1. Identify the slow step

The given mechanism is:

$A_2 \overset{k_1}{\underset{k_{-1}}{\rightleftharpoons}} 2A \quad \text{(fast)}$ $A + B_2 \xrightarrow{k_2} AB + B \quad \text{(slow)}$ $A + B \rightarrow AB \quad \text{(fast)}$

Since the second step is slow, it is the rate-determining step.

So, $\text{Rate} = k_2[A][B_2]$

2. Use the fast pre-equilibrium

From the first fast reversible step:

$A_2 \rightleftharpoons 2A$

For this equilibrium, $K = \frac{[A]^2}{[A_2]} = \frac{k_1}{k_{-1}}$

Hence, $[A]^2 = K[A_2]$ $[A] = \sqrt{K[A_2]}$

Thus, $[A] \propto [A_2]^{1/2}$

3. Substitute into the rate law

$\text{Rate} = k_2[A][B_2] = k_2\sqrt{K[A_2]}[B_2]$

So, $\text{Rate} = k'[A_2]^{1/2}[B_2]$

where $k' = k_2\sqrt{K}$

4. Find overall order

Order with respect to $A_2$ is $\frac{1}{2}$.

Order with respect to $B_2$ is $1$.

Therefore, overall order is $\frac{1}{2} + 1 = \frac{3}{2} = 1.5$

5. Check options

• A: $3$ ❌
• B: $2.5$ ❌
• C: $1.5$ ✅
• D: $2$ ❌

Therefore, the correct option is C.