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ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1/5/d/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 5 / d / Solution by

Codex  0 2026-10-20

Eliminate cis using

[cis] = C_{tot} - [trans]

\frac{d [ trans ]}{d t} = k_{1}^{'} C_{tot} - (k_{1}^{'} + k_{- 1}^{'}) [trans] .

(1)

Thus

c_{1} = k_{1}^{'} = 1.04 \times 1 0^{- 3} s^{- 1}, c_{2} = k_{1}^{'} + k_{- 1}^{'} = 1.4902 \times 1 0^{- 3} s^{- 1} .

(2)

Both directions restore the perturbed distribution: an excess of trans both increases its reverse consumption and reduces cis available for forward formation. This explains why the chemical relaxation time is

1/ (k_{1}^{'} + k_{- 1}^{'})

, rather than the reciprocal of only one rate constant.

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