1. Write the mechanism

The reaction occurs in two steps:

$\mathrm{NO}+\mathrm{Br}_2 \rightleftharpoons \mathrm{NOBr}_2 \qquad \text{(fast, equilibrium)}$ $\mathrm{NOBr}_2 + \mathrm{NO} \rightarrow 2\mathrm{NOBr} \qquad \text{(slow)}$

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

2. Rate law for the slow step

For the slow elementary step,

$\text{Rate} = k_2[\mathrm{NOBr}_2][\mathrm{NO}]$

But $\mathrm{NOBr}_2$ is an intermediate, so we must eliminate it using the first fast equilibrium.

3. Use the fast equilibrium

For the first step,

$\mathrm{NO}+\mathrm{Br}_2 \rightleftharpoons \mathrm{NOBr}_2$

The equilibrium constant is

$K = \frac{[\mathrm{NOBr}_2]}{[\mathrm{NO}][\mathrm{Br}_2]}$

So,

$[\mathrm{NOBr}_2] = K[\mathrm{NO}][\mathrm{Br}_2]$

4. Substitute into the rate law

$\text{Rate} = k_2(K[\mathrm{NO}][\mathrm{Br}_2])[\mathrm{NO}]$

$\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Br}_2]$

where $k = k_2K$.

5. Find overall order

From

$\text{Rate} = k[\mathrm{NO}]^2[\mathrm{Br}_2]$

• Order with respect to $\mathrm{NO}$ = $2$
• Order with respect to $\mathrm{Br}_2$ = $1$

Therefore, overall order

$2+1=3$

Hence, the overall order of the reaction is 3.