Reaction Rate.........

Question

When the concentration of A is doubled, the rate for the reaction: quadruples.

When the concentration of B is doubled the rate remains the same. Which mechanism below is consistent with the experimental observations?

a. Step 1: A B → D (slow)

Step 2: A D 2 C (fast equilibrium)

b. Step 1: 2 A → D (slow)

Step 2: B D → E (fast)

Step 3: E → 2 C (fast)

c. Step 1: 2 A D (fast equilibrium)

Step 2: B D → E (slow)

Step 3: E → 2 C (fast)

d. Step 1: A B D (fast equilibrium)

Step 2: A D → 2 C (slow)

How is this question answered??

Answer 1

Actually

on wikipedia

Answer 2

The rate and rate law of a prospective mechanism is derived from the slowest, or "rate controlling" step. The fact that the reaction is 2nd order in A means that the rate controlling step will either have 2 A's in it or one/two transient intermediates that are derived from an A. The fact that it is zero order in B means the rate controlling step has neither B nor a transient intermediate derived from B.

a. fails because the slow step has both A and B; it would be first order in both.

b. is second order in A and zero order in B. It could be the answer.

c. fails because B is in the slow step, and would be first order in B and 1/2 order in A (the intermediate D is derived from A)

d. fails because it would be 1 1/2 order in A and 1/2 order in B (D is derived from both A and B).

The answer is b., but there is not enough information to confirm the accuracy of Steps 2 and 3; that is, we are given no experimental or theoretical evidence of the existence of intermediate E.