= Solution
Combining $\Delta_rG^\circ=\Delta_rH^\circ-T\Delta_rS^\circ$ with the equilibrium relation gives
$$
\boxed{\ln K=-\frac{\Delta_rH^\circ}{R}\frac1T+\frac{\Delta_rS^\circ}{R}.}
$$
If <reaction enthalpy> and <reaction entropy> are approximately constant over the measurement interval, a plot of $y=\ln K$ against $x=1/T$ is straight, with \b[slope $-\Delta_rH^\circ/R$ and intercept $\Delta_rS^\circ/R$]. An exothermic reaction therefore has a positive slope. In the more general local form, the <Gibbs–Helmholtz equation> gives
$$
\frac{\mathrm d\ln K}{\mathrm d(1/T)}=-\frac{\Delta_rH^\circ(T)}R,
$$
which is the <Van't Hoff equation> without assuming a constant <enthalpy>.
Use a dimensionless <equilibrium constant> built from <thermodynamic activities>. Measure sufficiently separated temperatures after verifying true equilibrium, and avoid phase changes or a changing chemical reaction. Temperature-dependent <heat capacities> cause curvature: the fitted straight-line slope then estimates an interval-dependent <enthalpy>. Small errors in the equilibrium measurements can produce large slope errors over a narrow $1/T$ range. The <entropy> intercept is particularly sensitive because $1/T=0$ requires distant extrapolation, and slope/intercept errors are correlated. A numerical line through a few points is not, by itself, evidence that all these assumptions hold.
\Image[/past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2022/paper-1-vanthoff-plot.png]
{title=Original schematic van't Hoff plot showing how slope and intercept encode enthalpy and entropy}
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