For example, for the dissolution of a salt in water that is exothermic, heating the solution would drive the reaction towards the solid form of the salt according to Le Châtelier's principle.
However, according to the simplified Gibbs equation $(\mathrm dG = \mathrm dH - T\mathrm dS),$ the reaction would be spontaneous towards the products due to the increase in entropy and the negative enthalpy.
To me, these seem to contradict each other. How is this reconciled? Am I making a mistake somewhere?
Answer
The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ >0$ and $\Delta G^\circ<0$ must have a larger $K$ at a higher temperature because $\Delta G^\circ$ is more negative. If that were true, we would have a relationship of the form $\Delta G^\circ = -cK$, where $c$ is a constant. Instead, the key relationship is
$$\Delta G^\circ = -RT\ln K.$$
So $\Delta G^\circ$ is proportional to $-T\ln K$. In the case above where $\Delta G^\circ <0$ and $\Delta S^\circ>0$, as T increases, $\Delta G^\circ$ increases in magnitude (becomes more negative), but so does $-RT$, so we don't necessarily need to have a larger $K$ to satisfy the equation. To figure out the temperature dependence of $K$, we need to substitute $\Delta G^\circ$ with $\Delta H^\circ - T\Delta S^\circ$ and then rearrange things:
$$\Delta H^\circ - T\Delta S^\circ=-RT\ln K$$
$$\frac{\Delta H^\circ}{T}-\Delta S^\circ=-R\ln K$$
From that equation, hopefully it is clear that if $T$ increases (which reduces the magnitude of the $\frac{\Delta H^\circ}{T}$ term), $K$ will only increase if $\Delta H^\circ > 0$. If $\Delta H^\circ < 0$, K will have to decrease with increasing $T$ to maintain the equality. $\Delta S^\circ$ is a constant term that does not affect the change in $K$. Thus, our result is completely consistent with both Le Chatelier's principle and with the van't Hoff equation analysis.
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