Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature?
For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$

So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$