1.4 rate of reaction(1.2d)

Effect of Catalyst on the rate of Effect of Catalyst on the rate of reactionreaction

Experiment 1.4: To study the effect of a catalyst on the rate of decomposition of hydrogen peroxide


Problem statement How do catalyst affect the rate of decomposition of

hydrogen peroxide


Hypothesis Manganese (IV) oxide speed up the

decomposition of hydrogen peroxide


Variables: (a) Manipulated variable: The presence of

manganese (IV) oxide (b) Responding variable: The release of oxygen

gas (c) Fixed (controlled) variables: Volume and

concentration of hydrogen peroxide


Apparatus Boiling tube and wooden splinter

Material Hydrogen peroxide and manganese (IV) oxide


Procedure 1 A boiling tube is half-filled with hydrogen

peroxide. 2 A glowing splinter is placed at the mouth of

the boiling tube to test for the gas evolved (Figure 1.25).


Procedure 3 The changes that take place inside the

boiling tube and on the glowing splinter are recorded.

4 0.5 g of manganese (IV) oxide, MnO2 is added to hydrogen peroxide and shaken. The changes that take place in the boiling tube and on the glowing splinter are recorded.


Results


Discussion 1 Hydrogen peroxide decomposes to oxygen

gas and water according to the equation: 2H2O2 (aq) 2H2O(l) + O2 (g)


Discussion 2 The glowing splinter is rekindled in the

presence of oxygen gas.


Conclusion The rate of evolution of oxygen gas increases

when manganese (IV) oxide is added to hydrogen peroxide. This proves that manganese

(IV) oxide acts as a catalyst and speeds up the decomposition of hydrogen peroxide to water and oxygen. The hypothesis is accepted.

The effect of concentration of hydrogen The effect of concentration of hydrogen peroxide on the rate of reactionperoxide on the rate of reaction

1 The graph in Figure 1.21 shows the effect of concentration of hydrogen peroxide on the rate of decomposition of hydrogen peroxide.


In Experiment I, 50 cm3 of 0.14 mol dm-3 of hydrogen peroxide and 0.2 g of manganese (IV) oxide are used.

In Experiment II, a solution containing 25 cm3 of the same hydrogen peroxide mixed with 25 cm3 of water and 0.2 g of manganese (IV) oxide are used. For both the experiments, the temperature is kept constant.


2 (a) For Experiment I Concentration of H2O2 = 0.14 mol dm-3

For experiment II, hydrogen peroxide is diluted. (M1V1)before dilution =(M2V2)after dilution Concentration of H2O2 after dilution

307.050

2514.0

moldm


2 (b) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is faster than the rate of reaction in experiment II. We can therefore conclude that the higher the concentration of hydrogen peroxide, the faster the rate of reaction.

Factors that affect the rate of Factors that affect the rate of reactionreaction

2 (c) The maximum volume of oxygen gas produced in Experiment I is twice that produced in Experiment II. This is because the number of moles of hydrogen peroxide used in Experiment I is twice that used in Experiment II.

Explaining the effectiveness of different Explaining the effectiveness of different catalysts on the rate of decomposition of catalysts on the rate of decomposition of

hydrogen peroxidehydrogen peroxide

1 Figure 1.22 shows the results of an experiment carried out to study the effect of different catalysts (of the same mass) on the rate of decomposition of hydrogen peroxide.

Explaining the effectiveness of different catalysts on Explaining the effectiveness of different catalysts on the rate of decomposition of hydrogen peroxidethe rate of decomposition of hydrogen peroxide

In Experiment I, 50 cm3 of hydrogen peroxide and 0.5 g of manganese(IV) oxide are used.

In Experiment II, 50 cm3 of hydrogen peroxide and 0.5 g of iron (III) oxide are used.

For both the experiments, the concentration and volume of hydrogen peroxide as well as the temperature are kept constant.


2 Analysis of the reaction rate curve in Figure 1.22

(a) At any particular instant, the gradient of graph I is greater than the gradient of graph II. This means that the rate of reaction in Experiment I is faster than the rate of reaction in Experiment II. Thus, the experiment proves that manganese(IV) oxide is a more effective catalyst than iron(III) oxide in the decomposition of hydrogen peroxide.


2 Analysis of the reaction rate curve in Figure 1.22

(b) The maximum volumes of oxygen gas collected in both the experiments are the same because the volume and concentration of hydrogen peroxide used are the same. This experiment shows that a catalyst does not change the yield of the products.

Amount of catalysts on the rate of decomposition of Amount of catalysts on the rate of decomposition of hydrogen peroxidehydrogen peroxide

Experiment 1.5: To Invstigate the effect of the amount of the catalyst, manganese (IV) oxide on the decomposition of hydrogen peroxide


Problem statement How does the amount of manganese(IV)oxide

affect on the decomposition of hydrogen peroxide?


Hypothesis The rate of the decomposition of hydrogen

peroxide increases when the amount of the catalyst used is increased


Variables (a) Manipulated variable: Amount of the catalyst

used (b) Responding variable: The volume of oxygen

given off at half-minute intervals (c) Fixed (controlled) variables: Volume and

concentration of hydrogen peroxide, temperature of the experiment and type of the catalyst

Amount of catalysts on the rate of Amount of catalysts on the rate of decomposition of hydrogen peroxidedecomposition of hydrogen peroxide

Apparatus Measuring cylinder, conical flask, delivery tube,

rubber stopper, retort stand clamp and burette.

Materials 0.2 mol dm-3 hydrogen peroxide and

manganese(IV) oxide.


Experiment 1.5: To Invstigate the effect of the amount of the catalyst, manganese (IV) oxide on the decomposition of hydrogen peroxide


Procedure 1 Using a measuring cylinder, 25 cm3 of 0.2

mol dm-3 hydrogen peroxide is measured into a conical flask and 0.5 g of manganese(IV) oxide is added to the hydrogen peroxide.


Procedure 2 The conical flask is immediately closed with

a stopper fitted with a delivery tube (Figure 1.28) and the stopwatch is started simultaneously. The conical flask is swirled gently.


Procedure 3 The total volume of oxygen gas given off is

determined from the burette reading at intervals of ½ minute for 4 minutes.


Procedure 4 The experiment is repeated using 0.20 g of

manganese(IV) oxide instead of 0.50 g of manganese(IV) oxide.


Results Experiment l. Decomposition of hydrogen peroxide in the

presence of 0.5 g of manganese(IV) oxide


Results Experiment l. Decomposition of hydrogen peroxide in the

presence of 0.2 g of manganese(IV) oxide


Discussion 1 Based on the results of Experiments I and II, two

graphs of total volume of oxygen gas against time for the decomposition of hydrogen peroxide are plotted on the same axes (Figure 1.29).


Discussion 1 Graph I refers to the decomposition of hydrogen

peroxide catalysed by 0.5 g of manganese(IV) oxide, while graph II refers to the decomposition of hydrogen peroxide catalysed by 0.2 g of manganese(IV) oxide.


Discussion 2 The gradient of graph I is steeper than the

gradient of graph II, This shows that the rate of reaction I is faster than the rate of reaction II.


Discussion 3 If the decomposition of hydrogen peroxide in

both the experiments is allowed to complete, the maximum volumes of oxygen gas collected for both the experiments will be the same.


Discussion


Discussion 4 The quantity of catalyst does not affect

the amount of products formed.


Conclusion The larger the amount of the catalyst

manganese(IV) oxide used, the higher the rate of decomposition of hydrogen peroxide.

Applications of factors that affect rates Applications of factors that affect rates of reaction in daily life and in industrial of reaction in daily life and in industrial

processesprocesses Combustion of charcoal 1 Large pieces of charcoal will not catch fire

easily because the total surface area exposed to oxygen is small.


processesprocesses Combustion of charcoal 2 If small pieces of charcoal are used, they can

burn easily. This is because the total surface area exposed to the air increases. Thus, the rate of reaction with oxygen (combustion) increases.


processesprocesses Storing food in refrigerators 1 The decomposition and decay of food is a

chemical reaction caused by the action of microorganisms such as bacteria and fungi. These microorganisms multiply very rapidly at the temperature range of 10-60 °C.


processesprocesses Storing food in refrigerators 2 Room temperature is the optimum

temperature for the breeding of microorganisms in food. As a result, food turns bad quickly at room temperature.


processesprocesses Storing food in refrigerators

3 At low temperatures, for example, 5 °C (the normal temperature of a refrigerator), the activities of bacteria are slowed down. Hence, food that is kept in a refrigerator will last longer because the decaying reaction that destroys the food can be slowed down.


processesprocesses Storing food in refrigerators 4 In the supermarkets, fish, meat and other

types of fresh foods are kept in deep-freeze compartments where the temperature is about -20 °C. This keeps the food fresh for a few months because the very low temperature slows down the chemical reactions that cause the food to decay.


processesprocesses Cooking food in pressure cookers 1 Pressure cookers are used to speed up

cooking.


processesprocesses Cooking food in pressure cookers 2 In the pressure cooker, the higher pressure

enables water or oil to boil at a temperature higher than their normal boiling points. Furthermore, an increase in pressure causes an increase in the number of water molecules or cooking oil molecules coming into contact and colliding with the food particles.


processesprocesses Cooking food in pressure cookers 3 At a higher temperature and pressure, the

rate of reaction becomes faster. Thus, food cook faster in pressure cookers.


processesprocesses Uses of catalysts in industry 1 From the economic point of view, catalysts

play a vital role in industrial processes.


processesprocesses Uses of catalysts in industry 2 Catalysts do not increase the yields of

reactions. However, catalysts are used widely in industrial processes to speed up the rates of reactions so that the same amount of products can be obtained in a shorter time. As a result, the use of catalysts brings down the cost of production.


processesprocesses Uses of catalysts in industry 3 In the chemical industry, small pellets of

solid catalysts are used instead of big lumps. This is to give a larger surface for catalytic reaction to occur and hence a faster reaction will result.


processesprocesses The manufacture of ammonia (Haber process) 1 The Haber process is an industrial process

for the manufacture of ammonia from nitrogen and hydrogen.


processesprocesses The manufacture of ammonia (Haber process) 2 Nitrogen and hydrogen do not react at room

temperature and pressure. High temperature and pressure and the presence of a catalyst are required for nitrogen to react with hydrogen.


processesprocesses The manufacture of ammonia (Haber process) 3 The optimum conditions for obtaining a

maximum yield of ammonia in the Haber process are as follows:

(a) Temperature: 450-550 oC (b) Pressure : 200-500 atmospheres (c) Catalyst: Finely divided iron (Fe)


processesprocesses The manufacture of ammonia (Haber process) 4 In terms of industrial processes, a

temperature of 450 °C is considered as moderately high but the rate of reaction is slow at this temperature. Thus, a catalyst is required to increase the rate of reaction.


processesprocesses 5 In the Haber process, ammonia is produced

when a mixture of nitrogen and hydrogen (in the ratio of 1:3 by volume) is passed over finely divided iron as catalyst at 450-500 °C and 200-500 atmospheres. Under these conditions, about 10% yield of ammonia is obtained.


processesprocesses The manufacture of sulphuric add (Contact

process) 1 The contact process is the industrial process

for the manufacture of sulphuric acid from sulphur and oxygen.


processesprocesses The manufacture of sulphuric add (Contact process)

Raw materials required: sulphur, air and water. Conditions for the reaction of SO2 with O2 (from the

air): (a) Temperature: 450-500 °C (b) Pressure: 1-2 atmospheres (c) Catalyst: Vanadium(V) oxide, V2O5


processesprocesses 2 The following reaction scheme shows the steps

involved in the manufacture of sulphuric acid:


processesprocesses 3 In Step 2, sulphur dioxide is oxidised to sulphur

trioxide. The mixture of sulphur dioxide and oxygen is passed over vanadium(V) oxide, V2O5, as catalyst at 450-500 °C and a pressure of 1-2 atmospheres to form sulphur trioxide. Under these conditions, a yield of 98% of sulphur trioxide is obtained.


processesprocesses The manufacture of nitric acid (Ostwald process) 1. The Ostwald process is used to manufacture nitric

acid. Raw materials required: ammonia, air and water Conditions: (a) Temperature: 900 ° C (b) Pressure: 1-8 atmospheres (c) Catalyst: platinum


processesprocesses The manufacture of nitric acid (Ostwald process) The following reaction scheme shows the steps involved in

the manufacture of nitric acid.


processesprocesses The manufacture of nitric acid (Ostwald process) 2 In the Ostwald process, nitrogen monoxide, NO, is

produced (step 1) when ammonia gas is passed over the platinum (Pt) catalyst at about 900 ° C and 1-8 atmospheres.

In this reaction, ammonia is oxidised to nitrogen monoxide.


processesprocesses Example 5 Two experiments were carried out to determine

the rate of producing oxygen gas during the decomposition of hydrogen peroxide. In Experiment I, 20 cm3 of 2 moldm-3 hydrogen peroxide were used and the results of the experiment are shown on graph I in Figure 1.26.


processesprocesses Example 5 (a) Sketch a graph on the same axes to show

the results of the experiments that will be obtained if 5 cm3 of 4 mol dm-3 hydrogen peroxide were used for the reaction.


processesprocesses Example 5 Solution (a)


processesprocesses Example 5 (b) Explain your answer in (a).


processesprocesses Example 5 Solution (b) Differences in terms of rate of reaction Graph II is steeper than graph I because the rate of

reaction in Experiment II is expected to be faster than Experiment I. When the concentration of hydrogen peroxide is increased from 2 moldm-3 to 4 mol dm-3, the rate of reaction also increases


processesprocesses Example 5 Solution (b) Number of moles of H2O2 used in Experiment I

mol04.01000

202


processesprocesses Experiment 6 Solution (b) Number of moles of H2O2 used in Experiment I

Volume of oxygen collected at room temperature in Experiment I

mol04.01000

202

34802400004.02

1cm

2H2O2 (aq) 2H2O(l) + O2 (g)

2mol 1mol


processesprocesses Experiment 6 Solution (b) Number of moles of H2O2 used in

Experiment II= mol02.01000

54


processesprocesses Example 5 Solution (b) Volume of oxygen collected at room

temperature in Experiment II

=3240480

2

1cm


processesprocesses Example 5 (c) State the controlled variables for both the

experiments.


processesprocesses Example 5 Solution (c) Fixed (controlled) variables: In both the experiments, the same mass of the

catalyst and the same temperature of reaction are used.

1.4 rate of reaction(1.2d)

Documents

finely divided

raw materials

hydrogen peroxide

oxygen gas

reaction scheme

reaction rate

boiling tube

oxygen collected