Why is the equilibrium constant increasing in this exothermic reaction? A problem asks, if you have a positive $\Delta S$ (positive change in entropy) for an exothermic (meaning a negative $\Delta H$, or negative change in enthalpy) reaction and temperature increased, what would happen to $K_{\mathrm{eq}}$ and $K_{\mathrm{eq}}/Q$ (where $K_\mathrm{eq}$ is the equilibrium constant and $Q$ is the reaction quotient)?
The answer states that $K_{\mathrm{eq}}/Q$ and $K_{\mathrm{eq}}$ would increase. This is because if you use the thermodynamic equation, $\Delta G = \Delta H - T \Delta S$ (where $\Delta G$ is the change in Gibbs free energy and $T$ is for temperature), as you increase temperature you get a more negative number (since $\Delta S$ is positive and $T\Delta S$ is being subtracted). This would lead to a more negative $\Delta G$. Then, using another thermodynamic equation, $\Delta G = RT\ln(Q/K_{\mathrm{eq}})$ (where $R$ is the ideal gas constant), if $\Delta G$ is decreasing (becoming more negative), then the term $(Q/K_{\mathrm{eq}})$ must also be going down. This means that $K_{\mathrm{eq}}/Q$ must be going up and therefore $K_{\mathrm{eq}}$ increases, as temperature increases for this exothermic reaction.
So my issue is, I thought for an exothermic reaction, where heat can be represented as a product, I thought increasing the temperature would shift the reaction to the left and decrease $K_{\mathrm{eq}}$. Could someone clarify this for me? I tried looking this up with Chemistry Libretexts, but it didn't seem to clarify anything. The table seemed to suggest that we do use heat as a component of the reaction in endothermic or exothermic reactions to predict the $K_\mathrm{eq}$ change.
An exothermic reaction has a reduced equilibrium constant at higher $T$ because while the contribution of the change in the entropy of the system is a fixed quantity (for a small $T$ change), the effect of transferring heat to the surroundings is reduced at higher $T$ (because it causes a smaller change in the entropy of the surroundings). Mathematically this is described by the van't Hoff equation:
$$\left(\frac{\partial \ln K_\textrm{eq}}{\partial T}\right)_p = \frac{\Delta H^\circ}{RT^2}$$
On the other hand the change in $\Delta G$ with temperature is given by $\Delta S$:
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = -\Delta S$$
Since $\Delta G =\Delta G^\circ +RT\ln Q = -RT\ln (K/Q)$ one can write
$$\left(\frac{\partial\Delta G}{\partial T}\right)_p = \left(\frac{\partial\Delta G^\circ}{\partial T}\right)_p +R\ln Q\\ = -\Delta S^\circ+R\ln Q$$
Therefore if you want to see how $ K_\textrm{eq}$ depends on $T$ you should focus on $\Delta H^\circ$. If you want to see how $\Delta G$ depends on $T$ you should inspect $-\Delta S^\circ+R\ln Q$.
It is somehow unfortunate that the temperature dependence of $\Delta G$ and the equilibrium constant are different, because it leads to endless mix-ups, but it reminds you that they describe different things.