= Solution
Integrating the <Van't Hoff equation> with approximately constant <reaction enthalpy> gives
$$
\ln\!\left(\frac{K_{318}}{K_{298}}\right)=-\frac{\Delta_rH^\circ}{R}\left(\frac1{318}-\frac1{298}\right).
$$
Using the full calculated $K_{298}=2.4596538\ldots$ yields
$$
\boxed{\Delta_rH^\circ=-8.3145\frac{\ln(2.31/2.4596538\ldots)}{1/318-1/298}=-2.47\,\mathrm{kJ\,mol^{-1}}.}
$$
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 $-2.23\,\mathrm{kJ\,mol^{-1}}$ 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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