Q36 · CSIR-NET Chemistry, December 2019
Paper: CSIR-NET December 2019 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Hard
The oxidation of $\mathrm{NO}$ to $\mathrm{NO_2}$ occurs via the mechanism given below. \[ \begin{array}{l} 2\,\mathrm{NO} \underset{k_{-1}}{\overset{k_1}{\rightleftharpoons}} \mathrm{N_2O_2} \\ \mathrm{N_2O_2+O_2} \xrightarrow{k_2} 2\,\mathrm{NO_2} \end{array} \]
$\dfrac{d[\mathrm{NO_2}]}{dt}$ in the presence of large excess of $\mathrm{O_2}$ can be written as (a)$2k_1(\mathrm{NO})^{2}$
(b)$2k_1k_2(\mathrm{NO})^{2}(\mathrm{O_2})$
(c)$\dfrac{k_1}{k_2}(\mathrm{NO})^{2}$
(d)$2k_2(\mathrm{NO})^{2}$
Answer
Answer: A ✓ checked by 4AB · confidence medium
Explanation
In large excess O₂, k₂[O₂] ≫ k₋₁, so every N₂O₂ formed goes on to product. Dimerisation is rate-limiting: d[NO₂]/dt = 2k₁[NO]² (a).
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