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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/a/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 / a / iii / Solution by

Codex  0 2026-10-20

For a first-order reaction with no appreciable reverse contribution in this decay experiment,

\frac{d c}{d t} = - k c, c (t) = c_{0} e^{- k t} .

(1)

With fixed path length and attenuation coefficient, and absorbance attributable to ONOO after blank/background correction, the Beer-Lambert law gives

A (t) = εl c_{0} e^{- k t}

. Therefore

ln A (t) = ln (εl c_{0}) - k t .

(2)

The gradient is $- k$ , in

s^{- 1}

when time is measured in seconds. Constant illumination wavelength, homogeneous sampling and negligible product absorbance at that wavelength are required for this direct interpretation.

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