Consider the following reaction, 2 H 2 ( g )+2 NO ( g ) longrightarrow N 2 ( g )+2 H 2 O ( g ) Which follows the mechanism given below? 2 NO ( g ) underset K -1 stackrel K 1 rightleftharpoons N 2 O 2 ( g ) (fast equilibrium) N 2 O 2 ( g )+ H 2 ( g ) xrightarrow k 2 N 2 O ( g )+ H 2 O ( g ) (slow reaction) N 2 O ( g )+ H 2 ( g ) xrightarrow K 3 N 2 ( g )+ H 2 O ( g ) (fast reaction) The order of the reaction is \ \ \ \ \ .

Chemistry · Chemical Equilibrium

JEE Advanced 2024 — Paper 1 — Question 39

Consider the following reaction,

2 \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{NO}(\mathrm{~g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g})

Which follows the mechanism given below? $2 \mathrm{NO}(\mathrm{g}) \underset{\mathrm{K}_{-1}}{\stackrel{\mathrm{~K}_{1}}{\rightleftharpoons}} \mathrm{~N}_{2} \mathrm{O}_{2}(\mathrm{~g})$ (fast equilibrium) $\mathrm{N}_{2} \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \xrightarrow{\mathrm{k}_{2}} \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ (slow reaction) $\mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g}) \xrightarrow{\mathrm{K}_{3}} \mathrm{~N}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ (fast reaction) The order of the reaction is $\_\_\_\_\_$ .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

Net reaction: $2 \mathrm{H}_{2}(\mathrm{~g})+2 \mathrm{NO}(\mathrm{g}) \longrightarrow \mathrm{N}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ (i) $2 \mathrm{NO}(\mathrm{g}) \stackrel{\mathrm{K}_{1}}{\stackrel{\mathrm{~K}_{-1}}{\rightleftharpoons}} \mathrm{~N}_{2} \mathrm{O}_{2}$ (g) (fast)

\frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}=\frac{\left[\mathrm{N}_{2} \mathrm{O}_{2}\right]}{[\mathrm{NO}]^{2}} \end{gathered}$$ (ii) $\mathrm{N}_{2} \mathrm{O}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) \xrightarrow{\mathrm{k}_{2}} \mathrm{~N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}$ (g) (slow reaction) $\mathrm{r}=\mathrm{k}_{2}\left[\mathrm{~N}_{2} \mathrm{O}_{2}\right]\left[\mathrm{H}_{2}\right]$ $=\mathrm{k}_{2} \times \frac{\mathrm{k}_{1}}{\mathrm{k}_{-1}}[\mathrm{NO}]^{2} \times\left[\mathrm{H}_{2}\right]$ $\mathrm{r}=\frac{\mathrm{k}_{2} \mathrm{k}_{1}}{\mathrm{k}_{-1}}[\mathrm{NO}]^{2}\left[\mathrm{H}_{2}\right]$ So, order of reaction $=3$.

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2024
Paper: Paper 1
Subject: Chemistry
Chapter: Chemical Equilibrium
Topic: Introduction to Equilibrium and Law of Mass Action