1. Write the proposed mechanism

   $\text{Step 1: } NO + Br_2 \rightleftharpoons NOBr_2$ $\text{Step 2: } NOBr_2 + NO \to 2NOBr$

   We are told that Step 2 is the rate-determining step (slow step).

2. Rate law from the slow step

   Since the second step is slow, the rate is:

   $\text{Rate} = k_2[NOBr_2][NO]$

3. Eliminate the intermediate $NOBr_2$ using the fast equilibrium

   From Step 1 (fast equilibrium):

   $NO + Br_2 \rightleftharpoons NOBr_2$

   Its equilibrium constant is:

   $K = \frac{[NOBr_2]}{[NO][Br_2]}$

   Therefore,

   $[NOBr_2] = K[NO][Br_2]$

4. Substitute into the rate law

   $\text{Rate} = k_2(K[NO][Br_2])[NO]$

   $\text{Rate} = k_2K[NO]^2[Br_2]$

   Let $k = k_2K$, then

   $\text{Rate} = k[NO]^2[Br_2]$

5. Find the order with respect to $NO$

   In the rate law, the power of $[NO]$ is $2$.

   Hence, the order with respect to $NO$ is 2.

6. Check options

   • A: 1 ❌
   • B: 0 ❌
   • C: 3 ❌
   • D: 2 ✅

Therefore, the correct answer is D.