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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

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  • $\begingroup$ "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 $\endgroup$ – Buck Thorn May 24 at 6:39
  • $\begingroup$ @NightWriter I just realised. I messed up equilibrium with rate. But that doesn't essentially changes the problem $\endgroup$ – user226375 May 24 at 7:14
  • $\begingroup$ 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 $\endgroup$ – user226375 May 24 at 7:17
  • $\begingroup$ 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. $\endgroup$ – Buck Thorn May 24 at 7:21
  • $\begingroup$ In your question's title you do not mean 'rate constant' (or rate coefficient) but equilibrium constant which is entirely different. $\endgroup$ – porphyrin May 24 at 7:23
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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).

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