Question

For the reaction,

2A+ B→product

A friend proposes the following mechanism:

$A+B\rightleftharpoons M$

$A+M\to \text{Products}$

(a) Assuming that the second step is the rate determining step and the first step is a fast equilibrium step, determining the rate law. Represent the rate constant in terms of$K,K_{-1},K_2$ .

(b) Using the steady state approximation, determine the rate law.

(c) Under what conditions of $\left[A\right]\text{and}\left[B\right]$do you get the same rate law in parts a and b?

Short answer

The rate law is$\text{K}\left[\text{A}\right]\left[\text{M}\right]$ and the overall rate law expression is$\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[\text{A}\right]\left[\text{B}\right]}{{\text{K}}_{\text{- 1}}+{\text{K}}_{\text{2}}\left[\text{A}\right]}$

Step-by-step solution

Determining the initial rate law and rate constant equation:

Since second step is the rate determining one:

Rate $=\text{K}\left[\text{A}\right]\left[\text{M}\right]$

${\text{K}}_1\left[\text{A}\right]\left[\text{B}\right]={\text{K}}_{\text{- 1}}\left[\text{M}\right]+{\text{K}}_{\text{2}}\left[\text{A}\right]\left[\text{M}\right]$

Applying steady state approximation:

Now we must introduce the steady state approximation for intermediate :

${\text{K}}_{\text{1}}\left[\text{A}\right]\left[\text{B}\right]={\text{K}}_{\text{- 1}}\left[\text{M}\right]+{\text{K}}_{\text{2}}\left[\text{A}\right]\left[\text{M}\right]$

$\text{M}=\frac{{\text{K}}_{\text{1}}\left[\text{A}\right]\left[\text{B}\right]}{{\text{K}}_{\text{- 1}}+{\text{K}}_{\text{2}}\left[\text{A}\right]}$

When we now introduce that to the overall rate law expression:

$\text{Rate}=\frac{{\text{K}}_{\text{1}}{\text{K}}_{\text{2}}\left[\text{A}\right]\left[\text{B}\right]}{{\text{K}}_{\text{- 1}}+{\text{K}}_{\text{2}}\text{[A]}}$