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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-3/42/c/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 3 / 42 / c / ii / Solution by

Codex  0 2026-10-20

Let

v_{i}

denote the rate of reaction

i

. For odd oxygen defined as O plus O

_{3}

, reactions 5 and 6 each consume one unit, while NO

_{2}

photodissociation in reaction 7 restores one. Thus the loss from reactions 4–8 is

L = v_{5} + v_{6} - v_{7}

. HO

_{2}

stationarity gives

v_{5} = v_{6} + v_{8} + v_{4}

, and NO

_{2}

stationarity within this restricted recycling loop gives

v_{7} = v_{8}

. Therefore

L = 2 v_{6} + v_{4} ≃ 2 k_{6} [H O_{2}] [O] .

(1)

The requested expression neglects the small termination correction

v_{4}

in the odd-oxygen budget, while retaining it to determine the much slower total-radical balance. This is consistent when cycling greatly exceeds termination. The net main catalytic cycle is O plus O

_{3}

forming two O

_{2}

molecules; the NO-assisted route mainly interconverts O and O

_{3}

and cancels from this restricted net loss after NO

_{2}

recycling.

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