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

At the higher temperature the same equilibrium rate balance givesThis is a reverse first-order rate constant, not the net relaxation rate, which includes both elementary directions.

The temperature increase shifts the equilibrium towards cis. The ratio $[\mathrm{trans}]/[\mathrm{cis}]$ decreases from about $2.46$ to $2.31$, so the trans fraction falls from $0.71095$ to $2.31/3.31=0.69789$. Trans remains the more abundant isomer; “favours cis” describes the direction of the change, rather than requiring cis to become the majority.

Integrating the Van't Hoff equation with approximately constant reaction enthalpy givesUsing the full calculated $K_{298}=2.4596538\dots$ yieldsThe 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 $-2.23\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$ 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.