Q74 · CSIR-NET Chemistry, June 2013

Paper: CSIR-NET June 2013 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Hard

The rate equation for the reaction $2 \mathrm{AB}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}_{2}$ is given byrate $=\mathrm{k}[\mathrm{AB}]\left[\mathrm{B}_{2}\right]$ A possible mechanism consistent with this rate law is
(a)$2 \mathrm{AB}+\mathrm{B}_{2} \xrightarrow{\text { slow }} 2 \mathrm{AB}_{2}$
(b)$\mathrm{AB}+\mathrm{AB} \leftrightharpoons \mathrm{A}_{2} \mathrm{B}_{2}$ (fast) $; \mathrm{A}_{2} \mathrm{B}_{2}+\mathrm{B}_{2} \xrightarrow{\text { slow }} 2 \mathrm{AB}_{2}$
(c)$\mathrm{AB}+\mathrm{B}_{2} \xrightarrow{\text { slow }} \mathrm{AB}_{3} ; \mathrm{AB}_{3}+\mathrm{AB} \xrightarrow{\text { fast }} 2 \mathrm{AB}_{2}$
(d)$\mathrm{AB}+\mathrm{B}_{2} \leftrightharpoons \mathrm{AB}_{3}$ (fast) $; \mathrm{AB}_{3}+\mathrm{AB} \xrightarrow{\text { slow }} 2 \mathrm{AB}_{2}$
Answer
Answer: C ✓ checked by 4AB · confidence high

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

Explanation
A slow bimolecular first step AB + B₂ → AB₃ followed by a fast step gives rate = k[AB][B₂]; (a) would be termolecular in [AB]², and pre-equilibria (b, d) give different orders.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

· Browse this chapter in the app