QJEEAdv24Chb965: JEE Advanced 2024 Chemistry - Class 12 JEEAdv24Chb965 Solution: Consider the following reaction, which follows the mechanism…

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.