Question

A hypothetical reaction, $A_2+B_2\to 2AB$ follows the mechanism given below
$A_2\rightleftharpoons A+A\left(\text{fast}\right)$
$A+B_2\to AB+B\left(\text{slow}\right)$
$A+B\to AB\left(\text{fast}\right)$
The order of the overall reaction is:

• A. $2$
• B. $1$
• C. $1\frac{1}{2}$
• D. $\text{Zero}$

Answer 1

The correct option is C $1\frac{1}{2}$

$A_2+B_2\to 2AB$
Mechanism:
$A_2\stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons }}A+A\left(\text{fast}\right)$
$A+B_2\stackrel{k_2}{\to }AB+B\left(\text{slow}\right)$
$A+B\stackrel{k_3}{\to }AB\left(\text{fast}\right)$

The slow step is the rate determining step.
$\therefore \text{Rate}=k_2\left[A\right]\left[B_2\right]$

Since $A$ is not in the overall reaction, it cannot be in the rate law equation.
$\frac{k_1}{k_{-1}}=K_{eq}=\frac{[A{]}^2}{\left[A_2\right]}\Rightarrow \left[A\right]=K_{eq}^{1/2}[A_2{]}^{1/2}$

$\therefore \text{Rate}=k_2{K}_{eq}^{1/2}\left[A_2{]}^{1/2}\right[B_2]=k^{\prime }[A_2{]}^{1/2}\left[B_2\right]\therefore \text{Order}=\frac{1}{2}+1=1\frac{1}{2}$