The following mechanism has been proposed for the reaction of NO with $$Br_2$$ to form NOBr : $$$NO(g) + Br_2(g) \rightleftharpoons NOBr_2(g)$$$ $$$NOBr_2(g) + NO(g) \longrightarrow 2NOBr(g)$$$ If the second step is the rate determining step, the order of the reaction with respect to $$NO(g)$$ is

The mechanism contains two elementary steps:

$$NO + Br_2 \;\rightleftharpoons\; NOBr_2 \quad\text{(fast equilibrium)}$$
$$NOBr_2 + NO \;\longrightarrow\; 2\,NOBr \quad\text{(slow - rate-determining)}$$

For an elementary step, the rate law is written directly from its molecularity. Because the second step is slow, the overall rate is controlled by this step:

$$\text{rate} = k_2\,[NOBr_2]\,[NO]$$

The concentration $$[NOBr_2]$$ is not an experimental variable, so we eliminate it using the fast equilibrium of step 1. For the reversible step 1, write the equilibrium constant:

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

Substitute this expression into the rate law obtained from the slow step:

$$\text{rate} = k_2\,[NO]\,(K\,[NO]\,[Br_2]) = (k_2K)\,[NO]^2\,[Br_2]$$

The exponent of $$[NO]$$ in the final rate law is $$2$$, so the reaction is second-order with respect to nitric oxide.

Option D which is: 2