Suggested Answers to Chemical Kinetics Revision Exercise. The results of some investigations of the rate of this reaction are shown below.

Transcription

1 Suggested Answers to Chemical Kinetics Revision Exercise 1 At 973K, nitrogen monoxide and hydrogen react as follows: 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) The results of some investigations of the rate of this reaction are shown below. Experiment Initial Initial Initial rate of number concentration of concentration of reaction nitrogen monoxide hydrogen / mol dm -3 s -1 / mol dm -3 / mol dm (a) (i) Use the below data to determine the order of the reaction with respect to 1 Nitrogen monoxide 2 Hydrogen Let the rate equation be rate = k [NO] m [H 2 ] n For nitrogen monoxide Comparing experiments 1 and 2, When [NO] is increased by 2 times while keeping the [H 2 ] constant, the rate is increased by 4 times. Hence, the order of reaction with respect to NO = 2 For Hydrogen, Comparing experiments 4 and 5, When [H 2 ] is increased by 2 times while keeping the [NO] constant, the rate is increased by 2 times. Hence, the order of reaction with respect to H 2 = 1 (ii) Write a rate equation for the above reaction. Rate = k [NO] 2 [H 2 ] (iii) Calculate the rate constant, giving its units. Using experiment 2 (or any other set), 1

2 0.013 = k [0.0040] 2 [0.012] k = 6.77 x 10 4 mol -2 dm 6 s -1 (b) Explain briefly why the initial reaction rate would be expected to increase by increasing each of the following: (i) the pressure An increase in pressure of a gaseous reaction: gaseous reactant particles are brought closer together number of gaseous reactant particles per unit volume increases number of effective collision increases frequency of effective collision increases rate of reaction is proportional to the frequency of effective collisions rate of reaction increases (ii) the temperature [Note: a different explanation is required for each of these factors] An increase in temperature: number of reactant particles with E E a increases number of effective collisions increases frequency of effective collisions increases rate of reaction is proportional to the frequency of effective collisions rate of reaction increases (c) Suggest, with reasons, whether you would expect the reaction between nitrogen monoxide and hydrogen to be endothermic or exothermic. Large amount of energy will be evolved when N 2 gas is formed since N N is a very strong triple bond. The energy released by forming N 2 and H 2 O will be greater than the energy absorbed to break the bonds in NO and H 2. Hence. the reaction is exothermic. 2

3 2 Iodine and propanone (CH 3 COCH 3 ) react together in an aqueous acidic solution according to the equation CH 3 COCH 3 + I 2 CH 3 COCH 2 I + H + + I - The rate of the reaction can be measured by recording the reduction of the concentration of the iodine by the decrease in the intensity of its colour as measure in a colorimeter. Three sets of separate experiments were performed in which the initial concentration of each of the reactants, iodine, propanone and acid was varied in turn, the other two being kept constant. The results are shown below in graphical form. Graph 1 Graph 2 [I 2 ] / mol dm [I 2 ] / mol dm [CH 3COCH 3] = 0.4 mol dm -3 [CH 3COCH 3] = 0.8 mol dm -3 [CH 3COCH 3] = 1.2 mol dm [I 2] = mol dm -3 [I 2] = mol dm -3 [I 2] = mol dm Time / min Time / min Graph 3 [I 2 ] / mol dm [H + ] = 0.3 mol dm - 3 [H + ] = 0.1 mol dm - 3 [H + ] = 0.2 mol dm Time / min 3

4 Answer: (a) Use the graphs to determine the order of reaction with respect to (i) propanone (ii) iodine (iii) acid (a) From the Graph 1. When concentration of propanone = 0.4 mol dm -3, Rate 1 gradient of the graph = 1.00 x 10-4 mol dm -3 min When concentration of propanone = 0.8 mol dm -3, Rate 2 gradient of the graph = 2.00 x 10-4 mol dm -3 min When concentration of propanone is increased by 2 times, while keeping the other two reagents concentrations constant, rate of reaction is also increased by 2 times. Hence, the order of reaction with respect to propanone is 1. (ii) From Graph 2 When concentration of I 2 = mol dm -3, Rate 1 gradient of the graph = 8.33 x 10-5 mol dm -3 min When concentration of I 2 = mol dm -3, 4

5 Rate 2 gradient of the graph = 8.33 x 10-5 mol dm -3 min When concentration of iodine is changed for the 3 experiments while keeping the other two reagents concentrations constant, the rate (determined by gradient of the graph) remains the unchanged, hence order of reaction with respect to iodine is zero. (iii) From Graph 3 When concentration of H + = 0.1 mol dm -3, Rate 1 gradient of the graph = 8.33 x 10-5 mol dm -3 min When concentration of H + = 0.2 mol dm -3, Rate 2 gradient of the graph = 1.67 x 10-4 mol dm -3 min When concentration of H + is increased by 2 times, while keeping the other two reagents concentrations constant, rate of reaction is also increased by 2 times. Hence, the order of reaction with respect to propanone is 1. (b) Write the rate equation for the reaction. Rate = k [CH 3 COCH 3 ] [H + ] (c) What is the total order of reaction? Overall order of reaction = 1+1 = 2 (d) What is the units of the rate constant, k? Units of k = mol -1 dm 3 s -1 5

6 3(a) A bromoalkane, RBr, is hydrolysed by aqueous sodium hydroxide. Write a balanced equation for the reaction and suggest what type of reaction it is. RBr + NaOH ROH + NaBr (b) The following results were obtained from two experiments on such a hydrolysis. In each experiment, the overall [NaOH(aq)] remained virtually constant at the value given at the top of the column. Time / min [RBr] / mol dm -3 when [OH-] = 0.10 mol dm -3 [RBr] / mol dm -3 when [OH-] = 0.15 mol dm Plot these data on suitable axes and use your graph to determine the following. 1 Use the half-life method to deduce the order of reaction with respect to the bromoalkane. 2 Use the initial rates method to deduce the order of reaction with respect to sodium hydroxide 3 Construct a rate equation for the reaction. Graph of [RBr] / mol dm-3 vs time / min [RBr] / mol dm g 2 g 1 [OH - ] = 0.10 mol dm [OH - ] = 0.15 mol dm t ½ t ½ time / min 6

7 1 Let the rate equation be Rate = k [RBr] m [OH - ] n Since [NaOH] is constant (it is present in great excess), the shape of the graph is dependent on the order of reaction with respect to RBr and not OH -. From the graph, when [OH-] = 0.15 mol dm -3, half-life is constant at 75 mins. Hence, the order of reaction w.r.t [Rbr] is 1 2 Let g1 and g2 be the gradients of the graph at t = 0 when when [OH-] = 0.10 mol dm -3 and 0.15 mol dm -3 respectively. From the graphs, Initial rate for [OH - ] = 0.10 mol dm -3 initial rate 1 g 1 ( ) / (40-0) = 5.25 x 10-5 mol dm -3 min -1 Initial rate for [OH - ] = 0.15 moldm -3 is ( ) / (0 50) 8.4 x 10-4 mol dm -3 s -1 From the graphs, Initial rate for [OH - ] = 0.15 mol dm -3 initial rate 2 g 2 ( ) / (40-0) = 8.00 x 10-5 mol dm -3 min -1 By inspection, When [OH-] is increased by 1.5 times (from 0.15 mol dm -3 to 0.10 mol dm -3 ) while keeping [RBr] constant, the rate is increased by = 1.5 times Hence, order of reaction with respect to OH - = 1 3 Rate = k [RBr] [OH - ] Using initial rate = 5.25 x 10-5 mol dm -3 min -1, [RBr] = mol dm -3 and [OH-] = 0.10 mol dm -3 into the above reaction. Rate = k [RBr] [OH - ] 5.25 x 10-5 = k x 0.10 x 0.10 k = mol -1 dm 3 min -1 7

8 4 Oxides of nitrogen are a major source of air pollution. (a) One reaction which occurs in such polluted air is shown below. 2NO (g) + O 2 (g) 2NO 2 (g) To investigate how the rate of the above reaction is dependent on nitrogen monoxide, different concentrations of nitrogen monoxide were used, keeping oxygen concentration constant. The time taken for a fixed amount of nitrogen dioxide gas to be produced was then measured as shown in the table below. Experiment Initial concentration of NO (g), C / mol dm -3 Time, t / s Product (C t) / mol dm -3 s Product (C 2 t) / mol 2 dm -6 s (i) What is the significance of the products (C t) and (C 2 t)? [2] If product (C t) is constant for all three experiments, the reaction is first order, If product (C 2 t) is constant for all three experiments, the reaction is second order with respect to NO. Reason: The time taken for a fixed amount of oxygen to be produced is inversely proportionate to the initial rate of the reaction. 1 rate time rate = k[no] n [NO] n time = const. If n =1, order of reaction wrt NO is 1 If n =2, order of reaction wrt NO is 2 Hence each product (C t) or (C 2 t) shows if the reaction is 1 st or 2 nd order. (ii) Hence, deduce the order of the reaction with respect to NO concentration.[1] Reaction is 2 nd order with respect to NO as the product (C 2 t) is constant for all three experiments. 8

9 (b) At 700 ºC nitrogen monoxide can be reduced by hydrogen via a two-step exothermic reaction as follows: Activation energy 2NO (g) + H 2 (g) N 2 O (g) + H 2 O (g) slow Ea 1 N 2 O (g) + H 2 (g) N 2 (g) + H 2 O (g) fast Ea 2 (i) [2] Write the rate equation for the above reaction, stating the units of the rate constant. rate=k[no] 2 [H 2 ] [2] (ii) units of k= mol -2 dm 6 s -1 Sketch and label a reaction pathway diagram for the reaction, indicating clearly all relevant energy changes. Energy TS TS E a (RDS) E a (fast step) Reactants Intermediates H=-ve Products Progress of reaction 9

10 (iii) Explain, as fully as possible, why a relatively small increase in temperature can cause a large increase in the rate of a chemical reaction. Maxwell-Boltzmann Distribution Curve area under curve is proportional to number of reactant particles shaded area is larger when curve is flattened at higher temperature number of reactant particles with E E a increases number of effective collisions with increases frequency of effective collisions increases rate of reaction is proportional to the frequency of effective collisions rate of reaction increases 10