The overall order of a reaction is determined by the slow (rate-determining) step.

Here, the slow step is: $ \text{NOBr}_2 + \text{NO} \rightarrow 2 \text{NOBr} $

This is a second-order reaction: first-order with respect to $\text{NOBr}_2$ and first-order with respect to NO.

However, $\text{NOBr}_2$ is not a reactant in the overall reaction. It's an intermediate. So, we need to express it in terms of the initial reactants.

From the first (fast) step, we get: $ \text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2 $

In the steady state approximation, the rate of formation of $\text{NOBr}_2$ equals the rate of its consumption. We can write this as:

$ k_1[\text{NO}][\text{Br}_2] = k_{-1}[\text{NOBr}_2] + k_2[\text{NOBr}_2][\text{NO}] $

Since the second reaction (slow step) is much slower than the first one, $k_2[\text{NOBr}_2][\text{NO}]$ term is negligible in comparison to $k_{-1}[\text{NOBr}_2]$.

So, we can simplify to:

$ k_1[\text{NO}][\text{Br}_2] \approx k_{-1}[\text{NOBr}_2] $

Solving for $[\text{NOBr}_2]$, we get: $ [\text{NOBr}_2] \approx \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] $

Substitute $[\text{NOBr}_2]$ into the rate equation for the slow step:

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

$ \text{Rate} = k_2[\text{NO}] \left( \frac{k_1}{k_{-1}} [\text{NO}][\text{Br}_2] \right) $

So the overall reaction order is 3 (2 with respect to NO and 1 with respect to $\text{Br}_2$).