Consider the following reaction,

$2{\mathrm{H}}_2(\mathrm{g})+2\mathrm{N}\mathrm{O}(\mathrm{g})\to \; <!--\; \to \; -->{\mathrm{N}}_2(\mathrm{g})+2{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

which follows the mechanism given below :

$2NO(g)\underset{k_{-<!--\; -\; -->1}}{\stackrel{k_1}{\rightleftharpoons <!--\; \rightleftharpoons \; -->}}{N}_2{O}_2(g) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (fast\; equlibrium)$

${\mathrm{N}}_2{\mathrm{O}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})\stackrel{{\mathrm{k}}_2}{\to }{\mathrm{N}}_2\mathrm{O}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g}) \; \; (slow\; reaction)$

${\mathrm{N}}_2\mathrm{O}(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})\stackrel{{\mathrm{k}}_3}{\to }{\mathrm{N}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g}) \; \; \; \; (fast\; reaction)$

The order of the reaction is _____.

To determine the overall order of the given chemical reaction, we analyze its multi-step mechanism. The overall reaction is:

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

The proposed reaction mechanism consists of the following steps:
1. Step 1 (Fast equilibrium):
$2\mathrm{NO}(\mathrm{g})\underset{k_{-1}}{\overset{k_1}{\rightleftharpoons }}{\mathrm{N}}_2{\mathrm{O}}_2(\mathrm{g})$
2. Step 2 (Slow, rate-determining step):
${\mathrm{N}}_2{\mathrm{O}}_2(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})\stackrel{k_2}{\to }{\mathrm{N}}_2\mathrm{O}(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$
3. Step 3 (Fast step):
${\mathrm{N}}_2\mathrm{O}(\mathrm{g})+{\mathrm{H}}_2(\mathrm{g})\stackrel{k_3}{\to }{\mathrm{N}}_2(\mathrm{g})+{\mathrm{H}}_2\mathrm{O}(\mathrm{g})$

The overall rate of the reaction is determined by the slowest step (Step 2), which is the rate-determining step. We can write the rate law for this elementary step as:

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

Since ${\mathrm{N}}_2{\mathrm{O}}_2$ is an intermediate, it cannot appear in the final rate law expression. We express its concentration in terms of the reactants using Step 1 (the fast equilibrium step):

$K_{\mathrm{eq}}=\frac{k_1}{k_{-1}}=\frac{\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]}{{\left[\mathrm{NO}\right]}^2}$

Solving for the concentration of the intermediate $\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]$:

$\left[{\mathrm{N}}_2{\mathrm{O}}_2\right]=K_{\mathrm{eq}}{\left[\mathrm{NO}\right]}^2$

Substitute this expression back into the rate law for Step 2:

$\text{Rate}=k_2\left(K_{\mathrm{eq}},{\left[\mathrm{NO}\right]}^2\right)\left[{\mathrm{H}}_2\right]$

Combining the constants into a single effective rate constant, $k=k_2{K}_{\mathrm{eq}}$, we get:

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

From the final rate law equation, the order of the reaction with respect to $\mathrm{NO}$ is 2, and the order of the reaction with respect to ${\mathrm{H}}_2$ is 1.
Therefore, the overall order of the reaction is:
$\text{Overall Order}=2+1=3$

The correct answer is 3.