Pen-Write E-Library - Private

SCHEME OF WORK
WEEK:
1. Revision/ Nuclear chemistry. (i) Types and nature of radiation. (ii) Half-life as a measure of the stability of the nucleus.
2. Nuclear reaction i.e nuclear fusion and nuclear fission with examples. (i) Effect and application of radioactivity. (ii) Comparison of nuclear reaction and ordinary chemical reaction.
3. Simple Molecules and their shapes. (i) Covalent molecules e.g methane, ammonia. (ii) Crystalline solids: Their network structure e.g diamond.
4. Metallic bonding: Properties ; (i) Factors affecting the formation of metals. (ii) Intermolecular bonding, Van-der-Waal forces and hydrogen bonds. (iii) Comparison of all types of bond.
5. (a). Metal and their compound. (i) Extraction of metals e.g (AI. Cu, Sn and Fe).(ii) Their properties and reactions. (iii) Their uses.
6. Alloys: Composition and uses.
7. Introduction to qualitative analysis. (i) Test for cations using NaOH and NH4OH. (ii). Confirmatory test for the cations.
8. Test for Anions: (i) Identification of gases e.g CO2, SO2, and O2. (ii) Charateristics tests for anions e.g SO2/4, SO2/3, CO2/3, NO3-CI etc.
9. &10. Volumetric Analysis : Calculations basd on (i). Percentages purity and impurity of substances. (ii). Percentage/amount of water of rytallization. (iii). Mola mass of acid or base. (iv). Solubility of substance. (v). Volume of gases. (vi). Mole ratio of base.
10. Revision

TOPIC: ACIDS, BASES AND SALTS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define and explain neutralization with the aid of equations
b. Define salt and list the types of salts, with their meaning and examples.
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)

CONTENT: ACIDS, BASES AND SALTS
NEUTRALIZATION is a process in which an acid reacts completely with an appropriate amount of an alkali (or any other base) to produce a salt and water only.
Or
Neutralization is the combination of hydrogen ions, H+, and hydroxide ions, OH-, to form water molecules, H2O; and a salt is formed at the same time.
Acid + Alkali (or base) Salt + water
e.g. HCl(aq) + NaOH(aq) Na+Cl(aq) +H2O(l)
H+(aq) + OH-(aq) H2O(l)
From this definition of neutralization, abase may be defined as follows:
A BASE is a substance which will neutralize an acid to produce a salt and water only.
H2SO4(aq) + 2KOH(aq) K2SO4(aq) + 2H2O(l)
HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)

SALT
A salt is the compound formed when all or part of the ionizable hydrogen of an acid is replaced by metallic or ammonium ions.

TYPES OF SALTS
There are five main types of salts: normal salts, acid salts, basic salts, double salts and complex salts.

1) NORMAL SALTS: These are salts formed when all the replaceable hydrogen ions in the acid have been completely replaced by metallic ions.
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H2SO4(aq) + ZnO(s) ZnSO4(aq) + H2O(l)
Normal salts are neutral to litmus. However, a few normal salts such as Na2CO3, AlCl3 and sodium sulphide will undergo hydrolysis in water to give an acidic or alkaline medium

2) ACID SALTS: are formed when the replaceable hydrogen ions in acids are only partially replaced by a metal. They show acidic properties (e.g. turn blue litmus red) e.g. NaHSO4, KHS etc.
H2SO4(aq) + KOH(aq) KHSO4(aq) + H2O(l)
Acid salts cannot be formed by monobasic acids. Dibasic acids can form only one acid salt, while tribasic acids can form two different acid salts.

EVALUATION:
1. Define neutralization
2. Define a salt and give examples
3. List the types of salts

ASSIGNMENT:
1. Define a base on the basis of neutralization
2. Give the difference between a normal salt and an acid salt, with the aid of examples.






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Explain basic salts with examples
b. Explain double salts with examples
c. Explain complex salts with examples
d. Describe the preparation of salts and list the uses of salts.

CONTENT: TYPES OF SALTS
BASIC SALTS: Basic salts contain the hydroxide ion, OH-. Basic salts are produced when excess base is used to neutralize an acid. Basic salts have the properties of a base, e.g. a basic salt will turn red litmus blue and will react with excess acid to form a normal salt and water in the following way.
Zn(OH)Cl + HCl ZnCl2(aq) + H2O(l)
Other examples of basic salts are Cu(OH)Cl, Zn(OH)cl etc.

DOUBLE SALTS: Double salts are salts which ionize to produce three different types of ions in sodium. Usually two of these are positively charged (metallic or ammonium ions), while the other is negatively charged. E.g alum, KAl(SO4)2, cryolite NaAlF6.

COMPLEX SALTS: Complex salts contain complex ions consisting of a charged group of atoms
- Sodium tetrahydroxozincate (II) and
- Potassium hexacyanoferrate (II) are
Complex salts which ionize as follows:
Na2Zn(OH)4(aq) 2Na+(aq) + [Zn(OH)4]2-(aq)
K4Fe(CN)6(aq) 4K+(aq) + [Fe(CN)6]4-(aq)

PREPARATION OF SALTS:
The method used in preparing a salt depends on the nature of salt (i.e either soluble or insoluble). Some of the methods are as follows:

1) NEUTRALIZATION REACTION: This is used to prepare soluble salts e.g NaCl, KNO3 etc

2) DILUTE ACID AND METAL (for soluble salts) metals are more reactive than hydrogen will displace it from an acid. Zn(s) + 2HCl(aq) ZnCl(aq) + H2(g)

3) DILUTE ACID AND TRIOXOCARBONATE (IV): (for soluble salts)
H2SO4(aq) + CuCO3(s) CuSO4(aq) + H2O(l) + CO2(g)

4) DOUBLE DECOMPOSITION (for soluble salts)
this is the reaction between two soluble salts to form one soluble and one insoluble salt.
e.g. Na2SO4 + BaCl2 BaSO4 + 2NaCl

5) DIRECT COMBINATION OF CONSTITUENT ELEMENTS (for insoluble salts)
e.g. Fe(s) + S(s) FeS(s)
2Fe(s) + 3Cl2(s) 2FeCl3(s)

USES OF SALTS
1. As an electrolyte in dry cells e.g. NH4Cl
2. As drying agent e.g. fused CaCl2
3. For preserving food e.g. NaCl
4. Used as a laxative e.g. MgSO4
These are just few of the uses of salt.

EVALUATION:
1. Explain basic salts with examples
2. Explain double salts and complex salts with examples
3. Explain how to prepare soluble salts and insoluble salts
4. List three uses of salts

ASSIGNMENT:
1. Give two examples of salts and list their uses.
2. Explain the difference between an acid salt and a basic salt

BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Explain how to measure the acidity or alkalinity of substances using the pH scale
b. Explain the pH scale with the aid of its diagram
c. Briefly explain acid-base indicators.

CONTENT: MEASUREMENT OF ACIDITY AND ALKALINITY
The acidity and alkalinity of substances are measured using the pH scale.
THE Ph SCALE is a scale of numbers from 0 to 14 which is used to measure the acidity and alkalinity of substances.
A solution with a pH value of less than 7 is acidic while one with a value of more than 7 is alkaline. A solution with a pH value of 7 is neutral. Acidity increases with decreasing pH values, while alkalinity increases with increasing pH values. Thus, pH measures acidity or alkalinity of substances.
Mathematically, according to Sorenson:
pH = -log10 [H+]
pH is the negative logarithm to base ten of hydrogen ion concentration.

ACID-BASE INDICATORS
Indicators are organic dyes which change colour according to the pH of the medium
Indicator Colour in acidic medium Colour in alkaline medium
Methyl orange Red Yellow
Litmus Red Blue
Phenolphthalein Colourless Pink

EVALUATION:
1. Define pH
2. Define pH scale
3. Draw a pH scale
4. Define indicators and give three examples

ASSIGNMENT:
1. List three indicators and the colours of each in an acidic medium and an alkaline medium
2. How can the acidity or alkalinity of a substance be measured?






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Explain efflorescence with examples of efflorescent substances.
b. Explain deliquescence with examples of deliquescent substances.
c. Explain hygroscopy with examples of hygroscopic substances.
d. Explain drying agents with examples and the gases they are used to. Describe a dessicator.

CONTENT: EFFLORESCENCE, DELIQUESCENCE AND HYGROSCOPY WITH DRYING AGENTS
EFFLORESCENCE is a phenomenon whereby some crystalline salts will lose part or all of their water of crystallization when they are exposed to the atmosphere to form a lower dehydrated or anhydrous salt. Such a salt that undergoes efflorescence is called an EFFLORESCENT e.g. sodium trioxocarbonate (IV)-decahydrate, Na2CO3.10H2O, which losses nine out of its ten molecules of water of crystallization when exposed to the atmosphere.
Na2CO3.10H2O(s) Na2CO3.H2O(s) + 9H2O(g)

DELIQUESCENCE:is a phenomenon whereby some compounds absorb moisture from the atmosphere and eventually turn into solutions. Such compounds are said to be DELIQUESCENT. Examples are NaOH, FeCl3, KOH, CaCl2, MgCl2 and P2O5.

HYGROSCOPY: is a phenomenon whereby some substances also absorb moisture on exposure to the atmosphere and if they are solids, they will not form solutions but merely become sticky or moist while a hygroscopic liquid like conc. H2SO4 will absorb moisture from the air, usually diluting itself up to about three times its original volume. Other examples of hygroscopic substances include sodium trioxonitrate (V), NaNO3, copper (II) oxide, CuO and quick lime, CaO. Hygroscopic substances are commonly used as drying agents in the laboratory.

DRYING AGENTS: are hygroscopic or deliquescent substances that have a strong affinity for moisture or water and are usually used to dry gases in the laboratory. They are also commonly used in dessicators.
A drying agent cannot be used if it reacts with the substance to be dried. For example, conc.H2SO4 cannot be used to dry ammonia tetraoxosulphate (VI).
2NH3(g) + H2SO4(aq) (NH4)2SO4(aq)

A DESSICATOR: is an all glass apparatus used for drying substances in the laboratoey.
S/N DRYING AGENTS GASES
1 Concentrated H2SO4 All gases except NH3 and H2S
2 Fused calcium chloride All gases except ammonia
3 Phosphorous (V) oxide All gases except ammonia
4 Calcium oxide (Quick lime) Suitable for NH3 in particular
5 Silica gel All gases

EVALUATION:
1. Define efflorescence with examples of efflorescent substances.
2. Define deliquescence with examples of deliquescent substances
3. Define hygroscopy with examples of hygroscopic substances
4. Explain drying agents with examples
5. Describe and draw a dessicator

ASSIGNMENT:
1. Give the difference between deliquescence and hygroscopy.
2. With the aid of an equation, explain why conc. H2SO4 is not suitable for drying ammonia gas.

TOPIC: SOLUBILITY
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define and explain saturated and unsaturated solutions
b. Explain and define supersaturated solutions with given examples
c. Define solubility
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)

CONTENT: SOLUBILITY
For substances that are soluble in water, some are more soluble than others i.e. they have different solubilities. Solubility is a means of comparing the extent to which different solutes can dissolve in a particular solvent at a definite temperature.

SATURATED AND UNSATURATED SOLUTIONS
A SATURATED SOLUTION of a solute at a particular temperature is one which contains as much solute as it can dissolve at a particular temperature in the presence of undissolved solute particles.
The concentration of a saturated solution varies with the
(1) Solute
(2) Solvent
(3) Temperature

AN UNSATURATED SOLUTION is one which can continue to dissolve more solutes, if added until the solution becomes saturated.

A SUPERSATURATED SOLUTION is one which contains more of the solute than it can normally hold at that temperature e.g. Na2S2O3.5H2O, sodium trioxothiosulphate (VI)-pentahydrate, and Na2SO4.10H2O, sodium tetraoxosulphate (VI) decahydrates.
This is done by cooling the saturated solutions of certain substances (e.g. Na2SO4.10H2O) without the excess solute crystallizing out, provided no undissolved solids or dust particles are present. This cooled solution which contains more of the solute than is present in a saturated solution at that particular temperature is said to be supersaturated.
Supersaturated solutions are unstable and the excess solute will separate out if the solution is disturbed slightly be shaking or if a tiny crystal of the solute or even a dust particle is dropped into it. Even scratches inside the container can serve as centres of crystallization for the excess solute.

DEFINITION OF SOLUBILITY
The SOLUBILITY of a solute in a solvent at a particular temperature is the maximum amount of the solute in moles or grams that will saturate one dm3 of the solvent at that temperature.

EVALUATION:
1. Define saturated solution
2. Explain the factors which affect the concentration of a saturated solution
3. Define an unsaturated solution
4. Define a supersaturated solution and give examples.
5. Explain solubility

ASSIGNMENT:
1. Explain how the excess solute of a supersaturated solution can separate out.
2. Differentiate between a saturated solution and an unsaturated solution.






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Explain how to determine solubility.
b. Define and explain solubility graph or curve
c. Plot solubility graph and interpret solubility graph.

CONTENT: DETERMINATION OF SOLUBILITY
Two stages are involved in the determination of the solubility of a given solute:
Stage 1: Preparing a saturated solution of the given solute at a particular temperature.

Stage 2: Taking a known mass of the saturated solution and heating it to dryness so that the exact mass of the solvent and the solute in the saturated solution at that temperature can be calculated.

SOLUBILITY GRAPHS OR CURVES:
A SOLUBILITY GRAPH OR CURVE is a plot of the solubilities of a solute in a given solvent against their respective temperatures. The solubility graph or curve shows the effect of temperature on the solubility of the substance for comparing the solubilities of different solutes in a given solvent, several curves can be plotted on the same axes.

EVALUATION:
1. Explain the methods employed in the determination of the solubility of a given solute.
2. Define solubility graph or curve and interpret it.

ASSIGNMENT:
1. Briefly explain how to determine the solubility of potassium trioxonitrate (V), KNO3, at 60℃.
2. Explain what causes the solubility of alcohol in water.

BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. List and explain applications of solubility curves.
b. Explain the solubility of gases.
c. Differentiate between the solubility of solids and the solubility of gases.

CONTENT: APPLICATION OF SOLUBILITY CURVES
1. Solubility curves enable pharmacists to know the amounts of solid drugs that must be dissolved in a given quantity of solvent to give a prescribed drug mixture.

2. The solubility curves enable chemists and research workers to determine the most suitable solvents to be used at various temperatures for the extraction of essential chemicals from various natural sources.

3. Solubility curves enable a given mixture of solutes to be separated or purified by fractional crystallization. When a saturated solution of such a mixture is cooled, those fractions with low solubilities will be the first to crystallize out of solution while those that are still within their limits of solubility will remain in solution.

SOLUBILITY OF GASES
Most solid solutes have solubility which increases with a rise in temperature. However, the solubility of most common gases decreases with a rise in temperature. Hydrogen chloride gas is an exception. The solubility of a gas is directly proportional to the pressure of the gas. Aerated drinks are made by dissolving carbon (IV) oxide under pressure when an aerated drink can or bottle is opened, the pressure is released and bubbles of CO2 gas are given off.

DIFFERENCES BETWEEN SOLUBILITIES OF SOLIDS AND GASES
S/N SOLUBILITY OF SOLIDS SOLUBILITY OF GASES
1 General increases with a rise in temperature Generally increases with a rise in temperature (except HCl gas)

2 Pressure has no effect on the solubility of gases Solubility of gases increases with increase in pressure

3 Increase in surface area increases the solubility of a solid Surface area has no effect on the solubility of a gas

EVALUATION:
a. List three applications of solubility curves
b. Explain the effects of temperature and pressure on the solubility of gases.
c. Differentiate between the solubilities of solids and gases

ASSIGNMENT:
a. Explain pressure effect on the solubility of gases.
b. How are aerated drinks made?






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Solve calculation problems involving solubility

CONTENT: CALCULATIONS INVOLVING SOLUBILITY
1) When 50cm3 of a saturated solution of sugar of molar mass 342g at 40℃ was evaporated to dryness, 34.2g of dry solid was obtained. Calculate the solubility of sugar at 40℃ in moldm-3
SOLUTION:
No. of moles = mass(g)
Molar mass (gmol-1)
= 34.2g
342gmol-1
= 0.1 mol
No of moles of sugar = 0.1 mol
50cm3 off solution contained 0.1 mol of sugar
1000cm3 of solution will contain 0.1 mol X 1000cm3
50cm3
= 2 mol of sugar
2 mol of sugar in 100cm3 or 1dm3 solution at 40℃
Therefore solubility of sugar at 40℃ = 2moldm-3.
2) Water was added to 120g of salt, MgCl2 to produce 60.0cm3 of a saturated solution at 25℃. Its solubility at 25℃ is 8.0moldm-3. Calculate the mass of the salt which remained undissolved. (m=24, Cl= 35.5)

SOLUTION
McL2 = M + 2Cl; 24 + 2 (35.5)gmol-1
24 + 71(gmol-1) = 95gmol-1
Molar mass of MCl2 = 95gmol-1
Mass concentration (gdm-3) = molar concentration (moldm-3) X molar mass (gmol-1)
= 8 moldm-3 X 95gmol-1
= 760gdm-3
Therefore solubility in gdm-3 = 760gdm-3
1000cm3= 760g
60cm3 = 760 X 60cm3
1000cm3
= 45.6g
Solubility in 60cm3 = 45.6g
Mass of the salt which remained undissolved
= 120g - 45.6g
= 74.4g
Therefore the mass of the salt which remained undissolved = 74.4g

EVALUATION:
a. Establish the formulas of mass-volume relationship used in solubility calculations
b. Solve calculation problems in solubility

ASSIGNMENT:
1. 117.0g of Sodium chloride was dissolved in 1.0dm3 of distilled water at 25℃. Determine the solubility in moldm-3 of sodium chloride at that temperature [Na =23, Cl=35.5]
2. At 80℃, the solubility of KCl salt is 7moldm-3 and at 30℃, it is 5moldm-3. when the salt is cooled from 80℃ to 30℃, calculate the mass of crystals deposited. [K=39, Cl=35.5]

TOPIC: RATES OF REACTIONS
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Explain the rates of chemical reactions in terms of fastness and slowness, as well as surface area and temperature.
b. Define the rate of reaction
c. Explain the meaning of reaction rate and how it is measured
d. Give the formulas for rate of reaction.
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)
CONTENT: RATES OF CHEMICAL REACTIONS
Some chemical reactions are fast while others are slow. The speeds or rates of chemical reactions differ markedly. A white precipitate of AgCl forms almost immediately when AgNO3 is added to HCl acid while the rusting of iron nail in air is a slow process which takes several days.
A given reaction can also occur at different rates if the conditions of the reaction are different. A big piece of marble reacts slowly with cold water, but more rapidly with warm water (i.e. effect of temperature).
DEFINITION OF RATE OF REACTION
The rate of a chemical reaction is the number of moles of reactant converted or product formed per unit time.
MEANING AND MEASUREMENT OF RATE OF REACTION
Mg(s) + 2HCl(aq) MgCl2(s) + H2O(l)
In the above reaction, as it proceeds, magnesium and the acid, i.e. the reactants, are used up while MgCl2 and H2, i.e. the products are formed. The rate of reaction is the speed at which reactants are used up or products are formed.
Rate of reaction, such as the one above, can be measured by
a) Taking a given mass of magnesium
b) Adding an excess of HCl acid to it, and
c) Noting the time taken for all the magnesium to react
Rate of reaction = Amount of magnesium (moldm-3)
Time taken (min)

OR
Rate of reaction = change in concentration of reactant/product
Time taken for the change
The concentration of the reactant or product is in moldm-3 or gdm-3. The time it takes to undergo the complete reaction can be in seconds, minutes or even years for slow reactions.
EVALUATION: The students should be able to:
1. Explain why some reactions are fast while others are slow.
2. Using marble and dilute acid reaction as an example, explain why a given reaction can occur at different rates under different conditions of reaction.
3. Define the rate of a chemical reaction.
ASSIGNMENT:
1) Give the formula for rate of reaction in terms of amount of reactants and amount of products.
2) Explain why a piece of calcium reacts slowly with cold water, but more rapidly with warm water.


PERIOD: 2
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Solve calculation problems on reaction rates.
b. Explain the ways of measuring reaction rates
c. Define rate curve and sketch the graphs of rate curves.
CONTENT: RATES OF REACTIONS
CALCULATIONS
1) When 4.0g of calcium trioxocarbonate (IV) was added to excess dilute hydrochloric acid, carbon (IV) oxide was evolved. The entire reaction took 40 minutes. What was the rate of reaction?
SOLUTION:
Rate of reaction = mass of reactant (i.e. CaCO3)
Time taken for the change
= 4.0g
40min
= 0.1gmin-1
Therefore rate of reaction = 0.1gmin-1
This means that 0.1g of CaCO3 was converted to the product in 1 minute
2) When a dilute solution of “20 volume” H2O2 was heated, the total volume of oxygen collected was 120cm3 in just 15 minutes. What was the rate of formation of oxygen?
SOLUTION:
2H2O2(aq) 2H2O(l) + O2(g)
Rate of reaction = volume of oxygen formed
Time taken for the change
= 120cm3
15min
Rate of reaction = 8cm3min-1
Therefore 8cm3 of oxygen was produced in 1 minute
WAYS OF MEASURING REACTION RATES
Rates of reactions can be determined by measuring the rate at which the mass or concentration of a
a) Reactant is decreasing , or
b) Product is increasing
RATE CURVE
A rate curve is the graph which shows the rate of reaction.
EVALUATION: The students should be able to:
a) Define rate curve
b) Calculate the rate of formation of oxygen when a dilute solution of “30 – volume” H2O2 was heated; the total volume of oxygen collected was 180cm3 in 22.5seconds
ASSIGNMENT:
1) Sketch the rate curve for a reaction based on the formation of product
2) Sketch the rate curve for a reaction based on the composition of its reactant.



PERIOD: 3
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define the collision theory
b. Define effective collision
c. Define and explain activation energy
d. Define energy barrier
e. Define transition state of a chemical reaction
f. Draw the energy profile diagrams for an exothermic reaction and an endothermic reaction.
CONTENT: COLLISION THEORY
THE COLLISION THEORY assumes that there must be collisions between reactant particles for a chemical reaction to occur.
EFFECTIVE COLLISIONS are a small fraction of collisions which result in chemical reaction.
ACTIVATION ENERGY is the minimum amount of energy needed by the colliding reactant particles for a chemical reaction to occur.
In all chemical reactions, existing bonds in the reactant particles have to be broken first before new bonds can be formed to produce product particles. The breaking of bonds needs energy and is possible only if the reactant particles collide with sufficient energy to overcome this ENERGY BARRIER.
Reaction occurs if the energy of the colliding reactant particles is equal to or more than the activation energy.
ENERGY BARRIER is the energy required to be overcome by the reactant particles for a chemical reaction to go to completion and form products.
TRANSITION STATE is the stage at which the reactants are completely changing to products.
CHANGES IN ENERGY CONTENT DURING THE COURSE OF A CHEMICAL REACTION [REACTION PROFILE DIAGRAMS]
a) Exothermic Reaction
b) Endothermic Reaction
EVALUATION: The students should be able to:
1. Define collision theory and effective collisions
2. Explain energy barrier and transition state
3. Sketch the energy profile diagram for an exothermic reaction
4. Sketch the energy profile diagram for an endothermic reaction
ASSIGNMENT:
1. When 4.5g of CaCO3 was added to excess dilute HCl, CO2 was given off. The entire reaction took 15minutes. What was the rate of reaction?
2. Write the balanced chemical equation for the reaction between calcium trioxocarbonate (IV) and dilute hydrochloric acid.


PERIOD: 4
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. List the factors affecting the rates of chemical reactions.
b. Explain the factors affecting the rates of chemical reactions
CONTENT: FACTORS AFFECTING RATES OF REACTIONS
From the Collision Theory, we observe that the reaction rate would depend on the frequency of effective collisions between reactant particles.
FACTORS AFFECTING RATES OF REACTION
1. Nature of reactants
2. Concentration/pressure (for gases) of reactants
3. Surface area of reactants
4. Temperature of reaction mixture
5. Presence of light
6. Presence of catalyst
EFFECT OF NATURE OF REACTANT
The rate of a chemical reaction is dependent on the chemical nature of the reactants as different substances have different energy contents. For example, when a piece of iron is placed in dilute HCl, there is a slow evolution of hydrogen gas; but with a piece of Zinc, hydrogen is evolved rapidly; and with a piece of gold, there is no evidence of a reaction
EFFECT OF CONCENTRATION OF REACTANTS
Reactant particles will collide more often if they are crowded in a small space, i.e. frequency of collision is dependent upon concentration. An increase (or decrease) in concentration/pressure will lead to a corresponding increase (or decrease) in effective collisions of the reactants and hence in the rate of reaction.
EFFECT OF SURFACE AREA OF CONTACT
When a dilute acid is added to marble chips and to powdered marble, we notice a marked difference in the rate of reaction; the acid reacts more vigorously and faster with the powdered marble than with the marble chips because powdered marble offers a greater area contact with the acid than the marble chips.
By increasing the surface area of contact, reaction occurs faster since the time to be spent in breaking the old bond is minimized and vice-versa.
EFFECT OF TEMPERATURE
Increase in temperature leads to increase in average kinetic energy and thus, increase in effective collisions resulting in increase in rate of reaction.
EFFECT OF LIGHT
Some reactions are influenced by light and are called PHOTOCHEMICAL REACTIONS. Such reactions will be very fast in bright light, moderate in daylight but may be very slow if at all it is going to occur in the dark. for example, the reaction between hydrogen and chlorine is negligible without light, moderate in day light and explosive in bright sunlight. Other example is decomposition of H2O2; reaction between CH4 and Cl2; photosynthesis in plants; and conversion of silver halides to grey metallic silver (used in photography)
EFFECT OF A CATALYST
A CATALYST is a substance which alters the rate of a reaction but itself remains chemically and quantitatively unchanged at the end of a reaction. Addition of a suitable catalyst quickens the rate of a chemical reaction. In the preparation of O2 by heating KCLO3, in the presence of a small amount of the catalyst, MnO2, only moderate heat is needed to decompose KCLO3. In the absence of the catalyst, the KCLO3 must be heated to much higher temperature and for a longer time in order to obtain similar results. A positive catalyst usually acts by lowering the energy barrier of a chemical reaction. Thus in the presence of a catalyst, more reactant particles are able to react when they collide.
EVALUATION: The students should be able to:
1. List six factors which affect the rates of reactions
2. Explain the factors which affect reaction rates.
ASSIGNMENT:
1. Define a photochemical reaction
2. Explain the effect of light on photosynthesis in plants
3. Distinguish between photocatalysis and chemocatalysis

TOPIC: REVERSIBLE REACTION
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define a reversible reaction
b. Define a chemical equilibrium and explain a closed system
c. Derive the mathematical expression for equilibrium constant.
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)
CONTENT: REVERSIBLE REACTIONS AND CHEMICAL EQUILIBRIUM
A REVERSIBLE REACTION is the one that proceeds in both forward and backward directions under suitable conditions.
a reaction which is reversible under the same condition will have both the forward and the reverse reactions occurring at the same time, if all the reactants and products are present in one container.
CHEMICAL EQUILIBRIUM
A chemical reaction is in dynamic equilibrium when both the forward and backward reactions in a reversible system proceeds at the same rate, thereby producing no net change in the concentrations of the reactants and products.
CLOSED SYSTEM
A reversible reaction can only attain dynamic equilibrium in a closed system.
A CLOSED SYSTEM is a closed container, where during the process of the reaction, all reacting species (reactants and products) are present. In an OPEN SYSTEM i.e. one in which one or more of the substances are being removed, the reaction cannot attain equilibrium even if it is a reversible one. E.g.
CaCO3(s) CaO(s) + CO2(g) [closed system]
The above reaction is only reversible in a closed where it can attain dynamic equilibrium. However, in an open systems, the CO2 gas will escape into the atmosphere and the reaction will behave in an irreversible way without being able to attain equilibrium (i.e. CaCO3(s) CaO(s) + CO2(g) [open system]
EQUILIBRIUM CONSTANT
Consider the reaction: mA + nB r products.
Rate of reaction, r α[A]m n
Where [A] and are concentrations of A and B respectively.
R = K. [A]m n
Where k = velocity constant of that reaction at that temperature
Now let us apply the Law of mass action to the reversible reaction represented by the following equation
mA + nB r1 pC +qD
r2
r1 = k1.[A]m n and r2 = k2 .[C]p [D]q
for a reversible reaction in dynamic equilibrium,
r1 = r2; k1[A]mn = k2 [C]p [D]q
k1 = [C]p[D]q ; k1 = K
k2 [A]m n k2
K = [C]p[D]q
[A]mn
K = EQUILIBRIUM CONSTANT of that reaction
At that temperature
K only varies with temperature and is not affected by the addition of catalyst, and the varying of concentrations (if there are more reactants, the equilibrium position will shift to the right to produce more products).
EVALUATION: The students should be able to:
a. Define a reversible reaction
b. Define chemical equilibrium
c. Explain a closed system
d. Derive the mathematical expression for equilibrium constant
ASSIGNMENT:
1. Differentiate between a closed system and an open system
2. Explain how equilibrium constant varies with temperature and is independent of concentration and addition of a catalyst


PERIOD: 2
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. State the conditions which affect the equilibrium position of a reversible reaction
b. State Le Chatelier’s Principle
c. Briefly explain the importance of Le Chatelier’s principle
CONTENT: LE CHATELIER’S PRINCIPLE
Every reversible reaction attains its own specific equilibrium under a given set of conditions. The equilibrium position id dependent on
• Temperature of the reacting system
• Pressure of the reacting system (for gases);
• Concentration of the reacting system
A change in any one of these factors alter the position of the equilibrium. These factors and their effects on chemical systems in equilibrium were studied by Le Chatelier (1850-1936) who formulated the Le Chatelier’s principle.
LE CHATELIER’S PRINCIPLE states that if a chemical system is in equilibrium and one of the factors affecting the equilibrium position is altered, the equilibrium will shift so as to annul the effect of the change.
IMPORTANCE OF LE CHATELIER’S PRINCIPLE IN CHEMICAL INDUSTRY
1) It can help to define the optimum conditions for the chemical processes employed in industry;
2) It can help to reduce undesirable reversibility.
3) It can help to predict the effect of an altered factor on the equilibrium position of an untried reaction.
Physical observation of a reversible reaction (colour change)
heat
N2O4(g) 2NO2(g)
Pale yellow cold reddish brown
EVALUATION: The students should be able to:
a. List the factors that affect the position of dynamic equilibrium in a reversible reaction.
b. State Le Chatelier’s principle
c. List three importance of Le Chatelier’s principle in chemical industry
ASSIGNMENT:
1. List the factors which affect equilibrium constant and equilibrium position respectively.
2. Give the difference between a reversible and an irreversible reaction.




PERIOD: 3
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. List the factors that affects the equilibrium position of reversible reactions
b. Explain the factors that affect the equilibrium position of reversible reaction
CONTENT: FACTORS WHICH AFFECT THE EQUILIBRIUM POSITION OF A REVERSIBLE REACTION
1. CHANGE IN TEMPERATURE: if change in temperature is positive (for an endothermic reaction) an increase in temperature of the reversible reaction in equilibrium favours the forward reaction and the equilibrium position shifts to the right on the product(s) side, in order to maintain the equilibrium and vice versa. If change is negative (an exothermic reaction), decrease in temperature favours the forward reaction while increase in temperature favours the backward reaction, in order to sustain the equilibrium and vice versa.
It should be noted that change in temperature in any reversible reaction is for the forward reaction. For example +ve change in temperature means that the forward reaction is endothermic while the backward of the same reversible system is an exothermic reaction.
N2O4(g) 2NO2(g)
Pale yellow reddish-brown
2. CHANGE IN CONCENTRATION : In an equilibrium mixture, there is balance in the concentrations of the reactants and the products, i.e. these concentrations are in a definite ratio which depends on the conditions of the reaction. If the concentration of any of the reactants (on the left hand side) of a reversible reaction is increased, the equilibrium position shifts to the right to produce more products; also if any of the products is removed, this favours formation of more products.
3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)
3. CHANGE IN PRESSURE: This factor only has effect on an equilibrium in which:
• One of the reactants or products in the reversible reaction must be gaseous;
• The total number of moles of gaseous molecules on the left side of the equation must be different from the total number of moles of gaseous molecules on the right side. Increase in pressure of a system in equilibrium favours the side with less gaseous volume and thus equilibrium shifts to such side while, decrease in pressure brings the equilibrium to the side with higher gaseous volume. If both sides have equal gaseous volume, pressure has no effect.
N2(g) + 3H2(g) 2NH3(g)
4 moles or volumes 2 moles or volumes
In the above example, increase in pressure favours the side with less volume, i.e. more NH3 product is formed while decrease in pressure favours the side with more volume (more N2 and H2 are formed).
4. EFFECT OF A CATALYST: Catalysts do not have any effect on the position of an equilibrium, but only allow the equilibrium to be attained in a short time. A positive catalyst increases the rates of both the forward and backward reactions to the same extent by lowering the activation energy of the reaction. A negative catalyst slows down the rate of reaction so that a longer time is needed to reach equilibrium [Negative catalyst is called INHIBITOR]
EVALUATION: The students should be able to:
a. Explain the effect of temperature change on the equilibrium position for both endothermic and exothermic reversible reactions.
b. Explain the effect of concentration on equilibrium position of a reversible reaction.
c. Explain the effect of pressure change on equilibrium position of a reversible reaction.
d. Elaborate on the effect of a catalyst on the equilibrium position of a reversible reaction.
ASSIGNMENT:
a. Distinguish between a catalyst and an inhibitor.
b. List the conditions necessary for pressure change to alter the position of equilibrium in a reversible reaction.






PERIOD: 4
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
A) Establish the relationship between free energy change and the equilibrium constant, k
B) Establish the relationship between electrode potential and free energy change of an electrochemical cell.
CONTENT: EQUILIBRIUM CONSTANT, FREE ENERGY AND ELECTRODE POTENTIALS
FREE ENERGY of a chemical system, G, is that energy which is available for doing work. This is the driving force that brings about a chemical change.
∆G = ∆H - T∆S system [∆G = T∆Stotal]
The relationship between free energy and equilibrium constant;
∆GӨ = -RTink
Where GӨ is the standard free energy change of the system.
K is the equilibrium constant,
R is the gas constant and
T is the temperature of the system in Kelvin
In an electrochemical cell, oxidation occurs at one electrode and reduction at the other. If such a reaction (as the one below) is at equilibrium,
Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)
Then the relationship between EӨ and K is given by EӨ = RTink
nF
where EӨ is the standard electrode potential of the reaction,
R is the gas constant,
T is the temperature in Kelvin,
F is the Faraday’s constant,
K is the equilibrium constant, and
The value of K is important, especially in the electroplating industry. For example, if a coating of lead is to be deposited onto an object, a higher value of K would mean that the forward reaction is favoured, resulting in a thicker coating of lead.
Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)
The relationship between electrode potential and free energy change of an electrochemical cell can be expresses thus:
∆GӨ = - RTInk; EӨ = RTInk
nF
nFEӨ = RTInk
Therefore ∆GӨ = - nFEӨ
Where ∆GӨ is the standard free energy change,
n is the number of moles of electrons,
EӨ is the standard electrode potential, and
F is the Faraday’s constant
A negative value of ∆GӨ in this equation means that work is obtained from the electro-chemical cell.
EVALUATION: The students should be able to:
(i) Define free energy of a chemical system.
(ii) Derive the mathematical relationship between standard free energy and standard electrode potential.
ASSIGNMENT:
(1) Define the standard electrode potential.
(2) With the aid of balanced chemical equation, use Le Chatelier’s principle to explain the factors that affect the equilibrium position which favours the optimum yield of ammonia in the Haber process, given that the process is exothermic.

BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define the collision theory
b. Define effective collision
c. Define and explain activation energy
d. Define energy barrier
e. Define transition state of a chemical reaction
f. Draw the energy profile diagrams for an exothermic reaction and an endothermic reaction.

CONTENT: COLLISION THEORY
THE COLLISION THEORY assumes that there must be collisions between reactant particles for a chemical reaction to occur.

EFFECTIVE COLLISIONS are a small fraction of collisions which result in chemical reaction.

ACTIVATION ENERGY is the minimum amount of energy needed by the colliding reactant particles for a chemical reaction to occur.
In all chemical reactions, existing bonds in the reactant particles have to be broken first before new bonds can be formed to produce product particles. The breaking of bonds needs energy and is possible only if the reactant particles collide with sufficient energy to overcome this ENERGY BARRIER.
Reaction occurs if the energy of the colliding reactant particles is equal to or more than the activation energy.

ENERGY BARRIER is the energy required to be overcome by the reactant particles for a chemical reaction to go to completion and form products.

TRANSITION STATE is the stage at which the reactants are completely changing to products.

CHANGES IN ENERGY CONTENT DURING THE COURSE OF A CHEMICAL REACTION [REACTION PROFILE DIAGRAMS]
a) Exothermic Reaction
b) Endothermic Reaction

EVALUATION:
1. Define collision theory and effective collisions
2. Explain energy barrier and transition state
3. Sketch the energy profile diagram for an exothermic reaction
4. Sketch the energy profile diagram for an endothermic reaction

ASSIGNMENT:
1. When 4.5g of CaCO3 was added to excess dilute HCl, CO2 was given off. The entire reaction took 15minutes. What was the rate of reaction?
2. Write the balanced chemical equation for the reaction between calcium trioxocarbonate (IV) and dilute hydrochloric acid.






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. List the factors affecting the rates of chemical reactions.
b. Explain the factors affecting the rates of chemical reactions

CONTENT: FACTORS AFFECTING RATES OF REACTIONS
From the Collision Theory, we observe that the reaction rate would depend on the frequency of effective collisions between reactant particles.

FACTORS AFFECTING RATES OF REACTION
1. Nature of reactants
2. Concentration/pressure (for gases) of reactants
3. Surface area of reactants
4. Temperature of reaction mixture
5. Presence of light
6. Presence of catalyst

EFFECT OF NATURE OF REACTANT
The rate of a chemical reaction is dependent on the chemical nature of the reactants as different substances have different energy contents. For example, when a piece of iron is placed in dilute HCl, there is a slow evolution of hydrogen gas; but with a piece of Zinc, hydrogen is evolved rapidly; and with a piece of gold, there is no evidence of a reaction.

EFFECT OF CONCENTRATION OF REACTANTS
Reactant particles will collide more often if they are crowded in a small space, i.e. frequency of collision is dependent upon concentration. An increase (or decrease) in concentration/pressure will lead to a corresponding increase (or decrease) in effective collisions of the reactants and hence in the rate of reaction.

EFFECT OF SURFACE AREA OF CONTACT
When a dilute acid is added to marble chips and to powdered marble, we notice a marked difference in the rate of reaction; the acid reacts more vigorously and faster with the powdered marble than with the marble chips because powdered marble offers a greater area contact with the acid than the marble chips.
By increasing the surface area of contact, reaction occurs faster since the time to be spent in breaking the old bond is minimized and vice-versa.

EFFECT OF TEMPERATURE
Increase in temperature leads to increase in average kinetic energy and thus, increase in effective collisions resulting in increase in rate of reaction.

EFFECT OF LIGHT
Some reactions are influenced by light and are called PHOTOCHEMICAL REACTIONS. Such reactions will be very fast in bright light, moderate in daylight but may be very slow if at all it is going to occur in the dark. for example, the reaction between hydrogen and chlorine is negligible without light, moderate in day light and explosive in bright sunlight. Other example is decomposition of H2O2; reaction between CH4 and Cl2; photosynthesis in plants; and conversion of silver halides to grey metallic silver (used in photography).

EFFECT OF A CATALYST
A CATALYST is a substance which alters the rate of a reaction but itself remains chemically and quantitatively unchanged at the end of a reaction. Addition of a suitable catalyst quickens the rate of a chemical reaction. In the preparation of O2 by heating KCLO3, in the presence of a small amount of the catalyst, MnO2, only moderate heat is needed to decompose KCLO3. In the absence of the catalyst, the KCLO3 must be heated to much higher temperature and for a longer time in order to obtain similar results. A positive catalyst usually acts by lowering the energy barrier of a chemical reaction. Thus in the presence of a catalyst, more reactant particles are able to react when they collide.

EVALUATION:
1. List six factors which affect the rates of reactions
2. Explain the factors which affect reaction rates.

ASSIGNMENT:
1. Define a photochemical reaction
2. Explain the effect of light on photosynthesis in plants
3. Distinguish between photocatalysis and chemocatalysis

TOPIC: REVERSIBLE REACTION
BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. Define a reversible reaction
b. Define a chemical equilibrium and explain a closed system
c. Derive the mathematical expression for equilibrium constant.
REFERENCE: NEW CERTIFICATE CHEMISTRY FOR SSCE by OSEI YAW ABABIO (NEW EDITION)

CONTENT: REVERSIBLE REACTIONS AND CHEMICAL EQUILIBRIUM
A REVERSIBLE REACTION is the one that proceeds in both forward and backward directions under suitable conditions.
a reaction which is reversible under the same condition will have both the forward and the reverse reactions occurring at the same time, if all the reactants and products are present in one container.

CHEMICAL EQUILIBRIUM
A chemical reaction is in dynamic equilibrium when both the forward and backward reactions in a reversible system proceeds at the same rate, thereby producing no net change in the concentrations of the reactants and products.

CLOSED SYSTEM
A reversible reaction can only attain dynamic equilibrium in a closed system.
A CLOSED SYSTEM is a closed container, where during the process of the reaction, all reacting species (reactants and products) are present. In an OPEN SYSTEM i.e. one in which one or more of the substances are being removed, the reaction cannot attain equilibrium even if it is a reversible one. E.g.
CaCO3(s) CaO(s) + CO2(g) [closed system]
The above reaction is only reversible in a closed where it can attain dynamic equilibrium. However, in an open systems, the CO2 gas will escape into the atmosphere and the reaction will behave in an irreversible way without being able to attain equilibrium (i.e. CaCO3(s) CaO(s) + CO2(g) [open system]

EQUILIBRIUM CONSTANT
Consider the reaction: mA + nB r products.
Rate of reaction, r α[A]m n
Where [A] and are concentrations of A and B respectively.
R = K. [A]m n
Where k = velocity constant of that reaction at that temperature
Now let us apply the Law of mass action to the reversible reaction represented by the following equation
mA + nB r1 pC +qD
r2
r1 = k1.[A]m n and r2 = k2 .[C]p [D]q
for a reversible reaction in dynamic equilibrium,
r1 = r2; k1[A]mn = k2 [C]p [D]q
k1 = [C]p[D]q ; k1 = K
k2 [A]m n k2
K = [C]p[D]q
[A]mn
K = EQUILIBRIUM CONSTANT of that reaction
At that temperature
K only varies with temperature and is not affected by the addition of catalyst, and the varying of concentrations (if there are more reactants, the equilibrium position will shift to the right to produce more products).

EVALUATION:
a. Define a reversible reaction
b. Define chemical equilibrium
c. Explain a closed system
d. Derive the mathematical expression for equilibrium constant

ASSIGNMENT:
1. Differentiate between a closed system and an open system
2. Explain how equilibrium constant varies with temperature and is independent of concentration and addition of a catalyst






BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. State the conditions which affect the equilibrium position of a reversible reaction
b. State Le Chatelier's Principle
c. Briefly explain the importance of Le Chatelier's principle

CONTENT: LE CHATELIER'S PRINCIPLE
Every reversible reaction attains its own specific equilibrium under a given set of conditions. The equilibrium position id dependent on
- Temperature of the reacting system
- Pressure of the reacting system (for gases);
- Concentration of the reacting system
A change in any one of these factors alter the position of the equilibrium. These factors and their effects on chemical systems in equilibrium were studied by Le Chatelier (1850-1936) who formulated the Le Chatelier's principle.

LE CHATELIER'S PRINCIPLE states that if a chemical system is in equilibrium and one of the factors affecting the equilibrium position is altered, the equilibrium will shift so as to annul the effect of the change.

IMPORTANCE OF LE CHATELIER'S PRINCIPLE IN CHEMICAL INDUSTRY
1) It can help to define the optimum conditions for the chemical processes employed in industry;
2) It can help to reduce undesirable reversibility.
3) It can help to predict the effect of an altered factor on the equilibrium position of an untried reaction.
Physical observation of a reversible reaction (colour change)
heat
N2O4(g) 2NO2(g)
Pale yellow cold reddish brown

EVALUATION:
a. List the factors that affect the position of dynamic equilibrium in a reversible reaction.
b. State Le Chatelier's principle
c. List three importance of Le Chatelier's principle in chemical industry

ASSIGNMENT:
1. List the factors which affect equilibrium constant and equilibrium position respectively.
2. Give the difference between a reversible and an irreversible reaction.

BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
a. List the factors that affects the equilibrium position of reversible reactions
b. Explain the factors that affect the equilibrium position of reversible reaction

CONTENT: FACTORS WHICH AFFECT THE EQUILIBRIUM POSITION OF A REVERSIBLE REACTION
1. CHANGE IN TEMPERATURE: if change in temperature is positive (for an endothermic reaction) an increase in temperature of the reversible reaction in equilibrium favours the forward reaction and the equilibrium position shifts to the right on the product(s) side, in order to maintain the equilibrium and vice versa. If change is negative (an exothermic reaction), decrease in temperature favours the forward reaction while increase in temperature favours the backward reaction, in order to sustain the equilibrium and vice versa.
It should be noted that change in temperature in any reversible reaction is for the forward reaction. For example +ve change in temperature means that the forward reaction is endothermic while the backward of the same reversible system is an exothermic reaction.
N2O4(g) 2NO2(g)
Pale yellow reddish-brown

2. CHANGE IN CONCENTRATION : In an equilibrium mixture, there is balance in the concentrations of the reactants and the products, i.e. these concentrations are in a definite ratio which depends on the conditions of the reaction. If the concentration of any of the reactants (on the left hand side) of a reversible reaction is increased, the equilibrium position shifts to the right to produce more products; also if any of the products is removed, this favours formation of more products.
3Fe(s) + 4H2O(g) Fe3O4(s) + 4H2(g)

3. CHANGE IN PRESSURE: This factor only has effect on an equilibrium in which:
- One of the reactants or products in the reversible reaction must be gaseous;
- The total number of moles of gaseous molecules on the left side of the equation must be different from the total number of moles of gaseous molecules on the right side. Increase in pressure of a system in equilibrium favours the side with less gaseous volume and thus equilibrium shifts to such side while, decrease in pressure brings the equilibrium to the side with higher gaseous volume. If both sides have equal gaseous volume, pressure has no effect.
N2(g) + 3H2(g) 2NH3(g)
4 moles or volumes 2 moles or volumes
In the above example, increase in pressure favours the side with less volume, i.e. more NH3 product is formed while decrease in pressure favours the side with more volume (more N2 and H2 are formed).

4. EFFECT OF A CATALYST: Catalysts do not have any effect on the position of an equilibrium, but only allow the equilibrium to be attained in a short time. A positive catalyst increases the rates of both the forward and backward reactions to the same extent by lowering the activation energy of the reaction. A negative catalyst slows down the rate of reaction so that a longer time is needed to reach equilibrium [Negative catalyst is called INHIBITOR]

EVALUATION:
a. Explain the effect of temperature change on the equilibrium position for both endothermic and exothermic reversible reactions.
b. Explain the effect of concentration on equilibrium position of a reversible reaction.
c. Explain the effect of pressure change on equilibrium position of a reversible reaction.
d. Elaborate on the effect of a catalyst on the equilibrium position of a reversible reaction.

ASSIGNMENT:
a. Distinguish between a catalyst and an inhibitor.
b. List the conditions necessary for pressure change to alter the position of equilibrium in a reversible reaction.







BEHAVIOURAL OBJECTIVES: BY THE END OF THE LESSON, THE STUDENTS SHOULD BE ABLE TO:
A) Establish the relationship between free energy change and the equilibrium constant, k
B) Establish the relationship between electrode potential and free energy change of an electrochemical cell.

CONTENT: EQUILIBRIUM CONSTANT, FREE ENERGY AND ELECTRODE POTENTIALS
FREE ENERGY of a chemical system, G, is that energy which is available for doing work. This is the driving force that brings about a chemical change.
∆G = ∆H - T∆S system [∆G = T∆Stotal]
The relationship between free energy and equilibrium constant;
∆GӨ = -RTink
Where GӨ is the standard free energy change of the system.
K is the equilibrium constant,
R is the gas constant and
T is the temperature of the system in Kelvin
In an electrochemical cell, oxidation occurs at one electrode and reduction at the other. If such a reaction (as the one below) is at equilibrium,
Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)
Then the relationship between EӨ and K is given by EӨ = RTink
nF
where EӨ is the standard electrode potential of the reaction,
R is the gas constant,
T is the temperature in Kelvin,
F is the Faraday's constant,
K is the equilibrium constant, and
The value of K is important, especially in the electroplating industry. For example, if a coating of lead is to be deposited onto an object, a higher value of K would mean that the forward reaction is favoured, resulting in a thicker coating of lead.
Zn(s) + Pb2+(aq) Zn2+(aq) + Pb(s)
The relationship between electrode potential and free energy change of an electrochemical cell can be expresses thus:
∆GӨ = - RTInk; EӨ = RTInk
nF
nFEӨ = RTInk
Therefore ∆GӨ = - nFEӨ
Where ∆GӨ is the standard free energy change,
n is the number of moles of electrons,
EӨ is the standard electrode potential, and
F is the Faraday's constant
A negative value of ∆GӨ in this equation means that work is obtained from the electro-chemical cell.

EVALUATION:
(i) Define free energy of a chemical system.
(ii) Derive the mathematical relationship between standard free energy and standard electrode potential.

ASSIGNMENT:
(1) Define the standard electrode potential.
(2) With the aid of balanced chemical equation, use Le Chatelier's principle to explain the factors that affect the equilibrium position which favours the optimum yield of ammonia in the Haber process, given that the process is exothermic.