Q26 · CSIR-NET Chemistry, December 2015

Paper: CSIR-NET December 2015 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Easy

For the following reaction,
\[ \mathrm{A} \underset{k_{-1}}{\overset{k_{1}}{\rightleftharpoons}} 2\,\mathrm{B} \ ; \quad \mathrm{B} \xrightarrow{k_{2}} \mathrm{C} \]
$\dfrac{d[B]}{dt}$ is given by
(a)$k_{1}[A]-k_{-1}[B]^{2}-2k_{2}[B]$
(b)$2k_{1}[A]-k_{-1}[B]^{2}-k_{2}[B]$
(c)$\dfrac{1}{2}k_{1}[A]-\dfrac{1}{2}k_{-1}[B]^{2}-k_{2}[B]$
(d)$2k_{1}[A]-2k_{-1}[B]^{1/2}-k_{2}[B]$
Answer
Answer: B ✓ checked by 4AB · confidence medium

The source book printed no answer; this one was worked out and checked — see the explanation.

Explanation
Each A → 2B event makes two B, so the forward step contributes +2k₁[A] to d[B]/dt. The step B → C removes B at k₂[B].
The reverse step 2B → A is second order in B. Option (b) writes its term as k₋₁[B]², which means k₋₁ is defined as the rate of loss of B (−d[B]/dt = k₋₁[B]²). If k₋₁ were defined per reaction event, the term would be 2k₋₁[B]², and that form is not offered.
d[B]/dt = 2k₁[A] − k₋₁[B]² − k₂[B]. Answer (b).

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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