How to explain disagreement between Le Châtelier's principle and the simplified Gibbs free energy equation?
The primary flaw in your reasoning is assuming that $K$ is proportional to $-\Delta G^\circ$, so that a reaction with $\Delta S^\circ &gt;0$ and $\Delta G^\circ&lt;0$ 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 $\Delta G^\circ &lt;0$ and $\Delta S^\circ&gt;0$, 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 $\Delta H^\circ &gt; 0$. If $\Delta H^\circ &lt; 0$, 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.