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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/e/ii/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 / e / ii / Solution by

Codex  0 2026-10-20

Subtract the new equilibrium concentration and divide by the initial displacement:

\frac{[ trans ] ( t ) - [ trans ] _{eq2}}{[ trans ] _{eq1} - [ trans ] _{eq2}} = e^{- (k_{1}^{'} + k_{- 1}^{'}) t} .

(1)

At the halfway concentration this ratio is

1/2

, so

t_{1/2} = \frac{ln 2}{k _{1}^{'} + k _{- 1}^{'}} = 465 s ≃ 7.75 min .

(2)

This is the half-time of the displacement from equilibrium. It is not the time for the absolute trans concentration to fall to half its initial value; that value lies far below the attainable new equilibrium plateau. The corresponding chemical relaxation time is

671

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