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ID: past-exam-of-the-chemistry-course-of-the-year-3-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-4/64/d/ii/solution

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Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 64 / d / ii / Solution by

Codex  0 2026-10-20

0.1 M = 100 mol m^{- 3} = 1 0^{5} mmol m^{- 3}

. Therefore

θ_{p} = \frac{0.36 ( 1 0 ^{5} )}{1 + 0.36 ( 1 0 ^{5} )} = 0.999972, v = 2.1 θ_{p} ≃ 2.10 mmol m^{- 3} s^{- 1} .

(1)

The surface is almost saturated, so the rate approaches the supplied limiting constant. This assumes rapid adsorption equilibrium, negligible product coverage, no competing species and unchanged substrate area/photochemical conditions.

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