Reaction Mechanisms

Question

18) Consider the following elementary steps

Step 1: (fast) A + B ⇌ C + D (--> is k1, <-- is k-1)

Step 2: (slow) C + B --> D (--> is k2)

What is the overall rate law?

I don't understand how to do this problem at all... The answer is given as k2k1/k-1 [A][B]^2[D]^-1.

Answer 1

First, add the steps:

A + B ⇌ C + D

C + B --> D

A+ 2B ==> 2D. (Note C is an intermediate and will not appear in the overall rate law)

The slow step determines the overall rate.

At equilibrium, step 1: k1[A][B] = k-1[C][D]. Since [C] cannot appear in final rate law, we must solve for it before using in step 2:

[C] = k1[A][B] / k-1[D]

Substituting into step 2 …

For step 2: rate = k2[C][B]

rate = [B]k2 k1[A][B] / k-1[D]

rearranging we get

rate = k2k1/k-1[A]B]² /[D]

NOTE: This is a complex reaction mechanism in which the rate is actually negatively affected by the concentration of one of the products, D. The rate law can also be written as

rate = k2k1/k-1[A]B]² [D]^-1