CHEMICAL KINETICS A LEARNING NOTE

UNIT: 6 CHEMICAL KINETICS 1. Form the rate expression for the following reactions determine their order of reaction and dimensions of the rate constants. a) 3NO(g) ( N2O (g) + NO2 (g) Rate = K [NO] 2 b) H2O2 (aq) + 3 I – (aq) + 2H+ ( 2H2O (l) +I-3 Rate =- k [H2O] [I-] c) CH3 CHO(g) ( CH4(g) + CO(g) Rate = k [CH3 CHO]3/2 d) CHCl3 (g) + Cl2 (g) ( CCl4 (g) + HCl (g): Rate = [CHCl3] [Cl2] 1/2 e) C2 H5 Cl(g) ( C2H4(g) = HCl(g) : Rate = k [C2H5Cl] a) Order = 2 Unit of rate constant is s-1 mol-1 L [M1s-1] b) Order = 2; Unit of rate constant is s-1 mol-1 L or M-1 s-1 c) Order = 3/2 ; Unit of rate constant is s-1 ; mol-1/2 L1/2 d) Order = 3/2 ; Unit of rate constant is s-1 mol-1/2 L1/2 e) Order = 1 Unit of rate constant is s-1 2. For the reaction 2A + B + C ( A2 B+C The rate = k[A] [B]2 with k = 2 x 10-6 M-2 S-1, calculate initial rate of the reaction when [A] = 0.1M [B] = 0.2M and [c] = 0.8 M. If the rate of reverse reaction is negligible, then calculate the rate of reaction after [A] is reduced to 0.06M. Rate = k [A] [B] 2 = 2 x 10-6 M-2 s-1 x 0.1 M x (0.2M) 2 = 8 x 10-9 5Ms-1. When [A] is reduced to0.06m? [B] = 0.2 – 0.02 = 0.18 M ( Rate = 2 x 10-6 x 0.06 x (0.18) 2 = 38.88 x 10-10 = 3.888 x 10-9 M s-1. 3. The rate of decomposition of NH3 on platinum surface is zero order. What are the rate of production of N2 and H2 if k = 2.5 x 10-4 M-1 s-1. 2NH3 ( N2 + 3H2 Rate of decomposition of NH3 = k[NH3]0 = k = 2.5 x 10-4 M s-1 - d[NH3] = d[N2] = d[H2] 2 dt dt 3dt Rate of formation of N2 = d[NH3] 2dt = ½ x 2.5 x 10-4 M s-1 = 1.25x10-4 Rate of formation of H2 = 3 x d[NH3] 2 dt = 3/2 x 2.5 x 10-4 M s-1 = 3.75 x 10-4 M s-1 4. The decomposition of dimethyl ether leads to the formation of CH4, H2 and CO and the reaction rate is given by Rate = K [CH3 OCH3] 3/2 The rate of reaction is followed by increase in pressure in a closed vessel and the rate can also be expressed in terms of the partial pressure of dimethyl ether ie, rate k (PCH3 OCH3) 3/2. If the pressure is measured in bar and time in minute, then what are the units of rate and rate constant? Unit of rate is bar minute-1 Unit of rate constant Rate = k [P] 3/2 bar min-1 = k bar 3/2 Unit of k = (bar min-1 / bar3/2) = bar -1/2min-1 5. Mention the factors that affect the rate of a chemical reaction. a) Nature of reactants b) Physical states and surface area. c) Temperature. d) Concentration e) Catalyst 6. A reaction is second order with respect to a reactant. How is the rate of reaction affected if the concentration of reactant is (I) doubled (ii) reduced to ½? Consider the reaction A ( products Rate = k [A] 2 If concentration is doubled, rate k [2A] 2 = 4 k [A] 2 The rate of reaction becomes four times .If concentration is reduced to half, the rate of reaction becomes ¼. 7. What is the effect of temperature on the rate constant of a reaction? How can this temperature effect on rate constant be represented quantitatively? The rate constant of a reaction increases with increase in temperature. The relation between rate constant and temperature is given by Arrhenius equation K = A. e -Ea/RT 8. In a pseudo first order hydrolysis of ester in water following results were obtained. T/s 0 30 60 90 [Ester]/M 0.55 0.31 0.17 0.085 i) Calculate the average rate of reaction between the time intervals 30 to 60 seconds. ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. i) Average rate = - (C2 – C1) / (t2-t1) = - (0.17-0.31) / (60-30) = 0.14 / 30 = 4.6 x 10-3 Ms-1 ii) k = (2.303/t) log ((R0) / (R)) = (2.303)/30) log (0.55/0.31) = (2.303/30) x 0.249 = 1.91 x 10-2 s-1 9. A reaction is first order in A and second order in B. i) Write differential rate equation. ii) How is the rate affected when concentration of B is tripled? iii) How is the rate affected when the concentration of both A and B is doubled? i) –d[A]/dt = k [A]1 [B] 2. ii) If concentration of B is tripled rate becomes 9 times. iii) When concentration of both A and B are doubled rate becomes 8 times. 10. In a reaction between A and B, the initial rate of reaction was measured for different initial concentration of A and B as given below. [A]/M 0.2 0.2 0.4 [B]/M 0.3 0.1 0.05 r0 / Ms-1 5.07 x 10-5 5.07 x 10-5 7.6 x 10-5 What is the order of reaction with respect to A and B? -d[A]/dt = rate = k [A] x [B]y 5.07 x 10-5 = k x 0 .2x x 0 .3y ( (1) 5.07 x 10-5 = k x 0 .2x x 0.1y ( (2) 7.6 x 10-5 = k x 0 .4x x 0.05y ( (3) Dividing (1) by (2) 1 = (0.3./0.1)y = 3y y = 0. Dividing equation (2) by (3) (7.6 x 10-5 / 5.07 x 10-5) = (0.4/0.2)x = ( 2)x 1.5 = 2x log1.5 = xlog2 x = 0.1761/0.3010 = 0.585 Order with respect of [A] is 0.585 (Note: In the textbook(page 380), the order with respect to A is given as 1.5. For this answer r0 for the third run should be read as 1.434 x 10-4 instead of 7.6 x 10-5) 11. Reaction between NO2 and F2 to give NO2 F takes place by the following mechanism. NO2 (g) + F2(g) slow > NO2 F(g) + F(g) NO2(g) + F(g) fast > NO2F(g) Write the rate expression for the reaction. The rate-determining step is the slow step. So rate law is Rate = k[NO2] [F2]. 12. The following result has been obtained during the kinetic studies of the reaction. 2A + B ( C+D Experiment [A]/M [B]/M initial rate of formation of D/Mmin-1 1 0.1 0.1 6 x 10-3 2 0.3 0.2 7.2 x 10-2 3 0.3 0.4 2.88 x 10-1 4 0.4 0.1 2.4 x 10-2 Determine the rate law and rate constant for the reaction. Rate = -d[A]/dt = k[A]x [B]y 6 x 10-3 = k x 0 .1x x 0.1y ( (1) 7.2 x 10-2 = k x 0.3x x 0.2y ( (2) 2.88 x 10-1 = k x 0 .3x x 0 .4y ( (3) 2.4 x 10-2 = k x 0 .4x x 0.1y ( (4) (4)/(1) ( (2.4 x 10-2 / 6 x 10-3) = (k x 0.4x x 0.1y) / (k x 0 .1x x 0.1y) 4 = (0.4/0.1)x = 4x x = 1. (3) / (2) ( (2.88 x 10-1 / 7.2 x 10-2 ) = ( k x 0.3x x 0.4y) / (k x 0 .3x x 0.2y ) 4 = (0.4y / 0.2y) = ( 2)y ie, ( 2 ) 2 = ( 2)y y = 2. Rate = k [A]1 [B]2 6.0 x 10-3 = k x 0 .11 x 0 .12 k = (6 x 10-3/ 1 x 10-3) = 6 M-2min-1.( In the textbook the answer is given as 6 M-2s-1. Since th unit of rate is given as M min-1, the rate constant is 6 M-2min-1 or 0.1M-2 s-1 13. The reaction between A and B is first order with respect to A and zero order with respect to B. fill in the blank in the following table. Experiment [A]/M [B]/M initial rate / M min-1 1 0.1 0.1 2 x 10-2 2 -- 0.2 4 x 10-2 3 0.4 0.4 -- 4 -- 0.2 2 x 10-2 Rate = k [A] 1 [B] 0 = k [A] 1 From (1) 2 x 10-2 = k x 0.11 ( (1) K = (2 x 10-2 / 0.1) = 2 x 10-1 min-1 Form (2) 4 x 10-2 = k [A] [A] = 4 x 10-2 / 2 x 10-1) = 0.2 M. Form (3) rate = k x 0.41 = 2 x 10-1 x 0.41 = 8 x 10-2 M min-1 From (4) 2 x 10-2 = k [A] 1 2 x 10-2 = 2 x 10-1 [A] [A] = (2 x 10-2 / 2 x 10-1) = 0.1M. 14. Calculate the half-life of a first order reaction from their rate constants given below. (a) 200s-1 (b)2min-1 (c)4 year-1 a) t ½ = (.693/k) k = 200 s-1 ie, t ½ = (.693 /200 s-1) = 3.465 x 10-3s b) t ½ = (0.693 / 2 min –1) = 3.465 x 10-1 minute c) t ½ = (0.693 / 4 year-1) = 0.1732 year. 15. The experimental data for decomposing of N2O5 [2N2O5 ( 4No2 + O2] in gas phase at 318k are given below. t/s 0 400 800 1200 1600 2000 2400 2800 3200 102x[N2O5]/M 1.63 1.36 1.14 0.93 0.78 0.64 0.53 0.43 0.35 (a) Plot [N2O5] against t. (b) Find the half-life period for the reaction. (c) Draw of graph between log [N2O5] and t. (d) What is rate law? (e) Calculate the rate constant. (f) Calculate the half-life period from k and compare it with (b). (a) The shape of graph will be (b) t ½ = (0.693/-2.303 x slope) (c) The shape of the graph is (e) k = -slope x 2.303 (f) t ½ = (0.693/k) 16. The rate constant for a first order reaction 60 s-1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value? k = 60 s-1 Initial concentration = R0 Concentration at time t = (1 / 16) R0 t = (2.303/60 s) x log (R0/ Rt) = (2.303 / 60 s-1) x log [R0 / (R0/16)] = (2.303/60 s-1) log 16 = (2.303/60 s-1) x 1.2041 = 4.62 x 10-2 s. 17. The rate of most of the reactions double when their temperature is raised from 298k to 308k. Calculate their activation energy. log (k2/k1) = (Ea / 2.303 R) [ (1/T1 ) – (1/T2)] Ea = (2.303 RT1T2) x log (k2/k1) /(T2 – T1) = (2.303 x 8.314 x 308 x 298) x log 2 / (308 – 298)) = 52.897 kJ mol -1. 18. The rate constants for the decomposition of N2O5 at various temperatures are given below. T/c 0 20 40 60 80 105k/5-1 0.0787 1.7 25.7 178 2140 Draw a graph between lnk and 1/T and calculate the values of A and Ea. Predict the rate constant at 300 and 500. The shape of graph will be Ea= -R x slope ln A = intercept 19. The half-life for radioactive decay of 14C is 5730 y. an archeological artefact contained wood had only 80% of the 14C found in a living tree. Estimate the age of sample. ( = (0.693 / t ½) = (0.693 / 5750 y) t = (2.303 / () x log (No/Nt) = ((2.303 x 5730 y) /0.693) X log (100/80) = 1845 years. 20. During nuclear explosion, one of the product is 90 Sr with half life 28.1 Y. 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? ( = (0.693 / t ½) = (0.693 / 28.1Years) t = (2.303 / () log (No/Nt) 10 years = (2.303 x 28.1 years / 0.693) log (1μg/ Nt) Log (1/Nt) = (10 x .693) / (2.303 x 28.2) = 0. 1071 1/Nt = 1.279 Nt = 1/1.279 = 0.7818 μg When t = 60 years Log 1/Nt = (60 x 0.693 / 2.303 x 28.1) = 0.6427 1/Nt = 4.392 Nt= 1 / 4.392 = 0.228 μg.