1. Unit 1 Reaction Rates

2. 2.The balanced equation for the decomposition of hydrogen peroxide into water and oxygen is: 2H 2 O 2 (l)  2H 2 O(l) + O 2 (g) The following graph is obtained for the volume of oxygen released over time. Calculate the average rate of reaction between 10 s and 20 s. Rate = 40-24 / 20 = 0.8 cm 3 s -1.

3. 3. The decomposition of hydrogen peroxide solution into water and oxygen can be catalysed by an enzyme. 2H 2 O 2 (l)  2H 2 O(l) + O 2 (g) The rate of reaction can be followed by recording the mass loss over a period of time. (a) The following graph was obtained from experiments to find the effect of pH on the efficiency of the enzyme. (i) Calculate the average rate of the reaction over the first 20 minutes, in g min -1, for the experiment at pH 10. (ii) From the results shown on the graph, what can be concluded about the efficiency of the enzyme over the pH range used in the experiment? (b) Give one other way of following the rate of this reaction (a) (i) rate = (60.01)-(59.68) / 20 = 0.165g min -1. (ii) The enzyme is most efficient between a pH of 7 and 10. (b) The rate could be measured by collecting the oxygen gas in a gas syringe and plotting the volume of gas collected against time.

4. 4. The grid shows the possible effect of temperature change on reaction rate. Identify the graph which shows how the rate of reaction varies with temperature in the decomposition of hydrogen peroxide solution using catalase, an enzyme, found in potatoes. A B D EF C Graph in box D.

5. 5. T he effect of temperature changes on reaction rate can be studied using the reaction between an organic acid solution and acidified potassium permanganate solution. 5(COOH) 2 (aq) + 6H + (aq) + 2MnO 4 - (aq)  2Mn 2+ (aq) + 10CO 2 (g) + 8H 2 O(l) The apparatus required is shown in the diagram. (a) Name the organic acid. (b) Describe how the reaction time can be measured. (c) The headings for a set of results are shown below. Temperature of reaction /°C Reaction time /s Reaction rate / Complete the headings by entering the correct unit in the third column. (a)Oxalic acid. (b)Start timing when solutions are mixed and stop timing when purple colour appears. (c)s -1.

6. 6. Methanoic acid, HCOOH, can break down to carbon monoxide and water by two different reactions, A and B. Reaction A (catalysed) HCOOH(aq) + H + (aq)  CO(g) + H 2 O(l) + H + (aq) Reaction B (uncatalysed) HCOOH(aq)  CO(g) + H 2 O(l) (a) (i) What is the evidence in the equation for Reaction A that the H + (aq) ion acts as a catalyst? (ii) Explain whether reaction A is an example of heterogeneous or homogeneous catalysis. ( b) The energy diagram for the catalysed reaction is:- Copy the diagram and draw a line to show the reaction pathway for the uncatalysed reaction. (a) (i) Hydrogen ions are not used up in the reaction. (ii) Homogeneous – the catalyst and reactant are in the same physical state. (b) The uncatalysed reaction has a higher activation energy.

7. 7. The effect of temperature on reaction rate can be studied using the reaction between oxalic acid and acidified potassium permanganate solution. (a) What colour change would indicate that the reaction was complete? (b) A students results are shown in the graph below. ( (i) Use the graph to calculate the reaction time, in s, at 40 o C. (ii) Why is it difficult to obtain an accurate titration time when the reaction is carried out at room temperature (a) The appearance of a purple colour. (b) (i) Rate = 0.12 Time = 1 / 0.12 = 8.33 s (ii) Rate is very slow at room temperature.

8. 8. The diagram below shows the energy distribution of molecules in a gas at a particular temperature. Draw a second curve on the diagram to show the energy distribution of the molecules in the gas at a higher temperature. Label the diagram to indicate why an increase in temperature has such a significant effect on reaction rate. EAEA The second curve (blue line) shows that the average kinetic energy of the molecules has increases. The orange line shows the activation energy (E A ) for the reaction. The increase in temperature has greatly increased the number of molecules that possess the activation energy.