4AB › Chemistry PYQ › CSIR-NET › June 2015 › Q83Q83 · CSIR-NET Chemistry, June 2015 Paper: CSIR-NET June 2015 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Easy
For the following reaction, \[ \mathrm{A} \underset{\mathrm{k}_{-1}}{\overset{\mathrm{k}_{1}}{\rightleftharpoons}} 2 \mathrm{B} ; \mathrm{B} \xrightarrow{\mathrm{k}_{2}} \mathrm{C} \]
$\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$ is given by (a) $\mathrm{k}_{1}[\mathrm{A}]-\mathrm{k}_{-1}[\mathrm{B}]^{2}-2 \mathrm{k}_{2}[\mathrm{B}]$
(b) $2 \mathrm{k}_{1}[\mathrm{A}]-\mathrm{k}_{-1}[\mathrm{B}]^{2}-\mathrm{k}_{2}[\mathrm{B}]$
(c) $\frac{1}{2} \mathrm{k}_{1}[\mathrm{A}]-\frac{1}{2} \mathrm{k}_{-1}[\mathrm{B}]^{2}-\mathrm{k}_{2}[\mathrm{B}]$
(d) $2 \mathrm{k}_{1}[\mathrm{A}]-2 \mathrm{k}_{-1}[\mathrm{B}]^{1 / 2}-\mathrm{k}_{2}[\mathrm{B}]$
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 A → 2B gives B at 2k₁[A]; the reverse 2B → A is second order in B (k₋₁[B]²) and B → C removes B at k₂[B]: d[B]/dt = 2k₁[A] − k₋₁[B]² − k₂[B] (b), with k₋₁ defined per B consumed.
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