Relation between equilibrium constant and entropy change

Question

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?

Consider the following specific case,

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

[A] With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive

[B] With increase in temperature, the value of K for endothermic reaction increases because unfavourable change in entropy of the surroundings decreases

[C] With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative

[D] With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases

Answer: (B) and (D)

My attempt:

Since equilibrium with surrounding is given, the reaction must be reversible.

This implies, $∆S=0, ∆S_s=∆S_{surr}$.

Also, at equilibrium $∆G=0$.

This implies, ∆$H=T∆S_s=-T∆S_{surr}$

(where the s subscript denotes system, and surr denotes surrounding)

Writing the van't Hoff equation in differential form, I got

$= d(lnK)=∆H(dT/RT^2)$

$=d(lnK)=-∆S_{surr}(dT/RT)$

Now, aren't options A and C true too?

As on integrating, we can see directly the variation of $K$ with $∆S$.

What is the correct way to solve this?

Edit: Where ever there is a big space, it implies ∆. Example, ' ' H=T ' ' ,

Means, ∆H=T∆S

Answer 1

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:

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

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 $T$:

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

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.