# 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?

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

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

• "the Van't Hoff equation relates change in enthalpy to rate constant" That wording sounds incorrect, depending on what you mean by "rate constant". See eg en.wikipedia.org/wiki/Van_%27t_Hoff_equation – Buck Thorn May 24 at 6:39
• @NightWriter I just realised. I messed up equilibrium with rate. But that doesn't essentially changes the problem – user226375 May 24 at 7:14
• The ∆ is missing in all statements. Whoever edited my question thanks to you, but can you please bring back the ∆? Without it the problem is meaningless – user226375 May 24 at 7:17
• Please try to use standard terminology and notation, otherwise you will confuse a lot of people. As it stands I don't really understand what you mean with "missing deltas" and with the meaning of extra spaces. The deltas are quite visible, just include additional ones where necessary if they are missing. – Buck Thorn May 24 at 7:21
• In your question's title you do not mean 'rate constant' (or rate coefficient) but equilibrium constant which is entirely different. – porphyrin May 24 at 7:23

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}