Q.The equilibrium constant $K_P$ for the reaction

[!FORMULA] $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$

is $8.19 \times 10^2$ at 298 K and $4.6 \times 10^{-1}$ at 498 K. Calculate $\Delta H^0$ for the reaction.

The Van't Hoff equation makes the temperature-dependence of the equilibrium constant K fully quantitative. It follows directly from two thermodynamic relations: $\Delta G^0=-RT\ln K$ and $\Delta G^0=\Delta H^0-T\Delta S^0$. Equating these and rearranging gives

Differentiating with respect to temperature (treating $\Delta H^0$ as constant over the interval of interest) gives the differential form,

Integrating between two temperatures $T_1,T_2$ with equilibrium constants $K_1,K_2$ gives the integrated form,

This single equation lets either $\Delta H^0$ or an unknown K at a new temperature be calculated from the other quantities. Given $K_1=8.19\times10^2$ at $T_1=298$ K and $K_2=4.6\times10^{-1}$ at $T_2=498$ K for ammonia synthesis: $\log(K_2/K_1)=\log(0.46/819)=-3.2505$; rearranging for $\Delta H^0$, $\Delta H^0=\dfrac{2.303R\log(K_2/K_1)}{(T_2-T_1)/(T_1T_2)}=\dfrac{19.147\times(-3.2505)}{200/148404}\approx-46{,}190$ J mol$^{-1}\approx-46.2$ kJ mol$^{-1}$ -- negative, confirming ammonia synthesis is exothermic (K falls as T rises, exactly as Le Chatelier's principle predicts for an exothermic forward reaction). Given $\Delta H$ instead, an unknown $K_2$ can be found the same way: for $K_{P1}=0.0260$ at $T_1=298$ K with $\Delta H=32.4$ kJ mol$^{-1}$, $K_{P2}$ at $T_2=310$ K works out to $0.0431$. …