Consider the following reaction, $2 \mathrm{H}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow…

Consider the following reaction, $2 \mathrm{H}_2(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \rightarrow \mathrm{N}_2(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ which follows the mechanism given below: $2 \mathrm{NO}(\mathrm{g}) \stackrel{k_1}{\underset{k_{-1}}{\rightleftharpoons}} \mathrm{N}_2 \mathrm{O}_2(\mathrm{~g})$ $\quad$ (fast equlibrium) $\mathrm{N}_2 \mathrm{O}_2(\mathrm{~g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_2} \mathrm{~N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ $\quad$(slow reaction) $\mathrm{N}_2 \mathrm{O}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g}) \xrightarrow{k_3} \mathrm{~N}_2(\mathrm{~g})+\mathrm{H}_2 \mathrm{O}(\mathrm{g})$ $\quad$ (fast reaction) The order of the reaction is_____

Solution

Rate law $=\mathrm{k}_2\left[\mathrm{~N}_2 \mathrm{O}_2\right]\left[\mathrm{H}_2\right] \quad[\because$ slowest step of reaction is RDS $]$ $\because \frac{\mathrm{k}_1}{\mathrm{k}_{-1}}=\frac{\left[\mathrm{N}_2 \mathrm{O}_2\right]}{[\mathrm{NO}]^2}$ $\therefore \quad\left[\mathrm{N}_2 \mathrm{O}_2\right]=\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}[\mathrm{NO}]^2$ $\therefore \quad$ Rate $=\mathrm{k}_2 \times \frac{\mathrm{k}_1}{\mathrm{k}_{-1}}[\mathrm{NO}]^2\left[\mathrm{H}_2\right]$ $\therefore$ Order of reaction is (3)

Asked in: JEE Advanced 2024 (Paper 1)

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