The order of a reaction is determined from its experimentally derived rate law. When a reaction mechanism is provided, the rate law is determined by the slowest step, also known as the rate-determining step (RDS).

1. Identify the Rate-Determining Step (RDS): The given mechanism consists of three elementary steps. The second step is explicitly stated as the 'slow reaction', making it the rate-determining step. $\mathrm{N}_2 \mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g}) \quad \text { (slow reaction)}$

2. Write the Rate Law for the RDS: For an elementary reaction, the rate law can be written directly from its stoichiometry. The rate of the overall reaction is approximately equal to the rate of the RDS. $\text{Rate} = k_2 [\mathrm{N}_2 \mathrm{O}_2] [\mathrm{H}_2]$

3. Eliminate the Reaction Intermediate from the Rate Law: The species $\mathrm{N}_2 \mathrm{O}_2$ is a reaction intermediate because it is produced in one step and consumed in a subsequent step. The final rate law must be expressed only in terms of reactants and products of the overall reaction. We can express the concentration of the intermediate, $[\mathrm{N}_2 \mathrm{O}_2]$, using the first step, which is a fast equilibrium. $2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g}) \quad \text { (fast equilibrium) }$ For a fast equilibrium, the rate of the forward reaction equals the rate of the reverse reaction: $\text{Rate}_{\text{forward}} = \text{Rate}_{\text{reverse}}$ $k_1 [\mathrm{NO}]^2 = k_{-1} [\mathrm{N}_2 \mathrm{O}_2]$ We can also use the equilibrium constant expression, $K_{\text{eq}}$: $K_{\text{eq}} = \frac{k_1}{k_{-1}} = \frac{[\mathrm{N}_2 \mathrm{O}_2]}{[\mathrm{NO}]^2}$ Solving for the concentration of the intermediate, $[\mathrm{N}_2 \mathrm{O}_2]$: $[\mathrm{N}_2 \mathrm{O}_2] = K_{\text{eq}} [\mathrm{NO}]^2$

4. Substitute the Intermediate's Concentration into the Rate Law: Now, substitute the expression for $[\mathrm{N}_2 \mathrm{O}_2]$ back into the rate law derived from the RDS: $\text{Rate} = k_2 (K_{\text{eq}} [\mathrm{NO}]^2) [\mathrm{H}_2]$ Combining the constants $k_2$ and $K_{\text{eq}}$ into a new observed rate constant, $k_{\text{obs}} = k_2 K_{\text{eq}}$: $\text{Rate} = k_{\text{obs}} [\mathrm{NO}]^2 [\mathrm{H}_2]^1$

5. Determine the Overall Order of the Reaction: The overall order of the reaction is the sum of the exponents of the concentration terms in the final rate law. Order with respect to NO = 2 Order with respect to $\mathrm{H}_2$ = 1 Overall Order = 2 + 1 = 3

Therefore, the order of the reaction is 3.