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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/2024/paper-1/4/b/iii/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 / 2024 / Paper 1 / 4 / b / iii / Solution by

Codex  0 2026-10-20

“Low NO” means

k_{2} [NO] ≪ k_{- 1}

, or

[NO] ≪ 6.5 \times 1 0^{- 6} M

. Then

\frac{d [ N O _{2} ]}{d t} ≃ \frac{2 k _{1} k _{2}}{k _{- 1}} [NO]^{2} [O_{2}] .

The effective orders are two in NO and one in oxygen, giving overall order three. For the production-rate convention,

k_{eff, prod} = \frac{2 ( 189 ) ( 1.0 \times 1 0 ^{9} )}{6.5 \times 1 0 ^{3}} = 5.82 \times 1 0^{7} M^{- 2} s^{- 1} .

If the conventional reaction rate

v = \frac{1}{2} d [N O_{2}] / d t

is used, its coefficient is

2.91 \times 1 0^{7} M^{- 2} s^{- 1}

. These are the same kinetics under explicitly different normalizations. At high NO the denominator changes, so the effective NO order approaches one rather than remaining two.

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