Al_(s) + H₂SO_4(aq) → Al₂(SO₄)_3(aq) + H_2(g).

UNIT 2: CHEMICAL FAMILIES; PATTERNS AND PROPERTIES.

Checklist.

– Meaning of chemical families.

– Main chemical families;

Alkali metals
- Meaning and members
- Trends down the group
  - Atomic radius
  - Ionic radius
  - Energy levels
- Physical properties
  - Appearance
  - Ease of cutting
  - Melting and boiling [points
  - Electrical conductivity
  - 1^st ionization energy
- Chemical properties
  - Burning in air
  - Exposure to air
  - Reaction with water
  - Reaction with chlorine
- Similarity of ions and formula of compounds of alkali metals
- Uses of alkali metals.
Alkaline earth metals
- Meaning and members
- Trends down the group
  - Atomic radius
  - Ionic radius
  - Energy levels
- Physical properties
  - Melting and boiling [points
  - Electrical conductivity
  - 1^st ionization energy
  - 2^nd ionization energy
- Chemical properties
  - Reaction with air and water
  - Reaction with steam
  - Reaction with chlorine
  - Reaction with dilute acids
- Similarity of ions and formula of compounds of alkaline earth metals
- Uses of alkaline earth metals.
The Halogens
- Meaning and members
- Trends down the group
  - Atomic radius
  - Ionic radius
  - Energy levels
- Physical properties
  - Preparation and properties of chlorine
  - Appearance and physical states of halogens at room temperature
  - Melting and boiling points
  - Electrical conductivity
- Chemical properties
  - Ion formation
  - Electron affinity
  - Reaction with metals
  - Reaction with water
  - Reactivity trend of halogens.
- Similarity of ions and formula of compounds of alkali metals
- Uses of alkali metals.
The noble gases
- Meaning and members
- Trends down the group
  - Atomic radius
  - Ionic radius
  - Energy levels
- Physical properties
  - Melting and boiling [points
  - Electrical conductivity
  - 1^st ionization energy
- Chemical properties
- Uses of the noble gases

Introduction:

– Elements are classified and hence positioned in the periodic table based on the number of valence electrons and the number of energy levels.

– The number of valence electrons is equal to the group to which the element belongs; while the number of energy levels is equal to the period to which the element belongs.

– Elements in the same group are said to belong to the same chemical family.

– Thus a chemical family refers to a group of elements in the same number of valence electrons ad hence in the same group of the periodic table.

Characteristics of a chemical family:

– have same number of valence electrons;

– Show a uniform gradation in physical properties;

– have similar chemical properties;

Main chemical families.

– Four main chemical families will be studies in this section.

The Alkali metals
The Alkaline earth metals
The halogens
The noble gases

The Alkali metals.

– Are the elements with one valence electron and hence in group I of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

Lithium
Sodium
Potassium
Rubidium
Caesium
Francium

Electron arrangement of the first three alkali metals.

Elements	Electron arrangement
Lithium	2.1
Sodium	2.8.1
Potassium	2.8.81

Diagram: Part of periodic table showing the alkali metals

Gradation in properties of alkali metals.

Atomic and ionic radius.

Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

Trend:

– The ionic radius and the atomic radius of alkali metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from lithium to Francium.

Illustration:

– Lithium (2.1) has only 2 energy levels; sodium (2.8.1) has 3 energy levels while potassium (2.8.8.1) has 4 energy levels.

– Thus the outermost electron in potassium is further from the nucleus than the outermost electron in sodium and lithium.

Atomic and ionic radius of the same element.

– For the same alkali metals the atomic radius is larger than the ionic radius.

Reason:

– Alkali metals form ions by losing the valence electron, lading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

Illustration

– Potassium atom has electron arrangement of 2.8.8.1 hence 4 energy levels.

– During ion formation potassium reacts by losing the single valence electron to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

Diagrammatically: Potassium atom and potassium ion

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Atomic radius (nm)	Ionic radius (nm)
Lithium	Li	3	0.133	0.060
Sodium	Na	11	0.157	0.095
Potassium	K	19	0.203	0.133

Physical properties of alkali metals.

Appearance.

– Alkali metals have metallic luster when freshly cut. This refers to a shiny appearance on th cut surface.

– This surface however tarnishes due to reaction on exposure to air.

Ease of cutting.

– They are soft and easy to cut.

– The softness and ease of cutting increase down the group.

Reason:

– Alkali metals have giant metallic structures held together by metallic bonds.

– Metallic bond is due to attraction between the positively charged nucleus of one atom and the electrons in the outermost energy level of the next atom.

– Thus the force of attraction is stronger is smaller atoms than in larger atoms.

– The increase in atomic radius down the group implies that the strength of metallic bonds also decrease down the group (hence ease of cutting and softness).

They have relatively low melting and boiling points (in comparison to other metals).

Reason: they have relatively weaker metallic bonds.

– The melting and boiling points decrease down the group.

Reason: – The size of the atoms increase down the group due to increasing number of energy levels hence decrease in the strength of the metallic bonds (down the group).

Electrical conductivity.

– Alkali metals are good conductors of heat ad electricity.

Reason: they have delocalized electrons in the outermost energy level.

– The electrical conductivity is similar for all alkali metals.

Reason: all alkali metals have the same number of delocalized electron (a single electron) in the outermost energy level.

Note:

– In metals the electrons in the outermost energy level do not remain in one fixed position. They move randomly throughout the metallic structure and are thus said to be delocalsised.

Ionization energy.

Ionization energy is the minimum energy required to remove an electron from the outermost energy level of an atom in its gaseous state.

– The number of ionization energies an element may have is equivalent to the number of valence electrons.

– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.

– The first ionization among alkali metals decreases down the group.

Reason:

– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.

– Note that the atomic radius increases down the group due to increase in the number of energy levels.

Summary on physical properties of Alkali metals.

Element	Appearance	Ease of cutting	Melting point (^oC)	Boiling point (^oC)	Electrical conductivity	Atomic radius (nm)	Ionic radius (nm)
Lithium	Silver white	Slightly hard	180	1330	Good	0.133	520
Sodium	Shiny white	Easy	98	890	Good	0.157	496
Potassium	Shiny grey	Easy	64	774	Good	0.203	419

Chemical properties of alkali metals.

Reaction with air.

– When exposed to air alkali metals react with atmospheric moisture to form the corresponding metal hydroxide and hydrogen gas.

General equation:

Metal + Water → Metal hydroxide + hydrogen gas.

– The metal hydroxide further reacts with atmospheric carbon (IV) oxide to form hydrated metal carbonate.

General equation:

Metal hydroxide + carbon (IV) oxide → Hydrated metal carbonate.

Examples:

Lithium

With moisture: 2Li_(s) + 2H₂O_(l) → 2LiOH_(aq) + H_2(g);

Then with carbon (IV) oxide:2LiOH_(aq) + CO_2(g) → Li₂CO₃.H₂O_(s);

Sodium

With moisture: 2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g);

Then with carbon (IV) oxide: 2NaOH_(aq) + CO_2(g) → Na₂CO₃.H₂O_(s);

Potassium

With moisture: 2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g);

Then with carbon (IV) oxide: 2KOH_(aq) + CO_2(g) → K₂CO₃.H₂O_(s);

Burning in air.

– Alkali metals burn in air with characteristic flame colours to form corresponding metal oxides.

Examples:

– Burns in air to form lithium oxide as the only product.

Equation:

4Li_(s) + O_2(g) → 2Li₂O_(s)

– Burns in air to with a yellow flame to form sodium oxide as the only product.

Equation:

4Na_(s) + O_2(g) → 2Na₂O_(s)

Note:

– When burned in air enriched with oxygen or pure oxygen sodium burns with a yellow flame to fom sodium peroxide (instead of sodium oxide).

Equation: 2Na_(s) + O_2(g) → Na₂O_2(s)

– Burns in air with a lilac flame to form potassium oxide as the only product.

Equation:

4K_(s) + O_2(g) → 2K₂O_(s)

Reaction with water.

– Alkali metals react with water to form the corresponding hydroxides and hydrogen gas.

Examples:

Procedure:
– A small piece of potassium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It explosively melts into a silvery ball then disappears because reaction between water and potassium is exothermic (produces heat). The resultant heat melts the potassium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal bursts into a lilac flame; because hydrogen explodes into a flame which then burns the small quantities potassium vapour produced during the reaction;

– The resultant solution turns blue; because potassium hydroxide solution formed is a strong base;

(b). Reaction equations.

Equation I

2K_(s) + 2H₂O_(l) → 2KOH_(aq) + H_2(g);

Equation II

4K_(s) + O_{2 (g)} → 2K₂O_(s);

Equation III:

K₂O_(s) + H₂O_(l) → 2KOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

(b). Reaction equations.

Equation I

2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g);

Equation II

4Na_(s) + O_{2 (g)} → 2Na₂O_(s);

Equation III:

Na₂O_(s) + H₂O_(l) → 2NaOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Procedure:
– A small piece of lithium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It reacts less vigorously than sodium and des not melt since the melting point of lithium is relatively higher.

– It darts on the surface; due to propulsion by hydrogen;

– The gas does not ignite spontaneously;

– The resultant solution turns blue; because lithium hydroxide solution formed is a strong base;

(b). Reaction equations.

Equation I

2Li_(s) + 2H₂O_(l) → 2LiOH_(aq) + H_2(g);

Equation II

4Li_(s) + O_{2 (g)} → 2Li₂O_(s);

Equation III:

Li₂O_(s) + H₂O_(l) → 2LiOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Summary on reaction rate.

Metal	Reaction rate
Lithium	Vigorous
Sodium	More vigorous
Potassium	Explosive

Reaction with chlorine

– All alkali metals react with chlorine to form corresponding metal chlorides.

General procedure:

– A small piece of the alkali metal is cut and placed in a deflagrating spoon;

– It is then warmed and quickly lowered into a gas jar containing chlorine.

– The experiment should be done in a working fume chamber or in the open

Reason: Chlorine gas is poisonous.

– This experiment should not be attempted in the laboratory with potassium

Reason: the reaction is too explosive and very dangerous.

Observations:

Sodium:

– The metal bursts into a yellow flame.

– White fumes of sodium chloride are formed.

Equation: 2Na_(s) + Cl_2(g) → 2NaCl_(s)

Lithium:

– The metal reacts less vigorously than sodium without bursting into a flame.

– White fumes of lithium chloride are formed.

Equation: 2Li_(s) + Cl_2(g) → 2LiCl_(s)

Potassium:

– The metal bursts into a lilac flame.

– White fumes of potassium chloride are formed.

Equation: 2K_(s) + Cl_2(g) → 2KCl_(s)

The general trend in reactivity of alkali metals.

– The reactivity of alkali metals increase down the group.

Reason:

– Alkali metals react by losing the single valence electron. The ease of loss of the valence electron increases down the group due to decrease in attraction of the valence electron towards the positive nucleus. This in turn is due to the increase in atomic radius down the group as a result of increase in number of energy levels.

Similarity of ions and formulae of some compounds of Alkali metals.

– Alkali metals have a similar charge on their ions since they all have a single valence electron.

– Thus they have the same general formula in their ions and compounds.

Examples.

Alkali metal ion	Hydroxide (OH^–)	Oxides (O^2-)	Chloride (Cl^–)	Sulphates (SO₄^2-)
Li⁺
Na⁺
K⁺

Note:

– Alkali metals are usually not found as free elements; but rather in their combined states in the earths crust.

Reason:

– They have high reactivities.

Uses of Alkali metals and their compounds.

Sodium is used in the manufacture of sodium cyanide for gold extraction.
Lithium is used in the manufacture of special high strength glasses ad ceramics.
Lithium compounds are used in the manufacture of dry cells for use in mobile phones, laptops, stopwatches, and zero emission electric vehicles.
A molten mixture of sodium and potassium is used as a coolant in nuclear reactors.
Sodium vapour is used to produce the yellow glow in street ad advertisement lights.
Molten sodium is used as a reducing agent in the extraction of titanium.

Equation: Na_(l) + TiCl_4(g) → Ti_(s)+ 4NaCl_(l)

Sodium chloride is used as a food additive.
A mixture of sodium hydroxide (caustic soda) and carbon disulphide is used in the manufacture of artificial silk called rayon.

The Alkaline Earth metals.

– Are the elements with two valence electrons and hence in group II of the periodic table.

– All are metallic in nature.

– The members of the family in order down the group is as follows:

Beryllium
Magnesium
Calcium
Strontium
Barium
Radium

Electron arrangement of the first three alkaline earth metals.

Elements	Electron arrangement
Beryllium	2.2
Magnesium	2.8.2
Calcium	2.8.8.2

Diagram: Part of periodic table showing the alkaline earth metals

Dot and cross diagrams for the first three Alkaline Earth metals

Gradation in properties of alkali metals.

Atomic and ionic radius.

Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

Trend:

– The ionic radius and the atomic radius of alkaline earth metals increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Beryllium to radium.

Illustration:

– Beryllium (2.2) has only 2 energy levels; Magnesium (2.8.2) has 3 energy levels while calcium (2.8.8.2) has 4 energy levels.

– Thus the outermost electron in calcium is further from the nucleus than the outermost electron in magnesium and beryllium.

Atomic and ionic radius of the same element.

– For the same alkaline earth metal the atomic radius is larger than the ionic radius.

Reason:

– Alkaline earth metals form ions by losing the valence electrons, leading to the loss of an entire outermost energy level. Thus the atoms have more energy levels than the corresponding ion hence a larger radius in the atom than I the ion.

Illustration

– Calcium atom has electron arrangement of 2.8.8.2 hence 4 energy levels.

– During ion formation it reacts by losing the 2 valence electrons to acquire a new electron arrangement of 2.8.8 hence 3 energy levels.

– Thus the ion has a smaller radius than the atom.

Diagrammatically: calcium atom and calcium ion

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Electron arrangement	Atomic radius (nm)	Ionic radius (nm)
Beryllium	Be	4	2.2	0.089	0.031
Magnesium	Mg	12	2.8.2	0.136	0.065
Calcium	Ca	20	2.8.8.2	0.174	0.099

Physical properties of alkali metals.

Appearance.

– Alkaline earth metals acquire a metallic luster when polished. This refers to a shiny appearance on the cut surface.

– They however lose this metallic luster when exposed to air due to oxidation.

Note:

– The purpose of polishing alkaline earth metals before using them in experiment is to remove the oxide coating that slows down and prevents them from reacting.

Ease of cutting.

– Magnesium is hard to cut with a knife; but is however malleable and ductile.

– Calcium cannot also be cut with a knife because it is brittle.

Note:

Malleability:
– Refers to the ability of a material to be hammered into sheets.

Example: Iron sheets are possible to be made because iron metal is malleable.

Ductility:

– The ability of a material to be rolled into wires.

Example: electric cables are made of aluminium because aluminium metal is ductile.

Brittle:

– Refers to a substance which is hard and likely to break.

They have relatively high melting and boiling points in comparison to alkali metals.

Reason: they have relatively stronger metallic bonds (than alkali metals).

– The melting and boiling points decrease down the group.

Reason: –

– The size of the atoms increase down the group due to increasing number of energy levels. As the atomic radius increase the force of attraction between the positive nucleus and the delocalized electrons decrease. This leads to a decrease in the strength of the metallic bonds (down the group).

Electrical conductivity.

– Alkaline earth metals are good conductors of heat and electricity.

Reason: they have delocalized electrons in the outermost energy level.

– The electrical conductivity is similar for all the alkaline earth metals.

Reason: all alkaline earth metals have the same number of delocalized electron (two valence electrons) in the outermost energy level.

Ionization energy.

Ionization energy is the minimum energy required to remove an electron from the outermost energy level of an atom in its gaseous state.

– The number of ionization energies an element may have is equivalent to the number of valence electrons.

– Thus alkaline earth metals have two ionization energies.

– The first ionization energy is the minimum energy required to remove the first electron from the outermost energy level of an atom in its gaseous state.

– The second ionization energy is the minimum energy required to remove the second electron from the outermost energy level of an atom in its gaseous state.

– The first and second ionization energies among alkaline earth metals decreases down the group.

Reason:

– The effective force of attraction of on the outermost electron by the positive nucleus decreases with increasing atomic size and distance from the nucleus.

– Note that the atomic radius increases down the group due to increase in the number of energy levels.

Variation between 1^st and 2^nd Ionization energies.

– The first ionization energy is always lower than the second ionization energy for the same element.

Reason:

– After losing the first electron from an atom, the overall positive charge holds the remaining electrons more firmly. Thus removing a second electron from the ion requires more energy than the first electron

Note:

– The third ionization energy will also be higher tha the second ionization energy for the same reason.

Example:
Magnesium.

First ionization energy: Mg_(g)→ Mg⁺_(g) + e- (1^stE = 736 kJ per mole)
Second ionization energy: Mg⁺_(g)→ Mg²⁺_(g) + e- (2^ndE = 1450 kJ per mole)

Summary on physical properties of alkaline earth metals.

Element	Atomic number	Melting point (^oC)	Boiling point (^oC)	Atomic radius (nm)	1^st I.E (kJmol-1)	2^nd I.E (kJmol-1)
Beryllium	4	1280	2450	0.089	900	1800
Magnesium	12	650	1110	0.136	736	1450
Calcium	20	850	1140	0.174	590	1150
Strontium	38	789	1330	0.210	550	1060
Barium	56	725	1140	0.220	503	970

Chemical properties of Alkaline earth metals.

Burning alkaline earth metals in air.

– Alkaline earth metals react burn in air to form corresponding oxides.

– More reactive alkaline earth metals may also react with atmospheric nitrogen to form corresponding nitrides.

Examples:

– Burns in air with a blinding brilliant flame forming a white solid.

– The white solid is a mixture of magnesium oxide ad magnesium nitride.

Equations:

Reacting with oxygen: 2Mg_(s) + O_2(g) → 2MgO_(s)

Reacting with nitrogen: 3Mg_(s) + N_2(g) → Mg₃N_2(s)

– Burns in air with a faint orange flame forming a white solid.

– The white solid is a mixture of calcium oxide ad calcium nitride.

Equations:

Reacting with oxygen: 2Ca_(s) + O_2(g) → 2CaO_(s)

Reacting with nitrogen: 3Ca_(s) + N_2(g) → Ca₃N_2(s)

Note:

– The trend in the reactivity of alkaline earth metals when burning in air is not clear; due to the oxide coating on the calcium that tends to slow down the reaction of calcium in air.

– For this reason it is important to polish the surfaces of alkaline earth metals before using them in experiments.

Reaction of alkaline earth metals with cold water.

– Alkaline earth metals react slowly with cold water to form corresponding hydroxides and hydrogen gas.

Examples

Magnesium:

– Reacts slowly with water to form magnesium hydroxide and hydrogen gas.

– The reaction is very slow and the amount f hydrogen gas evolved is very low hence the hydrogen gas bubbles stick on the surface of the metal.

– The magnesium hydroxide formed dissolves slightly in water to form magnesium hydroxide.

– Thus the resultant solution is slightly alkaline.

Equation:

Mg_(s) + 2H₂O_(l) → Mg(OH)_2(aq) + H_2(g).

Calcium:

Diagram of apparatus:

Observations and explanations:

– Calcium sinks to the bottom of the beaker; because it is denser than water;

– Slow effervescence of a colourless gas; due to slow evolution of hydrogen gas;

– Soapy solution formed; due to formation of alkaline calcium hydroxide;

– A white suspension is formed; because calcium hydroxide is slightly soluble in water;

Reaction equation:

Ca_(s) + H₂O_(l) → Ca (OH)_2(aq) + H_{2 (g)};

Effect of resultant solution on litmus paper;

– Litmus paper slowly turns blue; calcium hydroxide formed is slightly soluble in water; releasing a small number of hydroxyl ions which result into alkaline conditions // high pH;

Reaction with steam.

– Alkaline earth metals react with steam to produce corresponding metal oxide and hydrogen gas.

– The reactivity with stem is faster ad more vigorous for each alkaline earth metals as compared to reaction with cold water.

Examples:

Magnesium

Procedure:
– A small amount of wet sand is put at the bottom of a boiling tube;

– A small piece of magnesium ribbon is cleaned and put in the middle of the combustion tube;

– The wet sand is heated gently first then the magnesium ribbon is heated strongly until it glows.

Reason:

– To generate steam that drives out the air that would otherwise react with the magnesium (preventing reaction with steam)

– The delivery tube is removed from the water before heating stops.

Reason:

To prevent sucking back (of the gas) as the apparatus cools

– The gas produced is tested using a burning splint.

Diagram of apparatus:

Observations and explanations.

– Magnesium burns with a white blinding flame;

– Grey solid (magnesium) forms a white solid; due to formation of magnesium oxide;

– Evolution of a colourless gas that produces a pop sound when exposed to a burning splint; confirming it is hydrogen;

Reaction equation.

Magnesium + Steam → Magnesium oxide + Hydrogen gas;

Mg_(s) + H₂O_(g) → MgO_(s) + H_2(g);

Calcium

– Reaction between calcium and steam would produce calcium oxide and hydrogen gas.

– However the reaction is too explosive to be done under laboratory conditions.

Reaction with chlorine.

– Alkaline earth metals react with chlorine to form corresponding chlorides as the only products.

Condition: presence of heat, hence the metal must be heated first.

Precaution: reaction should be done in a working fume chamber because chlorine gas is poisonous).

Examples:

Procedure:
– A piece of burning magnesium is lowered into a gas jar containing chlorine.

Observations:
– The metal continues to burn with a brilliant white flame.

– They grey solid forms a white powder.

Explanation.

– Reaction between magnesium and chlorine is exothermic.

– The heat produced keeps the metal burning; thus facilitates the reaction between magnesium and chlorine to form magnesium chloride, which is the white powder.

Equation:
Mg_(s) + Cl_2(g) → MgCl_2(s)

Procedure:
– A piece of burning calcium is lowered into a gas jar containing chlorine.

Observations and explanations.

– The metal burns shortly with an orange flame but soon smolders off.

– There is no steady reaction between calcium and chlorine.

Reason:

– When calcium is heated a coating of the metal oxide is formed first which prevents furtheer reaction between the metal and chlorine.

– However under suitable conditions calcium reacts with chlorine to form a white powder of calcium chloride.

Equation:
Ca_(s) + Cl_2(g) → CaCl_2(s)

Reaction of alkaline earth metals and dilute acids

– Generally alkaline earth metals react with dilute acids to form salts and hydrogen gas.

Examples:

With hydrochloric acid.
- Beryllium:

– When a piece of beryllium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium chloride remains in the test tube // boiling tube.

Equation: Be_(s) + 2HCl_(aq) → BeCl_2(aq) + H_2(g)

Magnesium:

– When a piece of magnesium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium chloride remains in the test tube // boiling tube.

Equation: Mg_(s) + 2HCl_(aq) → MgCl_2(aq) + H_2(g)

Calcium:

– When a piece of calcium is dropped into a beaker containing hydrochloric acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium chloride remains in the test tube // boiling tube.

Equation: Ca_(s) + 2HCl_(aq) → CaCl_2(aq) + H_2(g)

With sulphuric (VI) acid.
- Beryllium:

– When a piece of beryllium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium sulphate remains in the test tube // boiling tube.

Equation: Be_(s) + H₂SO_4(aq) → BeSO_4(aq) + H_2(g)

Magnesium:

– When a piece of magnesium is dropped into a beaker containing sulphuric (VI) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium sulphate remains in the test tube // boiling tube.

Equation: Mg_(s) + H₂SO_4(aq) → MgSO_4(aq) + H_2(g)

Calcium:

– When a piece of calcium is dropped into a beaker containing sulphuric (VI) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– The reaction however quickly stops and there is formation of a white precipitate in the test tube or boiling tube.

Reason:
– When calcium reacts with dilute sulphuric (VI) cid there is formation of calcium sulphate which is insoluble hence the formation of a white precipitate. The insoluble calcium sulphate coats the surface of the (calcium) metal stopping further reaction.

Equation: Ca_(s) + H₂SO_4(aq) → CaSO_4(s) + H_2(g)

With nitric (V) acid.
- Beryllium:

– When a piece of beryllium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of beryllium nitrate remains in the test tube // boiling tube.

Equation: Be_(s) + 2HNO_3(aq) → Be(NO₃)_2(aq) + H_2(g)

Magnesium:

– When a piece of magnesium is dropped into a beaker containing nitric (V) acid, there is effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of magnesium nitrate remains in the test tube // boiling tube.

Equation: Mg_(s) + 2HNO_3(aq) → Mg(NO₃)_2(aq) + H_2(g)

Calcium:

– When a piece of calcium is dropped into a beaker containing nitric (V) acid, there is slight effervescence of a colourless gas.

– The colourless gas produces a pop sound when exposed to a burning splint confirming that it is hydrogen gas.

– A colourless solution of calcium nitrate remains in the test tube // boiling tube.

Equation: Ca_(s) + 2HNO_3(aq) → Ca(NO₃)_2(aq) + H_2(g)

Trend in reactivity in alkaline earth metals.

– The reactivity of the alkaline earth metals increases down the group.

Reason:

– The atomic radius increases down the group due to increasing number of energy levels. The attraction of the two valence electrons towards the positive nucleus thus decreases down the group. Since alkaline earth metals react by losing electrons the ease with which the valence electrons are lost therefore decreases down the group

Similarity of ions and formulae of some compounds of Alkaline earth metals.

– Alkaline earth metals have the same valency (2) and a similar charge (2+) on their ions since they all have a single valence electron.

– Thus they have the same general formula in their ions and compounds.

Examples.

Alkali metal ion	Hydroxide (OH^–)	Oxides (O^2-)	Chloride (Cl^–)	Sulphates (SO₄^2-)
Be²⁺
Mg²⁺
Ca²⁺

Uses of some alkaline earth metals and their compounds

Magnesium is used in the manufacture of magnesium hydroxide which is used as an anti-acid medicine.

Reason: magnesium hydroxide is a non-toxic base.

A low-density alloy of magnesium and aluminium is used in aeroplanes construction.
Hydrated calcium sulphate (plaster of Paris) is used in hospitals to set fractured bones.
Cement is made by heating a mixture of calcium carbonate (limestone), clay and sand.
Calcium carbonate is used in the extraction of iron.
Calcium oxide (quicklime) is added tom acidic soils to raise pH for agricultural purposes.
Calcium nitrate is used as a nitrogenous fertilizer.
Magnesium oxide is used in the lining of furnaces.
Barium sulphate is used in the diagnosis of ulcers.
Barium nitrate is used to produce the green flame in fireworks.
Calcium carbonate is mixed with oil to make putty.

The Halogens.

– Are the elements with seven valence electrons and hence in group VII of the periodic table.

– All are non-metallic in nature.

– The members of the family in order down the group is as follows:

Chlorine
Bromine
Iodine
Astatine

Electron arrangement of the first three halogens.

Elements	Electron arrangement
Fluorine	2.7
Chlorine	2.8.7
Bromine	2.8.18.7

Note:

– Halogen is a derivative of two Greek words: halo and gen.

Diagram: Part of periodic table showing the halogens

Gradation in properties of halogens.

Atomic and ionic radius.

Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

Trend:

– The ionic radius and the atomic radius of halogens increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Fluorine to iodine.

Illustration:

– Fluorine (2.7) has only 2 energy levels; Chlorine (2.8.7) has 3 energy levels while bromine (2.8.18.7) has 4 energy levels.

– Thus the outermost electron in bromine is further from the nucleus than the outermost electron in chlorine and fluorine.

Atomic and ionic radius of the same element.

– For the same halogen the ionic radius is larger than the atomic radius.

Reason:

– Halogens form ions by gaining (an) electron into the outermost energy level. This increases the electron-electron repulsion in the outermost energy level hence decreasing the nuclear attraction of the outermost electrons towards the nucleus.

Illustration

– Chlorine atom has electron arrangement of 2.8.7 hence 3 energy levels.

– During ion formation it reacts by gaining an electron into the third energy level to acquire a new electron arrangement of 2.8.8.

– In the atom 17 protons are attracting 17 electrons; while in the chloride ion there are 17 protons attracting 18 electrons in the outermost energy level.

– Thus in the ion the effect of the positive nucleus is lower.

– This is due to increased repulsive effect between the existing electrons in the outermost energy level and the incoming electron (electron-electro repulsion)

Diagrammatically: chlorine atom and chloride ion

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Electron arrangement	Atomic radius (nm)	Ionic radius (nm)
Fluorine	F	9	2.7	0.064	0.136
Chlorine	Cl	17	2.8.7	0.099	0.181
Bromine	Br	35	2.8.18.7	0.114	0.195
Iodine	I	53	2.8.18.18.7	0.133	0.216

Laboratory preparation of chlorine.

Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.

Equation:

2HCl_(aq) + [O] Cl_2(g) + H₂O_(l)

– The [O] is from a substance containing oxygen (an oxidizing agent).

– The main oxidizing agents normally used for preparation of chlorine are:

Potassium manganate (VII); KMnO4.
Manganese (IV) oxide; MnO2

(a). Preparation of chlorine from MnO₂ and HCl.

(i). Apparatus:

(ii). Conditions:

– Heating;

– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.

(iii). Procedure:

– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);

Equation:

MnO_2(s) + 4HCl_(aq) ^Heat MnCl_2(aq) + 2H₂O_(l) + Cl_2(g)

(iv). Explanation:

– Manganese (IV) oxide oxidizes hydrochloric acid by removing hydrogen resulting into chlorine.

– The manganese (IV) oxide is reduced to water and manganese chloride.

– The resultant chlorine gas is passed through a bottle containing water.

Reason:

– To remove hydrogen chloride fumes (gas) which is very soluble in water.

– Next it is passed through concentrated sulphuric acid or anhydrous calcium chloride; to dry the gas.

(v). Collection:

(a). Wet chlorine is collected over brine (saturated sodium chloride solution) or hot water.

Reason:

– It does not dissolve in brine and is less soluble in water

(b). Dry chlorine is collected by downward delivery (upward displacement of air)

Reason:

– It is denser than air (2.5 times).

Note:

– Chlorine may also be dried by adding calcium chloride to the jar of chlorine.

(c). The first bottle must contain water and the second concentrated sulphuric acid.

Reason:

– If the gas is first passed through concentrated sulphuric acid in the first bottle then to the water; it will be made wet again.

Physical properties of Halogens.

Physical state and appearance at room temperature.

Halogen	State and appearance
Fluorine	Pale yellow gas
Chlorine	Green-yellow gas
Bromine	Volatile brown liquid
Iodine	Shiny dark grey solid.

Solubility

(a). In water

Experiment: To investigate solubility of halogens in water.

Procedure:

– A boiling tube containing chlorine gas is inverted into a beaker containing water.

– The experiment is repeated with fluorine, bromine and a few crystals of iodine.

Diagram:

Observations:
Fluorine:
– The level of solution rises in the boiling tube.

– The pale yellow colour of fluorine disappears.

Fluorine:
– The level of solution rises in the boiling tube.

– The green-yellow colour of fluorine disappears.

Bromine
– The level of solution rises in the boiling tube.

– The brown colour of fluorine disappears.

Iodine:
– The level of solution remains the same in the boiling tube.

– The shiny dark grey crystals remain in the beaker.

The rise in water level is higher in fluorine than in chlorine while the rise in chlorine is higher than in bromine.

Diagrams: observations after some time

Explanations:

– Fluorine, chlorine and bromine are all soluble in water, while iodine is insoluble in water.

– When a boiling tune containing the soluble halogens is inverted into a beaker containing water, the halogen dissolves in the water.

– This creates a partial vacuum and the water in the beaker thus rises to occupy the space left by the dissolved gas.

– The halogens dissolve in water to form acidic solutions.

– The more soluble the halogen, the higher the rise in water level in the boiling tube.

– Thus the solubility of halogens decreases down the group from fluorine to bromine.

Reason:

– All halogens have molecular structures with wreak van der waals forces between the molecules.

During the dissolution the Van der Waals must be broken. The strength of Van der Waals increase as the atomic size and hence the molecular size increases which occurs down the group.

(b). In tetrachloromethane.

– The same procedure (of dissolving halogens in water) is followed with tetrachloromethane.

Observations:

– All halogens are soluble in tetrachloromethane.

– The solubility of each halogen is higher in tetrachloromethane than in water.

Reason: Halogens are molecular thus non-polar and thus are more soluble in polar organic solvents like tetrachloromethane than in polar solvents like water.

Effect of heat.

– Fluorine and chlorine are gases at room temperature and simply expand and increase in volume when heated.

– Bromine is a brown liquid at room temperature and evolves brown fumes when heated.

– Iodine exists as shiny dark grey solid at room temperature and sublimes to give brown fumes when heated.

Electrical conductivity.

– All halogens are on conductors of heat and electricity.

Reason:

– Halogens are molecular and thus have neither delocalized electrons nor free mobile ions for electrical conductivity.

Trend in melting and boiling points.

– The melting and boiling points of halogens increase down the group.

Reason:

– Halogens exist as diatomic molecules and thus have molecular structures;

– The molecules are held together by intermolecular forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

Summary on some physical properties of halogens>

Halogen	Formula	Atomic number	Appearance	Melting point (^oC)	Boiling point (^oC)
Fluorine	F	9	Pale yellow gas	-238	-188
Chlorine	Cl	17	Green yellow gas	-101	-35
Bromine	Br	35	Brown liquid	-7	59
Iodine	I	53	Shiny dark grey solid	114	184

Chemical properties of Halogens.

Note: It is not easy for non-metals like halogens to form cations.

Reason: the ionization energy (amount of energy required to lose an electron(s) from the outermost energy level of a gaseous atom) is very large.

– Thus most non-metals react by forming anions (negatively charged ions) by electron loss.

Ion formation.

– Halogens react by gaining a single electron into the outermost energy level to form a stable electron configuration and corresponding anions.

– During ion formation by electron loss energy is released, and the energy change for this process is called electron affinity

Note: Definition.

– Electron affinity is thus the energy released when an atom in gaseous state gains an electron to form a negative ion (anion)

Trend in electron affinity of halogens:

– Generally the electron affinity of halogens decreases down the group.

Reason:

– The size of the atoms increases (due to increasing number of energy levels) down the group and thus the force of attraction of the electrons in the outermost energy level towards the nucleus decreases.

– Thus down the group the ease with which electrons are gained decreases and the faster the ease of electron gain, the more the energy released hence the more the electron affinity.

Summary:

Element	Ionization equation	Electron affinity (kJ per mole
Fluorine	F_(g) + e → F^–_(g)	-322
Chlorine	Cl_(g) + e → Cl^–_(g)	-349
Bromine	Br_(g) + e → Br^–_(g)	-325
Iodine	I_(g) + e → I^–_(g)	-295

Reaction with metals.

(a). Chlorine.

Reaction of chlorine with iron.

(i). Apparatus.

Note:

– The set up can also be modified by using sodium hydroxide to absorb excess chlorine gas as shown below.

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Chlorine gas is poisonous and will thus be harmful to the human body.

(iii). Procedure:

– A stream of dry chlorine gas is passed over heated iron wool as per the diagram.

(iv). Conditions.

Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)

Reason:

To prevent hydration hence oxidation of iron (which will then form Fe₂O₃.5H₂O) hence preventing reaction between iron and chlorine.

Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

Anhydrous calcium chloride.

– In the U-tube; to dry the chlorine gas.

– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.

Note: In the guard tube, calcium oxide is preferable to anhydrous calcium chloride.

Reason:

– Other than preventing atmospheric water vapour from getting into the flask with iron wool; it also absorbs excess chlorine thus preventing environmental pollution

(v). Observations:

– Iron metal glows red-hot.

– Red brown fumes (FeCl_3(g)) are formed in the combustion tube.

– A black solid (FeCl_3(s)) is collected in the flask.

Note:

– Iron (III) chloride cannot be easily collected in the combustion tube.

Reason:

– It sublimes when heated and hence the hotter combustion tube causes it to sublime and its vapour is collected on the cooler parts of the flask.

(vi). Reaction equation.

2Fe_(s) + 3Cl_2(g) → 2FeCl_3(g)

(vii). Conclusion.

– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.

Equation:

FeCl_3(s) FeCl_3(g)

(black) (Red-brown)

Reaction with aluminium.

– Chlorine reacts with aluminium I a similar manner to iron to from aluminium chloride.

Equation:

2Al_(s) + 3Cl_2(g) → 2AlCl_3(s)

2Al_(s) + 3Cl_2(g) → Al₂Cl_6(s)

Note:

– Aluminium chloride also sublimes on heating.

Equation:

AlCl_3(s) AlCl_3(g)

(White) (White)

Reaction with burning magnesium.

(i). Procedure:

– Burning magnesium is lowered into a gar jar of chlorine gas.

(ii). Observations:

– The magnesium continues to burn with a bright blinding flame;

– Formation of white fumes (MgCl₂); which cools into a white powder.

(iii). Equation:

Mg_(s) + Cl_2(g)→ MgCl_2(s)

– Generally chlorine reacts with most metals when hot top form corresponding chlorides.

Note:

Where a metal forms two chlorides when it reacts with chlorine, the higher chloride is usually formed.

Reason:

The higher chloride is stable. This explains why reactions of chlorine with iron results into iron (III) chloride and not iron (II) chloride.

(b). Bromine.

Reaction of bromine with iron

Apparatus

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Bromine gas is poisonous and will thus be harmful to the human body.

(iii). Procedure:

– Bromine liquid is heated to generate bromine vapour (fumes).

– The iron wool is then heated and a stream of the bromine gas is passed over heated iron wool as per the diagram.

(iv). Conditions.

Bromine must be heated to generate fumes before heating the iron.

Reason:

So that bromine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and bromine.

Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

(v). Observations:

– Brown fumes of bromine are produced on heating bromine liquid.

– The iron wool glows red-hot upon heating

– The iron wool forms dark red crystals (of iron (III) bromide)

(vi). Reaction equation.

Word: Iron + bromine → Iron (III) bromide.

Chemical: 2Fe_(s) + 3Br_2(g) → 2FeBr_3(s)

Dark red crystals.

Reaction of bromine with zinc

– Using the same set up bromine also reacts with zinc to form zinc bromide.

Equation: Zn_(s) + Br_2(g) → ZnBr_2(s).

Reaction of bromine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium bromide.

Equation: Mg_(s) + Br_2(g) → MgBr_2(s).

Reaction of bromine with sodium

– Using the same set up bromine also reacts with zinc to form sodium bromide.

Equation: 2Na_(s) + Br_2(g) → 2NaBr_(s).

(c). Iodine.

Reaction of iodine with iron

Apparatus

(ii). Precaution.

– Experiment should be done in a functional fume cupboard or in the open.

Reason:

– Iodine is poisonous and will thus be harmful to the human body.

(iii). Procedure:

– Iodine crystals are heated to generate iodine vapour (fumes).

– The iron wool is then heated and a stream of the iodine gas is passed over heated iron wool as per the diagram.

(iv). Conditions.

Iodine must be heated to generate fumes before heating the iron.

Reason:

So that iodine vapour will drive out air from the boiling tube to prevent oxidation of iron with oxygen which would otherwise prevent reaction between iron and iodine.

Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

(v). Observations:

– Purple vapour (fumes) of iodine is produced on heating iodine crystals.

– The iron wool glows red-hot upon heating

– The iron wool forms grey black crystals (of iron (II) iodide)

(vi). Reaction equation.

Word: Iron + iodine → Iron (II) iodide.

Chemical: Fe_(s) + I_2(g) → FeI_2(s)

Grey-black crystals.

Reaction of Iodine with zinc

– Using the same set up bromine also reacts with zinc to form zinc iodide.

Equation: Zn_(s) + I_2(g) → ZnI_2(s).

Reaction of Iodine with magnesium

– Using the same set up bromine also reacts with zinc to form magnesium iodide.

Equation: Mg_(s) + I_2(g) → MgI_2(s).

Reaction of Iodine with sodium

– Using the same set up bromine also reacts with zinc to form sodium iodide.

Equation: 2Na_(s) + I_2(g) → 2NaI_(s).

Note:

– The reactivity of chlorine with metals is more vigorous than that of bromine, which is more than that of iodine.

– Thus reactivity of halogens with metals decreases down the group.

Reason:

– Halogens react by gaining electrons; the ease of gaining electrons decrease down the group as the atomic size increases which leads to progressive decrease in the force of attraction of electrons in the outermost energy levels by the positively charged nucleus.

Reaction with water.

– Halogens that dissolve in water form a mixture of two acids.

Reaction of chlorine with water.

– Chlorine dissolves in water to form chlorine water, which is a mixture of two acids: chloric (I) acid and hydrochloric acid.

Equation:
Cl_2(g) + H₂O_(l) → HOCl_(aq) + HCl_(aq)

Chloric (I) acid Hydrochloric acid

Effects of sunlight on chlorine water.

(i). Procedure:

– Chlorine water is made by dissolving the gas in water.

– A long tube filled with chlorine water is inverted over a beaker containing water.

– It is then exposed to sunlight (bright light) as shown below.

(ii). Apparatus:

(iii). Observations:

– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.

(iv). Explanation:

– Chlorine water has two components.

Equation:

Cl_2(g) + H₂O_(l) ═ HCl_(aq) + HOCl_(aq)

– The HOCl being unstable will dissolve on exposure to sunlight, giving out oxygen.

Equation:

2HOCl_(aq) → 2HCl_(aq) + O_2(g) (slow reaction)

Overall reaction:

2H₂O_(l) + 2Cl_2(g) → 4HCl_(aq) + O_2(g)

Effect of chlorine water on litmus papers

(i). Procedure:

– A strip of blue ad a strip of red litmus papers are dropped into chlorine water in a beaker.

(ii). Observations:
– The blue litmus paper turns red; then both litmus papers are decolourised.

(iii). Explanations.

– Chlorine water contains a mixture of chloric (I) acid ad hydrochloric acid.

Equation:

Cl_2(g) + H₂O_(l) ═ HCl_(aq) + HOCl_(aq)

– The two acids cause the litmus paper to turn red (from blue) while the red litmus paper remained red.

– The litmus papers are then decolourised due to bleaching action of chlorine water, which is due to the activity of chloric (I) acid.

– The chloric (I) acid is unstable and thus decomposes to give hydrochloric acid and oxygen atom.

Equation:

HOCl(aq) → HCl_(aq) + [O]

– The oxygen atom combines with the chemical of the natural dye in the litmus thus decolourising it.

Equation:

Dye + HOCl_(aq) → HCl_(aq) + (Dye + [O])

Coloured Decolourised

Note: solutions of bromine and fluorine in water will behave in a similar manner.

Examples:

Fluorine dissolving in water

F_2(g) + H₂O_(l) ═ HF_(aq) + HOF_(aq)

Hydrofluoric acid Fluoric (I) acid

– The mixture is flourine water

Bromine dissolving in water

Br_2(g) + H₂O_(l) ═ HBr_(aq) + HOBr_(aq)

Hydrobromic acid Bromic (I) acid

– The mixture is bromine water.

Some uses of halogens and their compounds.

Fluorine is a raw material in the preparation of a synthetic fibre known as polytetrafluoroethane.
Some compounds of fluorine are added to water and some tooth pastes in small quantities to reduce tooth decay.
Fluorine is used in the manufacture of hydrogen fluoride used to engrave words or pictures on glass.
Chlorine is used to make bleaches used in paper and textile industries.
Chlorine is added to water to kill micro-organisms in water treatment works.
Chlorine is used in the manufacture of a plastic known as polyvinylchloride (PVC).
Chlorine is used in large scale manufacture of hydrochloric acid.
Bromine is used in the manufacture of silver bromide which is used to make the light sensitive photographic paper and films.
A solution of iodine in alcohol (tincture of iodine) is used as a disinfectant.

The Noble gases.

– Are the elements with the maximum possible number of valence electrons and hence in group VIII of the periodic table.

– All are gaseous in nature.

– The members of the family in order down the group is as follows:

Helium
Neon
Argon
Krypton
Xenon
radon

– They are found as free atoms in nature and form about 1% of air with argon being the most abundant

Electron arrangement of the first three noble gases.

Elements	Electron arrangement
Helium	2.
Neon	2.8.
Argon	2.8.8.

Note: – Helium with only two electrons has one occupied energy level; which is completely full and hence is said to have a duplet electron configuration

– The rest of the noble gases have eight electrons in their outermost occupied energy level hence are said to have the octet electron configuration.

Diagram: Part of periodic table showing the noble gases

Dot and cross diagrams for the first three Alkaline Earth metals

Physical properties of noble gases

Atomic and ionic radius.

Atomic radius:

– Refers to the distance between the centre of the nucleus of an atom and the outermost energy level occupied by electron(s)

Ionic radius:

– Refers to the distance between the centre of the nucleus of an ion and the outermost energy level occupied by electron(s)

Trend:

– The atomic radius of noble gases increase down the group.

Reason:

– There is an increase in the number of energy levels down the group from Helium to radon.

Illustration:

– Helium (2.) has only 1 energy level; Neon (2.8) has 2 energy levels while Argon (2.8.8) has 3 energy levels.

– Thus the outermost electron in Argon is further from the nucleus than the outermost electron in neon and helium.

Note:

– Under normal circumstance the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence comparative studies of ionic radii among noble gases are not feasible.

Ionization energies.

– Noble gases have very high ionization energies.

Reason:

– Noble gases are colourless monoatomic gases with very stable electron arrangements, either dupltet (helium) or octet (the rest).

– The nuclear attraction force of electrons in the outermost energy level towards the nucleus is thus very strong.

Trend:

– The ionization energy decrease down the noble gases group.

Reason:

– As the size of the atoms increase down the group (due to increase in the number of energy levels), the force with which the electros in the outermost energy levels are attracted towards the nucleus decrease hence increase in the ease of electron loss (from the outermost energy level).

Melting and boiling points.

– Generally noble gases have very low melting and boiling points.

Reason:

– They exist as monoatomic gases with simple molecular structures with weak van der waals forces that are easy to break.

Trend:

– Melting and boiling points among the noble gases increase down the group.

Reason:

– Noble gases exist as monoatomic molecules and thus have simple molecular structures;

– The molecules (atoms) are held together by intermolecular (inter-atomic) forces called the Van der Waals which have to be broken during melting and boiling;

– The strength of the Van der Waals increases as the size of the atoms and hence molecules increases which occurs down the group.

Summary: changes in atomic and ionic radius among alkali metals.

Element	Symbol	Atomic number	Atomic radius (nm)	1^st Ionization energy (kJmol^-1)	Melting point (^oC)	Boiling point (^oC)
Helium	He	2	0.128	2372	-270	-269
Neon	Ne	10	0.160	2080	-249	-246
Argon	Ar	18	0.192	1520	-189	-186
Krypton	Kr	36	0.197	1350	-157	-152
Xenon	Xe	54	0.217	1170	-112	-108

Chemical properties of Noble gases.

– Generally the noble gases neither lose nor gain electrons and are thus stable hence unreactive.

– Due to this they rarely form ions and hence have no feasible chemical reactions.

– However xenon and radon with very large atomic radii and smaller ionization energies take part in some reactions and thus display come chemical properties.

Uses of some noble gases.

Note: The application of noble gases is ironically centered on their inert nature.

Argon is used in light bulbs to provide an inert environment to prevent oxidation.
Argon is used as an insulator in arch-welding.
Neon gas is used in streets and advertisement lights.
Helium mixed with oxygen is used in deep sea diving and mountaineering.
Helium mixed with oxygen also is used in hospitals for breathing in patients with respiratory problems and those undergoing certain forms of surgery.
Helium can be used instead of hydrogen in balloons for meteorological research.

Reason:

– Hydrogen is explosive I presence of air and may explode causing serious accidents.

– Helium is less dense than hydrogen.

Helium is used in thermometers for measurements f very low temperatures.
Liquid helium is used to keep certain metal alloys at temperatures low enough for them to become superconductors.

UNIT 3: PROPERTIES AND TRENDS ACROSS A PERIOD.

Checklist.

Introduction
Trends in physical properties of elements in period 3

Electrical conductivity
Atomic and ionic radii
Melting and boiling points

Trends in chemical properties of elements in period 3

Reaction with oxygen
Reaction with water
Reaction with acids

Introduction:

– A period is a vertical row of elements in the periodic table.

– Elements in the same period have same number of energy levels.

– There are 7 periods in the periodic table except for lanthanides and actinides which are not assigned periods.

– Periods 1 3 have fewer elements because they lack the d-block elements and have only the s-block elements and the p-block elements.

Note:

s-block elements: group 1 and 2
d-block elements: transitional elements
p-block elements: groups III to VIII.

– To understand trends and properties across a typical period of the periodic table, we shall use period 3 as the reference.

The Period three of the periodic table.

– Consists of elements with three energy levels.

– Consists of the 8 elements from sodium to argon.

– It is only made of s-block and p-block elements ad lacks any element in the d-block group of elements.

Part of periodic table showing period 3 of the periodic table.

Trends in physical properties of elements in period 3.

Electrical conductivity.

– Sodium, magnesium and aluminium are good conductors of electricity.

Reason:

They all have giant metallic structures with delocalized electros which conduct electricity;

– The electrical conductivity increases from sodium to aluminium.

Reason:

Electrical conductivity increases with increase in the number of delocalized electrons; thus aluminium with the highest number of delocalized electrons (3) in each atom will have the highest electrical conductivity;

– In the metals the electrical conductivity decreases with increase in temperature.

Reason:

– Increase in temperature distorts the alignment of electrons thus preventing their easy flow and hence poor conductivity;

– Phosphorus, sulphur, chlorine and argon do not conduct electric current.

Reason:
They all have molecular structures and all the electrons in the atoms are used in bonding; thus they lack delocalized electron or free ions for electrical conductivity.

– Silicon conducts electric current, and its electrical conductivity increases with increase in temperatures.

Reason:

– It is a semi-conductor; making it a very unique element in this period.

Note:

– A semi-conductor is a crystalline material which only conducts electricity under certain conditions.

The atomic and ionic radii.

– The atomic radii of period 3 elements decrease across the period.

Reason.

– For the same number of energy levels the number of protons in the nucleus increases across the period; this leads to the increase in nuclear charge while the shielding effect remains the same hence decrease in atomic radius across the period.

Melting and boiling points

– Sodium, magnesium and aluminium have very high melting and boiling points.

Reason:

– They have giant metallic structures with strong metallic bonds which need a lot of energy to break.

– The boiling and melting points increase from sodium to aluminium.

Reason:

– As you move across the period from sodium to aluminium, the nuclear charge increases while the energy levels remain the same hence decrease in atomic radius; the smaller the atomic radius (for metals) the stronger the metallic bonds.

– Silicon, though a non-metal also has a very high melting and boiling points.

Reason:

– Silicon has a giant atomic structure with strong covalent bonds throughout the structure, which need a lot of heat energy to break.

– Phosphorus, Sulphur, Chlorine and argon have low melting and boiling points.

Reason:

– They all have molecular structures with strong covalent bonds between the atoms (except in argon) but weak van der waals forces between the molecules which are easy to break.

– Note that argon exist as atoms and thus a monoatomic molecule.

– The melting and boiling points decreases from phosphorus to argon.

– As we move across the period from phosphorus to argon, the size of the atoms decreases leading to smaller atoms and hence molecules, which lead to decrease in the strength of the van der Waals (across the period)

Note:

– Phosphorus and sulphur exists as solids at room temperature while chlorine and argon exists as gases at room temperature.

Reason: phosphorus and sulphur have giant molecular structures while chlorine and argon have simple molecular structures.

–
Summary: Some physical properties f elements in period 3.

Property	Na	Mg	Al	Si	P (white)	S (monoclinic)	Cl	Ar
Physical state and appearance	Silver	Silver solid	Silver solid	Black solid	White solid	Yellow solid	Green yellow gas	Colourless gas
Electron arrangement	2.8.1	2.8.2	2.8..3	2.8.4	2.8.5	2.8.6	2.8.7	2.8.8
Valency	1	2	3	4	5 or 5	6	7	8
Atomic radius	0.157	0.136	0.125	0.117	0.110	0.104	0.09	0.192
MP (^oC)	98	650	660	1410	44	119	-101	-189
BP (^oC)	890	1110	2470	2360	280	445	-35	-186
Structure	Giant metallic	Giant metallic	Giant metallic	Giant atomic	Molecular	Molecular	Simple molecular	Simple molecular /atomic
Bond type	metallic	metallic	metallic	Covalent	Covalent	Covalent	Covalent	Van der waals

Trends in chemical properties of the elements in period 3.

Note: Trends in reactivity.

– The reactivity among the metallic elements decreases across the period from sodium to aluminium.

Reason:

– There is a continuous increase in nuclear charge from sodium to aluminium which leads to increase in ionization energies hence increasing difficulty in removing an electron from the outermost energy level.

– Among the non-metallic elements, the reactivity increases across the period from phosphorus to chlorine.

Reason:

– There is increase in nuclear charge from phosphorus to chlorine, hence an increase in ease of electron gain (electro affinity) since non-metals react by gaining electrons.

– Argo is unreactive and can only react under very special conditions.

Reason:

– Its a noble gas with a stable octet configuration.

Reaction of period 3 elements with oxygen.

– All period three elements react with (burn in) oxygen with the exception of argon.

Experiment: To investigate the reactions between period 3 elements and oxygen

Procedure:

– A small piece of the element is placed in a deflagrating spoon and warmed gently until it catches fire.

– It is then lowered into a gas jar full of oxygen.

– The flame colour and the colour of the product are noted.

– 10 cm³ of water containing universal or litmus indicator is added into the gas jar with the products.

Sodium:

– Burns vigorously in oxygen with a golden yellow flame; to produce white solid of sodium oxide.

Equation:

Na_(s) + O_2(s) → Na₂O_(s)

– The resultant sodium oxide dissolves in water to form sodium hydroxide.

Equation:

Na₂O_(s) + H₂O_(l) → 2NaOH_(aq);

– The sodium hydroxide is alkaline in nature and thus turns litmus indicator blue;

Magnesium:

– Burns vigorously in oxygen with a bright white light; to produce white solid of magnesium oxide.

Equation:

2Mg_(s) + O_2(s) → 2MgO_(s)

– The resultant magnesium oxide is slightly soluble in water to form magnesium hydroxide.

Equation:

MgO_(s) + H₂O_(l) → Mg(OH)_2(aq);

– The magnesium hydroxide is alkaline in nature and thus turns litmus indicator blue;

Aluminium:

Note:

– Aluminium (foil) is usually coated with a thin layer of aluminium oxide, Al₂O₃; which prevents reaction with the oxygen.

– When polished, it reacts slowly with oxygen to form a white solid of aluminium oxide.

Equation:

Al_(s) + 3O_2(s) → Al₂O_3(s)

– The resultant aluminium (III) oxide is insoluble in water.

Silicon:

– Silicon powder can only bur in oxygen at very high temperatures of about 450^oC to form solid silicon (IV) oxide.

Equation:

Si_(s) + O_2(s) → SiO_2(s)

– The resultant silicon (IV) oxide is insoluble in water.

– Burns readily in oxygen with a bright orange flame; to produce dense white fumes (solid) of phosphorus (V) oxide

Equation:

P_4(s) + 5O_2(s) → 2P₂O_5(s)

Note: sulphur exists and therefore reacts as molecules of P₄.

– The resultant phosphorus (V) oxide readily dissolves in water to form phosphoric (V) acid.

Equation:

P₂O_5(s) + 3H₂O_(l) → 2H₃PO_4(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

– Burns in oxygen with a blue flame; to form a colourless gas of sulphur (IV) oxide

Equation:

S_(s) + O_2(s) → SO_2(s)

– The resultant sulphur (IV) oxide readily dissolves in water to form sulphuric (IV) acid.

Equation:

SO_2(s) + H₂O_(l) → H₂SO_3(aq);

– The phosphoric acid is acidic in nature and thus turns litmus indicator pink / red;

– The sulphuric (IV) acid is unstable and thus easily gets oxidized by oxygen to the stable sulphuric (VI) acid.

Equation:

H₂SO_3(aq) + O_2(g) → H₂SO_4(aq);

– Burns in oxygen only under certain conditions to form acidic oxides.

Equation:

2Cl_2(s) + O_2(s) → 2Cl₂O_(s)

– Argon is unreactive.

Conclusion:

– Metallic elements burn in oxygen to form basic oxides.

– Non-metallic oxides burn in oxygen to form acidic oxides.

Reaction of period 3 elements with water.

Procedure:
– A small piece of sodium metal is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal floats on the water surface; because it is less dense than water;

– A hissing sound is produced; due to production of hydrogen gas;

– It vigorously melts into a silvery ball then disappears because reaction between water and sodium is exothermic (produces heat). The resultant heat melts the sodium due to its low melting point.

– It darts on the surface; due to propulsion by hydrogen;

– The metal may burst into a golden yellow flame; because hydrogen may explode into a flame which then burns the sodium;

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

Reaction equations.

Equation I

2Na_(s) + 2H₂O_(l) → 2NaOH_(aq) + H_2(g);

Equation II

4Na_(s) + O_{2 (g)} → 2Na₂O_(s);

Equation III:

Na₂O_(s) + H₂O_(l) → 2NaOH_(aq)

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; sodium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

Procedure:
– A small piece of magnesium ribbon is cut and dropped into a trough containing water;

– The resultant solution is tested with litmus paper;

Diagram of apparatus:

Observations and explanations:

– The metal sinks into the water surface; because it is denser than water;

– It reacts slowly with water leading to slow evolution of hydrogen gas.

– The resultant solution turns blue; because sodium hydroxide solution formed is a strong base;

Reaction equation.

Mg_(s) + 2H₂O_(l) → Mg(OH)_2(aq) + H_2(g);

Effect of resultant solution on litmus paper;

– Litmus paper turns blue; magnesium hydroxide formed is highly soluble in water; releasing a large number of hydroxyl ions which result into alkaline conditions // high pH;

– Aluminium does not normally react with cold water or steam.

Reason:

– Aluminium is usually coated with a thin coating of aluminium oxide which prevents further reaction.

– However at very high temperatures of about 700^oCsteam can react with aluminium to form aluminium oxide and hydrogen gas.

Equation:

2Al_(s) + 3H₂O_(g) → Al₂O_3(s) + 3H_2(g);

Note:

– Due to its inability t react with water and air aluminium is preferable for making cooking utensils like sufurias and coking pans.

Silicon, phosphorus and sulphur.

– These non-metals do not displace hydrogen from water and thus do not react with water.

– Dissolves in water t form chlorine water, which is a mixture of chloric (I) acid ad hydrochloric acid.

Equation:
Cl_2(g) + H₂O_(l) → HOCl_(aq) + HCl_(aq)

Chloric (I) acid Hydrochloric acid

Reaction of period 3 elements with acids

Procedure:

– A piece of the element is dropped into 5 cm³ of an acid in a test tube.

– Any gas produced is tested.

Sodium:

– Reacts explosively with acids to form salts and hydrogen and thus reactions of sodium with acids should not be tried in the laboratory.

Magnesium:

– Reacts with both dilute hydrochloric and dilute sulphuric acid to form magnbsium salts and hydrogen gas.

With hydrochloric acid:

Mg_(s) + 2HCl_(aq)→ MgCl_2(aq) + H_2(g)

With sulphuric acid:

Mg_(s) + H₂SO_4(aq)→ MgSO_4(aq) + H_2(g)

Aluminium:

Note:

– It does not readily react with dilute acids.

Reason:

– This is due to presence of a thin aluminium oxide coating that prevents contact hence reaction with the acids.

– When polished t remove the oxide coating it reacts with both dilute hydrochloric and dilute sulphuric acid to form aluminium salts and hydrogen gas.

With hydrochloric acid:

2Al_(s) + 6HCl_(aq)→ 3AlCl_3(aq) + 3H_2(g)

With sulphuric acid:

2Al_(s) + 3H₂SO_4(aq)→ Al₂(SO₄)_2(aq) + 3H_2(g)

Silicon, phosphorus, sulphur and chlorine.

– They do not react with dilute acids.

Summary: Chemical properties of period 3 elements.

Element	Na	Mg	Al	Si	P	S	Cl
Reaction with air or oxygen	Readily reacts with air. – Burns brightly in oxygen with a golden yellow flame to form Na₂O	Reacts slowly with air. – Burns in oxygen with a bright white flame to form MgO	– Forms a protective coating of Al₂O₃ when it burns in oxygen.	Si powder burns at temperatures above 950^oC to form SiO₂	White phosphorus smolders in air; – P burns in air with a bright orange flame to form P₂O₃ and P₂O₅	Burns in air or oxygen with a blue flame to form SO₂ gas	N reaction with air or oxygen under normal conditions.
Reaction with water	Reacts vigorously to produce H₂ and NaOH	Slow reaction with cold water to form Mg(OH)₂ and H₂; – Reacts faster with steam to form MgO and H₂	No reaction	No reaction	No reaction	No reaction	Dissolves in water to form chlorine water
Reaction with dilute acids	Violent reaction giving out H₂ and a sodium salt	Rapid evolution of H₂ gas and a Mg salt is formed	Reacts slowly to give H₂ and an Al salt	No reaction	No reaction	No reaction	No reaction

UNIT4: STRUCTURE AND BONDING.

Meaning of structure and bond.
Nature of the chemical bond.
Types of bonds.

Ionic bonds
- Meaning
- Formation
- Dot ad cross diagrams.
- Giant Ionic structures
  - Examples
  - Meaning
  - Properties of giant ionic structures
- Covalent and dative bonds
  - Meaning
  - Formation
  - Dot ad cross diagrams.
  - Giant covalent (atomic) structures
    - Examples
    - Meaning
    - Properties of giant covalent structures
    - Details of Diamond, graphite and silicon (IV) oxide
  - Molecular (simple and giant) structures
    - Examples
    - Meaning
    - Hydrogen bond and Van der waals
    - Properties of molecular structures
  - Metallic bonds
    - Meaning
    - Formation
    - Dot ad cross diagrams.
    - Giant Metallic structures
      - Examples
      - Meaning
      - Properties of giant metallic structures

Bond types across period 3

Oxides of period 3 elements
- Nature
- Melting and boiling points
- General trends in bond types
Chlorides of period 3 elements
- Nature
- Melting and boiling points
- General trends in bond types

Summary on bond characteristics

Meaning of structure and bond.

Bond:

– The mutual force of attraction that holds particles together when atoms (similar or different) combine during chemical reactions.

Structure:

– A regular pattern of particles in a substance held together by chemical bonds.

Nature of the chemical bond.

– Atoms are made up of energy levels and nucleus.

– The energy levels contain electrons while the nucleus contains protons and neutrons.

– The electrons are negatively charged, protons are positively charged while neutrons are electrically neutral.

– The noble gases are chemically inert since their outermost energy levels are completely filled with the maximum possible number of electrons.

– Thus noble gases have either a stable duplet state (2) like in helium or a table octet configuration (2.8 or 2.8.8) as in neon and argon respectively.

– Other atoms are unstable because the outermost energy levels are not yet completely filled with the maximum possible number of electrons.

– To attain the stable duplet or octet noble gas configuration, such atoms lose, gain or share their valance electrons.

– It is the act of losing, gaining or sharing valence electrons that lead to the chemical bonds.

– When atoms gain or lose valence electron(s) they become charged forming anions and cations respectively.

– Particles of the same charge repel each other while particles of different charges attract one another.

Types of bonds.

– There are three main types of chemical bonds:

Ionic bonds
Covalent bond
Metallic bond

(a). Ionic bonds / electrovalent bond.

Meaning

– Is a bond formed due to complete transfer of electrons from one atom to another resulting into two oppositely charged ions.

Formation of an ionic bond.

– Formed due to complete transfer of electrons from one atom to another; and mainly formed between a metal and a non-metal.

– This occurs in a bid for both atoms to acquire a stable noble gas configuration.

– One atom loses all its valence electrons thus forming a cation (positively charged ions).

– The other atom gains all the lost valence electrons forming an anion (negatively charged ion).

– The cation and the anion are oppositely charged and thus develop a mutual force of attraction between them which is the ionic / electrovalent bond.

Illustration: formation of ionic bond between sodium and chlorine to form sodium chloride.

– Sodium metal (atomic number 11) has electronic configuration 2.8.1 and thus unstable with 1 valence electron.

– Chlorine gas (atomic number 17) has electronic configuration 2.8.7 and thus unstable with 7 valence electrons.

– Sodium is more electropositive (the tendency to lose electrons to form cations) than chlorine while chlorine is more electronegative (the tendency to gain electrons to form anions).

– Sodium loses its single valance electron to form sodium ion with electronic configuration 2.8 and a net charge of +1 (Na⁺)

– Chlorine atom accepts the single electron lost by sodium to form a chloride ion with electronic configuration 2.8.8 with a net charge of -1 (Cl^–).

– The positively charged sodium ion and the negatively charged chloride ion attract each other.

– The electrostatic forces of attraction develop between the two oppositely charged ions and this constitutes the ionic bond.

– Compounds formed due to ionic bonding are thus called ionic compounds.

Diagrammatically:

Examples of ionic compounds.

Sodium chloride.
Potassium fluoride.
Magnesium oxide
Aluminium (III) oxide.

Dot and cross diagrams for ionic compounds

Potassium fluoride.

Magnesium oxide

Sodium oxide.

Magnesium chloride.

Aluminium (III) oxide.

Giant Ionic structures

– Ionic bonding results into one type of structure, the giant ionic structure.

– This is a type of structure in which all ions are bonded with strong ionic bonds throughout the structure.

– Each ion in the giant ionic structure is surrounded by several others resulting into giant pattern of several ions, hence giant ionic structure.

– Most ionic substances with the giant ionic structure are crystalline in nature, made up crystals.

Note:

A crystal is a solid form of a substance in which the particles are arranged in a definite pattern repeated regularly in 3 dimensions.

Illustration of the giant ionic structure: sodium chloride structure.

– The NaCl structure consists of many Na⁺ and Cl^– arranged and packed in a regular pattern.

– Each Na⁺ is surrounded by six Cl^– that are equidistant from it.

– Similarly each Cl^– is surrounded by six Na⁺ that are equidistant from it.

– This pattern occurs repeatedly in all directions.

– The result is a giant of ions in all directions hence giant ionic structure.

Diagram of the cubic structure of sodium chloride.

Properties of giant ionic structures.

They are hard and brittle.

Reason:

– Ionic solids are hard because each ion is held in the crystal by strong attractions from the oppositely charged ions around it.

– They are brittle and thus may be split cleanly (cleaved) using a sharp-edged razor.

Explanation:

– When a crystal is tapped sharply along a particular plane it is possible to displace one layer of ions relative to the next.

– Due to the displacement, ions of similar charge come together leading to repulsive forces between the portions of the crystals.

– This forces the two portions of the crystals to split apart.

Diagram: crystal cleavage in ionic compounds.

They have high melting and boiling points.

Reason:

– They have strong electrostatic forces / ionic bonds / electrovalent bonds between the oppositely charged ions throughout the structure which require large amounts of energy to break.

Solubility.

(i). They are soluble in polar solvents like water, ethanol and acetone (propanone)

Reason:

– Water contains highly polar molecules.

– The positive ends of the polar water molecules are attracted to the negative ions in the crystal, and the negative ends of the water molecules are attracted to the positive ions in the crystal.

– This results to the formation of ion-solvent bonds which leads to release of energy.

– This energy is sufficient to cause the detachment of ions from the crystal lattice hence dissolution.

Note:
– This detachment of ions is called solvation, and the energy required for this is called solvation energy.

– Where the solvent is water the ions are said to have been hydrated, and the energy involved in the process is called hydration energy.

Diagrams: hydrated positive and negative ions.

(ii). They are insoluble in non-polar organic solvents like tetrachloromethane, benzene and hexane.

Reason:

– Non-polar molecules are held together by weak intermolecular forces, the Van der Waals forces.

– The Van der Waals are much smaller in magnitude compared to the ionic bonds in the ionic solid crystal lattice.

– Thus the ion-ion interactions in the ionic solid are stronger than the solvent-solvent interactions in the solvent or the solvent-ion interactions between the solid and the solvent.

– Thus the non-polar solvent molecules cannot penetrate the ionic lattice to cause salvation.

Electrical conductivity.

– Ionic substances do not conduct electric current in solid state.

Reason:

The ions are held in static positions in the solid crystal lattice and thus cannot move to conduct electric current.

– They conduct electric current in molten and solution (aqueous) states.

Reason:

– In molten and aqueous states the ions are free and mobile and thus move about conducting electric current.

Gradation in properties of some ionic compounds of sodium.

Property of compound		Compound of sodium
Property of compound		Sodium fluoride	Sodium chloride	Sodium bromide	Sodium iodide
Solubility in water		Soluble	Soluble	Soluble	Soluble
Melting point (^oC)		993	801	747	661
Boiling point (^oC)		1695	1413	1390	1304
Electrical conductivity	Solid	Does not	Does not	Does not	Does not
Electrical conductivity	Molten / solution	Conducts	Conducts	Conducts	Conducts

Note:

– Solubility of the compounds decrease from sodium fluoride to sodium iodide.

– Melting and boiling points decrease from sodium fluoride to sodium iodide.

The covalent bond.

Meaning:

– Refers to a bond formed when two atoms of the same or of different elements share electrons to become stable.

– Formation of covalent bond between atoms (similar or dissimilar) result to the formation of a molecule.

– Covalent bonds are usually formed by the association of non-metals.

Note:

– A molecule is a group of atoms (two or more) of the same or different elements that are held together by strong covalent bonds.

Formation of a covalent bond.

– Covalent bonding is brought about by the facts that the electro-positivity and the electro-negativity of the elements involved are very close.

– For that reason, none of the atoms can completely lose its valence electrons to the next atom.

– For this reason, both atoms donate electrons which are then shared between them.

– Both atoms thus attain a stable noble gas (duplet or octet) configuration.

Illustrations:

Formation of chlorine molecule.

– Each chloride atom has electronic configuration 2.8.7 and thus need to gain a single electron in the outermost energy level to attain a stable noble gas configuration.

– Since both chloride atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 1), which they share between them.

– Thus chlorine molecule is formed by sharing 2 electrons between two chlorine atoms, hence a single covalent bond.

Diagram: formation of chlorine molecule.

Dot and cross diagram for chlorine molecule.

Formation of oxygen molecule.

– Each oxygen atom has electronic configuration 2.8.6 and thus need to gain 2 electrons into the outermost energy level to attain a stable noble gas configuration.

– Since both oxygen atoms have same electro-negativities, none will easily lose an electron to the other.
– For this reason, both donate the number of electrons required by the other atom (in this case 2), which they share between them.

– Thus oxygen molecule is formed by sharing 4 electrons (2 from each atom) between two chlorine atoms, hence a double covalent bond.

Diagram: the oxygen molecule.

Note:

– A single covalent bond is represented in dot (.) and cross (x) diagrams using two dots, two crosses, a dot and a cross or a single line () between the atoms involved in the bond.

– Thus a single covalent bond like in chlorine can be represented as Cl Cl, a double covalent bond like in oxygen can be represented as O = O while a triple covalent bond like in nitrogen can be represented as N ≡ N.

Dot (.) and cross (x) diagrams for various covalent compounds.

Hydrogen, H₂

Hydrogen chloride, HCl

Nitrogen, N₂

Water

Carbon (IV) oxide, CO₂

Ammonia gas, NH₃

Phosphene, PH₃

Methane, CH₄

Ethane. CH₂CH₂

Ethyne, C₂H₂

Ethanol, C₂H₅OH

Bromoethane, C₂H₄Br.

The coordinate bond.

– Refers to a type of covalent bond in which the shared pair of electrons forming the bond is contributed by only one of the atoms forming the bond.

– It is also called the dative bond.

Examples:

Formation of ammonium ion.

– Occurs when an ammonia gas molecule combines with a hydrogen ion (proton).

– All the atoms in the ammonia molecule have a stable noble gas configuration and thus the molecule is stable.

– However the nitrogen in the ammonia molecule has a lone pair of electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts bonds with the lone pair (2) of electrons in the nitrogen of the ammonia molecule forming a dative bond.

– The total number of electrons in the ammonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

Note:

– In a dot (.) and cross (x) diagram where the covalent bond is represented by horizontal lines (), the dative // coordinate bond is represented by an arrow (→) pointing the atom that “accepts” the electrons.

Diagram: formation of ammonium ion.

Formation of hydroxonium ion (H₃O⁺)

– Occurs when a water molecule combines with a hydrogen ion (proton).

– All the atoms in the water molecule have a stable noble gas configuration and thus the molecule is stable.

– However the oxygen in the ammonia molecule has 2 lone pairs of electrons // four electrons (electrons that have not yet been used in bond formation)

– The hydrogen ion has lost its outermost single valence electron to form the hydrogen ion.

– Thus the hydrogen ion has no electron(s) in its outermost energy level.

– To be stable the hydrogen needs two electrons in its outermost energy level.

– The hydrogen ion thus accepts and bonds with two of the four electrons in the oxygen of the water molecule forming a dative bond // coordinate bond.

– The total number of electrons in the hydroxonium ion is 11 while the total number of electrons is 10 leading to a net positive charge of +1

Diagram: formation of the hydroxonium ion.

Note:

– The hydroxonium ion (H₃O⁺) can still further react with another hydrogen ion to form another ion of the formula H₄O²⁺.

– This is due to the presence of a single lone pair of electrons in the structure of the hydroxonium ion (H₃O⁺).

Diagram: formation of H₄O²⁺ from H₃O⁺ and H⁺

Note:

– The H₄O²⁺ cannot however react further since all the valence electrons in all its atoms have been used in bonding leaving no lone pairs.

Further examples of dative covalent bonds in other compounds.

Carbon (IV) oxide

Formation of PH₄⁺

Aluminium chlorine dimer (Al₂Cl₆).

Ammonia-aluminium chloride vapours complex, AlCl₃.NH₃

Note:

– Substances with covalent bonds form two main types of structures:

Molecular structures
Giant atomic // Giant covalent structures.

The Molecular structures.

– Refers to a structure in which covalent bonds holds atoms together to form molecules and the resultant molecules are held together by intermolecular forces.

– Substance with molecular structures are usually gases or liquids at room temperature.

– Few solids such as sulphur, iodine, fats, sugar, naphthalene and paraffin wax also have molecular structures.

– The intermolecular forces in molecular structures are of two types:

Van der Waals
Hydrogen bonds

(a). The Van der Waals.

– Are the weakest form of intermolecular forces due to induced dipole- induced dipole attractions between molecules.

– As the size of the molecule increases, the number of constituent electrons increases leading to increase in strength of the induced dipole induced dipole interactions.

– The strength of the Van der Waals thus increases as the molecular size increases.

Note:

This explains the trend in boiling points for the halogens, which increase from fluorine to iodine.

Reason:

– From fluorine to iodine the size of the atoms and hence the molecules increases leading to increase in molecular masses that lead to stronger induced dipole induced dipole interactions hence increasing strength of the Van der Waals (from F₂ to I₂)

– This also applies for the increase in boiling points for the homologous series of alkanes.

Diagram: Illustration of Van der Waals forces

(i). Iodine

(ii). Graphite.

(b). The hydrogen bonds.

– Is an intermolecular force in which the electropositive hydrogen atom of one molecule is attracted to an electronegative atom of another molecule.

– The essential requirements for the formation of a hydrogen bond (H bond) are:

A hydrogen atom attached to a highly electronegative atom.
An unshared pair of electrons on the electronegative atom.

Note:

This explains why hydrogen bonds are common in molecules in which hydrogen are bonded to highly electronegative atoms like nitrogen, oxygen and fluorine.

The formation of hydrogen bonds

– Occurs when hydrogen atom is bonded to a highly electronegative atom like nitrogen, oxygen and fluorine.

– The electrons in the covalent bond (between hydrogen and the more electronegative atom) are drawn towards the electronegative atom.

– Hydrogen atom has no electrons other than the one it contributes to the covalent bond, which is also being pulled away from it (by the more electronegative atom).

– Hydrogen atom has no outer energy level of electrons making the single proton in the nucleus unusually bare.

– The proton is thus readily available for any form of dipole-dipole attractions.

– The bare proton of the hydrogen atom thus attracts the more electronegative atom (e.g. N, O and F) on either side.

– It thus exerts an attractive force on the more electronegative atom hence bonding them together.

– The two (more electronegative) larger atoms are drawn closer with a hydrogen atom effectively buried in their electron clouds.

– This constitutes the Hydrogen bond.

Illustration: formation of hydrogen bonds in water.

– In water two hydrogen atoms are bonded to an oxygen atom which highly electronegative.

– The electrons in the covalent bond (between each hydrogen and oxygen atom) are drawn towards the more electronegative oxygen atom.

– The hydrogen atoms have no electrons other than their share of those in the covalent bond, which are also being pulled away from them by the oxygen atom.

– Hydrogen atoms have no outer energy level of electrons making the single proton in their nucleus unusually bare.

– The proton is thus readily available for any form of dipole-dipole attractions (with oxygen in this case).

– The bare proton of each of the hydrogen atoms in one molecule thus attracts the more electronegative oxygen atom of the neighboring molecule.

– Each hydrogen atom thus exerts an attractive force on the oxygen atom of the next molecule hence bonding them together.

– The electronegative oxygen atom of one molecule is drawn to the electropositive hydrogen atom of the next molecule with the hydrogen atoms effectively buried in the electron clouds of oxygen.

– This constitutes the Hydrogen bond in the water molecule.

Diagram: hydrogen bonds in water molecules.

Note:

– Other compounds with hydrogen bonds include ethanol, ammonia, hydrogen fluoride etc.

– Hydrogen bonds are much stronger than the weak Van der Waals forces but still weaker than the covalent bonds.

Effect of hydrogen bonding on properties of molecular substances.

– Hydrogen bonding tends to disrupt the gradation in physical properties of molecular substances in relation to molecular weights.

– The effect on molecular masses on the melting and boiling points only apply when the intermolecular force is the same.

Examples.

Both ethanol (C₂H₅OH) and dimethyl ether (C₂H₆O) have the same relative molecular mass of 46. However the boiling point of ethanol is higher at 78.5oC than that of dimethyl ether at only -24oC.

Reason:

– Even though both have molecular structures with covalent bonds between the atoms, the intermolecular forces in ethanol are hydrogen bonds which are much stronger than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

Ethanol (C₂H₅OH) has a molecular mass of 46 while butane (C₄H₁₀) has a molecular mass of 58, yet the boiling point of ethanol is higher than that of butane.

Reason:

– Both have molecular structures with covalent bonds between the atoms. However, the intermolecular forces in ethanol are hydrogen bonds which are much stronger (and require more energy to break) than the intermolecular forces in dimethyl ether which are weak Van der Waals forces.

– Molecular substances are generally insoluble in polar solvents like water. However those with hydrogen bonding as the intermolecular forces are soluble in water since the hydrogen bonding confers them some polarity.

Examples:

– Sugar, ethanol, ethanoic acid etc

Properties of molecular structures.

Melting and boiling points.

– The generally have low melting and boiling points.

Reason:

– Although the atoms forming the molecules are held together by strong covalent bonds, the intermolecular forces are weak Van der Waals forces which require low amounts of energy to break.

Heat and electrical conductivity

– They are poor conductors of heat and electricity at any state.

Reason:

– They have neither delocalized electrons nor free mobile ions for electrical conductivity.

Solubility

– Molecular substances are generally insoluble in polar solvents like water but soluble in non-polar organic solvents like benzene.

Reason:

– In polar solvents like water there are strong water water attractions which are considerable stronger than the intermolecular forces (Van der Waals) attractions or molecule water (solvent) attractions, making the molecules unable to penetrate the water (solvent) structure for dissolution to occur.

– In non-polar solvents like benzene, the benzene-benzene attractions are similar in strength to the intermolecular forces or the molecule benzene (solvent) attractions, enabling the molecules to penetrate the solvent thus allowing dissolution.

Note:

– Molecular substances with hydrogen bonds as the intermolecular forces are soluble in polar solvents like water.

Reason:

– The hydrogen bonds in the molecules are equal in strength to the water water interactions which are also hydrogen bonds, thus the molecules are able to penetrate the structure of water leading to salvation // dissolution // hydration i.e. they are polar like the water molecules.

Summary: Properties of some molecular substances.

Property of compound		Molecular substance
Property of compound		Sugar (Sucrose)	Naphthalene	Iodine	Rhombic sulphur	Water	Hydrogen sulphide
Solubility in water		Soluble	Insoluble	Insoluble	Insoluble	–	Slightly soluble
Molecular mass		183	128	186	256	18	34
Melting point (^oC)		186	82	113	114	0	-85
Boiling point (^oC)		–	218	183	444	100	-60
Electrical conductivity	Solid	Does not	Does not	Does not	Does not	Does not	Does not
Electrical conductivity	Molten / solution	Does not	Does not	Does not	Does not	Does not	Does not

Giant covalent structures // Giant atomic structures.

– Are molecular substances in which atoms are linked throughout the whole structure by very strong covalent bonds from one atom to the next.

– The result is an indefinite number of atoms which are all covalently bonded together.

– This pattern occurs repeatedly throughout the structure leading to a giant of atoms all covalently bonded.

Examples:

(a). Diamond.

– Is an allotrope of carbon.

Note: allotropes are different crystalline forms of the same element in the same physical state.

– In diamond, each carbon atom is bonded to four other carbon atoms by strong covalent bonds.

– The carbon atoms in diamond are covalently bonded into an octahedral pattern, which repeats itself in all directions resulting into a giant atomic structure.

– Since each carbon atom is bonded to four others, all the four valence electrons in each carbon are used in bonding hence no delocalized electrons in the structure of diamond.

– Diamond is the hardest substance known due to the fact that all the atoms are covalently bonded together and are closely packed together.

Diagram: Structure of diamond:

Properties of diamond.

Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the structure which require large amounts of energy to break.

It is insoluble in water.

Reason:

– It is non-polar and thus cannot dissolve in polar water molecules since there are no intermolecular interactions which would facilitate penetration into the water molecules for dissolution to occur.

iii. Does not conduct heat and electricity.

Reason:

– Each carbon atom in the structure of diamond is bonded to four others hence uses all its four valence electrons in bonding and thus lacks any delocalized electrons for electrical conductivity.

It is the hardest substance known.

Reason:

– All carbon atoms are compactly bonded in a continuous octahedral pattern with strong covalent bonds throughout the structure which are very difficult to break.

(b). Graphite.

– Is also an allotrope of carbon.

– In graphite, each carbon atom is bonded to three other carbon atoms by strong covalent bonds.

– Since each carbon atom is bonded to only three others, only three of the four valence electrons in each carbon are used in bonding hence presence delocalized electrons in the structure of graphite.

– This explains the electrical conductivity of graphite.

– The carbon atoms in graphite are covalently bonded into hexagonal layers, which are joined to each other by weak Van der Waals forces.

– The presence of weak Van der Waals forces explains the slippery nature of graphite.

Diagram: Structure of graphite.

Properties of Graphite.

Have high melting and boiling points.

Reason:

– It has a giant atomic structure with strong covalent bonds throughout the hexagonal layers which require large amounts of energy to break. Even though there are Van der Waals between the layers the effect of the large number of covalent bonds still contribute to high melting and boiling points in graphite

Insoluble in water.

Reason:

iii. It is a good conductor heat and electricity.

Reason:

– Each carbon atom in the structure of graphite is bonded to three others hence uses only three of its four valence electrons in bonding. This leads to presence of delocalized electrons in the structure of graphite which conducts heat and electricity.

It is soft and slippery.

Reason:

– The carbon atoms in graphite are covalently bonded into hexagonal layers which are joined to each other by weak Van der Waals forces.

– The weak Van der Waals forces easily slide over each other when pressed hence the soft and slippery feel.

(c). Silicon (IV) oxide.

– Is a covalent compound of silicon and oxygen.

– Silicon has four electrons in its outermost energy level while oxygen has six.

– Each silicon atom is bonded to four oxygen atoms by strong covalent bonds.

– Each oxygen atom is bonded to two silicon atoms by strong covalent bonds.

– This means there are no delocalized electrons in the structure of silicon (IV) oxide making it unable to conduct electric current.

– The silicon and oxygen atoms are all covalently bonded together (by strong covalent bonds) in a repeated manner leading to a giant of covalent bonds throughout the structure.

– The extra ordinarily strong covalent bonds in silicon (IV) oxide throuout the structure contribute to the very high melting (1728^oC) and boiling (2231^oC) points.

Diagram: structure of silicon (IV) oxide.

Properties of giant atomic // giant covalent structures.

– Have high melting and boiling points
– They are non-conductors of heat and electricity with the exception of graphite

iii. – They are insoluble in water

– Most are generally very hard, with the exception of graphite

Note: for explanations of the above properties, refer to individual explanations of each compound.

The metallic bond.

– Is a bond formed due to electrostatic attraction between the positively charged nuclei and the negatively charged delocalized electrons that hold atoms together.

Formation of a metallic bond.

– In a metal there are usually many atoms surrounding any one atom.

– The valence electrons of any one atom are therefore mutually attracted to many nuclei.

– This leads to a situation in which the positive nuclei appear to be immersed in a sea of mobile electrons.

– The sea of mobile electrons are said to be delocalized which explains the ability of substances with metallic bonds to conduct electric current.

– This pattern of positive nuclei in a sea of electrons is repeated many times throughout the structure leading to a giant metallic structure, the only structure due to metallic bonds.

Diagram: model of the giant metallic structure.

Properties of the giant metallic structure.

They have high melting and boiling points.

Reason:

– They have strong metallic bonds throughout the structure which need large amounts of energy to break.

– The strength of the metallic bond increases with decrease in atomic size, as well as with increase in the number of delocalized electrons.

– Thus metals with smaller atoms and more delocalized electrons tend to have stronger metallic bonds hence higher melting and boiling points.

– This explains why the melting and boiling points of metallic elements in period three increase from sodium to aluminium.

They are good conductors of heat and electricity.

Reason:

– The have delocalized electrons which heat and electric current.

Note:

– The electrical conductivity of metals increases with increase in the number of delocalized electrons in each atom in the structure.

– This explains why aluminium metal is a better conductor of heat and electricity than both magnesium and sodium.

They are insoluble in water.

Reason:

– There are no dipoles in the giant metallic structure and are thus non-polar, so cannot dissolve in polar water molecules.

Summary: some physical properties of metals

Metal	Valency	Melting point (^oC)	Boiling point (^oC)	Atomic radii (nm)	Electrical conductivity
Lithium	1	180	1330	0.133	Good
Sodium	1	98	890	0.155	Good
Potassium	1	64	774	0.203	Good
Magnesium	2	651	1110	0.136	Good
Aluminium	3	1083	2582	0.125	Good

Summary: Comparing various types of structures.

Attribute	Giant metallic	Giant atomic / giant covalent	Giant ionic	Giant metallic
1. Structure i. Examples	Na, Fe, Cu.	Diamond, SiC, SiO₂.	Ca²⁺O^2-, (K⁺)₂SO₄^2-, Na⁺Cl^–,	I₂, S₈, C₁₀H₈, HCl, CH₄.
ii. Constituent particles	Atoms	Atoms	Ions	Molecules
iii. Type of substance compound	Metal element with low electronegativity	Non-metal element in group IV or its compound.	Metal/non-metal compound (a compound of elements with a large difference in electronegativity)	Non-metal element or non-metal/non-metal compound (elements with high electronegativity)
2. Bonding: In the solid	Attraction of outer mobile electrons for positive nuclei binds atoms together by strong metallic bonds	Atoms are linked through the whole structure by very strong covalent bonds from one atom to the next.	Attraction of positive ions for negative ions results in strong ionic bonds	Strong covalent bonds hold atoms together within the separate molecules; separate molecules are held together by weak intermolecular forces.
3. Properties. i. Volatility: State at room temp.	Non-volatile. Very high melting and very high boiling points. Usually solid	Non-volatile. Very high melting and very high boiling points. Solid	Non-volatile. Very high melting and very high boiling points. Solid	Volatile. Low melting and low boiling points. Usually gases or volatile liquids
ii. Hardness // malleability	Hard, yet malleable.	Very hard and brittle	Hard and brittle	Soft
iii. Conductivity:	Good conductors when solid or molten.	Non-conductors in any state (except graphite)	Non-conductors when solid; good conductors when molten or in aqueous solution (electrolytes)	Non-conductors when solid, molten and in aqueous solution. (A few like HCl react with water to form electrolytes)
iv. Solubility:	Insoluble in polar and non-polar solvents but soluble in molten metals	Insoluble in all solvents	Soluble in polar solvents (e.g. H₂O), insoluble in polar solvents like tetrachloromethane, CCl₄.	Polar molecules e.g. HCl are soluble in polar solvents like water; but insoluble in non-polar solvents like CCl₄ and vise-versa.

Types of bonds across a period.

– The number of valence electrons play an important role in determination of chemical bonding.

– Across a period in the periodic table, the nature of the bonds varies from metallic to covalent.

– The structure also thus varies from giant metallic to simple molecular.

– Thus similar compounds of the elements in period 3 will also exhibit variation in bond types, structures and properties.

Variation in bond types in oxides of period three elements.

Oxide	Na₂O	MgO	Al₂O₃	SiO₂	P₂O₅	SO₂	Cl₂O₇
Physical state	Solid	Solid	Solid	Solid	Solid	Gas	Gas
M.P (^oC)	1193	3075	2045	1728	563	-76	-60
B.P (^oC)	1278	3601	2980	2231	301	-10	-9
Structure	Giant ionic	Giant ionic	Giant ionic	Giant atomic	Molecular	Molecular	Molecular
Bonding	Ionic	Ionic	Ionic	Covalent	Covalent with Van der Waals	Covalent with Van der Waals	Covalent with Van der Waals
Nature of oxide	Basic (alkaline)	Basic (weakly alkaline)	Amphoteric	Acidic	Acidic	Acidic	Acidic
Solubility in water	Dissolves to forms an alkaline solution	Dissolves to forms an alkaline solution	Insoluble	Insoluble	Dissolves in water to form acidic solution.	Dissolves in water to form acidic solution.	Dissolves in water to form acidic solution.
Reaction with acids	Reacts to form salt and water.	Reacts to form salt and water.	Reacts to form salt and water.	No reaction.	No reaction.	No reaction	No reaction

Note:

– The melting and boiling points of the magnesium oxide is higher than that of sodium oxide

Reason:

– Both have giant ionic structures. However the electrostatic forces of attraction between magnesium ions and oxide ions are stronger due to the fact that magnesium ion has a charge of +2 and is smaller in size than the sodium ion.

Effect of bond types of properties of chlorides of period 3 elements

– Most period 3 elements form stable chlorides.

– The trend in bond types, structures and properties of chlorides of period 3 elements show variation across the period.

Properties:

Reaction with water.

Procedure:

– A test tube is half filled with water and initial temperature of the water recorded.

– A spatula end full of sodium chloride is added and stirred until it dissolves.

– The hiughest temperature attained when all the solid dissolves is recored and the temperature change calculated.

– Two three drops of universal indicator are added and the pH of the solution noted and recorded.

Observations:

Chloride	Observations
Chloride	Solubility	Temperature change (^oC)	pH of solution
Sodium chloride	Dissolves	Drop in temperature	7
Magnesium chloride	Dissolves	Slight increase	6.5
Aluminium chloride	Hydrolyzed	Increases	3
Silicon (IV) oxide	Hydrolyzed	Increases	2
Phosphorus (III) chloride	Hydrolyzed	Increases	2
Phosphorus (V) chloride	Hydrolyzed	Increases	2

Explanations.

Sodium and magnesium chlorides.

– Sodium chloride dissolves in water causing a slight drop in temperature.

– Magnesium dissolves readily with a small increase in temperature.

– Both chlorides are ionic and when added to water there is an immediate attraction of polar water molecules for ions in the chlorides.

– The solid thus readily dissolves forming aquated ions such as Na⁺_(aq) and Cl^–_(aq).

– These are separate metal and non-metal ions surrounded by polar water molecules.

– Since there is no production of either H⁺ or OH^– ions the solutions are neutral.

Equations:

NaCl_(s) + H₂O_(l) → Na⁺_(aq) + Cl^–_(aq) + H₂O_(l).

MgCl_2(s) + H₂O_(l) → Mg²⁺_(aq) + 2Cl^–_(aq) + H₂O_(l).

Anhydrous aluminium chloride.

– It exits in molecular form as a dimeric molecule of Al₂Cl₆.

– The dimeric molecule is formed when aluminium chloride (AlCl₃) molecules vapour condense and combine forming larger molecules of Al₂Cl₆.

Diagram: Formation of a dimer in aluminium chloride.

– When added to water aluminium chloride is hydrolyzed to form an acidic solution.

– By so doing it behaves like a covalent chloride rather than an ionic chloride.

– The hydrolysis is an exothermic reaction accompanied by release of heat hence the increase in temperature.

Explanations.

– The hydrolysis is due to the very small but highly charged aluminium ion, Al³⁺.

– The Al³⁺ draws electrons away from its surrounding water molecules and causes them to give up H+ ions.

– This reaction usually involves aluminium ions combining with six water molecules to form hexa-aqua-aluminium (III) ions which dissociate to give H⁺.

Equation:

[Al(H₂O)₆]³⁺_(aq) → [Al(H₂O)₅(OH)]²⁺_(aq) + H⁺_(aq)

Note:

This reaction can be simplified as follows:

– Aluminium chloride reacting with water and hydrolyzing to give HCl_(aq) as one of the products.

Equation: Al₂Cl_6(s) + 6H₂O_(l) → 2Al(OH)_3(aq) + 6HCl_(aq)

– Then the HCl dissociates to give H+ and Cl-

Equation: 6HCl_(aq) → H⁺_(aq) + Cl^–_(aq)

– It is the resultant H⁺ that confer the solution its acidic properties.

Silicon (IV) chloride

– Undergoes hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are silicon (IV) oxide solid and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H⁺ which leads to acidic conditions.

Equations:

SiCl_4(s) + 2H₂O_(l) → SiO_2(s) + 4HCl_(aq)

Then: 4HCl_(aq) → 4H⁺_(aq) + 4Cl^–_(aq).

Phosphorus (III) chloride and phosphorus (V) chloride

– Both undergo hydrolysis in water in an exothermic reaction producing a lot of heat.

– The products of the hydrolysis are phosphorus (III) acid and phosphoric (V) acid respectively, and hydrogen chloride gas.

– The hydrogen chloride gas immediately dissolves in the water to form hydrochloric acid.

– The hydrochloric acid dissociated to liberate H⁺ which leads to acidic conditions.

Equations:

With phosphorus (III) chloride:

PCl_3(s) + 2H₂O_(l) → H₃PO_3(s) + 3HCl_(aq)

Then: 3HCl_(aq) → 3H⁺_(aq) + 3Cl^–_(aq).

With phosphorus (V) chloride:

PCl_5(s) + 4H₂O_(l) → H₃PO_4(s) + 5HCl_(aq)

Then: 5HCl_(aq) → 5H⁺_(aq) + 5Cl^–_(aq).

Trends in bond types and properties of chlorides of elements in period 3

Property	Period 3 chloride
Formula	NaCl	MgCl₂	Al₂Cl₆	SiCl₄	PCl₃ & PCl₅	SCl₂
Physical state at RT	Solid	Solid	Solid	Liquid	Liquid	Liquid
M.P (^oC)	801	714	Sublimes at 180^oC	-70	Sublimes at -94^oC	-78
B.P (^oC)	1467	1437	–	57	–	Decomposes at 57^oC
Conductivity	Good	Good	V. poor	nil	nil	Nil
Structure	Giant ionic	Giant ionic	Molecular dimer	Molecular	Molecular	Molecular
Bond type	Ionic	Ionic	Ionic / covalent	Covalent	Covalent	Covalent
Effect on water	Readily dissolves	Readily dissolves	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes	Hydrolyzed to give HCl fumes
pH of solution	7	6.5	3	2	2	2

Summary: Characteristics of bonds

Property	Substances with
Property	Covalent bonds	Ionic bonds	Metallic bonds
Electrical conductivity	Non-conductors except graphite	– Solids do not conduct. – Aqueous solutions and molten state conduct	– Conducts
Thermal conductivity	– Non-conductors except graphite	– Do not conduct	– Conducts
Melting point (oC)	– Low for molecular substances – High for giant atomic structures	– Usually high.	– Generally high
Boiling point (oC)	– Low for molecular substances – High for giant atomic structures	– Usually high	– Generally high
Solubility	– Generally insoluble in water but soluble in organic solvents	– Generally soluble in water	– Some metals react with water

UNIT 5: SALTS.

Meaning.
Types of salts:

Normal salts
Acid salts
Basic salts
Double salts.

Solubility of salts in water

Sulphates
Chlorides
Nitrates
Carbonates
Sodium, potassium and ammonium salts

Solubility of bases in water.
Obtaining crystals
Preparation of salts

Insoluble salt
- Double decomposition
Soluble salts

A sodium, potassium and ammonium salts

A non-potassium, sodium and ammonium salt.

That reacts with water

Direct synthesis

That does not react with water

Acid and metal method

Acid + base method

Neutralization

Uses of salts
Action of heat on salts

Carbonates
Nitrates

Teachers' Resources

FORM 3 CHEMISTRY NOTES

October 5, 2025 Hillary Kangwana Leave a comment

UNIT 2: NITROGEN AND ITS COMPOUNDS.

Unit checklist.

Introduction
Preparation of nitrogen

Isolation from air
Isolation from liquid air
Laboratory preparation
Preparation from ammonia
Properties of nitrogen
Oxides of nitrogen
- Nitrogen (I) oxide
- Nitrogen (II) oxide
- Nitrogen (IV) oxide

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Action of heat on nitrates.
Ammonia gas

Preparation
Laboratory preparation
Preparation from caustic soda
Test for ammonia
Fountain experiment
Properties and reactions of ammonia
Large scale manufacture of ammonia gas: the Haber process
Uses of ammonia

Nitric (V) acid

Laboratory preparation
Industrial manufacture of nitric (V) acid: The Otswalds process.
Reactions of dilute nitric acid
Reactions of concentrated nitric acid
Uses of nitric acid

Test for nitrates.
Pollution effects of nitrogen and its compounds
Reducing pollution environmental pollution by nitrogen compounds.

Introduction:

– About 78% of air is nitrogen, existing as N₂ molecules.

– The two atoms in the molecules are firmly held together.

– Nitrogen does not take part in many chemical reactions due to its low reactivity.

– Its presence in air dilutes oxygen and slows down respiration, burning and rusting.

Preparation of nitrogen.

(a). Isolation from air.

(i). Apparatus.

(ii). Procedure.

– Air is driven out of the aspirator by passing water into the aspirator from a tap.

– The air is the passed through a wash bottle containing concentrated potassium hydroxide solution.

Reason:

– To remove carbon (IV) oxide from air.

Equations:

2KOH_(aq) + CO_2(g) K₂CO_3(aq) + H₂O_(l)

Then

K₂CO_3(aq) + H₂O_(l) + CO_2(g) 2KHCO_3(aq)

Thus;

KOH_(aq) + CO_2(g) KHCO_3(aq)

– The carbon (IV) oxide-free air is then passed into a combustion tube with heated copper metal.

Reason:

– To remove oxygen from the air.

Note:

In this reaction the brown copper metal is oxidized to black copper (II) oxide.

Equation:

2Cu_(s) + O_2(g) 2CuO_(s)

Brown Black

Note:

– Alternatively oxygen can be removed by passing the carbon (IV) oxide-free air through pyrogallic acid.

– The remaining part of air is mainly nitrogen and is collected over water.

Note:

– Nitrogen obtained by this method contains noble gases like xenon, argon etc as impurities.

– Purer nitrogen may be obtained by heating ammonium nitrite.

Equation:

NH₄NO_3(s) ^Heat N_2(g) + 2H₂O_(g)

Summary.

(b). Removal from liquid air.

– Liquid air is primarily a mixture of nitrogen and oxygen with small amounts of noble gases.

– This method involves manufacture of liquid air and consequent fractional distillation.

The chemical process.

Step 1: removal of dust particles.

– Dust particles are first removed by either of the two processes:

Electrostatic precipitation

(i). Electrostatic precipitation:

– Air is passed through oppositely charged plates hence an electric field.

– Dust particles (charged) are consequently attracted to plates of opposite charges.

Diagram: electrostatic precipitation:

(ii). Filtration:

– The air is passed through a series of filters which traps dust particles as the air is forced through.

Step 2: removal of carbon (IV) oxide.

– The dust-free air is passed through a solution of potassium hydroxide; to remove carbon (IV) oxide.

Equations:

2KOH_(aq) + CO_2(g) K₂CO_3(aq) + H₂O_(l)

Then:

K₂CO_3(aq) + H₂O_(l) + CO_2(g) 2KHCO_3(aq)

(Excess)

– Alternatively, sodium hydroxide may be used in place of potassium hydroxide.

Step 3: Removal of water vapour.

– The dustless, carbon (IV) oxide-free air is next passed into a chamber with concentrated sulphuric acid or anhydrous calcium chloride in which water vapour is separated and removed.

Note:

To remove water vapour, air may be alternatively passed into a freezing chamber where it is condensed at -25^oC.

– The water vapour solidifies and is then absorbed by silica gel and separated out.

– Air is freed from carbon (IV) oxide, water vapour and dust particles (before compression) so as to prevent blockage of the pipes caused by solid materials at liquefaction temperatures i.e. carbon (IV) oxide and water vapour form solids which may block the collection pipes.

Step 4: Liquification of air.

– The air free from dust, carbon (IV) oxide and water vapour is then compressed at about 200 atmospheres, cooled and allowed to expand through fine jet.

– This sudden expansion causes further cooling and the gases eventually liquefy.

– The liquid is tapped off through a valve while gas which has escaped liquefaction returns to the compressor.

– Liquid air is a transparent pale blue liquid.

– This liquid is then fractionally distilled.

Step 5: Fractional distillation of liquid air.

– The boiling point of nitrogen is -196^oC (77K) and that of oxygen is -183^oC (90K).

– Consequently when liquid air is allowed to warm up, the nitrogen boils off first and the remaining liquid becomes richer in oxygen.

– The top of the fractionating column is a few degrees cooler than the bottom.

– Oxygen, the liquid with the higher boiling point (-183^oC) collects at the bottom as the liquid.

– The gas at the top of the column is nitrogen which ahs a lower boiling point.

– The more easily vapourised nitrogen is taken off.

– This way about 99.57% nitrogen is obtained.

Note:

– The separation of nitrogen and oxygen from air is a proof that air is a mixture and not a compound.

Summary: Fractional distillation of liquid air.

AIR

Step 1: Elimination of dust by Filtration

and electrostatic precipitation

Step 2: CO₂ removal, pass dust free air

through KOH or NaOH

Step 3: Removal of water vapour; through

condensation -25^oC) or conc. H₂SO₄

Recycling Step 4: Compression at approximately 200

atmospheres; cooling and expansion of air

Step 5: Fractional distillation

(c). Laboratory preparation method.

(i). Apparatus.

(ii). Procedure:

– Concentrated solutions of sodium nitrite and ammonium chloride are heated together in a round bottomed flask.

(iii). Observations.

– Colourless gas (nitrogen) is evolved rapidly and is collected over water.

(iv). Equation.

NaNO_2(aq) + NH₄Cl_(aq) ^heat NaCl_(aq) + N_2(g) + 2H₂O_(l).

Note: the resultant gas is less dense than that isolated from air.

Reason:

– It does not contain impurities.

(d). Preparation from ammonia.

(i). Apparatus.

(ii). Procedure:

– Dry ammonia gas is passed over a heated metal oxide e.g. copper metal.

– The metal oxide is reduced to the metal while ammonia gas is itself oxidized to nitrogen and water.

– Water is condensed and collected in a u-tube immersed in ice cubes.

– Nitrogen produced is collected over water.

(iii). Observations and explanations.

Copper (II) oxide:

3CuO_(s) + 2NH_3(g) 3Cu_(s) + N_2(g) + 3H₂O_(l)

(Black) (Brown) (Colourless)

Zinc (II) oxide

3ZnO_(s) + 2NH_3(g) 3Zn_(s) + N_2(g) + 3H₂O_(l)

(Yellow-hot) (Grey) (Colourless)

(White-cold)

Lead (II) oxide

3PbO_(s) + 2NH_3(g) 3Cu_(s) + N_2(g) + 3H₂O_(l)

(Red-hot) (Grey) (Colourless)

(Yellow-cold)

Properties of nitrogen.

(a). Physical properties.

It is a colourless, odourless and tasteless gas; almost completely insoluble in water.
Slightly lighter than air.

(b). Chemical properties.

It is inert (unreactive)

Reason:

– The inert nature of nitrogen is due to the strong covalent bonds between the two nitrogen atoms in the molecule; N₂.

Structurally;

– In air, it neither burns nor supports combustion and acts mainly as a diluent for the oxygen; slowing down the rate of burning.

Chemical test for nitrogen.

– A gas is proved to be nitrogen by elimination: –

It extinguishes a lighted splint and dos not burn; hence it is not oxygen, hydrogen or carbon (II) oxide.
It has neither smell nor colour; and therefore is not chlorine, ammonia, sulphur (IV) oxide or hydrogen chloride.
It does not form a white precipitate in lime water, and so it is not carbon (IV) oxide.
It is neutral to litmus and therefore cannot be carbon (IV) oxide, hydrogen sulphide, ammonia, hydrogen chloride

Reaction with hydrogen.

– Under special conditions (i.e. high pressure, low temperatures and presence of iron catalyst), nitrogen combines with hydrogen to produce ammonia.

Equation:

N_2(g) + 3H_2(g) 2NH_3(g)

– This reaction forms the basis of Haber process used in the manufacture of ammonia.

Reaction with burning magnesium.

(i). Apparatus.

(ii). Procedure:

– A piece of burning magnesium ribbon is introduced into a gas jar full of nitrogen.

(iii). Observations:

– The magnesium ribbon continues to burn and a white solid; magnesium nitride is formed.

Equation:

3Mg_(s) + N_2(g) ^Heat Mg₃N_2(s)

Note:

– When magnesium nitride is treated with water or a solution of sodium hydroxide; the characteristic pungent smell of ammonia can be detected.

Equations:

In water

Mg₃N_2(s) + 6H₂O_(l) 2NH_3(g) + 3Mg(OH)_2(aq)

In sodium hydroxide:

Mg₃N_2(s) + NaOH_(aq)

Reaction with oxygen.

– When nitrogen and oxygen in air are passed through an electric arc small quantities of nitrogen (II) oxide are formed.

Equation:

N_2(g) + O_2(g) 2NO_(g)

Note:

– Nitrogen reacts with oxygen under various conditions to give different types of nitrogen oxides.

Uses of nitrogen

Used in the Haber process in the manufacture of ammonia.
Due to its inert nature, it is mixed with argon to fill electric bulbs (to avoid soot formation).
In liquid state it is used as an inert refrigerant e.g. storage of semen for artificial insemination.
Due to its inert nature, it is used in food preservation particularly for canned products i.e. it prevents combination of oxygen and oil which tends to enhance rusting.
It is used in oil field operation called enhanced oil recovery where it helps to force oil from subterranean deposits.

Oxides of nitrogen.

– The three main oxides of nitrogen are:

Nitrogen (I) oxide, N₂O
Nitrogen (II) oxide, NO
Nitrogen (IV) oxide, NO₂

Nitrogen (I) oxide.

Preparation of nitrogen (I) oxide, N2O

(i). Apparatus.

(ii). Procedure:

– Ammonium nitrate is gently heated in a boiling tube and gas produced collected over warm water.

– Heating is stopped while excess ammonium nitrate still remains.

Reason:

– To avoid chances of an explosion.

(iii). Observations:

– The solid (ammonium nitrate) melts and gives off nitrogen (I) oxide which is collected over warm water.

Reasons:

– Nitrogen (I) oxide is slightly soluble in cold water.

(iv). Equation:

NH₄NO_3(s) ^Heat NO_2(g) + 2H₂O_(l)

Properties:

It is a colourless gas, denser than air, fairly soluble in cold water and neutral to litmus.
It supports combustion by oxidizing elements like sulphur, magnesium and phosphorus under strong heat.

Equations:

N₂O_(g) + Mg_(s) ^Heat MgO_(s) + N_2(g)

2N₂O_(g) + S_(s) ^Heat SO_2(g) + 2N_2(g)

2N₂O_(g) + C_(s) ^Heat CO_2(g) + 2N_2(g)

5N₂O_(g) + 2P_(s) ^Heat P₂O_5(g) + 5N_2(g)

Magnesium decomposes the gas and continues to burn in it.

Equation:

N₂O_(g) + Mg_(s) ^Heat MgO_(s) + N_2(g)

When exposed over red-hot finely divided copper it is reduced to nitrogen.

Equation:

N₂O_(g) + Cu_(s) ^Heat CuO_(s) + N_2(g)

Chemical test.

It relights a glowing splint.

Note:

It can be distinguished from oxygen by the following tests:

It has a sweet sickly smell; oxygen is odourless.
It will not give brown fumes (NO₂) with nitrogen (II) oxide; oxygen does.
It is fairly soluble in cold water; oxygen is insoluble.
It extinguishes feebly burning sulphur; oxygen does not.

Uses of nitrogen (I) oxide.

– It was formerly used in hospitals as an aesthetic for dental surgery but has since been discontinued due to availability of more efficient anaesthetics.

Note:

– Nitrogen (I) oxide is also called laughing gas; because patients regaining consciousness from its effects may laugh hysterically.

Nitrogen (II) oxide, NO.

Preparation:

(i). Apparatus.

(ii). Procedure:

– Action of heat on 50% concentrated nitric acid on copper turnings.

– Not any heat is required.

Equation:

3Cu_(s) + 8HNO_3(aq) 3Cu(NO₃)_2(aq) + 4H₂O_(l) + 2NO_(g)

(iii). Observations:

– An effervescence occurs in the flask; with brown fumes because the nitrogen (II) oxide produced reacts with oxygen of the air in the flask to form a brown gas, nitrogen (IV) oxide.

Equation:

2NO_(g) + O_2(g) 2NO_2(g)

Colourless Colourless Brown

– The brown fumes eventually disappear and the gas collected over water.

– The NO₂ fumes dissolve in the water in the trough, resulting into an acidic solution of nitrous acid.

– The residue in the flask is a green solution of copper (II) nitrate.

– Industrially, the gas is obtained when ammonia reacts with oxygen in the presence of platinum catalyst.

– This is the first stage in the production of nitric acid.

(v). Properties.

It is a colourless, insoluble and neutral to litmus. It is also slightly denser than air.
Readily combines with oxygen in air and forms brown fumes of nitrogen (IV) oxide.
Does not support combustion except in the case of strongly burning magnesium and phosphorus; which continues to burn in it, thereby reducing it i.e. it is an oxidizing agent.

Example:

2Mg_(s) + 2NO_(g) 2MgO_(s) + N_2(g)

4P_(s) + 10NO_(g) 2P₂O_5(s) + 5N_2(g)

When passed over red-hot finely divided copper, it is reduced to nitrogen gas.

Equation:

2Cu_(s) + 2NO_(g) 2CuO_(s) + N_2(g)

Reaction with iron (II) sulphate.

– When iron (II) sulphate solution (freshly prepared) is poured into a gas jar of nitrogen (II) oxide, a dark brown colouration of Nitroso-iron (II) sulphate is obtained.

Equation:

FeSO_4(aq) + NO_(g) FeSO4.NO_(aq)

Green solution Dark brown

(Nitroso-iron (II) sulphate/ nitrogen (II) oxide iron (II) sulphate complex)

It is also a reducing agent.

Equation:

Cl_2(g) + 2NO_(g) 2ClNO_(l)

Chloro nitrogen (II) oxide.

Reaction with hydrogen.

– When electrically sparked with hydrogen, NO is reduced to nitrogen.

Equation:

2H_2(g) + 2NO_(g) 2H₂O_(l) + N_2(g)

Chemical test:

– When exposed to air, nitrogen (II) oxide forms brown fumes of nitrogen (IV) oxide.

Uses of Nitrogen (II) oxide.

Note: –It is not easy to handle owing to its ease of oxidation.

It is an intermediate material in the manufacture of nitric acid

Nitrogen (IV) oxide.

Preparation:

(i). Apparatus.

(ii). Procedure:

– Action of conc. Nitric acid on copper metal.

Equation:

Cu_(s) + 4HNO_3(l) Cu(NO₃)_2(aq) + 2NO_2(g) + 2H₂O_(l)

Note:

– NO₂ is also prepared by the action of heat on nitrates of heavy metals like lead nitrate.

– NO₂ is given off together with oxygen when nitrates of heavy metals are heated.

– It is best prepared by heating lead (II) nitrate in a hard glass test tube.

Lead (II) nitrate is the most suitable because it crystallizes without water of crystallization (like other nitrates) which would interfere with preparation of nitrogen (IV) oxide that is soluble in water.

– The gas evolved passes into a U-tube immersed in an ice-salt mixture.

Apparatus:

Equation:

2Pb(NO₃)_2(s) 2PbO_(s) + 4NO_2(g) + O_2(g)

Observations:

– The heated white lead (II) nitrate crystals decompose and decrepitates (cracking sound) to yield red lead (II) oxide; which turns yellow on cooling.

– A colourless gas, oxygen is liberated, followed immediately by brown fumes of nitrogen (IV) oxide.

– Nitrogen (IV) oxide is condensed as a yellow liquid; dinitrogen tetroxide (N₂O₄); and is collected in the U-tube.

Note:

– At room temperature, nitrogen (IV) oxide consists of nitrogen (IV) oxide and dinitrogen tetroxide in equilibrium with each other.

Equation:
2NO_2(g) N₂O_4(g)

(Nitrogen (IV) oxide) (Dinitrogen tetroxide)

– The oxygen being liberated does not condense because it ahs a low boiling point of -183oC.

Properties of nitrogen (IV) oxide.

Red-brown gas with a pungent chocking smell
It is extremely poisonous.
It is acidic, hence turns moist litmus paper red.
When reacted with water, the brown fumes dissolve showing that it is readily soluble in water.

Equation:

2NO_2(g) + H₂O_(l) HNO_3(aq) + HNO_2(aq)

(Nitric (V) acid) (Nitrous (III) acid)When liquid nitrogen

– Like carbonic (IV) acid, nitrous (III) acid could not be isolated. It is easily oxidized to nitric (V) acid.

Equation:

2NHO_2(aq) + O_2(g) 2NHO_3(aq)

(Nitric (III) acid) (Nitrous (V) acid)

Reaction with magnesium.

– Nitrogen (IV) oxide does not support combustion.

– However burning magnesium continues to burn in it.

Reason:

– The high heat of combustion of burning magnesium decomposes the nitrogen (IV) oxide to nitrogen and oxygen; the oxygen then supports the burning of the magnesium.

Equation:

4MgO_(s) + 2NO_2(g) 4MgO_(s) + 2N_2(g)

Note:

– Generally nitrogen (IV) oxide oxidizes hot metals and non-metals to oxides and itself reduced to nitrogen gas.

Examples:

(i). Copper:

4Cu_(s) + 2NO_2(g) 4CuO_(s) + N_2(g)

(ii). Phosphorus:

8P_(s) + 10NO_2(g) 4P₂O_5(s) + 5N_2(g)

(iii). Sulphur:

2S_(s) + 2NO_2(g) 2SO_2(g) + N_2(g)

Note:

– NO₂ reacts with burning substances because the heat decomposes it to NO₂ and O₂.

Equation:

2NO_2(g) ^Heat 2NO_(g) + O_2(g)

– This is the oxidizing property of nitrogen (IV) oxide.

– The resultant oxygen supports the burning.

Effects of heat:

– On heating, nitrogen (IV) oxide dissociates to nitrogen (II) oxide and oxygen and will support a burning splint.

Equation:

2NO_2(g) ^Heat 2NO_(g) + O_2(g)

– When liquid nitrogen (IV) oxide or dinitrogen tetroxide is warmed, it produces a pale brown vapour.

– This is due to the reversible set of reactions:

Heat Heat

N₂O_4(l) 2NO_2(g) 2NO_(g) + O_2(g)

(Dinitrogen tetroxide) Cool (Nitrogen (IV) oxide) Cool (Nitrogen (II) oxide) (Oxygen)

Pale yellow Brown

Colourless

– Percentage of each in the equilibrium depends on temperature.

– At low temperatures, percentage of N₂O₄ is high and the mixture is pale yellow in colour.

– Percentage of nitrogen (IV) oxide increases with increase in temperature and the colour darkens till at 150^oC when the gas is entirely NO₂ and is almost black.

– Still at higher temperatures, nitrogen (IV) oxide dissociates into colourless gas (NO and O₂).

Reaction with alkalis.

– A solution of aqueous sodium hydroxide is added to a gas jar of nitrogen (IV) oxide and shaken.

Observation:

– The brown fumes disappear.

Explanation:

– Formation of sodium nitrate and sodium nitrite.

Equation:

2NaOH_(aq) + 2NO_2(g) 2NaNO_3(g) + NaNO_2(aq) + H₂O_(l)

Ionically:

2OH^–_(aq) + 2NO_2(g) NO₃^–_(aq) + NO₂^–_(aq) + H₂O_(l)

Conclusion:

Nitrogen (IV) oxide is an acidic gas because it can react with an alkali.

Uses of nitrogen (IV) oxide.

Mainly used in the manufacture of nitric (V) acid.

Summary on comparison between oxides of nitrogen.

	Nitrogen (I) oxide	Nitrogen (II) oxide	Nitrogen (IV) oxide
Colour	– Colourless gas – Sweet sickly smell	– Colourless; turns brown in air; – Odourless	– Red brown gas; – Choking pungent smell;
2. Solubility	– Fairly soluble in cold water; but less soluble in hot water;	– Almost insoluble in water	– Readily soluble in water to form nitric (V) acid and nitrous (III) acid;
3. Action on litmus	– Neutral to litmus	– Neutral to litmus	– Turns moist blue litmus paper red; i.e. acidic.
4. Combustion	– Supports combustion; relights a glowing splint;	– Does not support combustion;	– Does not support combustion.
5. Density	– Denser than air	– Slightly denser than air	– Denser than air;
6. Raw materials and conditions	– Ammonium nitrate and heat;	– Copper and 50% nitric acid;	– Copper metal and concentrated nitric acid;

Action of heat on nitrates.

– All nitrates except ammonium nitrate decompose on heating tom produce oxygen gas as one of the products.

– Nitrates can be categorized into 4 categories based on the products formed when they are heated.

– The ease with which nitrates decompose increases down the electrochemical series of metals.

Nitrates of metals higher in the electrochemical series like sodium and potassium decompose on heating to give the corresponding metal nitrite and oxygen.

Examples:

2NaNO_3(s) ^Heat 2NaNO_2(s) + O_2(g)

2KNO_3(s) ^Heat 2KNO_2(s) + O_2(g)

Nitrates of most other metals (heavy metals) that are average in the electrochemical series decompose on heating to give the metals oxide; nitrogen (IV) oxide and oxygen gas.

Example: action of heat on lead (II) nitrate.

(i). Apparatus:

(ii). Procedure:

– Solid white lead (II) nitrate crystals are strongly heated in a boiling (ignition) tube.

– Products are passed into a U- tube immerse in ice.

– Excess gases are channeled out to a fume chamber.

(iii). Observations:

– The white crystalline solid decrepitates.

– A colourless gas (oxygen) is liberated and immediately followed by a red brown fumes/ gas (nitrogen (IV) oxide).

– A pale yellow liquid (dinitrogen tetroxide) condenses in the U-tube in the ice cubes.

– This is due to condensation of nitrogen (IV) oxide.

– A residue which is red when hot and yellow on cooling remains in the boiling (ignition) tube

Equation:

2Pb(NO₃)_2(s) ^Heat 2PbO_(s) + 4NO_2(g) + O_2(g)

(White crystalline solid) (Red-hot Brown Fumes Colourless gas
yellow-cold)

Further examples:

2Ca(NO₃)_2(s) ^Heat 2CaO_(s) + 4NO_2(g) + O_2(g)

(White solid) (White solid) Brown Fumes Colourless gas

2Mg(NO₃)_2(s) ^Heat 2MgO_(s) + 4NO_2(g) + O_2(g)

(White solid) (White solid) Brown Fumes Colourless gas

2Zn(NO₃)_2(s) ^Heat 2ZnO_(s) + 4NO_2(g) + O_2(g)

(White solid) (Yellow-hot Brown Fumes Colourless gas
White-cold)

2Cu(NO₃)_2(s) ^Heat 2CuO_(s) + 4NO_2(g) + O_2(g)

(Blue solid) (Black solid) Brown Fumes Colourless gas

Note:

– Some nitrates are hydrated and when heated first give out their water of crystallization; and then proceed to as usual on further heating.

Examples:

Ca(NO₃)₂.4H₂O_(s) ^Heat Ca(NO₃)_2(s) + 4H₂O_(g)

(White solid) (White solid) Colourless gas

On further heating;

2Ca(NO₃)_2(s) ^Heat 2CaO_(s) + 4NO_2(g) + O_2(g)

(White solid) (White solid) Brown Fumes Colourless gas

Nitrates of metals lower in the reactivity series e.g. mercury and silver decompose on heating to give the metal, nitrogen (IV) oxide and oxygen.

Example:

Hg(NO₃)_2(s) ^Heat Hg_(s) + 2NO_2(g) + O_2(g)

(White solid) Brown Fumes Colourless gas

2AgNO_3(s) ^Heat 2Ag_(s) + 2NO_2(g) + O_2(g)

(White solid) Brown Fumes Colourless gas

Ammonium nitrate decomposes to nitrogen (I) oxide and water vapour.

Example:

NH₄NO_3(s) ^Heat N₂O_(g) + O_2(g)

Colourless gas Colourless gas
Note:

This reaction is potentially dangerous as ammonium nitrate explodes on strong heating.

Ammonia.

– Is a compound of nitrogen and hydrogen and is the most important hydride of nitrogen.

– It is formed when any ammonium salt is heated with an alkali whether in solid or solution form.

– It is a colourless gas with a pungent smell of urine.

– It is alkaline and turns moist red litmus paper to blue when introduced to it.

Laboratory preparation of ammonia.

(i). Reagents.

Base + ammonium salt NH_3(g) + H₂O_(l)

(ii). Apparatus.

(iii). Procedure.

– Ammonium chloride (NH₄Cl)/ sal-ammoniac is mixed with a little dry slaked lime i.e. Ca(OH)₂ and the mixture thoroughly ground in a mortar.

Reason:

– To increase surface area for the reactions.

– The mixture is then heated in a round-bottomed flask.

Note:

– A round-bottomed flask ensures uniform distribution of heat while heating the reagents.

– The flask should not be thin-walled.

Reason:

The pressure of ammonia gas liberated during heating may easily crack or break it.

– The flask is positioned slanting downwards.

Reason:

– So that as water condenses from the reaction, it does not run back to the hot parts of the flask and crack it.

– The mixture on heating produces ammonia, calcium chloride and water.

Equation:

Ca(OH)_2(s) + NH₄Cl_(s) CaCl_2(aq) + 2NH_3(g) + 2H₂O_(g)

(Slaked lime)

(iv). Drying:

– Ammonia is dried by passing it through a tower or U-tube filled with quicklime (calcium oxide) or pellets of caustic potash but not caustic soda which is deliquescent.

Note:

Ammonia cannot be dried with the usual drying agents; concentrated sulphuric acid and calcium chloride as it reacts with them.

With concentrated sulphuric acid.

2NH_3(g) + H₂SO_4(l) (NH₄)₂SO_4(aq)

With fused calcium chloride:

CaCl_2(aq) + 4NH_3(g) CaCl₂.4NH_3(s)

– i.e. ammonia reacts forming complex ammonium salt.

(v). Collection:

– Dry ammonia gas is collected by upward delivery.

Reasons:

– It is lighter than air.

– It is soluble in water.

Other methods of preparing ammonia.

(b). Ammonia from caustic soda (sodium hydroxide) or caustic potash (potassium hydroxide)

Note:

– The slaked lime is replaced by either of the above solutions.

– Thus the solid reactant is ammonium chloride and the liquid reactant is potassium hydroxide.

(i). Apparatus:

(ii). Procedure:

– The flask is not slanted. It is vertical and heated on a tripod stand and wire gauze.

Reason:

– No need of slanting since water produced is in liquid form and not gaseous. Thus there is no possibility of condensation of water on hotter parts.

Equations:

(i). With caustic soda:

NaOH_(aq) + NH₄Cl_(s) NaCl_(aq) + H₂O_(l) + NH_3(g)

Ionically;

Na⁺_(aq) + OH^–_(aq) + NH₄Cl_(s) Na⁺_(aq) + Cl^–_(aq) + H₂O_(l) + NH_3(g)

Hence; NH₄Cl_(s) + OH^–_(aq) Cl^–_(aq) + H₂O_(l) + NH_3(g)

(ii). With caustic potash:

KOH_(aq) + NH₄Cl_(s) KCl_(aq) + H₂O_(l) + NH_3(g)

Ionically;

K⁺_(aq) + OH^–_(aq) + NH₄Cl_(s) K⁺_(aq) + Cl^–_(aq) + H₂O_(l) + NH_3(g)

Hence; NH₄Cl_(s) + OH^–_(aq) Cl^–_(aq) + H₂O_(l) + NH_3(g)

Note:

Ammonium sulphate could be used in place of ammonium chloride in either case.

Equations:

(i). With caustic soda:

2NaOH_(aq) + (NH₄)₂SO_4(s) Na₂SO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

2Na⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) 2Na⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(g)

(ii). With caustic potash:

2KOH_(aq) + (NH₄)₂SO_4(s) K₂SO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

2K⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) 2K⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(g)

(iii). With calcium hydroxide:

Ca(OH)_2(aq) + (NH₄)₂SO_4(s) CaSO_4(aq) + 2H₂O_(l) + 2NH_3(g)

Ionically;

Ca²⁺_(aq) + 2OH^–_(aq) + (NH₄)₂SO_4(s) Ca²⁺_(aq) + SO₄^2-_(aq) + H₂O_(l) + NH_3(g)

Hence; (NH₄)₂SO_4(s) + 2OH^–_(aq) SO₄^2-_(aq) + 2H₂O_(l) + 2NH_3(g)

Note:

Reaction with calcium hydroxide however stops prematurely, almost as soon as the reaction starts.

Reason;

– Formation of insoluble calcium sulphate which coats the ammonium sulphate preventing further reaction.

Preparation of ammonium solution.

(i). Apparatus.

(ii). Procedure:

– The apparatus is altered as above.

– The drying tower is removed and the gas produced is directly passed into water by an inverted funnel.

Reasons for the inverted broad funnel.

– It increases the surface area for the dissolution of thereby preventing water from sucking back into the hot flask and hence prevents chances of an explosion.

(iii). Equation.

NH_3(g) + H₂O_(l) NH₄OH_(aq)

Note:

– The solution cannot be prepared by leading the gas directly to water by the delivery tube.

Reason:

– Ammonia gas is very soluble in water and so water would rush up the delivery tube and into the hot flask causing it to crack.

– The rim of the inverted funnel is just below the water surface.

Tests for ammonia.

It is a colourless gas with a pungent smell.
It is the only common gas that is alkaline as it turns moist red litmus paper blue.
When ammonia is brought into contact with hydrogen chloride gas, dense white fumes of ammonium chloride are formed.

Equation:

NH_3(g) + HCl_(g) NH₄Cl_(s)

Fountain experiment.

(i). Diagram:

(ii). Procedure:

– Dry ammonia is collected in a round-bottomed flask and set up as above.

– The clip is open and solution let to rise up the tube.

– The clip is closed when the solution reaches the top of the tube after which it is again opened fro a while.

(iii). Observations and explanations.

– When a drop of water gets to the jet, it dissolves so much of the ammonia gas that a partial vacuum is created inside the flask.

– As the ammonia in the flask dissolves, the pressure in the flask is greatly reduced.

– The atmospheric pressure on the water surface in the beaker forces water into the flask vigorously.

– The drawn-out jet of the tube causes a fountain to be produced.

– The fountain appears blue due to the alkaline nature of ammonia.

(iv). Caution:

– Ammonia is highly soluble in water forming an alkaline solution of ammonium hydroxide.

Note:

1 volume of water dissolves about 750 volumes of ammonia at room temperature.

Properties and reactions of ammonia.

Smell: has a characteristic pungent smell.
Solubility: it is highly soluble in water. The dissolved ammonia molecule reacts partially with water to form ammonium ions (NH₄⁺) and hydroxyl ions (OH^–)

Equation:

NH_3(g) + H₂O_(l) NH4⁺_(aq) + OH^–_(aq)

– Formation of hydroxyl ions means that the aqueous solution of ammonia is (weakly) alkaline and turns universal indicator purple.

Reaction with acids.

– Sulphuric acid and concentrated ammonia solution are put in a dish and heated slowly.

– The mixture is evaporated to dryness.

Observations:

– A white solid is formed.

Equation:

2NH₄OH_(aq) + H₂SO_4(aq) (NH₄)₂SO_4(aq) + H₂O_(l)

Ionically:

2NH₄⁺_(aq) + 2OH^–_(aq) + 2H⁺_(aq) + SO₄^2-_(aq) 2NH₄⁺_(aq) + SO₄^2-_(aq) + 2H⁺_(aq) + 2OH^–_(aq) + H₂O_(l).

Then;

2H⁺_(aq) + 2OH^–_(aq) 2H₂O_(l)

– To some of the resultant white solid, a little NaOH_(aq) was added and the mixture warmed.

– The gas evolved was tested fro ammonia.

Observation:

– The resultant gas tested positive for ammonia.

Equation:

(NH₄)₂SO_4(s) + 2NaOH_(aq) Na₂SO_4(aq) + 2NH_3(g) + 2H₂O_(l).

Explanations:

– Evolution of ammonia shows that the white solid formed is an ammonium salt.

– The ammonia reacts with acids to from ammonium salt and water only.

Further examples:
HCl_(aq)+ NH₄OH_(aq) NH₄Cl_(aq) + H₂O_(l)

HNO_3(aq) + NH₄OH_(aq) NH₄NO_3(aq)+ H₂O_(l)

Ionic equation:

NH_3(g) + H⁺_(aq) NH₄⁺_(aq)

Reaction of ammonia with oxygen.

– Ammonia extinguishes a lighted taper because it dos not support burning.

– It is non-combustible.

– However it burns in air enriched with oxygen with a green-yellow flame.

Experiment: Burning ammonia in oxygen.

(i). Apparatus.

(ii). Procedure:

– Dry oxygen is passed in the U-tube for a while to drive out air.

– Dry ammonia gas is then passed into the tube.

– A lighted splint is then passed into the tube.

(iii). Observations:

– A colourless gas is liberated.

– Droplets of a colourless liquid collect on cooler parts of the tube.

(iv). Explanations:

– The conditions for the reactions are:

Dry ammonia and oxygen gas i.e. the gases must be dry.
All air must be driven out of the tube.

– Ammonia burns continuously in oxygen (air enriched with oxygen) forming nitrogen and water vapour i.e. ammonia is oxidized as hydrogen is removed from it leaving nitrogen.

Equation:

4NH_3(g) + 3O_2(g) 2N_2(g) + 6H₂O_(g)

Sample question:

Suggest the role of glass wool in the tube.

Solution:

– To slow down the escape of oxygen in the combustion tube, thus providing more time for combustion of ammonia.

Ammonia as a reducing agent.

– It reduces oxides of metals below iron in the reactivity series.

Experiment: reaction between ammonia and copper (II) oxide.

(i). Apparatus.

Ice cubes

(ii). Procedure:

– Copper (II) oxide is heated strongly and dry ammonia is passed over it.

– The products are then passed through a U-tube immersed in cold water (ice cubes).

(iii). Observations.

– The copper (II) oxide glows as the reaction is exothermic.

– A colourless liquid collects in the U-tube.

– A colourless gas is collected over water.

– The black copper (II) oxide changes to brown copper metal.

(iv). Explanations.

– Ammonia gas reduces copper (II) oxide to copper and is itself oxidized to nitrogen and water.

Equation:

3CuO_(s) + 2NH_3(g) 3Cu_(s) + 3H₂O_(l) + N_2(g)

Black red-brown (colourless)

– The water produced condenses in the U-tube immersed in cold (ice) water.

– The resultant nitrogen is collected by downward displacement of water.

– The nitrogen gas collected is ascertained indirectly as follows:

A lighted splint is extinguished and the gas does not burn; thus it is not oxygen, hydrogen, or carbon (II) oxide.
It has neither smell nor colour; it is not ammonia, chlorine, sulphur (IV) oxide or nitrogen (IV) oxide.
It is not carbon (II) oxide because it does not turn lime water into a white precipitate.

Note:

– This experiment proves that ammonia contains nitrogen.

Reaction with chlorine.

(i). Procedure:

– Ammonia gas is passed into a bell jar containing chlorine.

(ii). Apparatus:

(iii). Observations:

– The ammonia catches fire and burns for a while at the end of the tube.

– The flame then goes out and the jar then gets filled with dense white fumes of ammonium chloride.

Equations:

2NH_3(g) + 3Cl_2(g) 6HCl_(g) + N_2(g)

Then;

6HCl_(g) + 6NH_3(g) 6NH₄Cl_(s)

Overall equation:

8NH_3(g)+ 3Cl_2(g) 6NH₄Cl_(s) + N_2(g)

Ammonia solution as an alkali.

– Solution of ammonia in water contains hydroxyl ions.

Equation:

NH_3(g) + H₂O_(l) NH₄⁺_(aq) + OH^–_(aq)

– Thus it has many properties of a typical alkali.

– Ammonia salts are similar to metallic salts.

– The group (NH₄⁺) precipitates in the reaction as a whole without splitting in any way.

– It exhibits unit valency in its compounds and therefore called a basic radical.

Note:

– It cannot exist freely as ammonia gas (NH₃) which is a compound.

– Like other alkalis, ammonia solution precipitates insoluble metallic hydroxides by double decomposition when mixed with solution of salts of the metals.

Reaction with air in the presence of platinum wire.

(i). Apparatus:

(ii). Procedure:

– Concentrated ammonia solution is put in a conical flask.

– The platinum (or even copper) wire is heated until white-hot.

– Oxygen gas or air is then passed through the ammonia solution.

– The red-hot platinum (copper) wire is then put into the flask containing the concentrated ammonia.

(iii). Observations:

– The hot platinum wire glows.

– Red-brown fumes are evolved.

(iv). Explanations:

– The hot platinum coil glows when it comes into contact with the ammonia fumes, which come from the concentrated ammonia solution.

– Reaction between ammonia and oxygen takes place on the surface of the platinum wire that acts a s a catalyst.

– A lot of heat is produced in the reaction that enables the platinum coil to continue glowing.

– Ammonia is oxidized to nitrogen (IV) oxide.

Equation:

4NH_3(g) + 5O_2(g)^{Platinum catalyst} 4NO_(g) + 6H₂O_(l)

– Red-brown fumes of nitrogen (IV) oxide are produced due to further oxidation of the nitrogen (II) oxide to from nitrogen (IV) oxide.

Equation:

2NO_(g) + O_2(g) 2NO_2(g)

Action of aqueous ammonia on solution of metallic salts

(i). Procedure:

– To about 2cm³ of solutions containing ions of calcium, magnesium, aluminium, zinc, iron, lead, copper etc in separate test tubes; aqueous ammonia is added dropwise till in excess.

(ii). Observations:

The various metal ions reacted as summarized in the table below.

Metal ions in solution	Observations on addition of ammonia
Metal ions in solution	Few drops of ammonia	Excess drops of ammonia
Ca²⁺	White precipitate	White precipitate persists;
Mg²⁺	White precipitate	Precipitate persists;
Al³⁺	White precipitate	Precipitate persists;
Zn²⁺	White precipitate	Precipitate dissolves;
Fe²⁺	Pale green precipitate	Precipitate persists; slowly turns red-brown on exposure to air;
Fe³⁺	Red-brown precipitate	Precipitate persists;
Pb²⁺	White precipitate	Precipitate persists;
Cu²⁺	Pale blue precipitate	Precipitate dissolves forming a deep blue solution;

(iii). Explanations:

– Most metal ions in solution react with ammonia solution to form insoluble metal hydroxides.

– In excess ammonia, some of the so formed hydroxides dissolve forming complex ions.

(iv). Equations:

Mg²⁺_(aq) from MgCl₂;

MgCl_2(aq) + 2NH₄OH_(aq) Mg(OH)_2(s) + 2NH₄Cl_(aq)

Ionically:

Mg²⁺_(aq) + 2OH^–_(aq) Mg(OH)_2(s)

(White ppt)

Fe²⁺ from Fe(NO₃)₂;

Fe(NO₃)_2(aq) + 2NH₄OH_(aq) Fe(OH)_2(s) + 2NH₄NO_3(aq)

Ionically:

Fe²⁺_(aq) + 2OH^–_(aq) Fe(OH)_2(s)

(Pale green ppt)

Fe³⁺ from FeCl₃;

Ionically:

Fe³⁺_(aq) + 3OH^–_(aq) Fe(OH)_3(s)

(Red brown ppt)

Note:

Zn²⁺_(aq)and Cu²⁺_(aq) dissolve in excess ammonia solution forming complex ions.

Zinc ions and ammonia solution.

With little ammonia:

ZnCl_2(aq) + 2NH₄OH_(aq) Zn(OH)_2(s) + 2NH₄Cl_(aq)

Ionically:

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

(White ppt.)

In excess ammonia:

– The white precipitate of Zn(OH)_2(s) dissolves in excess ammonia to form a colourless solution; proof that solution has Zn²⁺ions;

– The colourless solution is a complex salt of tetra-amine zinc (II) ions.

Equation:

Zn(OH)_2(s) + 4NH_3(aq) [Zn(NH₃)₄]²⁺_(aq) + 2OH^–_(aq)

(White ppt.) (Colourless solution-tetra amine zinc (II) ions)

Copper (II) ions.

With little ammonia:

– A pale blue precipitate is formed.

Ionically:

Cu²⁺_(aq) + 2OH^–_(aq) Cu(OH)_2(s)

(Pale blue ppt.)

In excess ammonia:

– The pale blue precipitate of Cu(OH)_2(s) dissolves in excess ammonia to form a deep blue solution; proof that solution has Cu²⁺ions;

– The deep blue solution is a complex salt of tetra-amine copper (II) ions.

Equation:

Cu(OH)_2(s) + 4NH_3(aq) [Cu(NH₃)₄]²⁺_(aq) + 2OH^–_(aq)

(Pale blue ppt.) (Deep blue solution-tetra amine copper (II) ions)

Uses of ammonia gas and its solution:

Ammonia gas is used in the manufacture of nitric acid and nylon.
Ammonia gas is important in the preparation of ammonium salts used as fertilizers.
It liquefies fairly easily (B.P is -33^oC) and the liquid is used as a refrigerant in large cold storages and ice cream factories.
Liquid ammonia is injected directly into the soil as a high nitrogen content fertilizer.
Ammonia solution is used in laundry work as a water softener and a cleansing agent (stain remover)
Ammonia is used in the manufacture of sodium carbonate in the Solvay process.
Ammonia is used in smelling salts. It has a slightly stimulating effect on the action of the heart and so may prevent fainting

Qualitative analysis for cations using sodium hydroxide solution

(i). Procedure:

– To about 2cm³ of solutions containing ions of calcium, magnesium, aluminium, zinc, iron, lead, copper etc in separate test tubes; aqueous sodium hydroxide is added dropwise till in excess.

(ii). Observations:

The various metal ions reacted as summarized in the table below.

Metal ions in solution	Observations on addition of ammonia
Metal ions in solution	Few drops of ammonia	Excess drops of ammonia
Ca²⁺	White precipitate	White precipitate persists
Mg²⁺	White precipitate	Precipitate persists;
Al³⁺	White precipitate	Precipitate dissolves;
Zn²⁺	White precipitate	Precipitate dissolves;
Fe²⁺	Pale green precipitate	Precipitate persists; slowly turns red-brown on exposure to air;
Fe³⁺	Red-brown precipitate	Precipitate persists;
Pb²⁺	White precipitate	Precipitate dissolves;
Cu²⁺	Pale blue precipitate	Precipitate dissolves forming a deep blue solution;

(iii). Explanations:

– Most metal ions in solution react with sodium hydroxide solution to form insoluble metal hydroxides.

– In excess sodium hydroxide, some of the so formed hydroxides (hydroxides of Zn, Al, Pb and Cu) dissolve forming complex ions.

(iv). Equations:

Ca²⁺_(aq) + 2OH^–_(aq) Ca(OH)_2(s)

(White ppt)

Mg²⁺_(aq) + 2OH^–_(aq) Mg(OH)_2(s)

(White ppt)

Al³⁺_(aq) + 3OH^–_(aq) Al(OH)_3(s)

(White ppt)

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

(White ppt)

Pb²⁺_(aq) + 2OH^–_(aq) Pb(OH)_2(s)

(White ppt)

Cu²⁺_(aq) + 2OH^–_(aq) Cu(OH)_2(s)

(Pale blue ppt)

Fe²⁺_(aq) + 2OH^–_(aq) Fe(OH)_2(s)

(Pale green ppt)

Fe³⁺_(aq) + 3OH^–_(aq) Fe(OH)_3(s)

(Red brown ppt)

Note:

Hydroxides of Zn²⁺_(aq); Pb²⁺_(aq); and Al³⁺_(aq) dissolve in excess ammonia solution forming complex ions.

Zinc ions and sodium hydroxide solution.

With little sodium hydroxide:

Zn²⁺_(aq) + 2OH^–_(aq) Zn(OH)_2(s)

(White ppt.)

In excess sodium hydroxide:

– The white precipitate of Zn(OH)_2(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo zinc (II) ions (zincate ion).

Equation:

Zn(OH)_2(s) + 2OH^–_(aq) [Zn(OH)₄]^2-_(aq)

(White ppt.) (Colourless solution-tetra hydroxo- zinc (II) ion/ zincate ion)

Aluminium ions and sodium hydroxide solution.

With little sodium hydroxide:

Al³⁺_(aq) + 3OH^–_(aq) Al(OH)_3(s)

(White ppt.)

In excess sodium hydroxide:

– The white precipitate of Al(OH)_3(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo aluminium (III) ions (aluminate ion).

Equation:

Al(OH)_3(s) + OH^–_(aq) [Al(OH)₄]^–_(aq)

(White ppt.) (Colourless solution-tetra hydroxo- aluminium (III) ion/aluminate ion

Lead (II) ions and sodium hydroxide solution.

With little sodium hydroxide:

Pb²⁺_(aq) + 2OH^–_(aq) Pb(OH)_2(s)

(White ppt.)

In excess sodium hydroxide:

– The white precipitate of Pb(OH)_2(s) dissolves in excess sodium hydroxide to form a colourless solution;

– The colourless solution is a complex salt of tetra-hydroxo lead (II) ions (plumbate ions).

Equation:

Zn(OH)_2(s) + 2OH^–_(aq) [Zn(OH)₄]^2-_(aq)

(White ppt.) (Colourless solution-tetra hydroxo- lead (II) ion/ plumbate ion)

Summary and useful information on qualitative analysis:

Colours of substances in solids and solutions in water.

COLOUR		IDENTITY
SOLID	AQUESOUS SOLUTION (IF SOLUBLE)	IDENTITY
1. White	Colourless	Compound of K⁺; Na⁺, Ca²⁺; Mg²⁺; Al³⁺; Zn²⁺; Pb²⁺; NH₄⁺
2. Yellow	Insoluble	Zinc oxide, ZnO (turns white on cooling); Lead oxide, PbO (remains yellow on cooling, red when hot)
2. Yellow	Yellow	Potassium or sodium chromate;
3. Blue	Blue	Copper (II) compound, Cu²⁺
4. Pale green Green	Pale green (almost colourless) Green	Iron (II) compounds,Fe²⁺ Nickel (II) compound, Ni²⁺; Chromium (II) compounds, Cr³⁺; (Sometimes copper (II) compound, Cu²⁺)
5. Brown	Brown (sometimes yellow) Insoluble	Iron (III) compounds, Fe³⁺; Lead (IV) oxide, PbO₂
6. Pink	Pink (almost colourless) Insoluble	Manganese (II) compounds, Mn²⁺; Copper metal as element (sometimes brown but will turn black on heating in air)
7. Orange	Insoluble	Red lead, Pb₃O₄ (could also be mercury (II) oxide, HgO)
8. Black	Purple Brown Insoluble	Manganate (VII) ions (MnO^–) as in KMnO₄; Iodine (element)-purple vapour Manganese (IV) oxide, MnO₂ Copper (II) oxide, CuO Carbon powder (element) Various metal powders (elements)

Reactions of cations with common laboratory reagents and solubilities of some salts in water

CATION	SOLUBLE COMPOUNDS (IN WATER)	INSUOLUBLE COMPOUNDS (IN WATER)	REACTION WITH AQUEOUS SODIUM HYDROXIDE	REACTION WITH AQUEOUS AMMONIA SOLUTION
Na⁺	All	None	No reaction	No reaction
K⁺	All	None	No reaction	No reaction
Ca²⁺	Cl^–; NO₃^–;	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate insoluble in excess	White precipitate insoluble in excess, on standing;
Al³⁺	Cl^–; NO₃^–; SO₄^2-	O^2-; OH^–;	White precipitate soluble in excess	White precipitate insoluble in excess
Pb²⁺	NO₃^–; ethanoate;	All others;	White precipitate soluble in excess	White precipitate insoluble in excess
Zn²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; SO₄^2-; OH^–;	White precipitate soluble in excess	White precipitate soluble in excess
Fe²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Dark) pale green precipitate insoluble in excess	(Dark) pale green precipitate insoluble in excess
Fe³⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	(Red) brown precipitate insoluble in excess	(Red) brown precipitate insoluble in excess
Cu²⁺	Cl^–; NO₃^–; SO₄^2-	CO₃^2-; O^2-; OH^–;	Pale blue precipitate insoluble in excess	Pale blue precipitate soluble in excess forming a deep blue solution
NH₄⁺	All	None;	Ammonias gas on warming	Not applicable.

Qualitative analysis for common anions.

	SO₄^2-_(aq)	Cl^–_(aq)	NO₃^–_(aq)	CO₃^2-_(aq)
TEST	Add Ba²⁺_(aq) ions from Ba(NO₃)_2(aq); acidify with dilute HNO_3(aq)	Add Ag⁺_(aq) from AgNO_3(aq). Acidify with dilute HNO₃ Alternatively; Add Pb²⁺ from Pb(NO₃)₂ and warm	Add FeSO_4(aq); Tilt the tube and carefully add 1-2 cm³ of concentrated H₂SO_4(aq)	Add dilute HNO_3(aq); bubble gas through lime water;
OBSERVATION	The formation of a white precipitate shows presence of SO₄^2- ion;	The formation of a white precipitate shows presence of Cl^– ion; Formation of a white precipitate that dissolves on warming shown presence of Cl^–_(aq) ions	The formation of a brown ring shows the presence of NO₃^– ions	Evolution of a colourless gas that forma a white precipitate with lime water, turns moist blue litmus paper red; and extinguishes a glowing splint shows presence of CO₃^2- ions
EXPLANATION	Only BaSO₄ and BaCO₃ can be formed as white precipitates. BaCO₃ is soluble in dilute acids and so BaSO₄ will remain on adding dilute nitric acid	Only AgCl and AgCO₃ can be formed as white precipitates. AgCO₃ is soluble in dilute acids but AgCl is not; – PbCl₂ is the only white precipitate that dissolves on warming	Concentrated H₂SO₄ forms nitrogen (II) oxide with NO₃^–_(aq) and this forms brown ring complex (FeSO₄.NO) with FeSO₄;	All CO₃^2- or HCO₃^– will liberate carbon (IV) oxide with dilute acids

Checklist:

Why is it not possible to use dilute sulphuric acid in the test for SO₄^2- ions;
Why is it not possible to use dilute hydrochloric acid in the test for chloride ions?
Why is it best to use dilute nitric acid instead of the other two mineral acids in the test for CO₃^2-ions?
How would you distinguish two white solids, Na₂CO₃ and NaHCO₃?

What to look for when a substance is heated.

1. Sublimation	White solids on cool, parts of a test tube indicates NH₄⁺ compounds; Purple vapour condensing to black solid indicates iodine crystals;
2. Water vapour (condensed)	Colourless droplets on cool parts of the test tube indicate water of crystallization or HCO₃^– (see below)
3. Carbon (IV) oxide	CO₃^2- of Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺; Cu²⁺;
4. Carbon (IV) oxide and water vapour (condensed)	HCO₃^–
5. Nitrogen (IV) oxide	NO₃^–of Cu²⁺; Al³⁺; Zn²⁺; Pb²⁺; Fe²⁺; Fe³⁺
6. Oxygen	NO₃^– or BaO₂; MnO₂; PbO₂;

Industrial manufacture of ammonia-The Haber process.

– Most of the worlds supply of ammonia is from the synthesis of Nitrogen and hydrogen in the Haber process.

(i). Raw materials

Nitrogen

– Usually obtained from liquid air by fractional distillation

Hydrogen

– Obtained from water gas in the Bosch process.

– Also from crude oil (cracking)

(ii). General equation

N_2(g) + 3 H_2(g) 2NH_3(g) + heat;

Note:

– Nitrogen and hydrogen combine in the ratio 1:3 respectively to form two volumes of ammonia gas plus heat.

-The reaction is exothermic releasing heat to the surrounding.

(iii). Conditions

High pressures

– The process is favoured by high pressures and thus a pressure of approximately 200 to 300 atmospheres is used.

Reason:

– The volume of gaseous reactants from equation is higher than volume of gaseous products. Thus increased pressure shifts the equilibrium to the right; favoring the production of more ammonia.

Note:

Such high pressures are however uneconomical.

Low temperatures

– Low temperatures favour production of ammonia;

Reason:

– The reaction is exothermic (releases heat to the surrounding) hence lower temperature will favour the forward reaction (shift the equilibrium to the right), producing more ammonia.

Catalyst

– The low temperatures make the reaction slow and therefore a catalyst is used to increase the rate of reaction

– The catalyst used is finely divided iron; impregnated with Aluminium oxide (Al₂O₃)

(iv). The chemical processes

Step 1: Purification

-The raw materials, nitrogen and hydrogen are passed through a purification chamber in which impurities are removed.

-The main impurities are CO₂, water vapour, dust particles, SO₂, CO₂ and O₂;

Reason:

The impurities would poison the catalyst

Step 2: Compression

– The purified Nitrogen and Hydrogen gases are compressed in a compressor at 500 atmospheres.

Reasons:

To increase chances of molecules reacting;
To increase rate of collision of the reacting particles.
To increase pressure (attain desired pressures); and hence increase concentration of reacting particles.

Step 3: Heat exchanger reactions

– Upon compression the gaseous mixture, nitrogen and hydrogen are channeled into a heat exchanger; which heats them increasing their temperature.

– This enables the reactants (hydrogen and nitrogen) to attain the optimum temperatures for the succeeding reactions (in the catalytic chamber)

– From the heat exchanger the gases go to the catalyst chamber.

Step 4: Catalytic chamber.

– The gases then combine in the ratio of 1:3 (N₂:H₂respectively), to form ammonia.

– This reaction occurs in presence of a catalyst; which speeds up the rate of ammonia formation;

– The catalyst is finely divided iron impregnated with aluminium oxide (Al₂O₃ increases the catalytic activity of iron).

Equation in catalytic chamber

N_2(g) + 3H_2(g) 2NH_3(g) + Heat (-92kjmol^–)

– Only about 6-10% of the gases combine.

– Due to the high heat evolution involved, the products are again taken back to the heat exchanger; to cool the gases coming from the catalytic chamber.

Step 5: Heat exchanger

– The gases from the catalytic chamber enter the heat exchanger a second time.

Reason:

– To cool the gases coming from the catalytic chamber, thus reduce cost of condensation.

-The gaseous mixture; ammonia and uncombined nitrogen and hydrogen are the passed through a condenser.

Step 6: The condenser reactions (cooling chamber)

– The pressure and the low temperatures in this chamber liquefy ammonia, which is then drawn off.

– The uncombined (unreacted) gases are recirculated back to the compressor, from where they repeat the entire process.

Summary: flow chart of Haber process.

Fractional distillation of air

Nitrogen

Hydrogen

Crude oil cracking; or water gas in Bosch process

Purifier: removal of duct particles; CO₂; H₂O vapour etc

Unreacted gases

(recycling)

6-10% ammonia + air;

LIQUID AMMONIA

Citing a Haber process plant

– When choosing a site for this industrial plant, the following factors are considered:

Availability of raw materials (natural gas and crude oil)
Presence of cheap sources of energy.
Availability of transport and marketing.
Availability of appropriate technology and labour force.

Ammonium salts as fertilizers

– Ammonium salts are prepared by the reaction between ammonia and the appropriate acid in dilute solution followed by evaporation and crystallization

(a). Ammonium sulphate

– Is prepared by absorbing ammonia in sulphuric acid.

Equation:

2NH_3(g) + H₂SO_4(aq) (NH₄)₂SO_4(aq)

Note: It is a cheap fertilizer.

(b). Ammonium nitrate

– Is prepared by neutralization nitric acid by ammonia.

Equation:

NH_3(g) + HNO_3(aq) NH₄NO_3(aq)

– As there is some danger of exploding during storage, ammonium nitrate is mixed with finely powdered limestone (CaCO₃).

-The mixture, sold as nitro-chalk is much safer.

(c). Ammonium phosphate

– It is particularly useful as it supplies both nitrogen and phosphorus to the soil.

– It is prepared by neutralizing othophosphoric acid by ammonia

Equation:

3NH_3(g) + H₃PO_4(aq) (NH₄)₃ PO_4(aq)

(d) Urea CO (NH₂)₂

– Is made from ammonia and carbon (IV) oxide

– Its nitrogen content by mass is very high; nearly 47%

Equation:

NH_3(g) +CO_2(g) CO (NH₂)_2(aq)+ H₂O_(l)

Nitric (V) acid

– Is a monobasic acid (has only one replaceable Hydrogen atom); and has been known as strong water (aqua forty).

– It is a compound of hydrogen, oxygen and nitrogen.

Laboratory preparation of nitric (V) acid

(i). Apparatus

(ii). Reagents

– Nitric acid is prepared in the laboratory by action of concentrated sulphuric acid on solid nitrates e.g. potassium nitrate (KNO₃) and sodium nitrate (NaNO₃)

(iii). Procedure

– 30-40 grams of small crystal of KNO₃ are put in a retort flask.

– Concentrated sulphuric acid is added just enough to cover the nitrate; and then heated (warmed) gently.

– The apparatus is all glass.

Reason:

– Nitric (V) acid would attack rubber connections.

– The neck of the retort flask is inserted into a flask that is kept cool continually under running water; this is where nitric acid is collected.

Note:

The cold water running over the collection flask is meant to cool (condense) the hot fumes of nitric (V) acid.

(iv). Observations and explanations

– Fumes of nitric are observed in the retort;

Equation

KNO_3(g) + H₂SO_4(aq) KHSO_4(aq) +HNO_3(g)

– If Lead (II) nitrate was used;

Pb(NO₃)_2(s) + H₂SO_4(aq) PbSO_4(s) + 2HNO_3(g)

Note: with lead (II) nitrate the reaction soon stops because the insoluble lead (II) sulphate coats the surface of the nitrate preventing further reaction; yield of nitric (V) acid is thus lower;

-These fumes of nitric acid appear brown.

Reason:

– Due to the presence of nitrogen (iv) oxide gas formed by thermal decomposition of nitric acid.

Equation:

4HNO_3(aq) 4NO_2(g) + O_2(g) + 2H₂O_(g)

– Pure nitric (V) acid is colourless but may appear yellow (brown) due to the presence of Nitrogen (IV) oxide.

– The brown colour can be removed by blowing air through the acid.

– Fuming nitric acid boils at 83^oC and is 99% pure; while concentrated nitric acid is only 70% acid and 30% water.

Note: Nitric acid is usually stored in dark bottles.

Reason:

– To avoid its decomposition by light to nitrogen (IV) oxide, oxygen and water.

– The reaction in the retort flask is a typical displacement reaction; in which the more volatile nitric (V) acid is displaced from nitrates by the less volatile sulphuric acid.

– The nitric acid distills over because it is more volatile than sulphuric acid.

Properties of concentrated nitric acid

– Nitric (V) acid readily gives oxygen and therefore is called an oxidizer.

– The acid is usually reduced to nitrogen (IV) oxide and water.

Effects of heat on concentrated nitric acid

(i) Apparatus

(ii) Observations

– Brown fumes are seen in the hard glass tube.

– Colourless gas is collected over water.

(ii). Explanations

– Sand soaked in concentrated nitric acid produces nitric solid vapour on heating.

– The hot glass wool catalyzes the decomposition of nitric acid to nitrogen (IV) oxide (brown fumes), water vapour and oxygen.

Equation

4HNO_3(l) 4NO_2(g) + 2H₂O_(l) + O_2(g)

(Brown fumes)

– The so formed nitrogen (IV) oxide dissolves in water forming both nitric and nitrous acids.

Equation:

2NO_2(g) + H₂O_(l) HNO_2(aq) + HNO_3(aq)

– The oxygen gas is collected over water; and with the solution becoming acidic.

Reaction with saw dust

– Saw dust contains compounds of carbon Hydrogen and oxygen.

Procedure:

– Some saw dust is heated in an evaporating dish and some few drops of concentrated nitric (V) acid on it (this is done in a fume cupboard)

Observation:

– A violent reaction occurs, the saw dust catches fire easily and a lot of brown fumes of nitrogen (IV) oxide given off.

– Nitric (V) acid oxidizes the compounds in saw dust to CO₂ and water; and itself it is reduced to nitrogen (IV) oxide and water.

Equation:

(C, H, O)_n(s) + HNO_3(l) NO_2(g) + CO_2(g) +H₂O_(g)

Saw dust

– Warm concentrated nitric (V) acid oxidizes pure carbon and many other compounds containing carbon.

Equation:

C_(s) + 4HNO_3(l) 2H₂O_(l) + 4NO_2(g) + CO_2(g)

Reaction with sulphur

Procedure:

– 2 cm³of concentrated nitric (V) acid is added to a little sulphur in a test tube and warmed.

– The mixture is filtered to remove excess sulphur and the filtrate diluted with distilled water.

– Drops of barium chloride are then added to the resultant solution.

Observations:

– Red brown gas, nitrogen (IV) oxide (NO₂) is evolved and the sulphur is oxidized to sulphuric acid.

Equation

S_(s) + 6HNO_3(l) H₂SO_4(aq) + 6NO_2(g) +2H₂O_(l)

– On addition of barium chloride to the solution, a white precipitate is formed.

– This is due to formation of barium sulphate and is a confirmation for the presence of SO₄²– ions.

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

(White precipitate)

Reaction with metals

– Concentrated nitric (V) acid reacts with metals except gold and platinum.

– Actual reaction depends on the concentration of the acid and the position of the metal in the reactivity series.

– The reaction results in a metal nitrate, NO₂and water.

– Copper, which is low in the reactivity series, reduces conc. HNO₃ to NO₂.

Equation:

C_u(s) + HNO_3(l) C_u(NO₃)_2(aq) + 2NO_2(g) + 2H₂O_(l)

– Metals more reactive than copper e.g. Magnesium may reduce nitric acid to dinitrogen monoxide (N₂O) or Nitrogen (N₂).

– Some metals like iron and aluminium form insoluble layers when reacted with nitric acid thus stopping any further reaction.

Reaction with iron (II) salts

Procedure:

– Few crystals of iron (II) sulphate are dissolved in dilute sulphuric acid.

– A little concentrated nitric (V) acid is added to the solution and mixture warmed.

Observation:

– Green solution turns brown.

Equation:

6FeSO_4(s) + 3H₂SO_4(aq) +3HNO_3(l) 4H₂O_(l) +2NO_(g) + 3Fe₂ (SO₄)_3(aq)

Explanation:

– Nitric acid oxidizes green iron (II) salts (Fe²⁺) to brown iron (III) salts (Fe³⁺) and itself is reduced to Nitrogen (II) Oxide.

Note:

– In air, nitrogen (II) oxide is readily oxidized to nitrogen (IV) oxide; resulting to brown fumes.

Equation:
2NO_(g) + O_2(g) 2NO_2(g)

Reduction of nitric (V) acid by hydrogen sulphide.

Procedure

– A few drops of conc. nitric (V) acid are added to a gas jar full of hydrogen sulphide and the jar then covered.

Observations

– Fumes (brown) of Nitrogen (IV) oxide and yellow deposits of sulphur;

Equation

– It is a REDOX reaction.

Oxidation

H₂S_(g) + 2HNO_3(l) 2H₂O_(l) + 2NO_2(g) +S_(s)

Reduction

Properties of dilute nitric (V) acid

Reaction with metals

– Dilute nitric (V) acid reacts with most metals to form nitrogen (II) oxide instead of hydrogen.

Example:

3Mg_(s) + 8HNO_3(aq) 3Mg(NO₃)_2(aq) +2NO_(g) + 4H₂O_(l)

– In fact HNO₃is reduced to NO and water but NO soon gets oxidized in air to form brown fumes of NO₂.

– However very dilute HNO₃(cold) reacts with more active metals such as Magnesium to produce Hydrogen.

Dilute nitric (V) acid as a typical acid

(a). It turns blue litmus paper red.

(b). It reacts with metal oxides and metal hydroxides to form a metal nitrate and water (Neutralization)

Examples

CuO_(s) + 2HNO_3(aq) Cu (NO₃)_2(aq)+ H₂O_(l)

(Black) (Blue)

Zn(OH)_2(s) + 2HNO_3(aq) Zn (NO₃)_2(aq) + 2H₂O_(l)

(White ppt) (Colourless)

KOH_(aq) + HNO_3(aq) KNO_3(aq) + H₂O_(l)

(Alkali) (Acid) (Salt) (Water)

Reaction with metal carbonates and hydrogen carbonates

– Dilute HNO₃reacts with metal carbonates and hydrogen carbonates to form a nitrate, CO₂and water.

Examples.

CuCO₃(s) + 2HNO_3(aq) Cu(NO₃)_2(aq) + CO_2(g) + H₂O_(l)

(Green) (Blue solution)

NaHCO_3(s) + HNO_3(aq) NaNO_3(aq) + CO_2(g) + H₂O_(l)

Test for nitrates/nitric acid

Oxidation of iron (ii) sulphate

– Concentrated HNO₃ oxidizes green Iron (II) sulphate in presence of sulphuric acid into Iron (III) sulphate (yellow/brown)

– However the solution turns dark brown due to formation of a compound, FeSO₄.NO

– NO is produced by reduction of nitrate to nitrogen monoxide by Fe²⁺

Ionically;

Fe²_+(aq) Fe³⁺_(aq) + e^– (oxidized)

NO₃^–_(aq) + 2H⁺_(aq) + e^– NO_2(g) + H₂O_(l) (reduced)

Brown ring test

Procedure.

– An unknown solid is dissolved then acidified using dilute H₂SO₄.

– Some FeSO₄ solution is then added.

– The test tube is then held at an angle and concentrated sulphuric (V) acid is added slowly (dropwise) to the mixture.

Observations

– The oily liquid (conc. H₂SO₄) is denser than water hence sinks to the bottom.

– A brown ring forms between the two liquid layers if the solid is a nitrate.

Diagrams:

Explanations:

– Suppose the solution tested isKNO₃, the conc. H₂SO₄and the KNO₃reacted to produce HNO₃.

Equation:

KNO₃(aq) +H₂SO_4(aq) KHSO_4(aq) + HNO_3(aq)

– The NO₃^– from nitric acid oxidizes some of the FeSO₄ to Fe₂(SO₄)₃ (Fe²⁺ toFe³⁺) and itself reduced to NO by the Fe²⁺

-The NO so formed reacts with more FeSO4 to give a brown compound (FeSO4 NO) which appears as a brown ring.

Equation:

FeSO_4(aq) + NO_(g) FeSO₄. NO_(aq)

(Green) (Brown)

Ionically:

Fe²⁺_(aq) + 5H₂O_(l) + NO_(g) [Fe(H₂O)₅NO]²⁺_(aq)

(Green) (Brown)

Heat

– Nitrates of less reactive metals decompose easily with gentle heating; clouds of brown NO₂ can be seen.

Equation:

2Cu(NO₃)₂ ^heat 2CuO_(s) + 4NO_2(g) + O_2(g)

(Brown, acidic)

– The nitrates of more reactive metals need much stronger heating and decompose in a different way.

Equation:

2Na NO_3(s) ^heat 2NaNO_2(s) + O_2(g)

Uses of nitric acid

– Large quantities are used in fertilizer manufacture.

– Manufacture of explosives (TNT)

– Manufacture of dyes

– Making nitrate salts

– Etching of metals.

– Manufacture of nylon and terylene

– Refining precious metals

– An oxidizing agent.

Industrial manufacture of nitric acid

The Otswalds process

(a). Introduction

– Nitric acid is manufactured by the catalyst oxidation of ammonia and dissolving the products in water.

(b). Raw materials

– Atmosphere air

– Ammonia from Haber process.

(c). Conditions

– Platinum-rhodium catalyst or platinum gauze.

– The ammonia-air mixture must be cleaned (purified) to remove dust particles which could otherwise poison the catalyst.

(d). Chemical reactions.

Step 1: Compressor reactions.

– Ammonia and excess air (oxygen) (1:10 by volume) is slightly compressed.

– The mixture is then cleaned to remove particles which would otherwise poison the catalyst.

– They are then passed to the heat exchanger.

Step 2: Heat exchanger and catalytic chamber.

– In the heat exchanger, the gaseous mixture is heated to about 900^oC and then passed over a platinum-rhodium catalytic chamber.

– An exothermic reaction occurs and ammonia is oxidized to nitrogen (II) oxide and steam.

Equation:

4NH_3(g) + 5O_2(g) 4NO_(g) + 6H₂O_(g) + Heat.

– The exothermic reaction once started, provides the heat necessary to maintain the required catalytic temperature.

-This is of economical advantage i.e. electrical heating of catalyst is not continued hence lowering production costs.

Step 3: Heat exchanger.

– The hot products from catalytic chamber are again passed back through the heat exchanger.

– The hot gases are cooled and then passed into the cooling chamber.

Step 4: Cooling chamber

– Once cooled, the NO is oxidized to NO₂ by reacting it with excess oxygen.

Equation:

2NO_(g) + O_2(g) 2NO_2(g)

Step 5: Absorption towers:

– The NO₂ in excess air is then passed through a series of absorption towers where they meet a stream of hot water and form nitric (V) acid and nitrous (III) acid.

Equations:

2NO_2(g) + H₂O_(l) HNO_3(aq) + HNO_2(aq) (blue solution)

Nitric Nitrous

– The so produced nitrous (III) acid is oxidized by oxygen in excess air to nitric (V) acid so that the concentration of nitric acid in the solution (liquid) gradually increases.

Equation:

2 HNO_2(aq) + O_2(g) 2HNO_3(aq)

– The resultant HNO₃ is only 55%-65% concentrated.

– It is made more concentrated by careful distillation of the solution.

The process of distillation (increasing the concentration).

– Concentrated sulphuric (VI) acid is added to the dilute nitric (V) acid.

– The heat produced (when dilute sulphuric acid reacts with water) vapourises the nitric (V) acid.

– The resultant nitric (V) acid vapour is condensed.

Note:

Nitric (V) acid is stored in dark bottles.

Reason:

– To prevent its decomposition since it undergoes slow decomposition when exposed to light.

Dilute nitric (V) acid has higher ions concentration than concentrated nitric (V) acid.

Reason.

– Dilute nitric (V) acid is a stronger acid hence ionizes fully to yield more hydrogen ions than concentrated nitric (V) acid.

– Dilute nitric (V) acid is ionic whereas concentrated nitric (V) acid is molecular;

– Dilute nitric (V) acid is more (highly) ionized than concentrated nitric (V) acid.

Flow diagram for the otswalds process.

Ammonia

HEAT EXCHANGER

CATALYTIC CHAMBER

Air

Water Unreacted NO_(g)

NO + air;

Nitric (V) acid

Pollution effects of nitrogen compounds.

Acid rain

– Nitrogen (II) oxide is produced in internal combustion engines on combination of nitrogen and oxygen.

– Nitrogen (II) oxide oxidized to nitrogen (IV) oxide which dissolves in water to form nitric (III) and nitric (V) acids.

– Nitric (v) acid eventually reaches ground as acid rain and causes:

Loss of chlorophyll (chlorosis) from leaves
Corrosion of stone buildings and metallic structures, weakening them and destroying beauty.
Leaching of vital minerals from soils. These are converted into soluble nitrates and washed away from top soil. This leads to poor crop yields.

Smog formation.

– Nitrogen (IV) oxide also undergoes series of chemical reactions in air to produce one of the major components of smog.

– Smog reduces visibility for motorists, irritates eyes and causes breathing problems.

Eutrophication:

– Refers to enrichment of water with excess nutrients for algal growth.

– Presence of nitrate ions from nitrogen fertilizers in a water mass encourages rapid growth of algae.

– This eventually leads to reduction of dissolved oxygen in water, killing aquatic animals like fish.

– Presence of nitrate ions in drinking water may also cause ill health to humans. This is because they are converted into carcinogenic compounds.

Prevention.

Recycling unreacted gases in manufacture of nitric acid to prevent release into environment.
Treating sewage and industrial effluents to remove nitrogen compounds before releasing to rivers and lakes.
Fitting exhausts systems of vehicles with catalytic converters which convert nitrogen oxides into harmless nitrogen gas.
Adding lime to lakes and soils in surrounding regions to reduce acidity.
Applying fertilizers at right and in the correct proportion to prevent them from being washed into water masses.

UNIT 3: SULPHUR AND ITS COMPOUNDS

Checklist:

Occurrence of sulphur
Extraction of sulphur

The Frasch pump
Extraction process

Properties of sulphur

Physical properties
Chemical properties

Uses of sulphur
Allotropes of sulphur

Rhombic sulphur
Monoclinic sulphur

Compounds of sulphur

Sulphur (IV) oxide
Laboratory preparation
Other preparation methods
Properties of sulphur (IV) oxide
- Physical properties
- Chemical properties
- Uses of sulphur (IV) oxide

Sulphur (VI) oxide

Laboratory preparation
Properties of sulphur (VI) oxide

Sulphuric (VI) acid

Large scale manufacture
- Raw materials
- The chemical process
- Pollution control
Properties of concentrated sulphuric (VI) acid
- Physical properties
- Chemical properties
Properties of dilute sulphuric (VI) acid
Uses of sulphuric (VI) acid

Hydrogen sulphide gas

Laboratory preparation
Properties of hydrogen sulphide
Physical properties of hydrogen sulphide
Chemical properties of hydrogen sulphide

Atmospheric pollution by sulphur compounds

Occurrence

– Occurs naturally as s free element in the underground deposits in Texas and Louisiana (USA) and Sicily (ITALY).

– It also occurs as a sulphate and sulphide ores.

Examples;

Metallic sulphides: iron pyrites (FeS₂); Zinc blende (ZnS) Copper pyrites (CuFeS₂)

Metallic sulphates e.g. Gypsum, CaSO₄

Hydrogen sulphide e.g. H₂S present in natural gas.

Extraction of sulphur: The Frasch process

– Is done using a set of 3 concentric pipes called Frasch pump; hence the name Frasch process.

(i). Apparatus: Frasch pump

Hot compressed air

Superheated water at 170^oC

Froth of molten sulphur

Cross section of the Frasch pump

Outermost pipe: brings superheated water at 170^oC

Innermost pipe: brings in hot compressed air;

Middle pipe: brings out a froth of molten sulphur

(ii). Chemical process

Note: Sulphur cannot be mined by conventional mining methods such as open cast, alluvial mining etc

Reasons:

– Sulphur deposits lie very deep under several layers of quicksand hence cannot be accessed easily.

– Sulphur deposits are associated with poisonous gases such as sulphur (IV) oxide gas which can cause massive pollution if exposed to open environment.

– Three concentric pipes, constituting the Frasch pump are drilled through the rock and soil down to the sulphur deposits.

(a). The outer tube (pipe)

– Is used to pump superheated water at 170^o c and 10 atmospheres down the deposits.

– The heat of the water melts the sulphur.

– By the time the water reaches the sulphur, its temperature drops to 120^oC, but this is enough to melt sulphur whose M.P is 114^oC.

(b). The innermost tube

– Is the smallest pipe and is used to blow or force a jet of hot compressed air down the sulphur deposits.

– This produces a light froth of molten sulphur (mixture of air, water and sulphur) which is forced up the middle pipe.

(c). The middle pipe.

– Allows the sulphur froth (mixture of molten sulphur, water and air) into the surface; where mixture is run into large tanks.

– The forth usually settles in two layers, the bottom layer is mainly water while the upper layer is mainly molten sulphur; due to differences in density.

– Once in the settling tanks, sulphur solidifies and separates out; giving 99% pure sulphur.

– The sulphur is removed, melted again and poured into moulds, to form roll sulphur in which form it is sold.

Properties of sulphur

Physical properties

– It is a yellow solid which exists in one amorphous form and 2 crystalline forms.

– A molecule of sulphur consists of a pluckered ring of 8 sulphur atoms covalently bonded.

Diagram: structure of a sulphur molecule.

Solubility

– It is insoluble in water but soluble in organic solvents like carbon disulphide, xylene and toluene.

It is a poor conductor of heat and electricity since it is a covalent element lacking free electrons or ions.

Effects of heat

– When sulphur is heated out of contact with air, it melts at low temperatures of about 113^oC to form an amber (orange) coloured mobile liquid.

Reason:

– The S₈ rings open up to form chains of S₈.

Diagrams:

The pluckered S₈ ring of sulphur molecule Chains of S₈ molecule

– On further heating, the liquid darkens in colour.

– At 160^oC, the liquid becomes much darker and very viscous (such that the test tube can be inverted without the sulphur pouring out.)

– The viscosity continues to increase until a temperature of about 195⁰C

Reason:

– The S₈ rings of sulphur are broken and they then join to form very long chains of sulphur atoms, with over 100,000 atoms (S_{100 000}).

Note: As the chains entangle with one another the viscosity increases and colour darkens.

– Near the boiling point, the liquid becomes less dark i.e. red-brown and more mobile (runny).

Reason

– The long chains are broken to shorter chains.

– At 444^oC (boiling point), sulphur vapourises to form a red-brown vapour consisting of S₈, S₆, S₄ and S₂ molecules.

Reason

– The sulphur liquid changes state to form sulphur vapour.

– The vapour is light brown in colour, and consists of a mixture of molecules of formula S₂-S₁₀

Note

– If heated further the larger sulphur vapour molecules (S₈, S₆ etc) dissociate and at 750^oC the vapour is mostly constituted of diatomic molecules (S₂)

– On exposure to cold surfaces the light brown vapour condenses to a yellow sublimate. The yellow sublimate is called flowers of sulphur.

Chemical properties

Burning in air

– It burns in air with a bright blue flame forming a misty gas with a choking smell.

– The gas is sulphur (IV) oxide, with traces of sulphur (VI) oxide, both of which are acidic.

Equation:

S_(s) + O_2(g) SO_2(g)

Note:

The SO₃ is formed due to further oxidation of some of the SO₂ gas

Equation:

2SO_2(s) + O_2(g) 2SO_3(g)

Reaction with acids.

– Dilute acids have no effect on sulphur.

– It is however easily oxidized by concentrated (VI) sulphuric acid and Nitric (VI) acid.

With conc. H₂SO₄

– When warmed with conc. H₂SO₄, sulphur is oxidized to sulphur (IV) oxide while the acid is reduced to the same gas.

Equation:

S_(s) + 2H₂SO_4(l) 3SO_2(g) + 2H₂O_(l)

With conc. HNO₃

– Sulphur is oxidized to sulphuric (VI) acid while acid itself is reduced to red-brown Nitrogen (IV) oxide.

Equation:

S_(s) + 6HNO_3(l) H₂SO_4(aq) + 6NO_2(g) + 2H₂O_(l)

Note:

– The resultant solution gives a white precipitate with a solution of Barium chloride.

Reason

– Due to presence of sulphate ions which combine with Ba²⁺ to form insoluble BaSO_4(s)

Ionically;

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

Reaction with other elements.

– It combines directly with many other elements to form sulphides.

– With metals, sulphur forms metal sulphides, most of which are black.

Examples.

(a). With metals

Iron metal

Fe_(s) + S_(s) FeS_(s) + Heat

(Grey) (Yellow) (Black)

Note:

– During the reaction, the mixture glows spontaneously; immediately the reaction has started.

Copper

2Cu_(s) + S_(s) Cu₂S

(Red-brown) (Yellow) (Black copper (I) sulphide))

(b). Non-metals

Carbon

C_(s) + 2S_(s) CS₂(s)

(Black) (Yellow) (Black Carbon disulphide)

Note.

– Carbon (IV) sulphide has a distinct smell.

– It is an excellent solvent and is used as a pesticide due to its poisonous nature.

Hydrogen

H_2(g) + S_(s) H₂S_(g)

Fluorine

S_(s) + F_2(g) SF_2(g)

Chlorine

S_(s) + Cl_2(g) SCl_2(g)

Bromine

2S_(s) +Br_2(g) S₂Br_2(g)

Phosphorous

10S_(s) + 4P_(s) P₄S_10(s)

Note:

– Sulphur does not react with inert gases, nitrogen and iodine.

Uses of sulphur

Industrial manufacture of sulphuric (VI) acid in the contact process.
It is used as a fungicide for treatment of fungal skin diseases.
It is used for vulcanization (hardening) of rubber
Manufacture of calcium hydrogen sulphite (Ca(HSO₃)₂ used for bleaching in paper and textile industries.
Manufacture of matches and fireworks.
Manufacture of dyes e.g. sulphur blacks that gives paint smooth texture.
Manufacture of sulphur ointments and drugs e.g. sulphur-guanidine for dysentery.
Manufacture of hair oil.
Small amounts of sulphur are added to concrete to prevent corrosion by acids.
Manufacture of fungicides for spraying crops against fungal infections e.g. ridomil, dithane for potato and tomato blights

Allotropes of sulphur

– Allotropy is the existence of an element in more than one form without change of state.

– Sulphur has 2 allotropes

Rhombic sulphur/ octahedral/ alpha-sulphur
Monoclinic/ prismatic sulphur/ beta-sulphur.

-Unlike carbon only the rhombic sulphur occurs naturally.

Comparison of rhombic and monoclinic sulphur.

Allotrope Characteristic	Rhombic sulphur	Monoclinic sulphur
Stability	– Stable below transitional temp. of 96^oC	– Stable above 96^oC
Colour	– Bright yellow crystalline solid	– Pale yellow crystalline solid
Melting point	– Melts at 113^oC;	– Melts at 119^oC;
Density	– About 2.06gcm^-3(heavier than monoclinic Sulphur)	– Lighter than 1.98gcm^-3 (lighter than rhombic sulphur)
Shape	– Octahedral shape Diagram:	– Needle-like/ prismatic Diagram:

Note.

96^oC is called transitional temperature; because both allotropes are stable.

Compounds of sulphur

Oxides of sulphur.

Sulphur (IV) oxide

Laboratory preparation of sulphur (IV) oxide

(i). Apparatus:

Dry sulphur (IV) oxide gas

Sodium sulphite

Dilute HCl

Conc. H₂SO_4(l)

(ii). Procedure

– Dilute HCl or H₂SO₄ is poured into sodium sulphite crystals in the flask.

– The gas produced is passed through conc. Sulphuric acid to dry it.

– If the reaction is slow, the round-bottomed flask is heated (warmed) gently.

– Dry gas is collected by downward delivery as it is denser than air.

(ii). Equation.

Na₂SO_3(aq) + 2HCl_(aq) H₂O_(l) + SO_2(g) + 2NaCl_(aq)

Ionically;

2H⁺_(aq) + SO₃^2-_(aq) H₂O_(l) + SO_2(g)

Note:

– Nitric (V) acid should not be used.

Reason:

– It is a strong oxidizing agent and cannot therefore reduce the metal sulphites.

– Instead it will oxidize the SO₂ produced to sulphuric (VI) acid

Equation:

2HNO_3(aq) + SO_2(g) 2NO_2(g) + H₂SO_4(l)

Other methods of preparing sulphur (IV) oxide.

(b). Preparation from concentrated sulphuric (VI) acid

(i). Apparatus

– As in (a) above

(ii). Procedure

– Copper turnings are covered with concentrated sulphuric (VI) acid and the mixture heated (a must in this case).

Note:

– Dilute sulphuric (VI) acid does not react with copper hence the need for concentrated acid.

– Cold concentrated sulphuric (VI) acid does not also react with copper hence warming.

(iii). Observation.

– When the solution becomes hot, there is evolution of sulphur (IV) oxide gas.

Equation.

C_u(s) +2H₂SO_4(l) CuSO_4(aq) + 2H₂O_(l) + SO_2(g)

Note:

– This reaction is in two stages.

Oxidation of Cu to CuO

– Concentrated sulphuric (VI) acid oxidizes copper to Copper (II) oxide

Equation:

Cu_(s) + H₂SO_4(l) CuO_(s) + H₂O_(l) + SO_2(g)

CuO further reacts with the acid to form salt and water.

Equation:

CuO_(s) + H₂SO_4(l) CuSO_4(aq) + H₂O_(l)

Overall equation:

Cu_(s) + H₂SO_4(l) CuSO_4(aq) + 2H₂O_(l) + SO_2(g)

(c). Roasting sulphur in air

– When sulphur is burnt in air, SO₂ is produced.

Equation:

S_(s) + O_2(g) SO_2(g)

Note:

This reaction is not suitable for preparing a pure sample of the gas in the lab.

Reason

– The gas is contaminated with traces of O₂; N₂; CO₂ and inert gases.

– There are higher chances of environmental pollution, due to escape of some of the gas into the atmosphere.

(d). Roasting metal sulphides in air

Examples:

2FeS_(g) + 3O_2(g) 2FeO_(s) + 2SO_2(g)

2ZnS_(g) + 3O_2(g) 2ZnO_(s) + 2SO_2(g)

Preparation of sulphur (IV) oxide solution.

(i). Apparatus

(ii). Procedure

– Gas is directly passed into water using an inverted funnel; to prevent “sucking back” by increasing surface area for dissolution.

Properties of sulphur (IV) oxide gas

Physical properties

It is a colourless gas with an irritating (pungent) characteristic smell.
It neither burns nor supports combustion i.e. when a lighted splint is introduced into a gas jar full of sulphur (IV) oxide, the splint is extinguished.
It has a low PH.

Chemical properties.

– It is a strong reducing agent.

– An aqueous solution of sulphur (IV) oxide, sulphurous acid is strong reducing agent.

– The sulphite radical, SO₃^2-, acts as a supplier of electrons; the overall reaction results into formation of sulphate ions.

Equations:

H₂SO_3(aq) 2H⁺_(aq) + SO₃^2-_(aq) then;

SO₃^2-_(aq) + H₂O_(l) SO₄^2-_(aq) + 2H⁺_(aq) + 2e-

– The resultant electrons supplied are accepted by an oxidizing agent, which consequently gets reduced.

Examples:

(i). Reduction of acidified potassium manganate (VII).

Procedure.

-To about 2 cm³ of sulphur (IV) oxide solution, 2 cm³ of dilute H₂SO₄was added followed by an equal volume of potassium manganate (VII) solution.

Observations

– Purple solution changes to colourless.

Explanation

– Purple manganate (VII) ions are reduced to colourless manganate (II) ions, while H₂SO₃ (sulphurous (IV) acid) is reduced to sulphate ions and water.

Equation:

5SO_2(g) + 2KMnO_4(aq) + 2H₂O K₂SO_4(aq) + 2MnSO_4(aq)+ H₂SO_4(aq)

Ionically;

2MnO^–_4(aq) + 5SO₃^2-_(aq) + 6H⁺_(aq) 2Mn²⁺_(aq) + 5SO₄^2-_(aq) + 3H₂O_(l)

(ii). Reduction of potassium chromate (IV) solution

Procedure

– To 2 cm³of Sulphur (IV) oxide solution, 2 cm³ of dilute H₂SO₄ was added followed by an equivalent volume of potassium chromate (VI) solution.

Observation

– Acidified potassium chromate (VI) solution change from orange to green.

Equation

K₂Cr₂O_7(aq) + 3SO_2(aq) + H₂SO_4(aq) K₂SO_4(aq) + H₂O_(l) + Cr₂(SO₄)_3(aq)

(Orange) (Green)

Ionically: Oxidation

Cr₂O₇^2-_(aq)+ 3SO₃^2-_(aq) + 8H⁺_(aq) 2Cr³⁺_(aq) + 3SO₄^2-_(aq)

Reduction

Note: this is the usual chemical test for sulphur (IV) oxide.

(iii). Reduction of Iron (III) ions to Iron (II) ions (Fe³⁺ to Fe²⁺)

Procedure

– About 3 cm³ of Iron (III) chloride solution are heated in a test tube and sulphur (IV) oxide gas bubbled into it.

Observations

– The brown solution turns green.

Explanation

– Aqueous sulphur (IV) oxide reduces to Fe³⁺ in FeCl₃ which are brown to green Fe²⁺ in FeCl_2(aq).

Ionically

2Fe³⁺_(aq) + SO₃^2-_(aq) + H₂O_(l) Fe²⁺_(aq) + SO₄^2-_(aq) + H⁺_(aq)

(iv). Reduction of bromine water

Procedure

– Bromine water (red brown) is added to a solution of sulphur (IV) oxide followed by HCl and BaCl₂solution.

Equation

Br_2(aq) + SO_2(g) + 2H₂O_(l) 2HBr_(aq) + H₂SO_4(aq)

Ionically: ^Oxidation

Br_2(aq) + H₂O_(l) + SO₃^2-_(aq) 2HBr_(aq) + SO₄^2-_(aq)

(Red-brown) (Colourless)

Reduction

On addition of barium chloride

– A white precipitate is formed, due to the formation of insoluble barium sulphate.

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

Note

– This test confirms presence of SO₄^2- since a white precipitate insoluble in dilute hydrochloric acid is formed.

– CO₃^2-_(aq) and SO₃^2- also forms a white precipitate with BaCl_2(aq) but the white precipitates dissolve in dilute HCl_(aq)

(v). Reduction of hydrogen peroxide

Procedure

– To 2 cm³ of aqueous sulphur (IV) oxide, an equal volume of hydrogen peroxide is added followed by 1 cm³of HCl, then a few drops BaCl₂ solution.

Observation and explanations:

– Bubbles of a colourless gas; that relights a glowing splint.

– Hydrogen peroxide is reduced to water; while the sulphite ion in aqueous sulphur (IV) oxide (H₂SO_3(aq)) is oxidized to SO₄^2-_(aq)

Equation

H₂O_2(l) +SO₃^2-_(aq) H₂O_(l) + SO₄^2-_(aq)

– On addition of BaCl₂, a white precipitate insoluble in dilute HCl.

– This confirms presence of sulphate ions.

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

(vi). Reduction of concentrated nitric (V) acid

Procedure

– Sulphur (IV) oxide is bubbled through (into) a solution of concentrated nitric (v) acid.

Observation

– Brown fumes (of NO₂) are liberated.

Explanation

– Sulphur (IV) oxide reduces nitric (V) acid to nitrogen (IV) oxide (brown) while it is itself oxidized by HNO₃ to form H₂SO₄.

– Thus while SO₂ is the reducing agent; HNO₃ is the oxidizing agent.

Equation:

2HNO_3(l) + SO_2(g) 2NO_2(g) + H₂SO_4(aq)

(Brown fumes)

(vii). Reaction with atmospheric oxygen in light.

Procedure:

– About 2 cm³ of Sulphur (IV) oxide solution is left in a test tube in light for 24 hours, dilute HCl is then added, followed by barium chloride.

Observations and explanations:

– Atmospheric oxygen in light oxidizes sulphite ion (SO₃^2-) into sulphate (SO₄^2-)

Equation:

2SO₃^2-_(aq) + O_2(g) 2SO₄^2-_(aq)

– On adding barium chloride, a white precipitate insoluble in dilute HCl results; confirming presence of sulphate ion.

Equation:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO4_(s)

(White ppt)

Sulphur (IV) oxide as oxidizing agent

– It reacts as an oxidizing agent with reducing agents more powerful than itself.

Examples

(a). Reaction with hydrogen sulphide

Procedure

– A test tube of dry hydrogen sulphide gas is inverted into a gas jar full of moist sulphur (IV) oxide, and the gases allowed to mix.

Observation

– Yellow deposits of sulphur is produced.

Examples:

Oxidation

2H₂S_(g) + SO_2(g) 2H₂O_(l) + 3S_(s)

Reduction

Explanations:

– H₂S is a stronger reducing agent than sulphur (IV) oxide.

– Thus sulphur (IV) oxide acts as an oxidizing agent supplying oxygen to the hydrogen sulphide.

Note

– Dry gases do not react and for this reaction to occur, the gases must be moist or at least one of them.

(b). Reaction with burning magnesium

Procedure

– Burning magnesium is lowered into a gas jar full of sulphur (IV) oxide.

Observation

– White fumes of magnesium oxide and yellow specks of sulphur.

Equation

2Mg_(s) + SO_2(g) 2MgO_(s) + S_(s)

Sulphur (IV) oxide as bleaching agent.

Procedure

– Coloured flower petals are placed in a test-tube full of sulphur (IV) oxide.

Observation

– The coloured (blue or red) petals are bleached (turned colorless);

Explanations:

– In presence of water, sulphur (IV) oxide acts as a bleaching agent. It bleaches by reduction (removal of oxygen form the dye)

– It first combines with water forming the sulphurous acid; which then reduces the dye to form a colourless product.

Equations:

SO_2(g) + H₂O_(l) H₂SO_3(aq)

H₂SO_3(aq) 2H⁺_(aq) + SO₃^2-_(aq)

Then;

SO₃^2-_(aq) + [O] SO₄^2-_(aq)

From dye

General equation

SO_2(g) + H₂O_(l) + [Dye + (O)] Dye + H₂SO_4(aq)

Coloured Colourless

Note

– The original colour may be restored by oxidation or prolonged exposure to air. This explains why old newspapers which were originally bleached white by sulphur (IV) oxide turn brown with time.

– Chlorine bleaches by oxidation hence its oxidation is permanent; SO₂ is however preferred because it is milder in action.

Reaction with sodium hydroxide (alkalis)

Procedure

– A gas jar full of sulphur (IV) oxide is inverted over sodium hydroxide solution in a trough and shaken.

Observations

– Solution seen rises up in the jar.

Explanation

– Sulphur (IV) oxide is acidic, hence easily absorbed by alkaline solutions such as sodium hydroxide solution.

– Sodium sulphite and sodium hydrogen sulphites are formed depending on amount of sulphur oxide.

Equations

With limited sulphur (IV) oxide:

2NaOH_(aq) + SO_2(g) Na₂SO_3(aq) + H₂O_(l)

With excess sulphur (IV) oxide:

NaOH_(aq) + SO_2(g) NaHSO_3(aq)

Reaction with chlorine:

– Sulphur (IV) oxide reacts with moist chlorine to form an acidic mixture of sulphuric (VI) acid and hydrochloric acid.

Equation:

SO_2(g) + SO_2(g) H₂O_(l) H₂SO_4(aq) + 2HCl_(aq)

Explanation:

– Sulphur (IV) oxide serves as the reducing agent reducing chlorine into hydrochloric acid;

– Chlorine acts as the oxidizing agent; oxidizing the sulphur (IV) oxide into sulphuric (VI) acid

Tests for sulphur (iv) oxide

Characteristic pungent smell.
Bleaches flower petals.
Decolourises purple potassium manganate (VII)
Turns filter paper soaked in acidified orange potassium dichromate (VI) solution to green

Sulphur (IV) oxide as a pollutant

– It is industrial waste in some chemical processes.

– The emission to the air it dissolves forming sulphurous acid.

Equation:

SO_2(g) + H₂O_(l) H₂SO_3(aq)

– Sulphurous acid is readily oxidized to sulphuric (VI) acid; which attacks stonework and metal structures causing them to corrode.

– If breathed in, SO₂ causes lung damage.

Uses of sulphur (VI) oxide

– Industrial manufacture of sulphuric (VI) acid.

– Fumigation in green houses for purposes of pest and disease control.

– Preservative in jam and fruit juices.

– Bleaching agent for wool, straw, paper pulp etc.

Sulphuric (VI) acid

Industrial manufacture of sulphuric (VI) acid: The contact process

Raw materials

– Sulphide ores or sulphur.

– Water

– Oxygen (air)

– Concentrated sulphuric (VI) acid.

The chemical process

Step 1: Production of sulphur (VI) oxide

– Sulphur (IV) oxide is obtained b burning the metal ores of sulphides or elemental sulphur in air.

Equation:

S_(s)+ O_2(g) SO_2(g)

– Obtaining sulphur (IV) oxide form pyrites is cheaper than form sulphur.

– Flowers of sulphur form pyrites is impure and contains dust; which involves extra expenses and time in purification.

Step 2: Purification and drying

– The Sulphur (IV) oxide and excess air are passed through a series of driers and purifiers.

– Purifiers remove dust particles, which would otherwise poison the catalyst used in this process by taking up the catalytic surface thus impairing the catalytic efficiency.

– Purification (removal of dust) is by electrostatic precipitation.

– Are dried through concentrated sulphuric acid then passed through heat exchanger.

Step 3: Heat exchanger reactions

– The pure dry SO₂ and excess air mixture are passed into heat exchanger reactions.

Reason:

– To lower their temperatures since reaction in the proceeding chamber (catalytic chamber) are exothermic hence requiring lower temperatures.

Step 4: Catalytic chamber

– Dry dust-free SO₂ is mixed with clean excess air, heated and passed into a catalytic chamber containing vanadium (V) oxide catalyst.

Equation V₂O₅

2SO_2(g) + O_2(g) 2SO_3(g) + Heat

450^oC

– The product is sulphur (VI) oxide, SO₃.

– Formation of sulphur (VI) oxide is accompanied by evolution of heat (exothermic reaction) and a reduction in volume.

Note:

– A good yield of SO₃ is favoured by the following conditions.

Temperature

– The forward reaction is exothermic hence the yield can be favourable in low temperatures.

– However, at such low temperatures the equilibrium is attained very slowly.

– At high temperatures, equilibrium is achieved very quickly but sulphur (VI) oxide decomposes considerably.

– Thus a compromise optimum temperature of about 450^oC is used in order to enable as much sulphur (VI) oxide as possible to be made in a reasonable time.

– From the graph, high SO₃ yield is favoured by relatively low temperatures.

Graph: %age yield of sulphur (VI) oxide against temperature.

Pressure

– High pressures favour production of more sulphur (VI) oxide.

Reason

– The volume of gaseous reactants is higher than volume of gaseous products.

– Since reaction involves reduction in volume, theoretically pressure used should be as high as is economically convenient.

Note:

– High pressures are however disadvantageous.

Reason

– The equipment required to generate high pressure would be expensive to maintain.

– The high pressure could also liquefy sulphur (VI) oxide.

– A pressure slightly above atmospheric pressure is used providing 98% conversion at low maintenance costs.

Catalyst

– A catalyst neither takes part in a reaction nor increases the yield.

– It merely speeds up the reaction i.e. reduces the time taken to react at equilibrium of 450^oC.

– Main catalyst is vanadium (V) oxide (V₂O₅).

– It is spread out (in trays) on silica gel to increase the surface area for combination of reactants.

– Dust settled in the catalyst may reduce its effective area.

– Dust may also react with the catalyst, poison it and further reduce its efficiency.

– This explains need to purify gases thoroughly.

– An effective catalyst is platinised asbestos.

– However, vanadium (V) oxide is preferred.

Reasons:

– It is not easily poisoned by dust particles.

– It is cheaper and readily available.

Note:

– The highest yield of sulphur (VI) oxide is obtained at optimum conditions of 450⁰C and pressure 2-3 atmospheres in presence of vanadium (V) oxide or platinised asbestos.

Step 5: Heat exchanger reactions

– Hot SO₃gas from catalytic chamber is again passed through heat exchanger for cooling after which the cooled gas is taken into an absorption chamber.

Step 6: Absorption chamber

– The SO₃ is not dissolved (passed) into water directly.

Reason

– It dissolves in water exothermically with a loud, hissing sound giving off corrosive vapour resulting into harmful sulphuric acid sprays or mist all around.

– The SO₃ is dissolved in conc. H₂SO₄ forming oleum (pyrosulphuric acid/ fuming sulphuric acid).

Equation:

SO_3(g) + H₂SO_4(l) H₂S₂O_7(l)

– Resultant Oleum is then channeled into a dilution chamber.

Step 7: Dilution chamber.

– Oleum is diluted with correct amounts of water to form concentrated sulphuric acid.

Equation:

H₂S₂O_7(l) + H₂O_(l) 2H₂SO_4(aq)

Summary: flow diagram for the contact process:

Pollution control in contact process.

– Main source of pollution is sulphur (IV) oxide.

– In catalyst chamber, SO₂ reacts with oxygen forming SO₃.

Equation: V₂O₅

2SO_2(g) + O_2(g) 2SO_3(g) + Heat

450^oC

– This is a reversible reaction and upto 98% conversion is possible and excess (unreacted) SO₂ warmed and released into atmosphere via long chimneys.

– However, SO₂ being a pollutant, little or none should be released into atmosphere.

– This is done by scrubbing the gas.

– This involves neutralizing the chimney gas by a solution of Calcium hydroxide forming a salt (calcium sulphite) and water.

Equation:

Ca(OH)_2(aq) + SO_2(g) CaSO_3(aq) + H₂O_(l)

Note:

– In certain cases, filters are also installed to remove any traces of acid spray or mist form the exhaust gases.

– The unreacted gases (SO₂ and SO₃) may also be recycled within the process.

Properties of concentrated sulphuric (VI) acid

Physical properties

– Colourless, odourless, oily liquid.
– Very dense; with density 1.84 gcm^-3.
– Soluble in water and gives out considerable heat when a solution is formed.
– It is hygroscopic absorbs atmospheric moisture to become wet.

Experiment: To show hygroscopic nature of conc. H₂SO₄.

(i). Procedure

– A small beaker half full of conc. H₂SO₄ is weighed.

– Level of acid in beaker is marked to the outside using gummed paper.

– Acid is left exposed to air for a week or so then weighed again and level also noted.

(ii). Observations

– There is an increase in weight of acid.

– Level of acid in beaker is now above the paper mark.

(iii). Explanations

– The increase in weight and size is due to water absorbed form the air by the conc. sulphuric (VI) acid.

Note:

– This explains why sulphuric (VI) acid is used as a drying agent.

Chemical properties

– It is a dehydrating agent.

Examples:

(a). Action on blue hydrated copper (II) sulphate (CuSO₄.5H₂O) crystals.

(i). Procedure

– A few crystals of hydrated CuSO₄.5H₂O were put in a test tube and enough concentrated sulphuric (VI) acid added, to cover them completely.

(ii). Observation:

– Blue copper (II) sulphate pentahydrate crystals turn to white powder of anhydrous CuSO₄.

Equation

Conc. H₂SO₄

CuSO₄.5H₂O_(s) CuSO_4(s) + 5H₂O_(l)

(Blue crystals) (White crystals)

Explanations:

– Conc.H₂SO₄ has a very strong affinity for water and hence removes water of crystallization from crystals hence dehydrating them.

(b). Action on white sugar (C₁₂H₂₂O₁₁)

(i). Procedure:

– A tablespoonful of sugar is put in an evaporating dish form a beaker and adequate volume of conc. H₂SO₄ is added.

(ii). Observations:

– Sugar turns form brown then yellow and finally to a charred black mass of carbon.

– A spongy black mass of charcoal (carbon) rises almost filling the dish.

– Steam is also give off and dish becomes very hot since reaction is exothermic.

Equation

Conc. H₂SO₄

C₁₂H₂₂O_11(s) 12C_(s) + 11H₂O_(l)

(White crystals) (Black solid)

Explanation

– The acid removed from the sugar elements of water (hydrogen and oxygen, ratio 2:1) to form water, leaving behind a black charred mass of carbon.

(c). Action on oxalic acid (ethanedioic acid (H₂C₂O₄)

– Conc. H₂SO dehydrates oxalic acid on heating to a mixture of carbon (II) oxide and carbon (IV) oxide.

Conc. H₂SO₄

Equation

H₂C₂O_4(s) CO_(g) + CO_2(g) + H₂O_(l)

Note:

– Conc. H₂SO₄ acid gives severe skin burns because it removes water and elements of water from skin tissue.

– Should the acid spill on skin, it is washed immediately with plenty of water followed with a solution of sodium hydrogen carbonate.

– Holes appear where the acid spills on clothes for same reason.

(d). Action on alcohols (alkanols)

– Conc. sulphuric (VI) acid dehydrates alcohols to corresponding alkenes.

Example: dehydration of ethanol to ethene

Equation:

Conc. H₂SO₄

CH₃CH₂OH_(s) C₂H_4(g) + H₂O_(l)

(Ethanol) (Ethene)

(e). Action on methanoic acid.

– Conc. sulphuric (VI) acid dehydrates methanoic acid to form CO.

Conc. H₂SO₄

Equation:

HCOOH_(s) CO_(g) + H₂O_(l)

Further reactions of conc. H₂SO₄ as an oxidizing agent.

– Hot concentrated Sulphuric acts as an oxidizing agent in which cases it is reduced to sulphur (IV) oxide and water.

Examples:

(a). Reaction with metals.

Copper

Cu_(s) + 2H₂SO_4(l) CuSO_4(aq) + SO_2(g) + 2H₂O_(l)

Note: the copper (II) sulphate formed is white since the conc. H₂SO₄ further dehydrates the hydrated CuSO₄.

Zinc

Zn_(s) + 2H₂SO_4(l) ZnSO_4(aq) + SO_2(g) + 2H₂O_(l)

(Hot acid)

Zn_(s) + H₂SO_4(l) ZnSO_4(aq) + H_2(g)

(Cold acid)

Lead

Pb_(s) + 2H₂SO_4(l) PbSO_4(aq) + SO_2(g) + 2H₂O_(l)

(Hot; conc.) (Insoluble)

Note:

– Dilute sulphuric (VI) acid doesnt have any action on copper.

Reason:

– Copper is below hydrogen in reactivity series hence cannot displace it from the acid.

– This acid (H₂SO₄) has very little effects on lead, and usually the amount of SO₂ liberated is very little.

Reason:

– Formation of an insoluble lead sulphate layer that forms a protective coating on the metal stopping further reaction.

(b). Reaction with non-metals.

– Concentrated sulphuric acid oxidizes non-metals such as sulphur and carbon to their respective oxides.

Equations:

Ø With carbon

C_(s) + 2H₂SO_4(l) CO_2(g) + 2SO_2(g) + 2H₂O_(l)

Ø With sulphur

S_(s) + 2H₂SO_4(l) 3SO_2(aq) + 2H₂O_(l)

It is a less volatile acid; and displaces more volatile acids (refer to lab preparation of HNO₃)

Reactions of dilute sulphuric acid

Reaction with metals

– It reacts with metals above hydrogen in the reactivity series to produce a salt and hydrogen.

– With potassium and sodium, reaction is violent.

Equations:

With magnesium:

Mg_(s) + H₂SO_4(aq) MgSO_4(aq) + H_2(g)

With zinc:

Zn_(s) + H₂SO_4(aq) ZnSO_4(aq) + H_2(g)

Note:

– Copper is below hydrogen in reactivity series hence cant displace hydrogen form dilute sulphuric (VI) acid.

Reaction with carbonates and hydrogen carbonates

– Dilute H₂SO_4(aq) reacts with carbonates and hydrogen carbonates to produce a salt, carbon (IV) oxide and water.

Equations

With sodium carbonate:

Na₂CO_3(s) + H₂SO_4(aq) Na₂SO_4(aq) + CO_2(g) + H₂O_(l)

With calcium hydrogen carbonate:

CaHCO_3(s) + H₂SO_4(aq) CaSO_4(aq) + CO_2(g) + H₂O_(l)

Note:

– Reaction with lead carbonate however stops soon after the reaction.

Reason:

– Formation of an insoluble coating of the lead (II) sulphate on the lead (II) carbonate which prevents further contact between acid and carbonate.

– The same logic applies for calcium carbonate.

Reaction with oxides and hydroxides

– Reacts to form salt and water.

– However, those metal oxides whose sulphates are insoluble react only for a while.

– Thus reaction between dilute sulphuric (VI) acid and lead (II) oxide stops almost immediately.

– This is due to formation of an insoluble layer of lead (II) sulphate which effectively prevents further contact between acid and oxide.

Equations:

With magnesium oxide:

MgO_(s) + H₂SO_4(aq) MgSO_4(aq) + H₂O_(g)

(White) (Colourless solution)

With copper (II) oxide:

CuO_(s) + H₂SO_4(aq) CuSO_4(aq) + H₂O_(g)

(Black) (Blue solution)

With sodium hydroxide:

NaOH_(s) + H₂SO_4(aq) Na₂SO_4(aq) + 2H₂O_(g)

(White) (Colourless solution)

With lead (II) oxide:

PbO_(s) + H₂SO_4(aq) PbSO_4(aq) + H₂O_(g)

(Red) (White ppt; reaction stops immediately)

Uses of sulphuric (VI) acid

Manufacture of fertilizers.
Processing of metal ores.
Manufacture of detergents.
Manufacture of plastics.
Manufacture of dyes and paints.
Manufacture of lead and accumulators.
Manufacture of polymers.
Manufacture of petroleum (petroleum refinery).
Drying agent in industrial processes.

Hydrogen sulphide gas

– It is a colourless gas with a characteristic rotten egg smell; and is usually given out by rotting cabbage and eggs.

Laboratory preparation

(i). Apparatus:

Warm water

H₂S_(g)

Iron (II) sulphide

Dil. HCl

Anhydrous Dry H₂S gas

Calcium chloride

Iron (II) sulphide

Dil HCl

(ii). Procedure:

– Dilute hydrochloric acid is poured into Iron (II) sulphide in a round-bottomed flask.

– Resultant gas is passed through U-tube with anhydrous calcium chloride to dry the gas.

– This can also be done with phosphorous (V) oxide.

Equation:

FeS_(s) + 2HCl_(aq) H₂S_(g) + FeCl_2(aq)

Ionically:

S^2-_(aq) + H⁺_(aq) H₂S_(g)

(iii). Collection of gas

– When dry, the gas is collected by downward delivery because it is denser than air.

– When wet is collected over warm water because it is more soluble in cold water.

Hydrogen sulphide test.

– When a strip of filter paper soaked in aqueous lead (II) ethanoate is put in hydrogen sulphide, the paper turns black or dark brown.

Reason:

– Due to the formation of lead (II) sulphide which is black.

Equation

H₂S_(g) + (CH₂COOH)₂Pb_(aq) PbS_(s) + 2CH₃COOH_(aq)

Properties of hydrogen sulphide gas

Physical properties

Colourless and very poisonous gas (similar to hydrogen cyanide)
Has a repulsive smell (similar to that of rotten eggs or decaying cabbages)
Soluble in water giving a weak acid (only slightly ionized)

Equation:

H2_S(g) + H₂O_(l) H₂S_(aq)

Then:

H₂S_(aq) H⁺_(aq) + HS^–_(aq) 2H⁺_(aq) + S^2-_(aq)

– The acid is dibasic hence forms hydrogen sulphides.

Equation:

2NaOH_(aq) + H₂S_(g) NaHS_(aq) + 2H₂O_(l)

Note:

– Potassium hydroxide reacts similarly like sodium hydroxide.

Chemical properties

Combustion

– Burns in a blue flame in a limited supply of oxygen (air) forming a yellow deposit of sulphur and steam.

Equation:

2H₂S_(g) + O_2(g) 2SO_2(s) + 2H₂O_(g)

– In plentiful supply (excess) of Oxygen (air) it burns with a blue flame forming SO₂ and steam.

Equation:

2H₂S_(g) +3O_2(g) 2S_(s) + 2H₂O_(g)

It is a reducing agent

– It supplies electrons which are accepted by the oxidizing agent and forms sulphur.

Ionically:

H₂S_(aq) + 2H⁺_(aq) + S^2-_(aq)

Then

S^2-_(aq) S_(s) + 2e^–_(aq)

H₂S_(aq) + [O] S_(s) + H₂O_(l); in terms of addition of oxygen.

Examples

(i). With acidified K₂Cr₂O₇ solution (potassium dichromate VI)

Equation:

Reduction:

Cr₂O₇^2-_(aq) + 3H₂S_(g) + 8H⁺_(aq) 2Cr³⁺_(aq) + 7H₂O_(l) + 3S_(s)

(Orange) (Green)

Oxidation

Observation: The orange solution turns green and H₂S oxidized to yellow sulphur.

(ii). Potassium manganate (VII) (KMnO₄)

Equation:

Reduction:

2MnO₄^–_(aq) + 5H₂S_(g) + 6H⁺_(aq) 2Mn²⁺_(aq) + 8H₂O_(l) + 5S_(s)

(Purple) (Colourless)

Oxidation

Observation:

– The Purple solution turns colourless

– Manganate (VII) ions are reduced to manganate (II) ions; H₂S oxidized to yellow sulphur.

(iii). Action on Iron (III) chloride ions

Equation:

FeCl_3(aq) + H₂S_(g) 2FeCl_2(aq) + 2HCl_(aq) + S_(s)

Ionically:

Reduction:

Fe³⁺_(aq) + S^2-_(g) Fe²⁺_(aq) + 3S_(s)

(Brown) (Pale green)

Oxidation

Observation:

– The brown solution turns pale green;

– The Fe³⁺_(aq) are reduced to Fe²⁺_(aq); while the S^2-_(aq) are oxidized to yellow sulphur.

(iv). Action with Conc. HNO₃

Equation:

2HNO_3(aq) + H₂S_(g) 2H₂O_(aq) + 2NO_2(aq) + S_(s)+ Heat

Ionically:

Reduction:

2H⁺_(aq) + 2NO₃^–_(aq) + H⁺_(aq) + S^2-_(aq) 2H₂O_(l) + 2NO_2(g) + S_(s)+ Heat

(Colourless solution) (Brown) (Yellow)

Oxidation

Observation:

– Evolution of brown fumes; and deposits of a yellow solid;

– HNO_3(aq) is reduced to brown NO_2(g); while S^2-_(aq) are oxidized to yellow sulphur;

Note: The solution also contains H₂SO₄ produced by the reaction:

Reduction

2HNO_3(aq) + H₂S_(g) H₂SO_4(aq) + 8NO_2(aq) + 4H₂O_(l);

Oxidation

(v). Action of air on H₂S

– The gas is dissolved in distilled water in a beaker and exposed to air; after a few days, a white disposal is formed.

Equation:

H₂S_(g) + O_2(g) 2H₂O_(l) + 2S_(s)

(vi). Action with concentrated sulphuric (VI) acid.

Equation

Reduction

H₂SO_4(aq) + 3H₂S_(g) 4S_(s) + 4H₂O_(l)

Oxidation

(vii). Action with halogen elements

Red-brown bromine water

– Red-brown bromine water is reduced forming colourless hydrogen bromide (Hydrobromic acid) and yellow deposits (suspension) of sulphur.

Equation:

Reduction

Br_2(aq) + H₂S_(g) 2HBr_(aq) + S_(s)

(Red-brown) (Colourless) (Yellow suspension)

Oxidation

(viii). Action with hydrogen peroxide.

Equation:

Reduction

H₂O_2(aq) + H₂S_(g) 2H₂O_(l) + S_(s)

(Red-brown) (Colourless) (Yellow suspension)

Oxidation

Preparation of metallic sulphides

– Hydrogen sulphide reacts with metal ions in solution to form precipitates of metal sulphides; majority of which are black in colour.

(i). Procedure

– The gas is bubbled through solutions of the following salts: Pb (NO₃)₂, CuSO₄, FeSO₄ etc.

(ii). Observations and equations

Lead ions:

Pb(NO₃)_2(aq) + H₂S_(aq) PbS_(s) + 2HNO_3(aq)

(Colourless) (Black)

Ionically:

Pb²⁺_(aq) + S^2-_(aq) PbS_(s)

Copper (II) ions:

CuSO_4(aq) + H₂S_(aq) CuS_(s) + H₂SO_4(aq)

(Blue) (Black)

Ionically:

Cu²⁺_(aq) + S^2-_(aq) CuS_(s)

Iron (II) ions:

FeSO_4(aq) + H₂S_(aq) FeS_(s) + H2SO_4(aq)

(Pal green) (Black)

Ionically:

Fe²⁺_(aq) + S^2-_(aq) FeS_(s)

Zinc ions:

Zn(NO₃)_2(aq) + H₂S_(aq) ZnS_(s) + 2HNO_3(aq)

(Colourless) (Black)

Ionically:

Zn²⁺_(aq) + S^2-_(aq) ZnS_(s)

Note:

– Most metal sulphides are insoluble in water except those of sodium, potassium and ammonium.

Sulphites

– Are compounds of the sulphite radical (SO₃^2-) and a metallic or ammonium cation

Effects of heat

– They decompose on heating, forming SO₂;

Example:

CuSO_3(s) ^Heat CuO_(s) + SO_2(g)

Test for sulphites

(i). Procedure

– To 2cm³ of the test solution, ad 2 cm³ of BaCl₂ or Ba (NO₃)₂; i.e. addition of barium ions.

– To the mixture add 2 cm³ of dilute HCl or HNO₃.

(ii). Observation

– A white precipitate (BaSO₃) is formed which dissolves on addition of acid.

– Production of a colourless gas that turns filter paper soaked in acidified orange potassium dichromate (VI) to green.

(iii). Explanations

– Only BaSO₃; BaCO₃ and BaSO₄ form white precipitates;

– The precipitates of BaSO₃ and BaCO₃ dissolve on addition of dilute acids; unlike BaSO₄;

– BaSO₃ produces SO_2(g) as it dissolves on addition of a dilute acid; SO₂ turns orange acidified potassium dichromate (VI) to green;

– BaCO₃ of the other hand dissolves in dilute acids producing CO₂; which has no effect on K₂Cr₂O₇; but forms a white precipitate in lime water;

Equations:

On addition of Ba²⁺:

Ba²⁺_(aq) + SO₃^2-_(aq) BaSO_3(s)

(White precipitate)

On addition of dilute HCl(aq):

BaSO_3(s) + 2HCl_(aq) BaCl_2(aq) + SO_2(g) + H₂O_(l)

(White precipitate) (Colourless)

Ionically:

BaSO_3(s) + 2H⁺_(aq) Ba²⁺_(aq) + SO_2(g) + H₂O_(l)

Sulphates

– Are compounds of the sulphate radical (SO₄^2-) and a metallic or ammonium cation.

Effects of heat.

– Decompose on heating and liberate SO₂ and SO₃or SO₃alone;

– However quite a number of sulphates do not decompose on heating; and thus require very strong heating in order to decompose.

Examples:

2FeSO_4(s) ^Heat Fe₂O_3(s) + SO_2(g) + SO_3(g)

(Pale green) (Brown) (Colourless gases)

CuSO_4(s) ^Heat CuO_(s) + SO_3(g)

(Blue) (Black) (Colourless)

Action of acids

Test for sulphates

– To about 2 cm³ of the test solution, 2 cm³ of BaCl₂ or Ba (NO₃)₂ solution is added.

– To the mixture, 2 cm₃ of dilute HCl or HNO₃ is added.

Observation

– A white precipitate is formed when Ba (NO₃)₂ is added; which is insoluble in excess acid.

Explanations.

– Only BaSO₃; BaCO₃ and BaSO₄ form white precipitates;

– The precipitates of BaSO₃ and BaCO₃ dissolve on addition of dilute acids; unlike BaSO₄;

– Thus the white precipitate insoluble in dilute HCl or HNO₃ could only be a sulphate; in this case barium sulphate.

Equations:

On addition of Ba²⁺:

Ba²⁺_(aq) + SO₄^2-_(aq) BaSO_4(s)

(white precipitate)

On addition of dilute acid:

BaSO_4(s) + 2HCl_(aq) BaSO_4(s) + 2HCl_(aq); i.e. no effect;

(White precipitate) (White precipitate)

Pollution by sulphur compounds.

– Main pollutants are sulphur (IV) Oxide and hydrogen sulphide.

(a). Sulphur (IV) oxide.

– SO₂ is emitted when sulphur-containing fuels are burnt; during extraction of metals like copper and in manufacture of sulphuric (VI) acid.

– SO₂ is oxidized to SO₃;

– SO₃ reacts with water in atmosphere to form sulphuric (VI) acid which comes down as acid rain or acid fog.

Acid rain (fog) has environmental effects:

Leaching of minerals in soil;
Erosion of stone work on buildings;
Corrosion of metallic structures;
Irritation of respiratory systems thus worsening respiratory illnesses;
Death of plants as a result of defoliation (falling of leaves);
Destruction of aquatic life in acidified lakes;
Stunted plant growth due to chlorosis;

(b). H₂S is very poisonous.

UNIT 5: CHLORINE AND ITS COMPOUNDS.

Unit Checklist:

About chlorine.
Preparation of chlorine.
Properties of chlorine.

Colour and smell
Solubility in water
Action on litmus paper
Bleaching action
Action on hot metals
Reaction with non-metals
Oxidation reactions
Reaction with alkalis
Effect of sunlight on chlorine water.

Industrial manufacture of chlorine (The mercury cathode cell)
Uses of chlorine and its compounds
Hydrogen chloride gas

Preparation
Properties

Test for chlorides.
Hydrochloric acid

Large scale manufacture
Uses of hydrochloric acid

Environmental pollution of chlorine and its compounds

Introduction:

– Chlorine is a molecular non-metallic element made up of diatomic molecules.

– Its electron arrangement is 2.8.7 and it belongs to the halogen family.

Preparation of chlorine.

Note: It is usually prepared by oxidation of concentrated hydrochloric acid by removal of hydrogen.

Equation:

2HCl_(aq) + [O] Cl_2(g) + H₂O_(l)

– The [O] is from a substance containing oxygen.

(a). Preparation of chlorine from MnO₂ and HCl.

(i). Apparatus:

(ii). Conditions:

– Heating;

– Presence of an oxidizing agent; in this case it is manganese (IV) oxide.

(iii). Procedure:

– Hydrochloric acid is reacted with manganese (IV) oxide (dropwise);

Equation:

MnO_2(s) + 4HCl_(aq) ^Heat MnCl_2(aq) + 2H₂O_(l) + Cl_2(g)

(iv). Explanation:

– Manganese (IV) oxide oxidizes hydrochloric acid by removing hydrogen resulting into chlorine.

– The manganese (IV) oxide is reduced to water and manganese chloride.

– The resultant chlorine gas is passed through a bottle containing water.

Reason:

– To remove hydrogen chloride fumes (gas) which is very soluble in water.

– Next it is passed through concentrated sulphuric acid or anhydrous calcium chloride; to dry the gas.

(v). Collection:

(a). Wet chlorine is collected over brine (saturated sodium chloride solution) or hot water.

Reason:

– It does not dissolve in brine and is less soluble in water

(b). Dry chlorine is collected by downward delivery (upward displacement of air)

Reason:

– It is denser than air (2.5 times).

Note:

– Chlorine may also be dried by adding calcium chloride to the jar of chlorine.

(c). The first bottle must contain water and the second concentrated sulphuric acid.

Reason:

– If the gas is first passed through concentrated sulphuric acid in the first bottle then to the water; it will be made wet again.

Properties of chlorine gas.

Colour and smell.

Caution: Chlorine is very poisonous.

– It is a green-yellow gas with an irritating pungent smell that attacks the nose and the lungs.

– It is 2.5 times denser than air, hence can be collected by downward delivery.

Solubility in water.

– It is fairly soluble in water forming green-yellow chlorine water.

Equation:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

– Chlorine water is composed of two acids; chloric (I) acid (hypochlorous acid) and hydrochloric acid.

Action on litmus paper.

– Moist chlorine turns litmus paper red then bleaches it.

– Dry chlorine turns damp blue litmus paper red then bleaches it.

– Moist chlorine bleaches red litmus paper; dry chlorine bleaches damp red litmus paper.

– Dry chlorine has no effect on dry litmus paper.

Reasons:

(i). In presence of moisture chlorine forms chlorine water which is acidic and hence turns blue litmus paper red.

(ii). Hypochlorous acid in the chlorine water is an oxidizing agent; thus adds oxygen (oxidizes) to the colour of most dyes; hence bleaching it.

Equations:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

Acidic solution

Then:

Dye + HOCl_(aq) HCl_(aq) + {Dye + [O]}

Coloured Colourless

Bleaching action.

– Moist chlorine bleaches dyes but not printers ink which is made of carbon.

– The colour change is due to oxidation by hypochlorous acid.

Equations:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

Acidic solution

Then:

Dye + HOCl_(aq) HCl_(aq) + {Dye + [O]}

Coloured Colourless

Action on a burning splint.

– The gas put out a glowing splint. It does not burn.

Action on hot metals.

(a). Preparation of iron (III) chloride.

(i). Apparatus.

(ii). Precaution.

– Experiment should be done in a fume cupboard or in the open.

Reason:

– Chlorine gas is poisonous and will thus be harmful to the human body.

(iii). Procedure:

– Dry chlorine gas is passed over iron wool as per the diagram.

(iv). Conditions.

Chlorine gas has to be dry (done by the anhydrous calcium chloride in the U-tube)

Reason:

To prevent hydration hence oxidation of iron (which will then form Fe₂O₃.5H₂O) hence preventing reaction between iron and chlorine.

Iron metal must be hot; and this is done by heating.

Reason:

To provide activation energy i.e. the minimum kinetic energy which the reactants must have to form products.

Anhydrous calcium chloride.

– In the U-tube; to dry the chlorine gas.

– In the thistle funnel; to prevent atmospheric water vapour (moisture) from getting into the apparatus and hence reacting with iron (III) chloride.

(v). Observations:

– Iron metal glows red-hot.

– Red brown fumes (FeCl_3(g)) are formed in the combustion tube.

– A black solid (FeCl_3(s)) is collected in the flask.

Note:

– Iron (III) chloride cannot be easily collected in the combustion tube.

Reason:

– It sublimes when heated and hence the hotter combustion tube causes it to sublime and its vapour is collected on the cooler parts of the flask.

(vi). Reaction equation.

2Fe_(s) + 3Cl_2(g) 2FeCl_3(g)

(vii). Conclusion.

– Iron (III) chloride sublimes on heating; the black solid changes to red-brown fumes on heating.

Equation:

FeCl_3(s) FeCl_3(g)

(black) (Red-brown)

(b). Aluminium chloride.

2Al_(s) + 3Cl_2(g) 2FeCl_2(s)

2Al_(s) + 3Cl_2(g) Al₂Cl_6(s)

Note:

– Aluminium chloride also sublimes on heating.

Equation:

AlCl_3(s) AlCl_3(g)

(White) (White)

(c). Reaction with burning magnesium.

(i). Procedure:

– Burning magnesium is lowered into a gar jar of chlorine gas.

(ii). Observations:

– The magnesium continues to burn with a bright blinding flame;

– Formation of white fumes (MgCl₂); which cools into a white powder.

(iii). Equation:

Mg_(s) + Cl_2(g) MgCl_2(s)

– Generally chlorine reacts with most metals when hot top form corresponding chlorides.

Note:

Where a metal forms two chlorides when it reacts with chlorine, the higher chloride is usually formed.

Reason:

The higher chloride is stable. This explains why reactions of chlorine with iron results into iron (III) chloride and not iron (II) chloride.

Reaction with non-metals.

– It reacts with hot metals; forming covalent molecular compounds.

(a). Reaction with phosphorus.

(i). Procedure:

– A piece of warm phosphorus is lowered into a gas jar of chlorine.

(ii). Observations:

– Phosphorus begins to smoulder and then ignites spontaneously.

– Evolution of white fumes (PbCl₃ and PCl₅)

(iv). Explanation.

– Chlorine reacts with warm dry phosphorus to form white fumes of phosphorus (III) and (V) chlorides.

Equations:

P_4(s) + 6Cl_2(g) 4PCl_3(s)

(With limited chlorine)

P_4(s) + 10Cl_2(g) 4PCl_5(s)

(With excess chlorine)

(b). Reaction with hydrogen.

(i). Conditions:

– Heating or presence of light; since chlorine and hydrogen do not react with each other at room temperature.

(ii). Precaution:

– The experiment is performed in a fume chamber (cupboard); since the reaction is explosive;

(iii). Procedure:

– Chlorine gas is mixed with hydrogen gas and the mixture heated or exposed to direct light; then aqueous ammonia brought near the mouth of the jar.

(iv). Observations:

– White fumes at the mouth of the jar.

(v). Explanations:

– Chlorine reacts explosively with hydrogen to form hydrogen chloride gas.

Equation:

Cl_2(g) + H_2(g)^{Heat/ Light} 2HCl_(g).

– The hydrogen chloride gas diffuses upwards and reacts with ammonia at the mouth of the test tube to form white fumes of ammonium chloride; NH₄Cl.

Equation:

HCl_(g) + NH_3(g) NH₄Cl_(g)

White fumes.

Chlorine as an oxidizing agent.

– Chlorine is a strong oxidizing agent and oxidizes many ions, by readily accepting electrons.

– During the process, chlorine itself undergoes reduction.

(a). Reaction with hydrogen sulphide gas.

(i). Procedure:

– A gas jar full of chlorine gas is inverted into another containing hydrogen sulphide gas.

(ii). Apparatus:

(iii). Observations:

– Yellow deposits (of sulphur)

– Misty fumes (hydrogen chloride gas)

(iv). Explanations:

– Chlorine oxidizes hydrogen sulphide gas to sulphur solid, while itself is reduced to hydrogen chloride gas.

Equation: _Oxidation

Cl_2(g) + H₂S_(g) 2HCl_(g)+ S_(s)

^Reduction

(v). Conditions:

– At least one of the gases must be moist; they do not react with each other in absence of moisture.

Note:

– In absence of moisture both gases are still in molecular form and hence cannot react; water facilitates their ionization hence ability to react.

– If aqueous hydrogen sulphide is used, then sulphur forms as a yellow suspension on the acidic solution.

Equations:

Stoichiometric:

Cl_2(g) + H₂S_(aq) 2HCl_(aq)+ S_(s)

Ionic:

Cl_2(g) + S^2-_(g) 2Cl^–_(g)+ S_(s)

(b). Reaction with sodium sulphite.

Procedure:

– Chlorine gas is bubbled through sodium sulphate in a beaker.

– Resulting solution is then divided into two portions.

– To the first portion, drops of dilute nitric acid are added followed by few drops of barium nitrate solution.

– To the second portion, few drops of lead (II) nitrate are added and the mixture warmed then cooled.

(ii). Observations:

1^st portion: White precipitate formed indicating presence of SO₄^2-;

Explanations:

– The white precipitate indicate presence of SO₄^2-; the precipitate is barium sulphate Ba(SO₄)₂;

– Chlorine oxidizes SO₃^2- in Na₂SO₃ to SO₄^2- while itself is reduced to chloride ions;

Equations:

H₂O_(l) + Cl_2(g) + Na₂SO_3(aq) Na₂SO_4(aq) + 2HCl_(aq)

Ionically:

Cl_2(g) + SO₃^2-_(aq) + H₂O_(l) SO₄^2-_(aq) +2H⁺_(aq) + 2Cl^–_(aq)

– On adding barium nitrate (Ba(NO₃)₂); the Ba²⁺ ions react with the SO₄^2-to form insoluble BaSO₄; the white precipitate.

Ionically;

Ba²⁺_(aq)+ SO₄^2-_(aq)BaSO_4(s)

(White precipitate)

Note:

– The solution is first acidified (with HNO₃) before addition of Ba(NO₃)₂ to prevent precipitation of BaSO_3(s)and BaCO_3(s).

2^nd portion:

Observation:

– Formation of a white precipitate on addition of Pb(NO₃)₂solution.

– On warming the white precipitate dissolves then recrystalizes back on cooling.

Explanations:

– The white precipitate shows presence of either Cl^–; SO₃^2-orSO₄^2-

– However the fact that it dissolves on warming confirms the presence of Cl^–_(aq) and not SO₃^2-_(aq) and SO₃^2-_(aq)

Equation:

Pb²⁺_(aq)+ Cl^–_(aq)PbCl_2(s)

(White precipitate soluble on warming)

(c). Reaction with ammonia.

(i). Procedure:

Chlorine gas is bubbled through aqueous ammonia.

(ii). Observations:

– Evolution of white fumes.

(iii). Explanation.

– Chlorine gas oxidizes ammonia to nitrogen, while is itself reduced to white fumes of ammonium chloride.

Equation: _Reduction

8NH_3(g) + 3Cl_2(g) 6NH₄Cl_(g)+ N_2(s)

^Oxidation

(d). Displacement reactions with other halogens.

(i). Procedure:

– Chlorine is bubbled through aqueous solutions of fluoride, bromide and iodide ions contained in separate test tubes.

(ii). Observations and explanations:

With fluoride ions.

– No observable change or no reaction; because chlorine is a weaker oxidizing agent than fluorine.

With bromide ions:

– If potassium bromide was used, the colourless solution turns red-brown.

Reason:

– Chlorine has a higher tendency to gain electrons than bromine.

– It readily oxidizes bromide ions (in KBr) to form potassium chloride and bromine which immediately dissolves to make the solution red-brown.

Equation: _Reduction

2KBr_(aq) + Cl_2(g) 2KCl_(aq)+ Br_2(l)

^Oxidation^{Red brown}

Ionically;

2Br^–_(aq) + Cl_2(g) 2Cl^–_(aq) + Br_2(l)

With iodide ions.

– Using potassium iodide the colourless solution would turn black.

Reason:

– Chlorine has a higher tendency to gain electrons that iodine.

– It readily oxidizes the I^– (in KI) to form iodine and potassium chloride.

– Iodine solid in the resulting solution makes it black.

Equation: _Reduction

2KI_(aq) + Cl_2(g) 2KCl_(aq)+ I_{2(l) (black)}

^Oxidation

Ionically;

2I^–_(aq) + Cl_2(g) 2l^–_(aq) + Br_2(l)

Reaction with alkalis.

(a). Reaction with sodium hydroxide solution.

(i). Procedure:

– Bubble chlorine slowly through cold dilute sodium hydroxide solution.

– Dip litmus paper.

(ii). Observation:

– Litmus paper is bleached; the product has the colour and smell of chlorine.

(iii). Explanation:

– Chlorine dissolves in sodium hydroxide to form a pale yellow solution of sodium chlorate (I) or sodium hypochlorite (NaClO);

– The sodium chlorate (I) bleaches dyes by oxidation.

Equation:

Cl_2(g)+ 2NaOH_(l) NaCl_(aq) + NaClO_(aq)+ H₂O_(l)

Pale yellow solution

Bleaching action of NaClO:

– The NaClO donates oxygen to the dye making it colourless; and thus it bleaches by oxidation.

Equation:

Dye + NaClO_(aq) NaCl_(aq) + {Dye + [O]}

Coloured Colourless

Note:

With hot concentrated sodium hydroxide, the chlorine forms sodium chlorate (III); NaClO₃.

Equation:

3Cl_2(g)+ 6NaOH_(l) 5NaCl_(aq) + NaClO_3(aq)+ 3H₂O_(l)

(b). Reaction with potassium hydroxide

– Follows the trend of sodium.

(c). Reaction with slaked lime {Ca(OH)_2(s)}

Equation:

Cl_2(g)+ Ca(OH)_2(l) CaOCl_2(aq)+ 3H₂O_(l)

Calcium chlorate I

Note:

Bleaching powder, CaOCl₂ always smells of strongly of chlorine because it reacts with carbon (IV) oxide present in the atmosphere to form chlorine.

Equation:

CaOCl_2(s) + CO_2(g) CaCO_3(s) + Cl_2(g)

Effects of chlorine gas on:

(a). A burning candle.

(i). Procedure:

– A burning candle is lowered into a gas jar of chlorine.

(ii). Observations:

– It burns with a small, red and sooty flame.

(iii). Explanations:

– Wax (in candles) consists of mainly hydrocarbons.

– The hydrogen of the hydrocarbon reacts with chlorine forming hydrogen chloride while leaving behind carbon.

(b). warm turpentine.

(i). Procedure:

– A little turpentine is warmed in a dish and a filter paper soaked (dipped) in it.

– The filter paper is then dropped into a gas jar of chlorine.

(ii). Observation:

– There is a red flash accompanied by a violent action whilst a black cloud of solid particles form.

(iii). Conclusion:

– Black cloud of slid is carbon.

– Turpentine (a hydrocarbon) consists of hydrogen and carbon combined together.

– The chlorine combines with hydrogen and leaves the black carbon behind.

Equation:

C₁₀H_16(l) + 8Cl_2(g) 16HCl_(g) + 10C_(s)

Effects of sunlight on chlorine water.

(i). Procedure:

– Chlorine water is made by dissolving the gas in water.

– A long tube filled with chlorine water is inverted over a beaker containing water.

– It is then exposed to sunlight (bright light) as shown below.

(ii). Apparatus:

(iii). Observations:

– After sometime a gas collects in the tube and on applying a glowing splint, the splint is rekindles showing that the gas collected is oxygen.

(iv). Explanation:

– Chlorine water has two components.

Equation:

Cl_2(g) + H₂O_(l) HCl_(aq) + HOCl_(aq)

– The HOCl being unstable will dissolve on exposure to sunlight, giving out oxygen.

Equation:

2HOCl_(aq) 2HCl_(aq) + O_2(g) (slow reaction)

Overall reaction:

2H₂O_(l) + 2Cl_2(g) 4HCl_(aq) + O_2(g)

Industrial manufacture of chlorine (the mercury cathode cell)

The electrolysis of brine

(i). Apparatus.

(ii). Electrolyte.

– Brine, concentrated sodium chloride solution, NaCl

(iii). Electrodes.

Anode: carbon (graphite)

Cathode: Flowing mercury;

(iv). Ions present:

NaCl_(aq) Na⁺_(aq) + Cl^–_(aq)

H₂O_(l) H⁺_(aq) + OH^–_(aq)

(v). Reactions:

Anode:

– Cl^– and OH^– migrate to the anode.

– Because of high concentration of Cl^–_(aq), they are discharged in preference to OH^– ions.

Equation:

2Cl^–_(aq) Cl_2(g) + 2e^–

(Green-yellow)

Cathode:

– H⁺_(aq) and Na⁺_(aq) migrate to the cathode.

– Because the cathode is made of mercury, Na⁺_(aq) is discharged in preference to H⁺_(aq) ions;

Equation:

2Na⁺_(aq) + 2e^– 2Na_(s)

Note:

– Sodium formed at the cathode dissolves in the flowing mercury cathode to form sodium amalgam (Na/Hg).

– Sodium amalgam is reacted with water to form sodium hydroxide and hydrogen.

– Mercury (in the sodium amalgam) remains unreacted.

Equation:

2Na/Hg_(l) + 2H₂O_(l) 2NaOH_(aq) + H_2(g) + 2Hg_(l)

– The unreacted mercury is recycled.

(vi). Products:

– Chlorine gas at the anode.

– Hydrogen and sodium hydroxide at the cathode.

Uses of chlorine gas and its compounds.

Manufacture of hydrochloric acid.
Used in form of bleaching powder in textile and paper industries.
For sterilization of water for both domestic and industrial use and in swimming pools.
Used in sewage treatment e.g. NaOClO₃ solution used in latrines.
Manufacture of plastics (polyvinyl chloride; PVC)
Manufacture of germicides, pesticides and fungicides e.g. DDT and some CFCs.
CFCs are used to manufacture aerosol propellants.
Manufacture of solvents such as trichloromethane and some chlorofluorocarbons (CFCs).
CFCs are commonly freons are used as refrigerants in fridges and air condition units due to their low boiling points.
Manufacture of chloroform, an aesthetic.

Hydrogen chloride gas.

Laboratory preparation of hydrogen chloride gas.

(i). Apparatus:

(ii). Procedure:

– Concentrated sulphuric acid is reacted with sodium chloride, and the mixture heated gently.

– Resultant gas is passed through conc. Sulphuric (VI) acid; to dry the gas.

(iii). Equation:

H₂SO_4(l) + NaCl_(aq) NaHSO_4(s) + HCl_(g)

Ionically;

H⁺_(aq) + Cl^–_(aq) HCl_(g)

Note:

– The reaction can proceed in the cold, but on large scale HCl(g) is produced by the same reaction but the heating is continued to re hot.

Properties of hydrogen chloride gas.

Colourless gas with a strong irritating pungent smell.
Slightly denser than air (1¼ times). This makes it possible to collect the gas by downward delivery.
Very soluble in water; and fumes strongly in moist air forming hydrochloric acid deposits.

Diagram:

– The aqueous solution is known as hydrochloric acid.

– It is almost completely ionized (a strong acid) in aqueous solution.

Equation:

HCl_(aq) H⁺_(aq) + Cl^–_(aq)

– This solution has the usual acidic properties:

Examples:

(i). turns blue litmus red.

(ii). Liberates hydrogen gas with certain metals e.g. zinc, Magnesium, iron etc.

Note:

Hydrochloric acid does not react with metals below hydrogen in the reactivity series.

Equations:

Zn_(s) + 2HCl_(aq) ZnCl_2(aq) + H_2(g)

Mg_(s) + 2HCl_(aq) MgCl_2(aq) + H_2(g)

Fe_(s) + 2HCl_(aq) FeCl_2(aq) + H_2(g)

(iii). Neutralizes bases to form salt and water.

Examples:

HCl(aq) + NaOH(aq) NaCl(aq) +H₂O(l)

2HCl(aq) + CuO(s) CuCl₂(aq) + H₂O(l)

(iv). Liberates carbon (IV) oxide from carbonates and hydrogen carbonates.

Examples:

CaCO_3(s) + 2HCl_(aq) CaCl_2(aq) + H₂O_(l) + CO_2(g)

ZnCO_3(s) + 2HCl_(aq) ZnCl_2(aq) + H₂O_(l) + CO_2(g)

NaHCO_3(s) + HCl_(aq) NaCl_(aq) + H₂O_(l) + CO_2(g)

Note:

As the hydrogen chloride gas very soluble in water, the solution must be prepared using a funnel arrangement; to prevent sucking back and increase the surface area for the dissolution of the gas;

Diagram: dissolution of hydrogen chloride gas

Dry hydrogen chloride is NOT particularly reactive at ordinary temperatures, although very reactive metals burn in it to form the chloride and hydrogen gas.

Equation:

2Na_(s) + 2HCl_(aq) 2NaCl_(s) + H_2(g)

Metals above hydrogen in the reactivity series react with hydrogen chloride gas when heated.

Note:

If reacted with some metals it forms 2 chlorides e.g. iron where iron (II) and iron (III) chlorides exist.

Hydrogen chloride gas forms white fumes of ammonium chloride when reacted with ammonia gas;

Equation:

NH_3(g) + HCl_(g) NH₄Cl_(s)

Note: This is the chemical test for hydrogen chloride gas.

Hydrogen chloride is decomposed by oxidizing agents, giving off chlorine.

Examples:

PbO_2(s) + 4HCl_(g) PbCl_2(s) + 2H₂O_(l) + Cl_2(g)

MnO_2(s) + 4HCl_(g) MnCl_2(s) + 2H₂O_(l) + Cl_2(g)

Diagram: reacting hydrogen chloride with an oxidizing agent.

Test for chlorides.

Test 1: Using silver ions:

Procedure:

– To the test solution, add silver ions from silver nitrate.

– Acidify with dilute nitric acid.

(ii). Observations and inference:

– Formation of a white precipitate shows presence of Cl^–_(aq)

(iii). Explanations:

– Only silver carbonate and silver chloride can be formed as white precipitates.

– Silver carbonate is soluble in dilute nitric acid but silver chloride is not.

Equations:

– Using Cl^– from NaCl as the test solution;

NaCl_(aq) + AgNO_3(aq) NaNO_3(aq) + AgCl_(s)

White ppt.

Ionically;

Ag⁺_(aq) + Cl^–_(aq) Ag_(s)

White ppt.

Note:

– This precipitate dissolves in excess ammonia.

– The white precipitate of silver chloride turns violet when exposed to light.

Test 2: Using lead ions

(i) Procedure:

– To the test solution, add lead ions from lead (II) nitrate, then warm

(ii). Observations and inference:

– Formation of a white precipitate that dissolves on warming shows presence of Cl^–_(aq)

(iii). Explanations:

– Only lead carbonate, lead sulphate, lead sulphite and lead chloride can be formed as white precipitates.

– Only lead chloride dissolves on warming; unlike the rest which are insoluble even on warming.

Equations:

Using Cl^– from NaCl as the test solution;

2NaCl_(aq) + Pb(NO₃)_2(aq) 2NaNO_3(aq) + PbCl_2(s)

White ppt.

Ionically;

Pb²⁺_(aq) + Cl^–_(aq) PbCl_2(s)

White ppt.

Hydrochloric acid.

Large scale manufacture of hydrochloric acid.

(i). Diagram:

(ii). Raw materials:

– Hydrogen obtained as a byproduct of petroleum industry; electrolysis of brine or from water by Bosch process;

– Chlorine obtained from the electrolysis of brine or as fused calcium chloride.

(iii). Procedure:

– A small sample of hydrogen gas is allowed through a jet and burnt in excess chlorine gas.

Equation:

H_2(g) + Cl_2(g) 2HCl_(g)

Precaution: A mixture of equal volumes of hydrogen and chlorine explodes when put in sunlight.

– The hydrogen chloride gas formed is dissolved in water over glass beads.

– The glass beads increase the surface area over which absorption takes place.

– Commercial hydrochloric acid is about 35% pure.

– Hydrochloric acid is transported in steel tanks lined inside with rubber.

– If the acid comes into contact with exposed parts of metal or with rust, it forms iron (III) chloride that makes the acid appear yellow.

Pollution in an industry manufacturing hydrochloric acid.

(i). Chlorine is poisonous.

(ii). Mixture of hydrogen and oxygen in air is explosive when ignited.

Uses of hydrochloric acid.

Sewage treatment.
Treatment of water (chlorination) at the waterworks.
Removing rust from metal e.g. descaling iron before it is galvanized or and other metals before they are electroplated.
Making dyes, drugs and photographic materials like silver chloride on photographic films.

Environmental pollution by chlorine and its compounds.

Chlorine may dissolve in rain and fall as acid rain, which has adverse effects on plants and animals, buildings and soil nutrients.
CFCs are non-biodegradable. Over time, they diffuse into the atmosphere breaking down to free chlorine and fluorine atoms. These atoms deplete the ozone layer. Chlorine is thus one of the greenhouse gases.
PVCs are non-biodegradable.
DDT is a pesticide containing chlorine and has a long life span, affecting plants and animal life.

Note: DDT is banned in Kenya; NEMA advises increased use of pyrethroids in mosquito control.

ORGANIC CHEMISTRY I

Contents checklist.

ORGANIC CHEMISTRY

Definition

– The chemistry of hydrogen carbon chain compounds.

– It the study of carbon compounds except the oxides of carbon i.e. CO, CO₂ and Carbons.

ORGANIC CHEMISTRY I: THE HYDROCARBONS

Hydrocarbons

Are compounds of hydrogen and carbon only; and are the simplest organic compounds.

Main groups of hydrocarbons

Are classified on the basis of the type of bonds found within the carbon atoms.

Alkanes: Are hydrocarbons in which carbon atoms are linked by single covalent bonds.
Alkenes: Carbon atoms are held by at least one double bond.
Alkynes: Have at least one triple bond between any tow carbon atoms.

Saturated and unsaturated hydrocarbons

(a). Saturated hydrocarbons

– Are hydrocarbons which the carbon atoms are bonded to the maximum number of other atoms possible.

– hydrocarbons which don react and hence cannot decolourise both Bromine water and acidified potassium manganate (VII).

– They are compounds in which each carbon atom has only single covalent bonds, throughout the structure.

(b). Unsaturated hydrocarbons

– Are hydrocarbons which contain at least one double or bond, between any two adjacent carbon atoms.

– The carbon atoms do not have maximum covalency.

– They can decolourise both bromine water and acidified potassium manganate (VII).

Examples: All alkenes and Alkynes.

Experiment: To verify saturated and unsaturated hydrocarbons.

Procedure:
– 3 to 4 drops of bromine wate are added to about 1 cm3 of the liquid under investigation.

– The mixture is then shaken thoroughly and the observations recorded;

– For gases the gas under investigation is bubbled ito 1 cm3 of bromine water;

– The procedures are then repeated with acidified potassium manganate (VII);

Observations:

COMPOUND	OBSERVATIONS
COMPOUND	With potassium permanganate	With Bromine water
Kerosene	No observable colour change	No colour change
Laboratory gas	No observable colour change	No observable colour change
Turpentine	Purple colour turns colourless	Solution is decolourised
Hexane	No observable colour change	No observable colour change
Pentene	Potassium permanganate is decolourised	Solution is decolourised

Conclusion

– Kerosene, laboratory gas and hexane are saturate hydrocarbons

– Turpentine and pentane are unsaturated hydrocarbons.

Homologous series

– Refers to a group of organic compounds that have the same general formula, whose consecutive members differ by a similar unit, and usually have similar chemical properties.

Characteristics of a Homologous series.

(i). Can be represented by a general formula;

(ii). Have similar chemical properties

(iii). Have similar structures and names

(iv). They show a steady gradation of physical properties

(v). Can usually be prepared by similar methods.

Structural and molecular formula

Molecular formulae

– Simply shows the number and type of elements (atoms) in the compound.

Structural formula

Shows how the different atoms in the molecules (of a compound) are bonded or joined together.

Example:

Methane

Molecular formula CH₄;

Structural formula

│

H – C – H

│

Alkanes

Are the simplest hydrocarbons with the general formula; C_nH_{2n + 2} where n = number of carbon atoms in the molecule.

Examples:

– For compound with only 1 carbon atom, formula = CH₄

– 2 carbon atoms; the formula = C₂H₆

Names and formulas of the first 10 Alkanes

Note:

Consecutive members of the alkane series differ by a CH₂-unit, hence a homologous series.

(a). General formula

– The Alkanes have a general formula C_nH_2n+2 where n is the number of carbon atoms in the molecule.

Example:

When n = 3, (2n + 2) = 8, and the alkane has the formula C₃H₈ (Propane)

(b). Structure

– In all Alkanes the distribution of bonds around each carbon atom is tetrahedral.

Example: Methane

(c). Homologous series

– The Alkanes differ from each other by a CH₂-.

– Thus methane, CH₄differs from ethane, C₂H₆ by CH₂-, and ethane in turn differs from propane C₃H₈ by C ₂-.

– They therefore form a homologous series.

(d). Functional groups

– A functional group is a part of a compound which has a characteristic set of properties.

– Thus when a bromine atom replaces a hydrogen atom in an alkane, it imparts to the compound new chemical and physical properties.

Examples: six important functional groups.

(e). Isomerism

– Is a situation whereby two or more compounds have similar molecular formulae but different structural formula.

– Such compounds are called isomers, i.e compounds with the same molecular formula but different structural formula.

Examples: For Butane, (C₄H₁₀) there are two possible structures.

Isomers have different physical and chemical properties.

Example: Ethanol and dimethyl ether.

– Molecular formula: both have C₂H₆O

Structural formula:

(i). Ethanol (ii). Dimethyl ether

Differences

Ethanol

Dimethyl ether

– A liquid of boiling point 78.4^oC

– Completely soluble in water

– Reacts with sodium ethoxide and liberates hydrogen gas

– A gas at room temperature (B.P 24⁰C).

– Slightly soluble in water.

– Does not react with sodium metal.

(f). Alkyl groups

– Is a group formed by the removal of a hydrogen atom form a hydrocarbon.

– Alkyl groups dont exist on their own but are always attached to another atom or group.

Naming of alkyl groups

– Is done by removing the ending -ane from the parent alkane and replacing it with yl.

Examples

Methane (CH₄) gives rise to Methyl -CH₃

Ethane (C₂H₆) gives rise to ethyl, – C₂H₅i.e. -CH₂CH₃

Propane (C₃H₈) gives rise to Propyl, – C₃H₇ // -CH₂CH₂CH₃;

(g). Nomenclature of Alkanes

– Generally all Alkanes end with the suffix -ane;

– Alkanes can either be straight chain or branched.

(i). Straight chain Alkanes

– The names of all Alkanes end with the suffix -ane;

Examples:

Methane, ethane, propane, butane.

– With the exception of the first 4 members of the series (i.e. the 4 listed above) the names of Alkanes begin with a Greek prefix indicating the number of carbon atoms in the main chain.

Examples: – Pentane 5 carbon atoms

Hexane 6 carbon atoms.

(ii). Branched Alkanes

The naming of branched chain Alkanes is based on the following rules:-

The largest continuous chain of carbon atoms in the molecule is used to deduce the parent name of the compound.
The carbon atoms of this chain are numbered such that the branching // substituents are attached to the carbon atom bearing the lowest number.
The substituent // branch is named e.g. methyl, ethyl etc and the name of the compound written as one word.

Examples

Further examples

H H H CH₂CH₂CHCH₂CH₃

│ │ │ │ │

H – C – C – C – H CH₃ CH₂

│ │

H – C – H CH₃

│ 3-ethylhexane;

2-methylpropane;

Further examples.

CH₃CH₂CH₂CH₃

│

CH₃

3-methylpentane;

CH₃

│

H₃C – C – CH₃

│

CH₃

2, 2-dimethylpropane;

Note: refer to course books and draw as many examples as possible.

Draw the structural isomers of:

Butane.

Pentane;

Hexane;

(f). Occurrence of Alkanes

– There are 3 known natural sources:

(i). Natural gas: this consists of mainly of methane;

(ii). Crude oil:

– Consists of a mixture of many Alkanes

– It can be separated into its components by fractional distillation.

Reason:

– The different components have different boiling points.

(iii). Biogas: This contains about 60-75% of methane gas/marshy gas.

Separation of the components of crude oil.

(i). Apparatus

(ii). Procedure

– The apparatus is arranged as shown above.

– The first distillate appears at about 120^oC and is collected, the of 40^oC intervals thereafter until the temperatures reach 350^oC.

(iii). Observations and explanations

– This method of separation is called fractional distillation, and depends on the fact that the various components of the mixture have different boiling points.

– The various fractions vary in properties as explained below.

(a). Appearance

– Intensity of the colour increases with increase in boiling point.

– Boiling point increases with increasing number of carbon atoms.

Reason:

– The higher the number of carbon atoms, the higher the number of covalent bonds.

– Thus the first fraction to be distilled (lab gas) is colourless while the last distillates (between) is dark black in colour.

(b). Viscosity

– Increases with increasing boiling point;

– The fractions with low boiling points are less viscous while the fraction with the highest boiling point is semi-solid;

(c). Inflammability:

– Decreases with increasing boiling points.

– The gaseous fractions, with least boiling points readily catches fire // burn, while the semi-solid fractions with very high boiling points are almost non-combustible.

Note: Some Hydrocarbons are found in more than one fraction of crude oil and more advanced chemical methods are necessary for complete separation.

Uses of the various fractions of crude oil.

No. f carbon atom per molecule	Fractions	Uses
1-4	Gases	Laboratory gases and gas cookers
5-12	Petrol	Fuel in petrol engines
9-16	Kerosene (paraffin)	Fuel for jet engines (aeroplanes) and domestic uses
15-18	Light diesel oils	Fuel for heavy diesel engines e.g. for ships
18-25	Diesel oils	Fuel for diesel engines
20-70	Lubricating oils	Used for smooth running of engine parts
>70	Bitumen	Road tarmacking

Changes // gradation of physical properties across the alkane homologous series

Name of alkane

Formula

State of room temperature (208K)

M.P (K)

B.P (K)

Density

(g cm^-3)

Solubility

Methane

Ethane

Propane

Butane

Pentane

Hexane

Heptane

Octane;

Nonane

Decane

CH₄

C₂H₆

C₃H₈

C₄H₁₀

C₅H₁₂

C₆H₁₄

C₇H₁₆

C₈H₁₀

C₉H₂₀

C₁₀H₂₂

↑

Gaseous

↓

↑

Liquid

↓

138

143

178

243

112

184

231

273

309

342

447

0.424

0.546

0.582

0.579

0.626

0.659

0.730

Preparation and chemical properties of Alkanes

Note:

– Alkanes, like any other Homologous series have similar chemical properties.

– Generally any alkane can be represented form the reaction represented by the following equation:

C_nH_{2n + 1}COONa + NaOH_(aq) → C_nH_{2n +2} + Na₂CO_3(aq);

Thus;

– Methane can be prepared form sodium ethanoate (CH₃COONa)

– Ethane can be prepared form sodium propanoate (CH₃CH₂COONa)

– Propane can be prepared form sodium Butanoate (CH₃CH₂CH₂COONa)

Laboratory Preparation of methane

(i). Apparatus

(ii). Procedure

– About 5g of odium ethanoate and an equal mass of soda lime is put in a hard glass test tube, upon mixing them thoroughly in a mortar.

– The mixture is heated thoroughly in the test-tube.

(iii). Observation

– A colourless gas collects over water

Reasons:

– Methane does not react with and is insoluble in water.

Equation

CH₃COONa + NaOH_(s) → CH_4(g) + Na₂CO_3(aq)

Sodium ethanoate sodalime Methane Sodium carbonate

Physical properties of methane

It is a non-poisonous, colourless gas.
It is slightly soluble in water, but quite soluble in organic solvents such as ethanol and ether.
II is less denser than air and when cooled under pressure, it liquefies.

Chemical properties

Burning

– It is flammable and burns in excess air // oxygen with a pale blue non-luminous flame to give carbon (IV) oxide ad water vapour.

Equation:

CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(g)

Note: In a limited supply of air, the flame is luminous.

Reason:

– This is due to incomplete combustion of the methane.

– A mixture of methane and air explodes violently when ignited if the volume ratio is approximately 1:10 and this is often the cause of fatal explosions in coal mines.

Reaction with Bromine water and acidified potassium permanganate

– When methane is bubbled through bromine water the red brown colour of bromine persists; and when bubbled through acidified potassium manganate (VII) solution; the purple colour of the solution remains;

– Thus it has no effect on either bromine water or acidified potassium permanganate.

Reason: It is a saturated hydrocarbon.

Substitution reactions

– A substitution reaction is one in which one atom replaces another atom in a molecule.

Example: The substitution of Bromine in methane.

Procedure:

– A sample of Methane (CH₄) is placed in a boiling tube and to it is added some bromine gas.

– The tube is stoppered, and the mixture shaken, then allowed to stand and exposed to ultra-violet lamp.

Observations

– The red colour of Bromine begins to fade, and the pungent smell of hydrogen bromide (HBr) gas is detectable when the stopper is removed.

– A moist blue litmus paper also turns red on dipping into the resultant mixture.

Equation CH_4(g) + Br_2(g) → CH₃Br_(g) + HBr_(g)

Explanation – For a chemical reaction to occur, bonds must be broken. – The light energy (V.V. light) splits the Bromine molecule into free atoms, which are very reactive species. – Similarly the energy breaks the weaker carbon hydrogen bonds, and not the stronger carbon carbon bonds. – The free bromine atoms can then substitute (replace one of the hydrogen atoms of methane, resulting unto bromomethane and hydrogen bromide gas.

Note: This process can be repeated until all hydrogen atoms in CH₄ are replaced.

Write all the equations to show the stepwise substitution of all hydrogen atoms in methane.

– The substitution reactions can also occur with chlorine, forming chloremethane dichloromethane, trichloromethane (chloroform) and tetrachloromethane (carbon tetrachloride) respectively.

Equations:

Uses of methane – It is used as a fuel – Used in the manufacture of carbon black which is used in printers ink and paints. – Used in the manufacture of methanol, methanal, chloromethane and ammonia.

Cracking of Alkanes – Is the breaking of large alkane molecules into smaller Alkanes, alkenes and often hydrogen. It occurs under elevated temperatures of about 400-700^oC

Equation

Example: Cracking of propane

Alkenes

– Are hydrocarbons with at least one carbon-carbon double bond, and have the general formula C_nH_2n.

– They thus form a homologous series with the simplest member behind ethane.

Names and formulae of the first six alkenes.

Name of alkene

Formula

Ethene

Propane

Pbut-l-ene

Pent-lene

Hex-tene

NOMENCLATURE OF ALKENES

Rules

The parent molecule is the longest carbon chain; and its prefix is followed by the suffix ene.
The carbon atoms in the chain are numbered such that the carbon atoms joined by the double bonds get the lowest possible numbers.
The position of the substituent groups is indicated by showing the position of the carbon atom to which they are attached.
In case of 2 double bonds in an alkene molecule, the carbon atom to which each double bond is attached must be identified.

Examples

Questions: For each of the following alkenes, draw the structural formula

Hex- l ene
Prop-l-ene

Hex-2-ene

Give the IUPAC names for:

Note: Branched alkenes:

Event for branched alkenes, the numbering of the longest carbon chain is done such that the carbon atoms joined by the double bonds gets the smallest numbers possible.

Isomerism in alkenes

Alkenes show two types of isomerism:-

Branching isomerism
Positional isomerism

i) Branching isomerism

Occurs when a substitutent groups is attached to one of the carbon atoms in the largest chain containing the double bond.

Positional isomerism; in alkenes

Is a situation whereby two or more unsaturated alkenes have same molecular formular but different structural formula; due to alteration of the position of the double bond.

Question: Draw all the possible isomers of Hexene , resulting from positional and branching isomerism.

Gradation of physical properties of Alkenes

Name of alkene

Formula

(MP⁰C)

B.P (0)

Density g/cm³

solubility

Ethene

Propene

But-l-ene

Pent-l-ene

Hex-l-ene

-169

-189

-185

-138

-98

-104

–47.7

-6.2

-3.0

-98

–

0.640

0.674

Note: the double bond is the reactive site in alkenes

Preparation and chemical properties of Ethene

i) Apparatus

Procedure

A mixture of ethanol and concentrated sulfuric acid in the ratio 1:2 respectively are heated in a flask to a temp. of 160⁰C 180⁰C.

Observation

A colourless gas results; and is collected over water.

Reasons: Its insoluble, unreactive and lighter than water.

Equation

Explanation

At 160⁰C 180⁰C the conc. H2SO4 dehydrates the ethanol, removing a water molecule form it and the remaining C and H atoms rearrange and combine to form Ethene which is collected as colourless gas.

Note: At temperature below 140⁰C, a different compound called ether is predominantly formed.

Ethene can also be prepared by passing hot aluminum oxide over ethanol. The later of which acts as a catalyst i.e.

Reactions of ethene/chemical properties

Burning/combustion

Just like an alkenes and alkanes, ethene burn in air, producing carbon dioxide and large quantities of heat.

Equation:

Caution: Mixtures of air and ethene can be explosive and must be handled very carefully.

Additional reactions:

Is a reaction in which are molecule adds to another to form a single product occur in alkenes due to presence of a double bond.

With oxidizing agents
i) Reaction with acidified potassium permanganate.

Procedure: Ethene is bubbled into a test tube containing acidified potassium permanganate.

Observation: The purple colour of the solution disappears.

Explanation: Ethene reduces the potassium permanganate.

The permanganate ion is reduced to Manganese (II) ion and water.

Equation

Note: The net effect of the above reaction is the addition of two OH groups to the double bond forming ethan-1, 2-dio(ethylene glycol).

In cold countries ethylene glycol is used as an antifreeze in car radiators.

Reaction with acidified potassium chromate (VI) (K2Cr2O7)

Halogenations is the addition of halogen atoms across a double bond.
i) Reaction with Bromine Br2(g)

Procedure: Ethene is mixed with Bromine liquid/gas

Observation: The reddish brown bromine gas is decoloursed/becomes colourless.

Explanation: Bromine is decoloursed due to the addition of Bromine atoms to the twocarbon atoms f the double bond forming 1.2 dibromethane.

ii) Reaction with chlorine

The Chlorine (greenish yellow) also gets decoloursied due the addition of its atoms on the double bond.

Note: Alkenes react with and decolourise halogens and potassium permanganate by additional reaction at room temperature and pressure.

The reaction site is the double bond and hence/all alkenes will react in a similar manner.

Example; Butene and Bromine

iii) Reaction with Bromine water

Bromine is dissolved in water and reacted with ethene.

Equation:

Further examples of additional reactions

Addition of hydrogen halides

With hydrobromic acid; HBr (aq)

With sulphuric acid

Addition of Ethene with sulphuric acid

Note: When ethylhydrogen sulphate is hydrolysed, ethanol is formed.

In this reaction, water is added to ehylhydrogen sulphate and the mixture warmed.

Ethene with Hydrogen i.e. Hydrogenation.

Is commonly termed hydrogenation though just a typical addition reaction.

Ethene is reacted with hydrogen, under special conditions.

Conditions; moderate temperature and pressure.

Nickel catalyst/palladure catalyst.

Equation:

Application: it is used industrially in the conversion f various oils into fats e.g. in the preparation of Margarine.

Polymerization reactions.

Also called self-addition reactions

Alkanes have the ability to link together (polymerise) to though the double bond to give a molecule of larger molecular mass (polymers)

Polymers: Are very large molecules formed when 2 or more (smaller) molecules link together to form a larger unit.

Polymers have properties different form those of the original constituent manners.

Examples: Polymerisation of ethene

i) Conditions

High temperatures of about 200⁰C
High/elevated pressures of approximately 1000 atmospheres
A trace of oxygen catalyst.

ii) Procedure: Ethene is heated at 200⁰C and 1000 atm. Pressure over a catalyst.

iii) Observation: Sticky white substance which hardens on cooling is formed. This solid is called polythene, commonly reffered to as polythene.

Equation:

Generally

Uses of polythene

Used for the manufacture of many domestic articles (bowls, buckets, water cans, and cold water pipes) e.t.c.

Note: Polythene pipes have a great advantage over metal pipes as they can be welded quickly and do not burst in frosty weather.

Manufacture of reagent bottles, droppers, stoppers etc. since polythene is unaffected by alkalis and acids.

Test for Alkenes

– They decolourise bromine water, acidified potassium manganate VII.

i.e. These addition reactions show the presence of a double bond.

Uses of Alkenes

Manufacture of plastics, through polymerization.
Manufacture of ethanol; through hydrolysis reactions
Ripening of fruits.
Manufacture of ethan 1, 2-diol(glyco) which is used as a coolant.

ALYKYNES

Are unsaturated hydrocarbons which form a homologous series of a general formula CnH2n-2, where n = 2 or more.

The functional groups of the alkyne series is the carbon carbon tripple bond.

They also undergo addition reactions because of High unsaturation and may be polymerised like the alkenes.

Examples

Name

Molecular formula

Structural formular

Ethyne

Propyne

But-l-yne

Pent-l-yne

C2H2

C3H4

C4H6

C5H8

CH CH

CH3C CH

CH3CH2C CH

CH3(CH2)2C CH

Nomenclature of alkynes

The largest chain with the tripple carbon carbon bond forms the parent molecule.
Numbering of the carbon atoms is done such that the carbon atom with the tripple bond acquires the lowest possible number.
The substituent branch if any is named, and the compound written as a single word.

Examples

Draw the structures of the following hydrocarbons
2,2 dimethyl-but-2-yne
propyne

4,4 diethyl-hex-2-yne.

Isomerism in alkynes

Positional isomerism

Isomerism commonly occurs in alkynes due to the fact that the position of the tripple bond can be altered.

Such isomers, as usual have same molecular but different structural formulas.

Examples

i) Isomers of Butyne

Branching isomerism occurs when alkyl group is present in the molecule.

Others

Gradation in physical properties of Alkynes

Name of Alkyne

Formula

M.P/⁰C

B.P/⁰C

Density/gcm-3

Ethyne

Propyne

Butyne

Pent-l-yne

Hex-l-yne

HC CH

CH3 CH

CH3CH2CC CH

CH3CH2CH2C CH

CH3(CH2)3C CH

-8108

-103

-122

-90

-132

-83.6

-23.2

8.1

39.3

–

0.695

0.716

Preparation and chemical properties of Ethyne.

Preparation
i) Apparatus

ii) Procedure:

Water is dripped over calcium carbide and is collected over water.

Reasons for over-water collection:-

Its insoluble in water
Unreactive and lighter than water.

Conditions
Room temperature

Equation

Properties of Ethyne
i) Physical

Colourless gas, with a sweet smell when pure.
Insoluble in water and can thus be collected over water.
Solubility is higher in non- solvents * Draw table on physical properties.

Chemical properties

Combustion

Ethyne burns with a luminous and very sooty flame; due to the high percentage of carbon content, some of which remains unburnt.

In excess air, the products are carbon dioxide and water.

Equation

In limited air, they undergoes incomplete combustion, forming a mixture of carbon and carbon dioxide.

Note: A sooty flame observed when a hydrocarbon burns in air is an indication of unsaturation in the hydrocarbon.

Addition reactions

During addition reactions of alkynes (Ethyne) the tripple bond breaks in stages;

Reaction with hydrogen (Hydrogenation)

Note: This reaction occurs under special conditions i.e. – Presence of a Nickel catalyst

Temperatures about 200⁰C

Reaction with halogens
i) Reaction with chlorine

With Bromine gas

The red-brown bromine vapour is decoloursed.

Equations

Note: In this reaction Cl2 should be diluted with an inert.

Reason: Pure Cl2 reacts explosively with Ethyne, forming carbon and HCl.

Reaction with Bromine liquid

When Ethyne reacts with Bromine water, the reddish brown colour of bromine water disappears.

Reason: The Bromine adds to the carbon tripple bond leading to the of 1;1,2,2 tetrabromoethane.

Equation

E; Ethyne also decolorizes acidified potassium permanganate.

Note: Decolourization of acidified potassium permanganate and bromine water are tests for unsaturated hydrocarbons (alkanes and alkynes)

Reaction with hydrogen halides

Uses of Ethyne

Industrial manufacture of compounds like adhesives and plastics
Its used in the oxy-acetylene flame which is used for welding and cutting metals.

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ORGANIC CHEMISTRY II: ALCOHOLS AND ALKANOIC ACIDS:

1. Alkanols

ALCOHOLS (ALKANOLS):

Nomenclature of alcohols

2,2 dimethylpropanol-1-ol; Propan-2-ol;

H H H H

H H OH H

H CH3 H

2-methyl propan-2-ol;

Preparation and properties of alkanols.

Starch molecule + water sucrose molecules

Properties of alcohols (Ethanol)

Then;

Ethanal H+ reaction Ethanoic acid

General equation

(b). Reaction with acidified potassium manganate (VII).

General equation:

Example: Catalytic oxidation of ethanol with hot copper metal

Then;

Ethanal H+ reaction Ethanoic acid

General equation

(a). Oils

– Oils occur naturally in plants and animals.

Examples: Whale oil, groundnut oil, corn oil and castor oil.

– Oils are liquids at room temperature.

– Oils can be converted to fats /hardened into fats by hydrogenation.

Soaps and soapless detergents

– They are similar in that they contain a long hydrocarbon chain ending in a carboxylate anion to which is attracted a sodium cation.

– A typical soap is sodium stearate; C17H35COO–Na+

Structurally

Preparation of soaps

Summary on soap preparation

Effect of hard water on soap

– Soap is wasted in this way until all the calcium (II) and magnesium (II) stearate has been removed.

3. Formation of kettle fur which make electrical appliances inefficient and increases running costs.

OR: – Acids are proton donors i.e. a substance which provides protons or hydrogen ions.

Measuring the strength of alkalis

(ii). Using PH values

Laboratory preparations of salts

(iii). Formation of kettle fur which makes electrical appliances inefficient hence increasing running costs.

UNIT 3: ENERGY CHNAGES IN CHEMICAL AND PHYSICAL PROCESSES (THERMOCHEMISTRY)

Unit Checklist.

1. Introduction

2. Specific heat capacity and enthalpy of a system

3. Endothermic and exothermic reactions.

4. Activation energy

5. Determination of heat changes

Ø Heat of combustion

Ø Fuels

Ø Heat of neutralization

Ø Enthalpy of solution

UNIT 4: RATES OF REACTION AND REVERSIBLE REACTIONS

Ø Measurements of reaction rates

Ø Apparatus for measuring reaction rates

§ Temperature

Ø Factors affecting position of equilibrium

§ Temperature

Measurements of reaction rates

Ø Reacting particles must collide before a chemical reaction occurs

Factors affecting the rate of reaction

Reversible reactions

– The Equilibrium state here is established when:

Characteristics of equilibrium systems.

(i) The concentrations of reactants and products do not change after the equilibrium has been reached unless the system is disturbed.

(ii). The equilibrium can be reached from either direction.

Factors affecting the position of equilibrium

Pale yellow dark brown

– Decrease in temperature favours exothermic reactions.

Example:

Pale yellow dark brown

Decrease in pressure:

Experiment 1:- Displacement reactions among metals

Oxidation and reduction in terms of electron loss and gain

QUANTITATIVE ASPECTS OF ELECTROLYSIS

Basic terminologies and concepts

H CH₃ H

Ethanal H⁺reaction Ethanoic acid

Ethanal H⁺reaction Ethanoic acid

– A typical soap is sodium stearate; C₁₇H₃₅COO^–Na⁺