Q119 · 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 NO to $\mathrm{NO}_{2}$ occurs via the mechanism given below \[ \begin{array}{l} 2\mathrm{NO} \underset{k_{-1}}{\overset{k_{1}}{\rightleftharpoons}} \mathrm{N_2O_2} \\ \mathrm{N_2O_2 + O_2} \xrightarrow{k_{2}} 2\mathrm{NO_2} \end{array} \]
$\frac{\mathrm{d}\left[\mathrm{NO}_{2}\right]}{\mathrm{dt}}$ in the presence of large excess of $\mathrm{O}_{2}$ can be written as (a)$2 \mathrm{k}_{1}(\mathrm{NO})_{2}$
(b)$2 \mathrm{k}_{1} \mathrm{k}_{2}(\mathrm{NO})_{2}\left(\mathrm{O}_{2}\right)$
(c)$\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}(\mathrm{NO})^{2}$
(d)$2 \mathrm{k}_{2}(\mathrm{NO})^{2}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
With excess O₂, every N₂O₂ formed goes on to product, so the dimerisation step limits the rate: d[NO₂]/dt = 2k₁[NO]² (a).
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