0.085<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n (i) Calculate the average rate of reaction between the time interval 30 to 60 seconds. \n(ii) Calculate the pseudo first order rate constant for the hydrolysis of ester. \nAnswer: \n(i) \n<\/p>\n (ii) For pseudo first order reaction, \nAverage rate = Rate constant \u00d7 Average concentration Average rate \n<\/p>\n Question 9. \nA reaction is first order in A and second order in B. \n(i) Write the differential rate equation. \n(ii) How is the rate affected on increasing the concentration of B three times? \n(iii) How is the rate affected when the concentrations of both A and B are doubled? \nAnswer: \n(i) \n \n(ii) When concentration of B is tripled, rate of reaction increases by 9 times. \n(iii) Rate of reaction increases by 8 times.<\/p>\n Question 10. \nIn a reaction between A and B, the initial rate reaction (r0<\/sub>) was measured for different initial concentrations of A and B as given below.<\/p>\n\n\n\nA\/ mol L-1<\/sup><\/td>\n0.20<\/td>\n | 0.20<\/td>\n | 0.40<\/td>\n<\/tr>\n | \nB\/ mol L-1<\/sup><\/td>\n0.30<\/td>\n | 0.10<\/td>\n | 0.05<\/td>\n<\/tr>\n | \nr\u03b8<\/sub>mol L-1<\/sup>S-5<\/sup><\/td>\n5.07×10-5<\/sup><\/td>\n5.07×10-5<\/sup><\/td>\n1.43 x 10-5<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n what is the order of the reaction with respect to A and B ? \nAnswer: \nLet the rate law be, r = K [A]x<\/sup> [B]y<\/sup> \nr1<\/sub> = K[0.2]x<\/sup>[0.3]y<\/sup> = 5.07 \u00d7 10-5<\/sup> -(1) . \nr2<\/sub> = K[0.2]x<\/sup> [0.1]y<\/sup> = 5.07 \u00d710-5<\/sup> -(2) \nr3<\/sub> = K [0.4]x<\/sup> [0.05]y<\/sup> = 7.6 \u00d7 10-5<\/sup> -(3) , \n \n\u2234 The order of reaction with respect to A is 0.585 and with respect to B is 0. \n\u2234 rate law = K[A]0.858<\/sup> [B]0<\/sup> = K [A]0.585<\/sup>.<\/p>\nQuestion 11. \nThe following results have been obtained during the kinetic studies of the reaction 2A + B \u2192 C + D<\/p>\n \n\n\nExperiment<\/strong><\/td>\n[A]\/mol L-1<\/sup><\/strong><\/td>\n[B]\/mol L-1<\/sup><\/strong><\/td>\nInitial rate of formation of D\/mol L-1<\/sup>min-1<\/sup><\/strong><\/td>\n<\/tr>\n\nI<\/td>\n | 0.1<\/td>\n | 0.1<\/td>\n | 6.0 x 10-3<\/sup><\/td>\n<\/tr>\n\nII<\/td>\n | 0.3<\/td>\n | 0.2<\/td>\n | 7.2 x 10-2<\/sup><\/td>\n<\/tr>\n\nIII<\/td>\n | 0.3<\/td>\n | 0.4<\/td>\n | 2.88 x 10-1<\/sup><\/td>\n<\/tr>\n\nIV<\/td>\n | 0.4<\/td>\n | 0.1<\/td>\n | 2.40 x 10-2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Determine the rate law and the rate constant for the reaction. \nAnswer: \nr = K [A]x<\/sup> [B]y<\/sup> \nr, = K[0.1]x<\/sup> [0.1]y<\/sup> = 6.0 \u00d7 10-3<\/sup> -(i) \nr2 = K [0.3]x<\/sup> [0.2]y<\/sup> = 7.2 \u00d7 10-2<\/sup> -(2) \nr3 = K [0.3]x<\/sup> [0.4]y<\/sup> = 2.88 \u00d7 10-1<\/sup> -(3) \nr4 = K [0.4]x<\/sup> [Q.l]y<\/sup> = 2.4 \u00d7 10-2<\/sup> -(4) \n \n4x<\/sup> = 4 \n\u21d2 x =1 \n \n2y<\/sup> = 4 \u21d2 y = 2.<\/p>\n\u2234 Rate law, r = K [A]1<\/sup> [B]2<\/sup> \nRate constant, \n \n= 6M-2<\/sup>min-1<\/sup><\/p>\n<\/p>\n Question 12. \nThe reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:<\/p>\n \n\n\nExperiment<\/strong><\/td>\n[A]\/mol L-1<\/sup><\/strong><\/td>\n[B]\/mol L-1<\/sup><\/strong><\/td>\nInitial rate \/mol L-1 <\/sup>min-1<\/sup><\/strong><\/td>\n<\/tr>\n\nI<\/td>\n | 0.1<\/td>\n | 0.1<\/td>\n | 2.0 x 10-2<\/sup><\/td>\n<\/tr>\n\nII<\/td>\n | x<\/td>\n | 0.2<\/td>\n | 4.0 x 10-2<\/sup><\/td>\n<\/tr>\n\nIII<\/td>\n | 0.4<\/td>\n | 0.4<\/td>\n | y<\/td>\n<\/tr>\n | \nIV<\/td>\n | z<\/td>\n | 0.2<\/td>\n | 2.0 x 10-2<\/sup><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n Answer: \nrate law = K [A] \nfrom I 2.0 \u00d7 10-2<\/sup> = K 0.1 \n\u21d2 K = 2.0 \u00d7 10-1<\/sup>min-1<\/sup> \nfrom II 4 .0 \u00d7 10-2<\/sup> = 2 .0 \u00d7 10-31<\/sup> x \n\u21d2 K = 2 .0 \u00d7 10-1<\/sup> mol L-1<\/sup> \nfrom III y = 2.0 \u00d7 10-1<\/sup> \u00d7 0.4 \n\u21d2 y = 8 \u00d7 10-2<\/sup> mol 1L \nfrom IV 2.0 \u00d7 10-2<\/sup> = 2.0 \u00d7 10-1<\/sup> \u00d7 z \n\u21d2 z = 0.1 mol L-1<\/sup>.<\/p>\nQuestion 13. \nCalculate the half – life of a first order reaction from their rate constants given below. \n(i) 200 s-1<\/sup> \n(ii) 2 min-1<\/sup> \n(iii) 4 years-1<\/sup> \nAnswer: \n<\/p>\nQuestion 14. \nThe half-life for radioactive decay of 14C is 5730 years. An archaeological artifact containing wood had only 80% of the 14C found in a living tree. Estimate the age of the sample. \nAnswer: \n[A]0<\/sub> = 100 \n[A] = 80 \n \n= 1845 years \n\u2234The age of the sample is 1845 years.<\/p>\nQuestion 15. \nThe rate constant for a first order reaction is 60 s-1<\/sup>. How much time will it take to reduce the initial concentration of the reactant to its 1\/16 th value? \nAnswer: \nK = 60 s-1<\/sup> \n<\/p>\n<\/p>\n Question 16. \nDuring nuclear explosion, one of the products is MSr with half-life of 28.1 years. If mg of 90<\/sup>Sr 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 ? \nAnswer: \nt1\/2<\/sup> = 28.1 years \n \n[A]0<\/sub> = 2.47 \u00d7 1\u03bcg \nFor 10 years \n \n[A] =0.78 \u00d7 10-6<\/sup>g = 0.78\u03bcg \nFor 60 years \n \nThe amount of 90<\/sup>Sr that will remain in the child\u2019s body after 10 years is 0.78\u03bcg and after 60 years is 0.229 \u03bcg.<\/p>\nQuestion 17. \nFor a first order reaction, show that time required for 99% completion is twice the time required for the completion of 90% of reaction. \nAnswer: \n \nfor 99% completion, \nif [A]0<\/sub> = 100 \n[A] = 100-99= 1 \nfor 90% completion \n[A] = 100-99 \n<\/p>\nQuestion 18. \nA first order reaction takes 40 min for 30% decomposition. Calculate t1\/2<\/sub>. | | | | | | | | | | | | | | | | | | | | | |