Chemical Kinetics Solved Problems

CHEMICAL KINETICS SOLVED PROBLEMS Q.1 Sol. For the reaction R P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes. Calculate the average rate of reaction using units of time both in minutes and seconds. R2  R1 0.02M  0.030 M R Average rate = t   t  t   25min 2 1 = = Q.2 Sol. 0.01M  4  10  4 Mmin1 25 min 0.01M  6.66  10 6 M s1 25  60 s The decomposition of NH3 on platinum surface is zero order. What are the rates of production of N2 and H2 if k = 2.5 ×10–4 mol–1 Ls–1 ? 2NH3(g) Pt N2(g) + 3H2(g) Rate of reaction dx  1 dNH3  dN2  1 dH2     dt 2 dt dt 3 dt Rate of production of N2 = Rate constant  Rate  K NH3 º  K  = 2.5 ×10–4 mol L–1 s–1 Rate of production of H2 = 3 × Rate constant = 3 ×2.5 ×10–4 mol L–1 s–1 = 7.5 ×10–4 mol L–1 s–1. Q.3 In a Pseudo first order hydrolysis of ester in water, the following results were obtained. t/s ester/ mole L1 Sol. 0 30 60 90 0.55 0.31 0.17 0.085 (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds (ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. (i) Average rate of reaction = Change in concentration 0.31  0.17   60  30  Change in time = 4.67 ×10–3L–1s–1 A  2.303 log 0 (ii) k= A  t Where [A0] = 0.55 M is initial concentration of ester ( t = 0 ) and [A] is the concentration of ester at time t. At 30 sec : k1 = 2.303 0.55 log = 1.91 ×10–2 s–1. 30 0.31 At 60 sec : k2 = 2.303 0.55 log = 1.96 ×10–2 s–1. 60 0.17 At 90 sec : k3 = 2.303 0.55 log = 2.07 ×10–2 s–1. 90 0.85 Thus, Average k= k1  k 2  k 3 1.91 10 2  1.96  10 2  2.07  10 2 = 3 3 k = 1.98 ×10–2 s–1. CHEMICAL KINETICS Q.4 Sol. A reaction is first order in A and second order in B. (i) Write differential rate equation. (ii) How is the rate affected on increasing the concentration of B three times ? (iii) How is the rate affected when the concentration of both A and B are doubled? (i) Rate = k [ A] 1[B]2 r0 = k [A]1[B]2 r1 = k [A]1[3B]2 = 9 k [A]1[B]2 = 9 × r0 The rate is increased by 9 times. (ii) r0 = k [A]1[B]2 r2 = k [2A]1[2B]2 = 8 k [A]1[B]2 = 8 × r0 The rate is increased by 8 times. Q.5 In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentration of A and B as given below: A / mol L1 B/ mol L1 Rate(r0 ) / mol L1S1 Sol. 0.20 0.20 0.40 0.30 0.10 0.05 5.07  10 5 5.07  10 5 1.43  10 4 What is the order of reaction with respect to A and B ? Let the rate of reaction, r = k [A]m [B]n (r1) 5.07 ×10–5 = k [0.20]m [0.30]n (r2) 5.07 ×10–5 = k [0.20]m [0.10]n (r3) 1.43 ×10–4 = k [0.40]m [0.05]n From experiment (i) and (ii) ......Exp (i) ......Exp (ii) ......Exp (iii)  r2  5.07  10 5 Ms 1 k 0.20 m 0.10 n    = 5.07 5 Ms 1 k 0.20 m 0.30 n  r1  n  0.10    =1  0.30  or 3n = 1  3n = 3º  n=o From experiment (i) and (iii)  r3  1.43  10 4 Ms 1 k 0.40 m 0.05 n    = 5 1 k 0.20 m 0.10 n  r1  5.07  10 Ms 2.8 = (2)m or m log2 = log 2.8 m × 0.3010 = 0.4472 n= Q.6 0.4472  1.49 0.3010 Thus, order of reaction with respect to A is 1.49 and order with respect to B is zero. So rate law is, Rate k [A]1.49 [B]º Calculate the half-life of a first order reaction from their rate constants given below: (a) 200 s–1 (b) 2 min–1 (c) 4 years–1 CHEMICAL KINETICS Sol. (a) (b) (c) Q.7 0.693 0.693  = 3.465 ×10–3 s k 200 0.693 0.693  t1/2 = = 3.465 ×10–1 min k 2 t1/2 = t1/2 = 0.693 0.693  = 0.17325 years k 4 The half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood has only 80% of the 14C found in a living tree. Estimate the age of the sample. Sol. t1/2 = If R0 is the initial 5730 years C activity in a living tree = 100 14 Activity in the dead wood R = 80 The nuclear radioactive decay follows the first order kinetics so Q.8 t= R  2.303 log 0 k R k= 0.693 t1/ 2 t= 2.303  t1/ 2 100 log = 1845.4 years 0.693 80 The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? Sol. k= R  2.303 log 0 t R R  2.303 log 0 k R [R0] = 1, then [R] = 1/16 t= 2.303 1 log = 0.0462 s–1 60 1/ 16 During nuclear explosion, one of the products is 90Sr with half-life of 28.1 years. If 1 g of 90Sr was absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years and 60 years if it is not lost metabolically? k= Q.9 Sol. T1/2 Sr90 = 28.1 years, k = We know, Now, k= t= R  2.303 log 0 R k R  2.303 log 0 R k At t = 10 year10 = R  2.303  28.1 log 0 R 0.693 log R0  R = 0.1070 0.693 0.693 t1/ 2 , k = 28.1 year–1 CHEMICAL KINETICS R0  R So, = antilog (0.1070) = 1.279 [R] = At t = 60 year60 = 1g = 0.7818 g 1.279 R  2.303  28.1 log 0 0.693 R log R0  R = antilog 0.9409 = 4.374 So, Q.10 [R] = 1g = 0.228 g. 4.374 For a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. Sol. t= R  2.303 log 0 k R 99% completion means If [R0] = 100, [R] = 100 – 99 = 1 t99% = 2.303 100 log k l = 2.303 2.303 log 102 ×2 k k 90% completion means If [R0] = 100, [R] = 100 – 90 = 10 t90% = R  2.303 log 0 k R = 2.303 100 log k 10 = 2.303 log 10 k = 2.303 k  t 99% 2.303  2 k   t 90% k 2.303 = 2 or t99% = 2 × t90% CHEMICAL KINETICS Q.11 A first order reaction takes 40 min for 30% decomposition. Calculate t1/2. Sol. t1/2 = k= 0.693 k R  2.303 log 0 t R 30% decomposition means that If [R0] = 100, [R] = 100 – 30 = 70 2.303 100 log k 70 k = k = 2.303  0.1549 40 t 1/2 = 0.693 k = 0.693 8.918  10 3 = 2.303 10 log k 7 = 8.918 ×10–3 min–1 min–1 = 77.7 min. Q.12 Sol. The following data were obtained during the first order thermal decomposition of SO2Cl2 at constant volume: SO2Cl2(g)  SO2(g) + Cl2(g) Experiment Time / s 1 Total Pr essure / atm 1 0 0.5 2 100 0.6 Calculate the rate of the reaction when total pressure is 0.65 atm. Suppose initial pressure of SO2Cl2 is Pi and its pressure decreases by x atm after time t then SO2Cl2(g)  SO2(g) + Cl2(g) Initial pressure (at t = 0) 0.5 atm 0 0 Pressure (at time = t) 0.05 – x x x Ptotal at time t = 0.5 – x + x + x = 0.5 + x atm i.e., 0.6 = 0.5 – x or x = 0.6 – 0.5 = 0.1 atm Now, Pi = 0.5 atm and P(at time 100 sec) = 0.5 – x = 0.5 – 0.1 = 0.4 atm (a) Evaluation of k : Initial pressure 2.303 k100 = log Pr essure at time t t = 2.303 0 .5 log t 0 .4 2.303 log 1.25 100 2.303 = 0.0969 = 2.23 ×10–3 s–1 100 Rate at Pt = 0.65 atm = (b) PSO2Cl2 at total pressure of 0.65 atm PSO2Cl2 = 2Pi – Pt = 1 – 0.65 = 0.35 atm Rate = k PSO2Cl2 = 2.23 ×10–3 ×0.35 = 7.8 ×10–4 atm s–1. CHEMICAL KINETICS Q.13 The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5 s–1 at 546 K. If the energy of activation is 179.9 kJ mol–1, what will be the value of pre-exponential factor? Sol. log A = logk + Ea 2.303 RT = log 2.418 × 10–5 s–1 + 179900 J mol 1 2.303  8.314 J k 1 mol 1  546 K = – 4.6184 + 17.21 log A = – 4.6184 + 17.21 = 12.5916 Taking antilog of both sides A = antilog 12.5916 = 3.9 × 10–12 s–1. Q.14 Sol. The decomposition of A into product has value of k as 4.5 ×103 s–1 at 10ºC and energy of activation 60 kJ mol–1. At what temperature would k be 1.5 ×104 s–1? k2 Ea  T2  T1  We know that, log k = 2.303 R  T T  1  1 2  log 1.5  10 4 4.5  10 3   T2  283  60000   2.303  8.314  283 T2   T2  283  0.5228 = 3133.62  283 T  2   or T2  233 0.5228 = 283 T 3133.62 2 T2  283 T2 1.67 ×10–4 × 283 = 0.0472 T2 = T2 – 283 T2 – 0.0472 T2 = 283 0.9528 T2 = 283 Hence, Q.15 T2 = 283 = 297.02 K . 0.9528 The time required for 10% completion of a first order reaction at 298 K is equal to that required for its 25% completion at 308. If the value of A is 4 × 1010s–1, calculate k at 308 K and Ea. Sol. k298 = Similarly, R  2.303 100 2.303 log 0 = log R k k 90 k308 = 2.303 100 log t 75 Dividing equation (ii) by equation (i), we get 2.303 100 log k 308 t 75  k 298 2.303 log 100 t 90 ..........(i) ..........(ii) CHEMICAL KINETICS As time required for 10% completion = time required for 25% completion. k 308 log100 / 75 0.1249   k 298 log100 / 90 0.0457 = 2.733  According to Arrhenius equation log Ea  1 1  k2     k1 2.303R  T1 T2  log 2.733  or Ea = = Ea  308  298  2.303  8.314  308  298  0436  2.303  8.314 1.089  10 4 8.36 1089  10  4 = 76658 J mol–1 Now, k308  A . e Ea / RT log k308 = log 4 × 1010  or = 10.602 – Ea 1  2.303 R 308 76658 2.303  8.314  308 = 10.602 – 12.998 = – 2.3 % or k308 = antilog (–2.396) = 4.02 × 10–3 s–1. Q.16 The rate of reaction quadruples when the temperature changes from 298 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature. Sol. T1 = 293 K ; T2 = 313 K According to Arrhenius equation log Ea  T2  T1  k2    k1 2.303 R  T1T2   T1T2  k2 Ea = 2.303 R ×  T  T  log k 1 1  2 Now, T1T2 293  313 T2  T1 = 313  293 = 4585.45 K k2 4  log  log 4  0.6021 k1 1 and log  Ea = 2.303  8314  JK 1mol1  293  313 4 log 313  293 1 Ea = 2.303  8314  293  313  0.6021 20 = 52863.33 J mol–1 = 52.86 kJ mol–1 CHEMICAL KINETICS Q.17 Express the rate of the following reactions in terms of the concentrations of reactants and products (i) 2NO2(g) + F2(g)  2NO2F(g) ( ii) 5Br –(aq) + BrO3–(aq) + 6H+ (aq)  3Br2(aq) + 3H2O(l) Sol. (i) For the given reaction, 2NO2(g) + F2(g)  2NO2F(g) Rate of reaction =  1 dNO2  dF  1 dNO2   2  2 dt dt 2 dt (ii) For given reaction, Rate of reaction = Q.18        1 d Br  BrO3  1 d H 1 dBr2   d   5 dt dt 6 dt 3 dt A first order reaction has specific rate constant 10–3 sec–1. How muct time time will it take for 10 g to reduce to half of the quantity? Sol. Half-life period for the first order reaction is given by the expression. t 0 .5  Now, 0.693 k k = 10–3 sec–1  t0,5 = 0.693 10 3 = 693 seconds. Q.19 60 per cent of a first order reaction was completed in 60 minutes. When was it half completed? Sol. Using first order rate constant equation. k= 2.303 a log t ax In first case: k= 2.303 100 log 60 min 100  60 in second case: k= 2.303 100 log t 100  50 Equating the two equations (i) and (ii), we get 2.303 100 2.303 100 log  log 60 min 40 t 50 or 60  log 2 60  0.3010  min 1  0.6020 t = log 10 4 = 45.4 min. ..........(i) ..........(ii) CHEMICAL KINETICS Q.20 A first orer reaction is 20% complete in 10 minutes. Calculates the time for 75% completion of the reaction. Sol. For a first order reaction, k = 2.303 a log a  x  t Substituting the given values, we get and, also k= 2.303 100 log 10 min 80 k= 2.303 100 log t 25 ............(i) ............(ii) Equating the two equation (i) and (ii) , we get 2.303 5 2.303 log  log 4 10 4 t or t= 10 log 4 log 5 / 4 or t= 10  0.6020 0.6990  0.6020 = 6.020 6020  min = 62.6 min 0.0970 97 Q.21 The half-life of a chemical reaction at a particular concentration is 25 minutes. When the concentration is doubled, the half-life becomes 50 minutes. Find the order of reaction. Sol. Suppose the initial concentration in the first case is a mol L–1. Then [A0]1 = a1(t1/2)1 = 25 minutes and Now, [A0]2 = 2a1(t1/2)2 = 50 minutes t1/ 2 1 t1/ 2 2  A     0 1  A 0 2  n1 n1 So,  a1     2a1   1=n-1  1   2 n1 n=2 Hence, the reaction is of second order or 1  1   2 2 n1