write the rate law of the rate determining step, and then use the equilibrium condition to eliminate any intermediates.

Question

Given a mechanism consisting of the following two elementary steps:

A (g)   2 B (g) fast equilibrium

B (g) ----> (K2) 2 C(g) slow step

Give the orders of the following substances which appear in the overall reaction:

{A} is ____ order.

{B} is ____ order.

{C} is ____ order.

For example, zero, half, first, second, third, etc.

(If it is negative, you would write, negative first, negative second, etc)

Answer 1

rxn1: A(g) <=k1=> 2B(g) fast

rxn2: B(g) =k2=> 2C(g) slow

Since the rate limiting step dictates the rate law, we can write...

Rate = k2[B] but since B is an intermediate, it cannot appear in the rate law.

So, we must manipulate the equations to eliminate B.

From rxn1, we have k1[A] = k-1[B]²

k1/k-1[A] = [B]2

[B] = sqrt k1/k-1[A] = sqrt k'[A]

Substituting this into Rate = k2[B], we have ...

Rate = k2 * sqrt k'[A] = k[A]^1/2

[A] is half

[B] is zero order

[C] is zero order