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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/a/i/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 / a / i / Solution by

Codex  0 2026-10-20

With the forward direction defined as cis-to-trans, the elementary steps have rates

k_{1} [cis]

and

k_{- 1} [trans]

. One trans molecule is gained per forward event and lost per reverse event, so

\frac{d [ trans ]}{d t} = k_{1} [cis] - k_{- 1} [trans] .

(1)

Each rate constant has units

s^{- 1}

, and the right side has units of concentration per unit time. This net rate can have either sign, even though the individual event rates are nonnegative.

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