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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/2022/paper-1/5/c/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 / 2022 / Paper 1 / 5 / c / iii / Solution by

Codex  0 2026-10-20

Integrating the Van't Hoff equation with approximately constant reaction enthalpy gives

ln (\frac{K _{318}}{K _{298}}) = - \frac{Δ _{r} H ^{\circ}}{R} (\frac{1}{318} - \frac{1}{298}) .

(1)

Using the full calculated

K_{298} = 2.4596538 \dots

yields

Δ_{r} H^{\circ} = - 8.3145 \frac{ln ( 2.31/2.4596538 \dots )}{1/318 - 1/298} = - 2.47 kJ mo l^{- 1} .

(2)

The negative sign agrees with the decrease of

K

on heating. This is the two-temperature estimate from the supplied equilibrium data. It is not exactly the

- 2.23 kJ mo l^{- 1}

implied by treating equal prefactors and a fixed activation-energy difference as exact throughout the interval: that stricter model predicts

K_{318} = 2.3243

rather than

2.31

. The data and idealizations are therefore only approximately mutually consistent; no rounding of intermediate calculations can make both numerical enthalpy estimates identical.

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