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The Gibbs energy change of reaction is given by $$\Delta G^0_{rx}=\Delta H^0_{rx}-T\Delta S^0_{rx}$$

From the look of it seems the direction of variation of equlibrium with $T$ is given by the sign of the entropy change $\Delta S^0_{rx}$.

However the Van't Hoff equation has $$\frac{d Ln K_a}{dT}=\frac{\Delta H^0_{rx}}{RT^2}$$where it now seems that the sign of the change in enthalpy is the relevant parameter. How are the two reconciled?

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up vote 1 down vote accepted

Consider the relationship between the Gibb's free energy and the equilibrium constant K $$\text{$\Delta $G}^0=-R T \log (K)$$ Note the explicit factor of T in this relationship. I believe this is the resolution of your question.

Combined with the expression you gave for change in free energy, we have $$-R T \log (K)=\text{$\Delta $H}^0-T \text{$\Delta $S}^0$$ or $$\log (K)=\frac{-\text{$\Delta $H}^0}{R T}+ \frac{\text{$\Delta $S}^0}{R}$$. Taking the derivative with respect to temperature yields the Van't Hoff equation $$\frac{\text{dlog}(K)}{\text{dT}}=\frac{\text{$\Delta $H}^0}{R T^2}$$

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