41 Working: Applying Hess law then: ∆Hf =∆Ha +∆Hi +∆Ha +∆He +∆Hl Substituting: -411= +108kJ + +500kJ + +121kJ +-364kJ + x -411 + -108kJ + -500kJ + -121kJ + +364kJ = x x= -776kJmole-1
2
20.0.0 REACTION RATES AND REVERSIBLE REACTIONS (15 LESSONS) A.THE RATE OF CHEMICAL REACTION (CHEMICAL KINETICS) 1.Introduction The rate of a chemical reaction is the time taken for a given mass/amount of products to be formed. The rate of a chemical reaction is also the time taken for a given mass/amount of reactant to be consumed /used up. Some reactions are too slow to be determined. e.g rusting ,decomposition of hydrogen peroxide and weathering. Some reactions are too fast and instantaneous e.g. neutralization of acid and bases/alkalis in aqueous solution and double decomposition/precipitation. Other reactions are explosive and very risky to carry out safely e.g. reaction of potassium with water and sodium with dilute acids. The study of the rate of chemical reaction is useful in knowing the factors that influence the reaction so that efficiency and profitability is maximized in industries. Theories of rates of reaction. The rate of a chemical reaction is defined as the rate of change of concentration/amount of reactants in unit time. It is also the rate of formation of given concentration of products in unit time. i.e. Rate of reaction = Change in concentration/amount of reactants Time taken for the change to occur Rate of reaction = Change in concentration/amount of products formed Time taken for the products to form For the above, therefore the rate of a chemical reaction is rate of decreasing reactants to form an increasing product. The SI unit of time is second(s) but minutes and hours are also used. (a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction.


(a)The collision theory The collision theory is an application of the Kinetic Theory of matter which assumes matter is made up of small/tiny/minute particles like ions atoms and molecules. The collision theory proposes that (i)for a reaction to occur, reacting particles must collide. (ii)not all collisions between reacting particles are successful in a reaction. Collisions that initiate a chemical reaction are called successful / fruitful/ effective collisions (iii)the speed at which particles collide is called collision frequency. The higher the collision frequency the higher the chances of successful / fruitful/ effective collisions to form products. (iv)the higher the chances of successful collisions, the faster the reaction. (v)the average distance between solid particles from one another is too big for them to meet and collide successfully. (vi)dissolving substances in a solvent ,make the solvent a medium for the reaction to take place. The solute particle distance is reduced as the particle ions are free to move in the solvent medium. (vii)successful collisions take place if the particles colliding have the required energy and right orientation which increases their vibration and intensity of successful / fruitful/ effective collisions to form products. (b)The Activation Energy(Ea) theory
The Enthalpy of activation(∆Ha) /Activation Energy(Ea) is the minimum amount of energy which the reactants must overcome before they react. Activation Energy(Ea) is usually required /needed in bond breaking of the reacting particles. Bond breaking is an endothermic process that require an energy input. The higher the bond energy the slower the reaction to start of. Activation energy does not influence whether a reaction is exothermic or endothermic. The energy level diagrams below shows the activation energy for exothermic and endothermic processes/reactions. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy.


Activated complex Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
The activated complex is a mixture of many intermediate possible products which may not exist under normal physical conditions ,but can theoretically exist. Exothermic reaction proceeds without further heating /external energy because it generates its own energy/heat to overcome activation energy. Endothermic reaction cannot proceed without further heating /external energy because it does not generates its own energy/heat to overcome activation energy. It generally therefore requires continuous supply of more energy/heat to sustain it to completion. 3. Measuring the rate of a chemical reaction. The rate of a chemical reaction can be measure as: (i)Volume of a gas in unit time; - if reaction is producing a gas as one of the products. - if reaction is using a gas as one reactants (ii)Change in mass of reactants/products for solid products/reactants in unit time. (iii)formation of a given mass of precipitate in unit time (iv)a certain mass of reactants to completely form products/diminish. Reactants may be homogenous or heterogenous. -Homogenous reactions involve reactants in the same phase/state e.g. solid-solid,gas-gas,liquid-liquid. -Heterogenous reactions involve reactants in the different phase/state e.g. solid-liquid,gas-liquid,solid-gas. 4. Factors influencing/altering/affecting/determining rate of reaction The following factors alter/influence/affect/determine the rate of a chemical reaction: (a)Concentration (b)Pressure (c) Temperature (d)Surface area
(e)Catalyst a) Influence of concentration on rate of reaction The higher the concentration, the higher the rate of a chemical reaction. An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker.


An increase in concentration of the reactants reduces the distance between the reacting particles increasing their collision frequency to form products. Practically an increase in concentration reduces the time taken for the reaction to take place. Practical determination of effect of concentration on reaction rate Method 1(a) Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the acid and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock.


Sample results:Table 1. Volume of acid(cm3) Volume of water(cm3) Volume of sodium thiosulphate(cm3) Time taken for mark ‘X’ to be invisible/obscured(seconds) Reciprocal of time 1 t 20.0 0.0 20.0 20.0 5.0 x 10-2 18.0 2.0 20.0 23.0 4.35 x 10-2 16.0 4.0 20.0 27.0 3.7 x 10-2 14.0 6.0 20.0 32.0 3.13 x 10-2 12.0 8.0 20.0 42.0 2.38 x 10-2 10.0 10.0 20.0 56.0 1.78 x 10-2 For most examining bodies/councils/boards the above results score for: (a) complete table as evidence for all the practical work done and completed. (b) (i)Consistent use of a decimal point on time as evidence of understanding/knowledge of the degree of accuracy of stop watches/clock. (ii)Consistent use of a minimum of four decimal points on inverse/reciprocal of time as evidence of understanding/knowledge of the degree of accuracy of scientific calculator. (c) accuracy against a school value based on candidate’s teachers-results submitted. (d) correct trend (time increase as more water is added/acid is diluted) in conformity with expected theoretical results. Sample questions 1. On separate graph papers plot a graph of: (i)volume of acid used(x-axis) against time. Label this graph I (ii) volume of acid used(x-axis) against 1/t. Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products.


Label this graph II 2. Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4.


Explain the shape of graph I Diluting/adding water is causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker.


Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it.


Sketch sample Graph I Time (seconds)
Sketch sample Graph II 3.From graph II ,determine the time taken for the cross to be obscured/invisible when the volume of the acid is: (i) 13cm3 From a correctly plotted graph 1/t at 13cm3 on the graph => 2.75 x 10-2 t = 1 / 2.75 x 10-2 = 36.3636 seconds (ii) 15cm3 From a correctly plotted graph 1/t at 15cm3 on the graph => 3.35 x 10-2 t = 1 / 3.35 x 10-2 = 29.8507 seconds (iii) 15cm3 From a correctly plotted graph 1/t Sec-1 x 10-2 Volume of acid (cm3) Volume of acid(cm3) Volume of acid(cm3)
1/t at 17cm3 on the graph => 4.0 x 10-2 t = 1 / 4.0 x 10-2 = 25.0 seconds (iv) 19cm3 From a correctly plotted graph 1/t at 19cm3 on the graph => 4.65 x 10-2 t = 1 / 4.65 x 10-2 = 21.5054 seconds 4.From graph II ,determine the volume of the acid used if the time taken for the cross to be obscured/invisible is: (i)25 seconds 1/t => 1/25 = 4.0 x 10-2 Reading from a correctly plotted graph; 4.0 x 10-2 correspond to 17.0 cm3 (ii)30 seconds 1/t => 1/30 = 3.33 x 10-2 Reading from a correctly plotted graph; 3.33 x 10-2 correspond to 14.7 cm3 (iii)40 seconds 1/t => 1/40 = 2.5 x 10-2 Reading from a correctly plotted graph; 2.5 x 10-2 correspond to 12.3 cm3 4. Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate.


Write the equation for the reaction taking place Na2S2O3 (aq) + 2HCl(aq) -> 2NaCl (aq)+ SO2 (g) + S(s) + H2O(l) Ionically: S2O32- (aq) + 2H+ (aq) -> SO2 (g) + S(s) + H2O(l) 5.Name the yellow precipitate Colloidal sulphur Method 1(b)
Reaction of sodium thisulphate with dilute hydrochloric acid You are provided with 2.0M Hydrochloric acid 0.4M sodium thiosulphate solution Procedure: Measure 10cm3 of sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Add 5.0cm3 of hydrochloric acid solution using a 10cm3 measuring cylinder into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1.


Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Repeat the procedure by measuring different volumes of the thiosulphate and adding the volumes of the distilled water to complete table 1. Sample results:Table 1. Volume of acid(cm3) Volume of water (cm3) Volume of sodium thiosulphate (cm3) Concentation of sodium thisulphate in molesdm-3 Time(T) taken for mark ‘X’ to be invisible/ obscured(seconds) T-1 5.0 0.0 25.0 0.4 20.0 5.0 x 10-2 5.0 5.0 20.0 0.32 23.0 4.35 x 10-2 5.0 10.0 15.0 0.24 27.0 3.7 x 10-2 5.0 15.0 10.0 0.16 32.0 3.13 x 10-2 Note concentration of diluted solution is got: C1V1=C2V2 => 0.4 x 25 = C2x 25 =0.4M C1V1=C2V2 => 0.4 x 20 = C2x 25 =0.32M C1V1=C2V2 => 0.4 x 15 = C2x 25 =0.24M C1V1=C2V2 => 0.4 x 10 = C2x 25 =0.16M Sample questions 1. On separate graph papers plot a graph of: (i)Concentration of sodium thiosulphate against time. Label this graph I
(ii)Concentration of sodium thiosulphate against against T-1.Label this graph II 2. Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water.


Explain the shape of graph I Diluting/adding water causes a decrease in concentration. Decrease in concentration reduces the rate of reaction by increasing the time taken for reacting particle to collide to form products. From graph II Determine the time taken if (i)12cm3 of sodium thisulphate is diluted with 13cm3 of water. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M From correct graph at concentration 0.192M => 2.4 x10-2 I/t = 2.4 x10-2 t = 41.6667seconds (ii)22cm3 of sodium thisulphate is diluted with 3cm3 of water. At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used.


At 22cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 22 = C2x 25 =0.352M From correct graph at concentration 0.352M => 3.6 x10-2 I/t = 3.6 x10-2 t = 27.7778seconds Determine the volume of water and sodium thiosulphate if T-1 is 3.0 x10-1 From correct graph at T-1 = 3.0 x10-1 => concentration = 0.65 M = C1V1=C2V2 => 0.4 x 25 = 0.65 M x V2 = 15.3846cm3 Volume of water = 25 - 15.3846cm3 = 9.6154cm3 Determine the concentration of hydrochloric acid if 12cm3 of sodium thiosulphate and 13cm3 of water was used. At 12cm3 concentration of sodium thisulphate = C1V1=C2V2 => 0.4 x 1 2 = C2x 25 =0.192M Mole ratio Na2S2 O3 :HCl =1:2 Moles of Na2S2 O3 = 0.192M x 12 => 2.304 x 10-3 moles 1000 Mole ratio HCl =2.304 x 10-1 moles = 1.152 x 10-3 moles 2 Molarity o f HCl = 1.152 x 10-3 moles x 1000 = 0.2304M 5.0
Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 10centimeter length of magnesium ribbon with sand paper/steel wool. Measure 40cm3 of 0.5M dilute hydrochloric acid into a flask .Fill a graduated gas jar with water and invert it into a trough. Stopper the flask and set up the apparatus to collect the gas produced as in the set up below: Carefully remove the stopper, carefully put the magnesium ribbon into the flask . cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar.


cork tightly. Add the acid into the flask. Connect the delivery tube into the gas jar. Immediately start off the stop watch and determine the volume of the gas produced after every 30 seconds to complete table II below. Sample results: Table II Time(seconds) 0 30 60 90 120 150 180 210 240 Volume of gas produced(cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 Sample practice questions 1.Plot a graph of volume of gas produced (y-axis) against time Magnesium ribbon Hydrochloric acid Graduated gas jar Hydrogen gas
2.Explain the shape of the graph. The rate of reaction is faster when the concentration of the acid is high . As time goes on, the concentration of the acid decreases and therefore less gas is produced. When all the acid has reacted, no more gas is produced after 210 seconds and the graph flattens. 3.Calculate the rate of reaction at 120 seconds From a tangent at 120 seconds rate of reaction = Change in volume of gas Change in time => From the tangent at 120seconds V2 - V1 = 96-84 = 12 = 0.2cm3sec-1 T2 - T1 150-90 60 4. Write an ionic equation for the reaction taking place. Mg2+(s) + 2H+(aq) -> Mg2+(aq) + H2 (g) 5. On the same axis sketch then explain the curve that would be obtained if: (i) 0.1 M hydrochloric acid is used –Label this curve I (ii)1.0 M hydrochloric acid is used –Label this curve II
Observation: Curve I is to the right Curve II is to the left Explanation A decrease in concentration shift the rate of reaction graph to the right as more time is taken for completion of the reaction. An increase in concentration shift the rate of reaction graph to the left as less time is taken for completion of the reaction. Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained.


Both graphs flatten after some time indicating the completion of the reaction. b)Influence of pressure on rate of reaction Pressure affects only gaseous reactants. An increase in pressure reduces the volume(Boyles law) in which the particles are contained. Decrease in volume of the container bring the reacting particles closer to each other which increases their chances of effective/successful/fruitful collision to form products. An increase in pressure therefore increases the rate of reaction by reducing the time for reacting particles of gases to react. At industrial level, the following are some reactions that are affected by pressure: (a)Haber process for manufacture of ammonia N2(g) + 3H2(g) -> 2NH3(g) (b)Contact process for manufacture of sulphuric(VI)acid 2SO2(g) + O2(g) -> 2SO3(g) (c)Ostwalds process for the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) -> 4NO (g) + 6H2O (l) The influence of pressure on reaction rate is not felt in solids and liquids. This is because the solid and liquid particles have fixed positions in their strong bonds and therefore no degree of freedom (Kinetic Theory of matter) c)Influence of temperature on rate of reaction
An increase in temperature increases the kinetic energy of the reacting particles by increasing their collision frequency. Increase in temperature increases the particles which can overcome the activation energy (Ea). A 10oC rise in temperature doubles the rate of reaction by reducing the time taken for the reaction to complete by a half. Practical determination of effect of Temperature on reaction rate Method 1 Reaction of sodium thisulphate with dilute hydrochloric acid Procedure: Measure 20cm3 of 0.05M sodium thisulphate into a 50cm3 glass beaker. Place the beaker on a white piece of filter paper with ink mark ‘X’ on it. Determine and record its temperature as room temperature in table 2 below. Measure 20cm3 of 0.1M hydrochloric acid solution using a 50cm3 measuring cylinder. Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above.


Put the acid into the beaker containing sodium thisulphate. Immediately start off the stop watch/clock. Determine the time taken for the ink mark ‘X’ to become invisible /obscured when viewed from above. Measure another 20cm3 separate portion of the thisulphate into a beaker, heat the solution to 30oC. Add the acid into the beaker and repeat the procedure above. Complete table 2 below using different temperatures of the thiosulphate. Sample results:Table 2. Temperature of Na2S2O3 Room temperature 30 40 50 60 Time taken for mark X to be obscured /invisible (seconds) 50.0 40.0 20.0 15.0 10.0 Reciprocal of time(1/t) 0.02 0.025 0.05 0.0667 0.1 Sample practice questions 1. Plot a graph of temperature(x-axis) against 1/t
2(a)From your graph determine the temperature at which: (i)1/t is ; I. 0.03 Reading directly from a correctly plotted graph = 32.25 oC II. 0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III.


0.07 Reading directly from a correctly plotted graph = 48.0 oC (ii) t is; I. 30 seconds 30 seconds => 1/t =1/30 =0.033 Reading directly from a correctly plotted graph 0.033 => 33.5 oC II. 45 seconds 45 seconds => 1/t =1/45 =0.022 Reading directly from a correctly plotted graph 0.022 => 29.0 oC III. 25 seconds 25 seconds => 1/t =1/25 =0.04 Reading directly from a correctly plotted graph 0.04 => 36.0 oC (b) From your graph determine the time taken for the cross to become invisible at: (i) 57.5 oC Reading directly from a correctly plotted graph at 57.5 oC= 0.094 =>1/t = 0.094 t= 1/0.094 => 10.6383 seconds
(ii) 45 oC Reading directly from a correctly plotted graph at 45 oC = 0.062 =>1/t = 0.062 t= 1/0.094 => 16.1290 seconds (iii) 35 oC Reading directly from a correctly plotted graph at 35 oC = 0.047 =>1/t = 0.047 t= 1/0.047 => 21.2766 seconds Method 2 Reaction of Magnesium with dilute hydrochloric acid Procedure Scub 5centimeter length of magnesium ribbon with sand paper/steel wool. Cut the piece into five equal one centimeter smaller pieces. Measure 20cm3 of 1.0M dilute hydrochloric acid into a glass beaker . Put one piece of the magnesium ribbon into the acid, swirl. Immediately start off the stop watch/clock. Determine the time taken for the effervescence/fizzing/bubbling to stop when viewed from above. Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above .


Record the time in table 2 at room temperature. Measure another 20cm3 portions of 1.0M dilute hydrochloric acid into a clean beaker. Heat separately one portion to 30oC, 40oC , 50oC and 60oC and adding 1cm length of the ribbon and determine the time taken for effervescence /fizzing /bubbling to stop when viewed from above . Record each time to complete table 2 below using different temperatures of the acid. Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions.


Sample results:Table 1. Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4.


Temperature of acid(oC) Room temperature 30 40 50 60 Time taken effervescence to stop (seconds) 80.0 50.0 21.0 13.5 10.0 Reciprocal of time(1/t) 0.0125 0.02 0.0476 0.0741 0.1 Sample practice questions
1. Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph.


Plot a graph of temperature(x-axis) against 1/t 2.(a)Calculate the number of moles of magnesium used given that 1cm of magnesium has a mass of 1g.(Mg= 24.0) Moles = Mass of magnesium => 1.0 = 4.167 x 10 -2 moles Molar mass of Mg 24 (b)Calculate the number of moles of hydrochloric acid used Moles of acid = molarity x volume of acid 1000 => 1.0 x 20 = 2.0 x 10 -2 moles 1000 (c)Calculate the mass of magnesium that remain unreacted Mole ratio Mg: HCl = 1:2 Moles Mg = ½ moles HCl => ½ x 2.0 x 10 -2 moles = 1.0 x 10 -2 moles Mass of reacted Mg = moles x molar mass => 1.0 x 10 -2 moles x 24 = 0.24 g Mass of unreacted Mg = Original total mass - Mass of reacted Mg => 1.0 g – 0.24 = 0.76 g (b)Calculate the total volume of hydrogen gas produced during the above reactions. 1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products.


1/t Temperature(oC)
Mole ratio Mg : H2 = 1:1 Moles of Mg that reacted per experiment = moles H2 =1.0 x 10 -2 moles Volume of Hydrogen at s.t.p produced per experiment = moles x 24 dm3 => 1.0 x 10 -2 moles x 24 dm3 = 0.24dm3 Volume of Hydrogen at s.t.p produced in 5 experiments =0.24 dm3 x 5 = 1.2 dm3 3.(a)At what temperature was the time taken for magnesium to react equal to: (i)70seconds 70 seconds => 1/t =1/70 =0.01429 Reading directly from a correctly plotted graph 0.01429 => 28.0 oC (ii)40seconds 40 seconds => 1/t =1/40 =0.025 Reading directly from a correctly plotted graph 0.025 => 32.0 oC (b)What is the time taken for magnesium to react if the reaction was done at: (i) 55.0 oC Reading directly from a correctly plotted graph at 55.0 oC=> 1/t = 8.0 x 10-2 => t = 1/8.0 x 10-2 = 12.5 seconds (ii) 47.0 oC Reading directly from a correctly plotted graph at 47.0 oC=> 1/t = 6.0 x 10-2 => t = 1/6.0 x 10-2 = 16.6667 seconds (iii) 33.0 oC Reading directly from a correctly plotted graph at 33.0 oC=> 1/t = 2.7 x 10-2 => t = 1/2.7 x 10-2 = 37.037 seconds 4. Explain the shape of the graph. Increase in temperature increases the rate of reaction as particles gain kinetic energy increasing their frequency and intensity of collision to form products. d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder.


d)Influence of surface area on rate of reaction Surface area is the area of contact. An increase in surface area is a decrease in particle size. Practically an increase in surface area involves chopping /cutting
solid lumps into smaller pieces/chips then crushing the chips into powder. Chips thus have a higher surface area than solid lumps but powder has a highest surface area. An increase in surface area of solids increases the area of contact with a liquid solution increasing the chances of successful/effective/fruitful collision to form products. The influence of surface area on rate of reaction is mainly in heterogeneous reactions. Reaction of chalk/calcium carbonate on dilute hydrochloric acid Procedure Measure 20cm3 of 1.0 M hydrochloric acid into three separate conical flasks labeled C1 C2 and C3 . Using a watch glass weigh three separate 2.5g a piece of white chalk. Place the conical flask C1 on an electronic balance. Reset the balance scale to 0.0. Put one weighed sample of the chalk into the acid in the conical flask. Determine the scale reading and record it at time =0.0. Simultaneously start of the stop watch. Determine and record the scale reading after every 30 seconds to complete Table I . Repeat all the above procedure separately with C2 and C3 to complete Table II and Table III by cutting the chalk into small pieces/chips for C2 and crushing the chalk to powder for C3 Sample results:Table 1. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 2.0 1.8 1.4 1.2 1.0 0.8 0.5 0.5 Loss in mass 0.0 0.5 0.7 1.1 1.3 1.5 1.7 2.0 2.0 Sample results:Table 1I.


Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 Mass of CaCO3 2.5 1.8 1.4 1.0 0.8 0.5 0.5 0.5 0.5 Loss in mass 0.0 0.7 1.1 1.5 1.7 2.0 2.0 2.0 2.0 Sample questions: 1.Calculate the loss in mass made at the end of each time from the original to complete table I,II and III 2.On the same axes plot a graph of total loss in mass against time (x-axes) and label them curve I, II, and III from Table I, II, and III. 3.Explain why there is a loss in mass in all experiments. Calcium carbonate react with the acid to form carbon(IV)oxide gas that escape to the atmosphere. 4.Write an ionic equation for the reaction that take place CaCO3(s) + 2H+(aq) -> Ca2+(aq) + H2O(l) + CO2(g) 5.Sulphuric(VI)acid cannot be used in the above reaction. On the same axes sketch the curve which would be obtained if the reaction was attempted by reacting a piece of a lump of chalk with 0.5M sulphuric(VI)acid. Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion.


Label it curve IV. Explain the shape of curve IV. Calcium carbonate would react with dilute 0.5M sulphuric(VI)acid to form insoluble calcium sulphate(VI) that coat /cover unreacted Calcium carbonate stopping the reaction from reaching completion. 6.Calculate the volume of carbon(IV)oxide evolved(molar gas volume at room temperature = 24 dm3, C= 12.0, O= 16.O Ca=40.0) Method I Mole ratio CaCO3(s) : CO2(g) = 1:1 Moles CaCO3(s) used = Mass CaCO3(s) = 0.025 moles Molar mass CaCO3(s) Moles CO2(g) = 0.025 moles Volume of CO2(g) = moles x molar gas volume =>0.025 moles x 24 dm3 = 0.600 dm3/600cm3
Method II Molar mass of CaCO3(s) = 100g produce 24 dm3 of CO2(g) Mass of CaCO3(s) =2.5 g produce 2.5 x 24 = 0.600dm3 100 7.From curve I ,determine the rate of reaction (loss in mass per second)at time 180 seconds on the curve. From tangent at 180 seconds on curve I Rate = M2-M1 => 2.08 – 1.375 = 0.625 = 0.006944g sec-1 T2- T1 222-132 90 8.What is the effect of particle size on the rate of reaction? A larger surface area is a reduction in particle size which increases the area of contact between reacting particles increasing their collision frequency. Theoretical examples 1. Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass?


Excess marble chips were put in a beaker containing 100cm3 of 0.2M hydrochloric acid. The beaker was then placed on a balance and total loss in mass recorded after every two minutes as in the table below. Time(minutes) 0.0 2.0 4.0 6.0 8.0 10.0 12.0 Loss in mass(g) 0.0 1.80 2.45 2.95 3.20 3.25 3.25 (a)Why was there a loss in mass? Carbon (IV) oxide gas was produced that escape to the surrounding (b)Calculate the average rate of loss in mass between: (i) 0 to 2 minutes Average rate =M2-M1 => 1.80 – 0.0 = 1.8 = 9.00g min-1 T2- T1 2.0 – 0.0 2 (i) 6 to 8 minutes Average rate =M2-M1 => 3.20 – 2.95 = 0.25 = 0.125g min-1 T2- T1 8.0 – 6.0 2 (iii) Explain the difference between the average rates of reaction in (i) and(ii) above. Between 0 and 2 minutes , the concentration of marble chips and hydrochloric acid is high therefore there is a higher collision frequency between the reacting particles leading to high successful rate of formation of products. Between 6 and 8 minutes , the concentration of marble chips and hydrochloric acid is low therefore there is low collision frequency between the reacting particles leading to less successful rate of formation of products. (c)Write the equation for the reaction that takes place. CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster.


CaCO3(s) + 2HCl (aq) -> CaCO3 (aq) + H2O(l) + CO2(g) (d)State and explain three ways in which the rate of reaction could be increased. (i)Heating the acid- increasing the temperature of the reacting particles increases their kinetic energy and thus collision frequency. (ii)Increasing the concentration of the acid-increasing in concentration reduces the distances between the reacting particles increasing their chances of effective/fruitful/successful collision to form products faster. (iii)Crushing the marble chips to powder-this reduces the particle size/increase surface area increasing the area of contact between reacting particles. (e)If the solution in the beaker was evaporated to dryness then left overnight in the open, explain what would happen. It becomes wet because calcium (II) chloride absorbs water from the atmosphere and form solution/is deliquescent. (f)When sodium sulphate (VI) was added to a portion of the contents in the beaker after the reaction , a white precipitate was formed . (i)Name the white precipitate. Calcium(II)sulphate(VI) (ii)Write an ionic equation for the formation of the white precipitate Ca2+(aq) + SO42-(aq)->CaSO4(s) (iii)State one use of the white precipitate -Making plaster for building -Manufacture of plaster of Paris
-Making sulphuric(VI)acid (g)(i) Plot a graph of total loss in mass(y-axes) against time (ii)From the graph, determine the rate of reaction at time 2 minutes. From a tangent/slope at 2 minutes; Rate of reaction = Average rate =M2-M1 => 2.25 – 1.30 = 0.95 = 0.3958g min-1 T2- T1 3.20 – 0.8 2.4 (iii)Sketch on the same axes the graph that would be obtained if 0.02M hydrochloric acid was used. Label it curve II e) Influence of catalyst on rate of reaction Catalyst is a substance that alter the rate /speed of a chemical reaction but remain chemically unchanged at the end of a reaction. Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder.


Biological catalysts are called enzymes. A catalyst does not alter the amount of products formed but itself may be altered physically e.g. from solid to powder to fine powder. Like biological enzymes, a catalyst only catalyse specific type of reactions Most industrial catalysts are transition metals or their compounds. Catalyst works by lowering the Enthalpy of activation(∆Ha)/activation energy (Ea) of the reactants .The catalyst lowers the Enthalpy of activation(∆Ha)/activation energy (Ea) by: (i) forming short lived intermediate compounds called activated complex that break up to form the final product/s (ii) being absorbed by the reactants thus providing the surface area on which reaction occurs. A catalyst has no effect on the enthalpy of reaction ∆Hr but only lowers the Enthalpy of activation(∆Ha)/activation energy (Ea)It thus do not affect/influence whether the reaction is exothermic or endothermic as shown in the energy level diagrams below. Energy level diagram showing the activation energy for exothermic processes /reactions. Activated complex Ea Catalysed A A B B Energy kJ Ea uncatalysed A-A B-B A-B A-B
Energy level diagram showing the activation energy for endothermic processes /reactions. Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities.


Activated complex The following are some catalysed reaction processes. (a)The contact process Vanadium(V) Oxide(V2O5) or platinum(Pt) catalyses the oxidation of sulphur(IV)oxide during the manufacture of sulphuric(VI) acid from contact process. SO2(g) + O2(g) ----V2O5--> SO3(g) To reduce industrial cost of manufacture of sulphuric (VI) acid from contact process Vanadium(V) Oxide(V2O5) is used because it is cheaper though it is easily poisoned by impurities. (b)Ostwalds process A A B B Energy kJ Reaction path/coordinate/path Ea A-A B-B A-B A-B ∆Hr
Platinum promoted with Rhodium catalyses the oxidation of ammonia to nitrogen(II)oxide and water during the manufacture of nitric(V)acid 4NH3(g) + 5O2(g) ----Pt/Rh--> 4NO (g) + 6H2O(l) (c)Haber process Platinum or iron catalyses the combination of nitrogen and hydrogen to form ammonia gas N2(g) + 3H2(g) ---Pt or Fe---> 2NH3(g) (d)Hydrogenation/Hardening of oil to fat Nickel (Ni) catalyses the hydrogenation of unsaturated compound containing - C=C- or –C=C- to saturated compounds without double or triple bond This process is used is used in hardening oil to fat. (e)Decomposition of hydrogen peroxide Manganese(IV)oxide speeds up the rate of decomposition of hydrogen peroxide to water and oxygen gas. This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen.


This process/reaction is used in the school laboratory preparation of Oxygen. 2H2O2 (g) ----MnO2--> O2(g) + 2H2O(l) (f)Reaction of metals with dilute sulphuric(VI)acid Copper(II)sulphate(VI) speeds up the rate of production of hydrogen gas from the reaction of Zinc and dilute sulphuric(VI)acid. This process/reaction is used in the school laboratory preparation of Hydrogen. H2 SO4 (aq) + Zn(s) ----CuSO4--> ZnSO4 (aq) + H2(g) (g) Substitution reactions When placed in bright sunlight or U.V /ultraviolet light , a mixture of a halogen and an alkane undergo substitution reactions explosively to form halogenoalkanes. When paced in diffused sunlight the reaction is very slow. e.g. CH4(g) + Cl2(g) ---u.v. light--> CH3Cl(g) + HCl(g) (h)Photosynthesis Plants convert carbon(IV)oxide gas from the atmosphere and water from the soil to form glucose and oxygen as a byproduct using sunlight / ultraviolet light. 6CO2(g) + 6H2O(l) ---u.v. light--> C6H12O6(g) + O2(g) (i)Photography
Photographic film contains silver bromide emulsion which decomposes to silver and bromine on exposure to sunlight. 2AgBr(s) ---u.v/sun light--> 2Ag(s) + Br2(l) When developed, the silver deposits give the picture of the object whose photograph was taken depending on intensity of light. A picture photographed in diffused light is therefore blurred. Practical determination of effect of catalyst on decomposition of hydrogen peroxide Measure 5cm3 of 20 volume hydrogen peroxide and then dilute to make 40cm3 in a measuring cylinder by adding distilled water. Divide it into two equal portions. (i)Transfer one 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask. Cork and swirl for 2 minutes. Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask.


Remove the cork. Test the gas produced using a glowing splint. Clean the conical/round bottomed/flat bottomed flask. (ii)Put 2.0g of Manganese (IV) oxide into the clean conical/round bottomed/flat bottomed flask. Stopper the flask. Transfer the second portion of the 20cm3volume hydrogen peroxide into a conical/round bottomed/flat bottomed flask through the dropping/thistle funnel. Connect the delivery tube to a calibrated/graduated gas jar as in the set up below. Start off the stop watch and determine the volume of gas in the calibrated/graduated gas jar after every 30 seconds to complete Table 1. (iii)Weigh a filter paper .Use the filter paper to filter the contents of the conical conical/round bottomed/flat bottomed flask. Put the residue on a sand bath to dry. Weigh the dry filter paper again .Determine the new mass Manganese (IV) oxide. Mass of MnO2 before reaction(g) Mass of MnO2 after reaction(g) 2.0 2.0 Plot a graph of volume of gas produced against time(x-axes) b) On the same axes, plot a graph of the uncatalysed reaction. (c) Explain the changes in mass of manganese(IV)oxide before and after the reaction. The mass of MnO2 before and after the reaction is the same but a more fine powder after the experiment. A catalyst therefore remains unchanged chemically but may physically change. Time(seconds) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0 240.0 270.0 Volume of gas (cm3) 0.0 20.0 40.0 60.0 80.0 90.0 95.0 96.0 96.0 96.0 Catalysed reaction Uncatalysed reaction
B.EQUILIBRIA (CHEMICAL CYBERNETICS) Equilibrium is a state of balance. Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e.


Chemical equilibrium is state of balance between the reactants and products. As reactants form products, some products form back the reactants. Reactions in which the reactants form products to completion are said to be reversible i.e. A + B -> C + D Reactions in which the reactants form products and the products can reform the reactants are said to be reversible. A + B C + D Reversible reactions may be: (a)Reversible physical changes (b)Reversible chemical changes (c)Dynamic equilibrium (a)Reversible physical changes Reversible physical change is one which involves: (i) change of state/phase from solid, liquid, gas or aqueous solutions. States of matter are interconvertible and a reaction involving a change from one state/phase can be reversed back to the original. (ii) colour changes. Some substances/compounds change their colours without change in chemical substance. Examples of reversible physical changes (i) colour change on heating and cooling: I. Zinc(II)Oxide changes from white when cool/cold to yellow when hot/heated and back. ZnO(s) ZnO(s) (white when cold) (yellow when hot) II. Lead(II)Oxide changes from yellow when cold/cool to brown when hot/heated and back. PbO(s) PbO(s)
(brown when hot) (yellow when cold) (ii)Sublimation I. Iodine sublimes from a grey crystalline solid on heating to purple vapour. Purple vapour undergoes deposition back to the grey crystalline solid. I2(s) I2(g) (grey crystalline solid (purple vapour undergo sublimation) undergo deposition) II. Carbon (IV)oxide gas undergoes deposition from a colourless gas to a white solid at very high pressures in a cylinder. It sublimes back to the colourless gas if pressure is reduced CO2(s) CO2(g) (white powdery solid (colourless/odourless gas undergo sublimation) undergo deposition) (iii)Melting/ freezing and boiling/condensation Ice on heating undergo melting to form a liquid/water. Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid.


Liquid/water on further heating boil/vaporizes to form gas/water vapour. Gas/water vapour on cooling, condenses/liquidifies to water/liquid. On further cooling, liquid water freezes to ice/solid. Melting boiling Freezing condensing (iv)Dissolving/ crystallization/distillation Solid crystals of soluble substances (solutes) dissolve in water /solvents to form a uniform mixture of the solute and solvent/solution. On crystallization /distillation /evaporation the solvent evaporate leaving a solute back. e.g. NaCl(s) + aq NaCl(aq) (b)Reversible chemical changes These are reactions that involve a chemical change of the reactants which can be reversed back by recombining the new substance formed/products. Examples of Reversible chemical changes (i)Heating Hydrated salts/adding water to anhydrous salts. H2O(s) H2O(l) H2O(s)
When hydrated salts are heated they lose some/all their water of crystallization and become anhydrous.Heating an unknown substance /compound that forms a colourless liquid droplets on the cooler parts of a dry test/boiling tube is in fact a confirmation inference that the substance/compound being heated is hydrated. When anhydrous salts are added (back) some water they form hydrated compound/salts. Heating Copper(II)sulphate(VI)pentahydrate and cobalt(II)chloride hexahydrate (i)Heat about 5.0g of Copper(II)sulphate(VI) pentahydrate in a clean dry test tube until there is no further colour change on a small Bunsen flame. Observe any changes on the side of the test/boiling tube. Allow the boiling tube to cool.Add about 10 drops of distilled water. Observe any changes. (ii)Dip a filter paper in a solution of cobalt(II)chloride hexahydrate. Pass one end the filter paper to a small Bunsen flame repeatedly. Observe any changes on the filter paper. Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white.


Dip the paper in a beaker containing distilled water. Observe any changes. Sample observations Hydrated compound Observation before heating Observation after heating Observation on adding water Copper(II)sulphate (VI) pentahydrate Blue crystalline solid (i)colour changes from blue to white. (ii)colourless liquid forms on the cooler parts of boiling / test tube (i)colour changes from white to blue (ii)boiling tube becomes warm /hot. Cobalt(II)chloride hexahydrate Pink crystalline solid/solution (i)colour changes from pink to blue. (ii) colourless liquid forms on the cooler parts of boiling / test tube (if crystal are used) (i)colour changes from blue to pink (ii)boiling tube becomes warm/hot. When blue Copper(II)sulphate (VI) pentahydrate is heated, it loses the five molecules of water of crystallization to form white anhydrous Copper(II)sulphate (VI).Water of crystallization form and condenses as colourless droplets on the cooler parts of a dry boiling/test tube. This is a chemical change that produces a new substance. On adding drops of water to an anhydrous white copper(II)sulphate(VI) the hydrated compound is
formed back. The change from hydrated to anhydrous and back is therefore reversible chemical change.Both anhydrous white copper(II)sulphate(VI) and blue cobalt(II)chloride hexahydrate are therefore used to test for the presence of water when they turn to blue and pink respectively. CuSO4(s) + 5H2 O(l) CuSO4.5H2 O(s/aq) (white/anhydrous) (blue/hydrated) CoCl2(s) + 6H2 O(l) CoCl2.6H2 O(s/aq) (blue/anhydrous) (pink/hydrated) (ii)Chemical sublimation Some compounds sublime from solid to gas by dissociating into new different compounds. e.g. Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made.


Heating ammonium chloride (i)Dip a glass rod containing concentrated hydrochloric acid. Bring it near the mouth of a bottle containing concentrated ammonia solution. Explain the observations made. When a glass rod containing hydrogen chloride gas is placed near ammonia gas, they react to form ammonium chloride solid that appear as white fumes. This experiment is used interchangeably to test for the presence of hydrogen chloride gas (and hence Cl- ions) and ammonia gas (and hence NH4+ ions) (ii)Put 2.0 g of ammonium chloride in a long dry boiling tube. Place wet / moist /damp blue and red litmus papers separately on the sides of the mouth of the boiling tube. Heat the boiling tube gently then strongly. Explain the observations made. When ammonium chloride is heated it dissociates into ammonia and hydrogen chloride gases. Since ammonia is less dense, it diffuses faster to turn both litmus papers blue before hydrogen chloride turn red because it is denser. The heating and cooling of ammonium chloride is therefore a reversible chemical change. NH3(g) + HCl(g) NH4Cl(s) (Turns moist (Turns moist (forms white fumes) litmus paper blue) litmus paper red)
(c)Dynamic equilibria For reversible reactions in a closed system: (i) at the beginning; -the reactants are decreasing in concentration with time -the products are increasing in concentration with time (ii) after some time a point is reached when as the reactants are forming products the products are forming reactants. This is called equilibrium. Sketch showing the changes in concentration of reactants and products in a closed system For a system in equilibrium: (i) a reaction from left to right (reactants to products) is called forward reaction. (ii) a reaction from right to left (products to reactants) is called backward reaction. (iii)a reaction in which the rate of forward reaction is equal to the rate of backward reaction is called a dynamic equilibrium. A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants.


A dynamic equilibrium is therefore a balance of the rate of formation of products and reactants. This balance continues until the reactants or products are disturbed/changed/ altered. Reactants concentration decreases to form products Products concentration increases from time=0.0 Equilibrium established /rate of formation of products equal to rate of formation of reactants. Concentration Mole dm-3 Reaction progress/path/coordinate
The influence of different factors on a dynamic equilibrium was first investigated from 1850-1936 by the French Chemist Louis Henry Le Chatellier. His findings were called Le Chatelliers Principle which states that: “if a stress/change is applied to a system in dynamic equilibrium, the system readjust/shift/move/behave so as to remove/ reduce/ counteract/ oppose the stress/change” Le Chatelliers Principle is applied in determining the effect/influence of several factors on systems in dynamic equilibrium. The following are the main factors that influence /alter/ affect systems in dynamic equilibrium: (a)Concentration (b)Pressure (c)Temperature (d)Catalyst (a)Influence of concentration on dynamic equilibrium An increase/decrease in concentration of reactants/products at equilibrium is a stress. From Le Chatelliers principle the system redjust so as to remove/add the excessreduced concentration. Examples of influence of concentration on dynamic equilibrium (i)Chromate(VI)/CrO42- ions in solution are yellow. Dichromate(VI)/Cr2O72- ions in solution are orange. The two solutions exist in equilibrium as in the equation: 2H+ (aq) + 2CrO42- (aq) Cr2O72- (aq) + H2O(l) (Yellow) (Orange) I. If an acid is/H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already H+ ions. The equilibrium shift forward to the right to remove/reduce the excess H+ ions added. Solution mixture becomes More Cr2O72- ions formed in the solution mixture make it to be more orange in colour. II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water.


II. If a base/OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side on the H+ ions. H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the H+ ions that have reacted with the OH- (aq) ions . More of the CrO42- ions formed in the solution mixture makes it to be more yellow in colour. 2OH- (aq) + 2Cr2O72- (aq) CrO42- (aq) + H2O(l) (Orange) (Yellow) I. If an acid/ H+ (aq) is added to the equilibrium mixture a stress is created on the reactant side on the OH- (aq). H+ ions react with OH- (aq) to form water. H+ (aq) +OH- (aq) -> H2O(l) The equilibrium shift backward to the left to add/replace the 2OH- (aq) that have reacted with the H+ (aq) ions . More Cr2O72- (aq)ions formed in the solution mixture makes it to be more Orange in colour. II. If a base /OH- (aq) is added to the equilibrium mixture a stress is created on the reactant side where there is already OH- (aq) ions. The equilibrium shift forward to the right to remove/reduce the excess OH- (aq) ions added. More of the Cr2O72- ions are formed in the solution mixture making it to be more orange in colour. (i)Practical determination of the influence of alkali/acid on Cr2O72- / CrO42- equilibrium mixture Measure about 2 cm3 of Potassium dichromate (VI) solution into a test tube. Note that the solution mixture is orange. Add three drops of 2M sulphuric(VI) acid. Shake the mixture carefully. Note that the solution mixture is remains orange. Add about six drops of 2M sodium hydroxide solution. Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium.


Shake carefully. Note that the solution mixture is turns yellow. Explanation The above observations can be explained from the fact that both the dichromate(VI)and chromate(VI) exist in equilibrium. Dichromate(VI) ions are stable in acidic solutions while chromate(VI)ions are stable in basic solutions. An equilibrium exist thus:
OH- H+ When an acid is added, the equilibrium shift forward to the right and the mixture become more orange as more Cr2O72- ions exist. When a base is added, the equilibrium shift backward to the left and the mixture become more yellow as more CrO42- ions exist. (ii)Practical determination of the influence of alkali/acid on bromine water in an equilibrium mixture Measure 2cm3 of bromine water into a boiling tube. Note its colour. Bromine water is yellow Add three drops of 2M sulphuric(VI)acid. Note any colour change Colour becomes more yellow Add seven drops of 2M sodium hydroxide solution. Note any colour change. Solution mixture becomes colourless/Bromine water is decolourized. Explanation When added distilled water,an equilibrium exist between bromine liquid (Br2(aq)) and the bromide ion(Br-), hydrobromite ion(OBr-) and hydrogen ion(H+) as in the equation: H2O(l) + Br2(aq) OBr- (aq) + H+ (aq) + Br- (aq) If an acid (H+)ions is added to the equilibrium mixture, it increases the concentration of the ions on the product side which shift backwards to the left to remove the excess H+ ions on the product side making the colour of the solution mixture more yellow. If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes.


If a base/alkali OH- is added to the equilibrium mixture, it reacts with H+ ions on the product side to form water. H+ (aq)+ OH-(aq) -> H2O(l) This decreases the concentration of the H+ ions on the product side which shift the equilibrium forward to the right to replace H+ ions making the solution mixture colourless/less yellow (Bromine water is decolorized) (iii)Practical determination of the influence of alkali/acid on common acidbase indicators. Cr2O72- CrO42-
Place 2cm3 of phenolphthalein ,methyl orange and litmus solutions each in three separate test tubes. To each test tube add two drops of water. Record your observations in Table 1 below. To the same test tubes, add three drops of 2M sulphuric(VI)acid. Record your observations in Table 1 below. To the same test tubes, add seven drops of 2M sodium hydroxide solution. Record your observations in Table 1 below. To the same test tubes, repeat adding four drops of 2M sulphuric(VI)acid. Table 1 Indicator Colour of indicator in Water Acid(2M sulphuric (VI) acid) Base(2M sodium hydroxide) Phenolphthalein Colourless Colourless Pink Methyl orange Yellow Red Orange Litmus solution Colourless Red Blue Explanation An indicator is a substance which shows whether another substance is an acid , base or neutral. Most indicators can be regarded as very weak acids that are partially dissociated into ions.An equilibrium exist between the undissociated molecules and the dissociated anions. Both the molecules and anions are coloured. i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water.


i.e. HIn(aq) H+ (aq) + In- (aq) (undissociated indicator (dissociated indicator molecule(coloured)) molecule(coloured)) When an acid H+ is added to an indicator, the H+ ions increase and equilibrium shift backward to remove excess H+ ions and therefore the colour of the undissociated (HIn) molecule shows/appears. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the H+ ion and therefore the colour of dissociated (In-) molecule shows/appears. The following examples illustrate the above. (i)Phenolphthalein indicator exist as: HPh H+ (aq) + Ph-(aq) (colourless molecule) (Pink anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore colourless. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The pink colour of dissociated (Ph-) molecule shows/appears. (ii)Methyl Orange indicator exists as: HMe H+ (aq) + Me-(aq) (Red molecule) (Yellow/Orange anion) On adding an acid ,equilibrium shift backward to the left to remove excess H+ ions and the solution mixture is therefore red. When a base/alkali OH- is added to the indicator, the OH- reacts with H+ ions from the dissociated indicator to form water. H+ (aq) + OH-(aq) -> H2O(l) (from indicator) (from alkali/base) The equilibrium shift forward to the right to replace the removed/reduced H+ ions. The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules.


The Orange colour of dissociated (Me-) molecule shows/appears. (b)Influence of Pressure on dynamic equilibrium Pressure affects gaseous reactants/products. Increase in pressure shift/favours the equilibrium towards the side with less volume/molecules. Decrease in pressure shift the equilibrium towards the side with more volume/molecules. More yield of products is obtained if high pressures produce less molecules / volume of products are formed. If the products and reactants have equal volume/molecules then pressure has no effect on the position of equilibrium The following examples show the influence of pressure on dynamic equilibrium: (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture
Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) Gay Lussacs law 2Volume 1Volume Avogadros law 2molecule 1molecule 2 volumes/molecules of Nitrogen(IV)oxide form 1 volumes/molecules of dinitrogen tetraoxide Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules.The equilibrium mixture become more yellow. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The equilibrium mixture become more brown. (ii)Iodine vapour-Hydrogen gas/Hydrogen Iodide mixture. Pure hydrogen gas reacts with Iodine vapour to form Hydrogen Iodide gas. Chemical equation : I2(g) + H2(g) ===== 2HI (g) Gay Lussacs law 1Volume 1Volume 2Volume Avogadros law 1molecule 1molecule 2molecule (1+1) 2 volumes/molecules of Iodine and Hydrogen gasform 2 volumes/molecules of Hydrogen Iodide gas. Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process.


Change in pressure thus has no effect on position of equilibrium. (iii)Haber process. Increase in pressure of the Nitrogen/Hydrogen mixture favours the formation of more molecules of Ammonia gas in Haber process. The yield of ammonia is thus favoured by high pressures Chemical equation : N2(g) + 3H2 (g) -> 2NH3 (g) Gay Lussacs law 1Volume 3Volume 2Volume Avogadros law 1molecule 3molecule 2molecule (1 + 3) 4 volumes/molecules of Nitrogen and Hydrogen react to form 2 volumes/molecules of ammonia. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of ammonia increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of ammonia decrease. (iv)Contact process. Increase in pressure of the Sulphur(IV)oxide/Oxygen mixture favours the formation of more molecules of Sulphur(VI)oxide gas in Contact process. The yield of Sulphur(VI)oxide gas is thus favoured by high pressures. Chemical equation : 2SO2(g) + O2 (g) -> 2SO3 (g) Gay Lussacs law 2Volume 1Volume 2Volume Avogadros law 2molecule 1molecule 2molecule (2 + 1) 3 volumes/molecules of Sulphur(IV)oxide/Oxygen mixture react to form 2 volumes/molecules of Sulphur(VI)oxide gas. Increase in pressure shift the equilibrium forward to the left where there is less volume/molecules. The yield of Sulphur(VI)oxide gas increase. Decrease in pressure shift the equilibrium backward to the right where there is more volume/molecules. The yield of Sulphur(VI)oxide gas decrease. (v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures.


(v)Ostwalds process. Increase in pressure of the Ammonia/Oxygen mixture favours the formation of more molecules of Nitrogen(II)oxide gas and water vapour in Ostwalds process. The yield of Nitrogen(II)oxide gas and water vapour is thus favoured by low pressures. Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (g) Gay Lussacs law 4Volume 5Volume 4Volume 6Volume Avogadros law 4molecule 5molecule 4molecule 6Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 10 volumes/molecules of Nitrogen(II)oxide gas and water vapour. Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour decrease. Decrease in pressure shift the equilibrium forward to the right where there is more volume/molecules. The yield of Nitrogen(II)oxide gas and water vapour increase. Note If the water vapour is condensed on cooling, then: Chemical equation : 4NH3(g) + 5O2 (g) -> 4NO(g) + 6H2O (l) Gay Lussacs law 4Volume 5Volume 4Volume 0Volume Avogadros law 4molecule 5molecule 4molecule 0Molecule (4 + 5) 9 volumes/molecules of Ammonia/Oxygen mixture react to form 4 volumes/molecules of Nitrogen(II)oxide gas and no vapour. Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules. The yield of Nitrogen(II)oxide gas increase. Decrease in pressure shift the equilibrium backward to the left where there is more volume/molecules. The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH).


The yield of Nitrogen(II)oxide gas decrease. (c)Influence of Temperature on dynamic equilibrium A decrease in temperature favours the reaction that liberate/generate more heat thus exothermic reaction(-ΔH). An increase in temperature favours the reaction that do not liberate /generate more heat thus endothermic reaction(+ΔH). Endothermic reaction are thus favoured by high temperature/heating Exothermic reaction are favoured by low temperature/cooling. If a reaction/equilibrium mixture is neither exothermic or endothermic, then a change in temperature/cooling/heating has no effect on the equilibrium position. (i)Nitrogen(IV)oxide /Dinitrogen tetroxide mixture Nitrogen(IV)oxide and dinitrogen tetraoxide can exist in dynamic equilibrium in a closed test tube. Nitrogen(IV)oxide is a brown gas. Dinitrogen tetraoxide is a yellow gas. Chemical equation : 2NO2(g) ===== N2 O4 (g) On heating /increasing temperature, the mixture becomes more brown. On cooling the mixture become more yellow. This show that (i)the forward reaction to the right is exothermic(-ΔH). On heating an exothermic process the equilibrium shifts to the side that generate /liberate less heat. (ii)the backward reaction to the right is endothermic(+ΔH). On cooling an endothermic process the equilibrium shifts to the side that do not generate /liberate heat. (c)Influence of Catalyst on dynamic equilibrium A catalyst has no effect on the position of equilibrium. It only speeds up the rate of attainment. e.g. Esterification of alkanols and alkanoic acids naturally take place in fruits.In the laboratory concentrated sulphuric(VI)acid catalyse the reaction.The equilibrium mixture forms the ester faster but the yield does not increase. CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount.


CH3CH2OH(l)+CH3COOH(l) ==Conc.H2SO4== CH3COOCH2CH3(aq) + H2O(l) (d)Influence of rate of reaction and dynamic equilibrium (Optimum conditions) on industrial processes Industrial processes are commercial profit oriented. All industrial processes take place in closed systems and thus in dynamic equilibrium. For manufacturers, obtaining the highest yield at minimum cost and shortest time is paramount. The conditions required to obtain the highest yield of products within the shortest time at minimum cost are called optimum conditions Optimum condition thus require understanding the effect of various factors on: (i)rate of reaction(Chemical kinetics) (ii)dynamic equilibrium(Chemical cybernetics)
1.Optimum condition in Haber process Chemical equation N2 (g) + 3H2 (g) ===Fe/Pt=== 2NH3 (g) ΔH = -92kJ Equilibrium/Reaction rate considerations (i)Removing ammonia gas once formed shift the equilibrium forward to the right to replace the ammonia. More/higher yield of ammonia is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of ammonia is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 500atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -92kJ) . Ammonia formed decomposes back to Nitrogen and Hydrogen to remove excess heat therefore a less yield of ammonia is attained. Very low temperature decrease the collision frequency of Nitrogen and Hydrogen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation.


(iv)Iron and platinum can be used as catalyst. Platinum is a better catalyst but more expensive and easily poisoned by impurities than Iron. Iron is promoted /impregnated with AluminiumOxide(Al2O3) to increase its surface area/area of contact with reactants and thus efficiency.The catalyst does not increase the yield of ammonia but it speed up its rate of formation. 2.Optimum condition in Contact process Chemical equation 2SO2 (g) + O2 (g) ===V2O5/Pt=== 2SO3 (g) ΔH = -197kJ Equilibrium/Reaction rate considerations
(i)Removing sulphur(VI)oxide gas once formed shift the equilibrium forward to the right to replace the sulphur(VI)oxide. More/higher yield of sulphur(VI) oxide is attained. (ii)Increase in pressure shift the equilibrium forward to the right where there is less volume/molecules . More/higher yield of sulphur(VI)oxide is attained. Very high pressures raises the cost of production because they are expensive to produce and maintain. An optimum pressure of about 1-2 atmospheres is normally used to attain about 96% yield of SO3. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -197kJ) . Sulphur(VI)oxide formed decomposes back to Sulphur(IV)oxide and Oxygen to remove excess heat therefore a less yield of Sulphur(VI)oxide is attained. Very low temperature decrease the collision frequency of Sulphur(IV)oxide and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 450oC is normally used. (iv)Vanadium(V)Oxide and platinum can be used as catalyst. Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation.


Platinum is a better catalyst and less easily poisoned by impurities but more expensive. Vanadium(V)Oxide is very cheap even if it is easily poisoned by impurities. The catalyst does not increase the yield of Sulphur (VI)Oxide but it speed up its rate of formation. 3.Optimum condition in Ostwalds process Chemical equation 4NH3 (g) + 5O2 (g) ===Pt/Rh=== 4NO (g) + 6H2O (g) ΔH = -950kJ Equilibrium/Reaction rate considerations (i)Removing Nitrogen(II)oxide gas once formed shift the equilibrium forward to the right to replace the Nitrogen(II)oxide. More/higher yield of Nitrogen(II) oxide is attained. (ii)Increase in pressure shift the equilibrium backward to the left where there is less volume/molecules . Less/lower yield of Nitrogen(II)oxide is attained. Very low pressures increases the distance between reacting NH3and O2 molecules. An optimum pressure of about 9 atmospheres is normally used. (iii)Increase in temperature shift the equilibrium backward to the left because the reaction is exothermic(ΔH = -950kJ) . Nitrogen(II)oxide and water vapour formed decomposes back to Ammonia and Oxygen to remove excess heat therefore a less yield of Nitrogen(II)oxide is attained. Very low temperature decrease the collision frequency of Ammonia and Oxygen and thus the rate of reaction too slow and uneconomical. An optimum temperature of about 900oC is normally used. (iv)Platinum can be used as catalyst. Platinum is very expensive.It is: -promoted with Rhodium to increase the surface area/area of contact. -added/coated on the surface of asbestos to form platinized –asbestos to reduce the amount/quantity used. The catalyst does not increase the yield of Nitrogen (II)Oxide but it speed up its rate of formation. C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2.


C.SAMPLE REVISION QUESTIONS 1.State two distinctive features of a dynamic equilibrium. (i)the rate of forward reaction is equal to the rate of forward reaction (ii)at equilibrium the concentrations of reactants and products do not change. 2. Explain the effect of increase in pressure on the following: (i) N2(g) + O2(g) ===== 2NO(g) Gay Lussacs law 1Volume 1Volume 2 Volume Avogadros law 1 molecule 1 molecule 2 molecule 2 volume on reactant side produce 2 volume on product side. Increase in pressure thus have no effect on position of equilibrium. (ii) 2H2(g) + CO(g) ===== CH3OH (g)
Gay Lussacs law 2Volume 1Volume 1 Volume Avogadros law 2 molecule 1 molecule 1 molecule 3 volume on reactant side produce 1 volume on product side. Increase in pressure shift the equilibrium forward to the left. More yield of CH3OH is formed. 4. Explain the effect of increasing temperature on the following: 2SO2(g) + O2 (g) ===== 2SO3 (g) ΔH = -189kJ Forward reaction is exothermic. Increase in temperature shift the equilibrium backward to reduce the excess heat. 5.120g of brass an alloy of copper and Zinc was put it a flask containing dilute hydrochloric acid. The flask was placed on an electric balance. The readings on the balance were recorded as in the table below Time(Seconds) Mass of flask(grams) Loss in mass(grams) 0 600 20 599.50 40 599.12 60 598.84 80 598.66 100 598.54 120 598.50 140 598.50 160 598.50 (a)Complete the table by calculating the loss in mass (b)What does the “600” gram reading on the balance represent The initial mass of brass and the acid before any reaction take place. (c)Plot a graph of Time (x-axes) against loss in mass. (d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant.


(d)Explain the shape of your graph The reaction produce hydrogen gas as one of the products that escape to the atmosphere. This decreases the mass of flask.After 120 seconds,the
react is complete. No more hydrogen is evolved.The mass of flask remain constant. (d)At what time was the loss in mass equal to: (i)1.20g Reading from a correctly plotted graph = (ii)1.30g Reading from a correctly plotted graph = (iii)1.40g Reading from a correctly plotted graph = (e)What was the loss in mass at: (i)50oC Reading from a correctly plotted graph = (ii) 70oC Reading from a correctly plotted graph = (iii) 90oC g Reading from a correctly plotted graph =
1 21.0.0 ELECTROCHEMISTRY (25 LESSONS) Electrochemistry can be defined as the study of the effects of electricity on a substance/ compound and how chemical reactions produce electricity. Electrochemistry therefore deals mainly with: i) Reduction and oxidation ii) Electrochemical (voltaic) cell iii) Electrolysis (electrolytic) cell (i)REDUCTION AND OXIDATION (REDOX) 1. In teams of oxygen transfer: i) Reduction is removal of oxygen. ii) Oxidation is addition of oxygen. iii) Redox is simultaneous addition and removal of oxygen. iv) Reducing agent is the species that undergoes oxidation, therefore gains oxygen. v) Oxidizing agent is the species that undergoes reduction, therefore looses/donates oxygen. e.g. When hydrogen is passed through heated copper (II) oxide, it is oxidised to copper metal as in the equation below: CuO (s) + H2 (g) -> Cu (s) + H2O (l)
2 (Oxidising agent) (Reducing agent) 2. In terms of hydrogen transfer: i) Oxidation is the removal of hydrogen. ii) Reduction is the addition of hydrogen. iii) Redox is simultaneous addition and removal of hydrogen. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates hydrogen. v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid).


v) Oxidizing agent is the species that undergoes reduction, therefore gains hydrogen. e.g. When hydrogen sulphide gas is bubbled into a gas jar containing chlorine gas it is oxidized (loose the hydrogen) to sulphur (yellow solid). The chlorine is reduced (gain hydrogen) to hydrogen chlorine gas. Cl2 (g) + H2S (g) -> S(S) + 2HCl (g) (Oxidizing agent) (Reducing agent) 3. In terms of electron transfer: i) Oxidation is donation/ loss/ removal of electrons. ii) Reduction is gain/ accept/ addition of electrons. iii) Redox is simultaneous gain/ accept/ addition and donation/ loss/ removal of electrons. iv) Reducing agent is the species that undergoes oxidation, therefore looses/ donates electrons. v) Oxidizing agent is the species that undergoes reduction, therefore gains/ accepts electrons. Example a) Displacement of metals from their solutions: Place 5cm3 each of Iron (II) sulphate (VI) solution into three different test tubes. Add about 1g of copper tunings / powder into one test tube then zinc and magnesium powders separately into the other test tubes. Shake thoroughly for 2 minutes each. Record any colour changes in the table below. Metal added to Iron (II) sulphate (VI) solution Colour changes Copper Solution remains green Zinc Green colour fades Magnesium Green colour fades Explanation -When a more reactive metal is added to a solution of less reactive metal, it displaces it from its solution. 3 -When a less reactive metal is added to a solution of a more reactive metal, it does not displace it from its solution. -Copper is less reactive than iron therefore cannot displace iron its solution. -Zinc is more reactive than iron therefore can displace iron from its solution. -Magnesium is more reactive than iron therefore can displace iron from its solution. In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1.


In terms of electron transfer: - the more reactive metal undergoes oxidation (reducing agent) by donating/loosing electrons to form ions -the less reactive metal undergoes reduction (oxidizing agent) by its ions in solution gaining /accepting/acquiring the electrons to form the metal. -displacement of metals involves therefore electron transfer from a more reactive metal to ions of another less reactive metal. Examples 1. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Zn(s) -> Zn2+(aq) + Fe(s) (redox/both donation and gain of electrons) 2. Mg(s) -> Mg2+(aq) + 2e (oxidation/donation of electrons) Fe2+(aq) + 2e -> Fe(s) (reduction/gain of electrons) Fe2+(aq) + Mg(s) -> Mg2+(aq) + Fe(s) (redox/both donation and gain of electrons) 3. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6.


Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cu2+(aq) + 2e -> Cu(s) (reduction/gain of electrons) Cu2+(aq) + Zn(s) -> Zn2+(aq) + Cu(s) (redox/both donation and gain of electrons) 4. Fe(s) -> Fe2+(aq) + 2e (oxidation/donation of electrons) 2Ag+(aq) + 2e -> 2Ag(s) (reduction/gain of electrons) 2Ag+(aq) + Fe(s) -> Fe2+(aq) + 2Ag(s) (redox/both donation and gain of electrons) 5. Zn(s) -> Zn2+(aq) + 2e (oxidation/donation of electrons) Cl2(g) + 2e -> 2Cl-(aq) (reduction/gain of electrons) Cl2(g) + Zn(s) -> Zn2+(aq) + 2Cl-(aq) (redox/both donation and gain of electrons) 6. 2Mg(s) -> 2Mg2+(aq) + 4e (oxidation/donation of electrons) O2(g) + 4e -> 2O2-(aq) (reduction/gain of electrons) O2(g) + 2Mg(s) -> 2Mg2+(aq) + 2O2-(aq) (redox/both donation and gain of electrons)
4 Note (i)The number of electrons donated/lost MUST be equal to the number of electrons gained/acquired. (i)During displacement reaction, the colour of ions /salts fades but does not if displacement does not take place. e.g a)Green colour of Fe2+(aq) fades if Fe2+(aq) ions are displaced from their solution. Green colour of Fe2+(aq) appear if Fe/iron displaces another salt/ions from their solution. b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution.


b)Blue colour of Cu2+(aq) fades if Cu2+(aq) ions are displaced from their solution and brown copper deposits appear. Blue colour of Cu2+(aq) appear if Cu/copper displaces another salt/ions from their solution. c)Brown colour of Fe3+(aq) fades if Fe3+(aq) ions are displaced from their solution. Brown colour of Fe3+(aq) appear if Fe/iron displaces another salt/ions from their solution to form Fe3+(aq). (iii)Displacement reactions also produce energy/heat. The closer/nearer the metals in the reactivity/electrochemical series the less energy/heat of displacement. (iv)The higher the metal in the reactivity series therefore the easier to loose/donate electrons and thus the stronger the reducing agent. 4. (a)In terms of oxidation number: i) Oxidation is increase in oxidation numbers. ii) Reduction is decrease in oxidation numbers. iii) Redox is simultaneous increase in oxidation numbers of one species/substance and a decrease in oxidation numbers of another species/substance. iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g.


iv) Reducing agent is the species that undergoes oxidation, therefore increases its oxidation number. v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g.


v) Oxidizing agent is the species that undergoes reduction, therefore increases its oxidation number. (b)The idea/concept of oxidation numbers uses/applies the following simple guideline rules: Guidelines /rules applied in assigning oxidation number 1.Oxidation number of combined Oxygen is always -2 except in peroxides (Na2O2/H2O2) where its Oxidation number is -1
5 2.Oxidation number of combined Hydrogen is always +1except in Hydrides (NaH/KH) where its Oxidation number is -1 3.All atoms and molecules of elements have oxidation number 0 (zero) Atom Oxidation number Molecule Oxidation number Na 0 Cl2 0 O 0 O2 0 H 0 H2 0 Al 0 N2 0 Ne 0 O3 0 K 0 P3 0 Cu 0 S8 0 4.All combined metals and non-metals have oxidation numbers equal to their valency /oxidation state e.g. Metal/non-metal ion Valency Oxidation state Oxidation number Fe2+ 2 -2 -2 Fe3+ 3 -3 -3 Cu2+ 2 -2 -2 Cu+ 1 +1 +1 Cl- 1 -1 -1 O2- 2 -2 -2 Na+ 1 +1 +1 Al3+ 3 +3 +3 P3- 3 -3 -3 Pb2+ 2 +2 +2 5.Sum of oxidation numbers of atoms of elements making a compound is equal zero(0) e.g. Using this rule ,an unknown oxidation number of an atom in a compound can be determined as below: a) CuSO4 has- -one atom of Cu with oxidation number +2( refer to Rule 4) -one atom of S with oxidation number +6 ( refer to Rule 4) -six atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in CuSO4 = (+2 + +6 + (-2 x 6)) = 0 b) H2SO4 has- -two atom of H each with oxidation number +1( refer to Rule 2) -one atom of S with oxidation number +6 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in H2SO4 = (+2 + +6 + (-2 x 4)) = 0
6 c) KMnO4 has- -one atom of K with oxidation number +1( refer to Rule 4) -one atom of Mn with oxidation number +7 ( refer to Rule 4) -four atoms of O each with oxidation number -2( refer to Rule 4) Sum of oxidation numbers of atoms in KMnO4 = (+1 + +7 + (-2 x 4)) = 0 Determine the oxidation number of: I.Nitrogen in; -NO => x + -2 = 0 thus x = 0 – (-2) = + 2 The chemical name of this compound is thus Nitrogen(II)oxide -NO2 => x + (-2 x2)= 0 thus x = 0 – (-4) = + 4 The chemical name of this compound is thus Nitrogen(IV)oxide -N2O => 2x + -2 = 0 thus 2x = 0 – (-2) = +2/2= +1 The chemical name of this compound is thus Nitrogen(I)oxide II.