Relation between equilibrium constant and entropy change
Like the Van't Hoff equation, which relates change in enthalpy to equilibrium constant, is there a similar equation for the relation between change in entropy and equilibrium constant?


A relation between $K$ and $\Delta S^\circ$ can be obtained as follows:

$$\begin{align}T\log K &amp;= -\frac{\Delta G^\circ}{R} \\ 
\left(\frac{\partial(T\log K)}{\partial T} \right)_p &amp;= -\frac{1}{R}\left(\frac{\partial \Delta G^\circ}{\partial T}\right)_p= \frac{\Delta S^\circ}{R}\end{align}$$

As regards the multiple choice question, the best way to answer it is to understand that $\Delta G$ represents a sum of the change in the entropy of the system and of the surroundings, scaled by $T$:

$$\begin{align}\Delta G &amp;= \Delta H - T\Delta S \\&amp;= -T\Delta S_{surr} - T\Delta S_{sys} \\&amp;= -T(\Delta S_{surr} + \Delta S_{sys}) \end{align}$$ 

Exchange of heat at constant pressure (equal to the enthalpy) alters the entropy of the surroundings, so when discussion turns to the heat exchanged at constant pressure, you don't need to consider the entropy of the system, only that of the surroundings (at least for this type of problem - this alone eliminates a and c as possible choices). Next you need to remember that endothermic means the surroundings lose heat to the system, which lowers the entropy of the surroundings. The opposite holds for an exothermic process (surroundings gain entropy).

To summarize: an endothermic reaction lowers the entropy of the surroundings. The decrease in the entropy of the surroundings decreases if you increase T. The result is therefore a net increase in the total entropy. Finally, since $$\Delta S ^\circ_{\text{total}} = -\frac{\Delta G^\circ}{T}=R \log K_{\text{eq}}$$
increasing the total entropy results in a higher equilibrium constant.