Q38 · CSIR-NET Chemistry, June 2021

Paper: CSIR-NET June 2021 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Chemical Kinetics – General · Marks: 2 · Difficulty: Hard

For the reaction,
\[ \left[\operatorname{cis}-\mathrm{M}(\mathrm{en})_{2}(\mathrm{OH})_{2}\right]^{+} \underset{\mathrm{k}_{2}}{\stackrel{\mathrm{k}_{1}}{\rightleftharpoons}}\left[\operatorname{trans}-\mathrm{M}(\mathrm{en})_{2}(\mathrm{OH})_{2}\right]^{+} \]
the equilibrium constant is 0.16 and $\mathrm{k}_{1}$ is $3.3 \times 10^{-4} \mathrm{s}^{-1}$. The experiment is started with pure cis form. The time taken for half the equilibrium amount of trans isomer to be formed is about
(a)290 s
(b)580 s
(c)190 s
(d)480 s
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
k₋₁ = k₁/K = 3.3×10⁻⁴/0.16 = 2.06×10⁻³ s⁻¹. Approach to equilibrium is first order with k₁ + k₋₁ = 2.39×10⁻³ s⁻¹; half of the equilibrium amount is reached at t = ln 2/(k₁ + k₋₁) = 290 s.

Study loop for Chemical Kinetics

1. Practise the PYQsPrevious-year questions, with answers

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