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Combining ${\Delta }_r{G}^{\circ }={\Delta }_r{H}^{\circ }-T{\Delta }_r{S}^{\circ }$ with the equilibrium relation givesIf 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 slope $-{\Delta }_r{H}^{\circ }/R$ and intercept ${\Delta }_r{S}^{\circ }/R$. An exothermic reaction therefore has a positive slope. In the more general local form, the Gibbs–Helmholtz equation giveswhich 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.

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.