Question

03 2 AG 2 B A B z cal FRIAL [A] [B ] Rate M/min 1 0.2 0.2 -32 2 0.4 0.2 1.28 3 0.4 0.4 2.56 40) Order of Rx with. Respect to [A] 4 b) ordea of Rx with Respect to [B] 4 c) Rate LAW 5 6 d) Rate Constant k (with units) - Proposed mechanism D A + A → Az - 2) A₂ + B A₂B E 3) AB+B A₂ B2 e) 2 intermedistes i mechanism Rate= 1 Are , of) which step is slow step (Ratedelenminus) narod on ette LAW

Answer 1

DATE: - 4= K CADA CBDO (Rate) 419) for trial L, we can write are - 1 = K 10:2) 10.270 7 0.32 = k10.2] (0.23 for trial 2, 1, 28 = K (0.4)*(0.270 Dividing @ by ③ * 4 = 10.41% % 4 = (21* 0. رون) *2 = و و درهم e Order wanit CAD is 2 (6) for frial 3 we can Write _ 2.56 = K (0.49% (0.498 - 3 Dividing 3 by 0 7 2=/0.4 10 7 2 = 20 10.2 ) o y = 1 e I order wort (B7 is 1 (o from part (W and (b) P Rate & CAD (B) 1 i.e. Rate= K[A] ² (B) 0. (d) Using trial #1 date in 6 2 M/min = K[0.25*10.2 M)? 7 K= 40 M² mint (e) As A2 and A₂ B appeared in the mechanism - are neither the reactants nor the products so A and AB are intermediates. l As rate is given by slowest step to according to ean 4 rate will be given by eg h 6 lie step 2) e second step i slow step.