Q57 · CSIR-NET Chemistry, December 2017

Paper: CSIR-NET December 2017 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Steady State Approximation · Marks: 2 · Difficulty: Medium

For a sequence of consecutive reactions, $\mathrm{A} \xrightarrow{\mathrm{k}_{1}} \mathrm{I} \xrightarrow{\mathrm{k}_{2}} \mathrm{P}$, the concentration of I would be, by steady state approximation.
(a)$\mathrm{k}_{1}[\mathrm{A}]$
(b)$\left(\mathrm{k}_{1}+\mathrm{k}_{2}\right)[\mathrm{A}]$
(c)$\mathrm{k}_{1} \mathrm{k}_{2}[\mathrm{A}]$
(d)$\frac{\mathrm{k}_{1}}{\mathrm{k}_{2}}[\mathrm{A}]$
Answer
Answer: A ✓ checked by 4AB · confidence medium

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

Explanation
Steady state d[I]/dt = k₁[A] − k₂[I] = 0 ⇒ [I] = k₁[A]/k₂.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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