Question

How to calculate the overall rate law base on a reaction mechanism that involves 2 steps: first step is fast step and the second step is an equilibrium ?

First Step: A --> B + C (fast)

Second Step: B --> D + E (slow equilibrium)   

Answer 1

We know that, Rate of a multistep reaction is determined by the slowest step of the reaction

First Step: A --> B + C (fast)

Since first step is a fast step we can say that this step is in equilibrium

Let,

Kep = Equilibrium constant for the first step

=> Keq = [B] [C] / [A] -------------- (1)

Second Step: B --> D + E (slow equilibrium)

Rate depends on the slowest step

=> Rate = K [B] ----------------- (2)

From (1)

Keq = [B] [C] / [A]

Assuming,

[B] = [C] at ant time

Keq = [B]^2 / [A]

=> [B]^2 = Keq [A]

Substituting this in equation (2) we get

Rate = K * sqrt (Keq x [A])

Let,

K x sqrt (Keq) = Kr

=> Rate = Kr [A]^0.5

where, Kr = Rate Constant