JEE Advanced · Chemistry · 18. Chemical Kinetics
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_____
- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(C) 3
Step-by-step Solution
Detailed explanation
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)
\(\because \frac{\mathrm{k}_1}{\mathrm{k}_{-1}}=\frac{\left[\mathrm{N}_2 \mathrm{O}_2\right]}{[\mathrm{NO}]^2}\)
\(\therefore \quad\left[\mathrm{N}_2 \mathrm{O}_2\right]=\frac{\mathrm{k}_1}{\mathrm{k}_{-1}}[\mathrm{NO}]^2\)
\(\therefore \quad\) 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)
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