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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/2023/paper-1/5/b/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 / 2023 / Paper 1 / 5 / b / ii / Solution by

Codex  0 2026-10-20

For forward and reverse elementary reactions at equilibrium, the two rates are equal. Both directions have two reactant particles and use the same concentration convention, so the conversion factors cancel and

K_{eq} = k_{3} / k_{- 3}

. Therefore

k_{- 3} = \frac{1.933 \times 1 0 ^{- 14}}{7.244 \times 1 0 ^{33}} = 2.668 \times 1 0^{- 48} d m^{3} molecul e^{- 1} s^{- 1} .

(1)

For molar rather than molecular number concentration, the bimolecular rate constant is multiplied by the Avogadro constant:

k_{- 3} = 2.668 \times 1 0^{- 48} (6.022 \times 1 0^{23}) = 1.607 \times 1 0^{- 24} d m^{3} mo l^{- 1} s^{- 1} .

(2)

Only one factor is needed: converting the two concentration factors contributes

N_{A}^{2}

, while converting the production rate from molecules to moles removes one factor.

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