The given reaction is:

$$2 \mathrm{H}_2(\mathrm{~g}) + 2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g}) + 2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$$

And it follows the mechanism below:

$$ \begin{array}{ll} 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) & \text { (fast equilibrum) } \\\\ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (slow reaction) } \\\\ \mathrm{N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_3} \mathrm{~N}_2(\mathrm{~g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) & \text { (fast reaction) } \end{array} $$

To determine the order of the reaction, we need to focus on the slow step, as it is the rate-determining step. The slow step is:

$$ \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) + \mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g}) + \mathrm{H}_2 \mathrm{O}(\mathrm{g}) $$

The rate law for this step can be written as:

$$ \text{Rate} = k_2 [\mathrm{N}_2\mathrm{O}_2][\mathrm{H}_2] $$

However, $$[\mathrm{N}_2 \mathrm{O}_2]$$ is an intermediate and its concentration can be expressed in terms of the reactants. From the fast equilibrium step:

$$ 2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) $$

The equilibrium constant for this step can be written as:

$\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}$ = $$ K = \frac{[\mathrm{N}_2 \mathrm{O}_2]}{[\mathrm{NO}]^2} $$

Hence, the concentration of the intermediate $$\mathrm{N}_2 \mathrm{O}_2$$ can be expressed as:

$$ [\mathrm{N}_2 \mathrm{O}_2] = K[\mathrm{NO}]^2 $$

Substituting this into the rate law for the slow step gives:

$$ \text{Rate} = k_2 K [\mathrm{NO}]^2[\mathrm{H}_2] $$

Let's combine $$k_2$$ and $$K$$ into a single constant, say $$k'$$:

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

This implies that the reaction is second-order in $$\mathrm{NO}$$ and first-order in $$\mathrm{H}_2$$. Therefore, the overall order of the reaction is:

$$ 2 + 1 = 3 $$

Thus, the order of the reaction is 3.