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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 the increase in temperature, the value of $K$ for exothermic reaction decreases because the entropy change of the system is positive

(B) With the increase in temperature, the value of $K$ for endothermic reaction increases because of an unfavorable change in entropy of the surroundings decreases

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

(D) With the increase in temperature, the value of $K$ for exothermic reaction decreases because the favorable change in entropy of the surroundings decreases

Now, I am having trouble understanding option (B) (which in fact, is one of the correct options).

Since the process is endothermic and the temperature is rising, the value of $K$ increases, so heat must flow from surroundings to the system to favor the reaction which implies increase in an unfavorable change in entropy of surroundings rather than decrease as given in the option.

Kindly clarify my doubt. Also, any insights (other than the ones already given in the links attached below) into other options are very appreciated.

This question might look Duplicate of the questions already posted here:

But since this question is not focusing on the same problem, so I hope, it will not be marked as duplicate.

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  • $\begingroup$ It's hard to understand exactly why this is not a duplicate. I tried making my answer a little clearer in case that helps. $\endgroup$ – Buck Thorn Apr 29 '20 at 18:55
  • $\begingroup$ @BuckThorn I believe it's not a duplicate b/c, at least based on my read of the question, the OP got himself confused about a side-issue that wasn't addressed in your answer -- he though $\Delta H$ is determined by the enthalpy change in going from reactants to the equilibrium mixture (as opposed to what it actually is: reactants to products), and thus was thinking that, if you increase T for an endothermic rxn, as it shifts to the right it's actually absorbing more heat (b/c you need to go further to the right to get to equilibrium). $\endgroup$ – theorist Apr 30 '20 at 2:55
  • $\begingroup$ @theorist That doesn't seem at all obvious from re-reading the post, but you are welcome to post an answer. Looks like a dupe to me. $\endgroup$ – Buck Thorn Apr 30 '20 at 6:20
  • $\begingroup$ @theorist Exactly, I was having that misconception. Thanks, I understand my mistake as well as the concept now. $\endgroup$ – Samarth May 4 '20 at 15:51
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I agree it's not a duplicate because, while you are asking about the same homework problem discussed in the other question you linked, the issue you are having is different. Hence, because the last answer didn't address your specific confusion, you are asking a different question about that same homework problem.

Here's where you've got things mixed up: Whether a reaction is exothermic or endothermic depends on the inherent nature of the reactants and products, not the equilibrium constant. I.e., $\Delta H^{o}_{rxn}$ is the enthalpy change for a complete conversion of reactants to products at standard state, irrespective of the equilibrium constant. It's not, as I believe you were thinking, the enthalpy change to go from pure reactants to the equilibrium mixture. Hence if you could, say, change $K_{eq}$ purely by tweaking $\Delta S^{o}_{rxn}$, then $\Delta H^{o}_{rxn}$ wouldn't change just because you've changed $K_{eq}$.

So B is correct. If you raise T (and here we're assuming $\Delta H^{o}_{rxn}$ is independent of T), and $\Delta H^{o}_{rxn} > 0$, the reason the reaction shifts to the right is this: In calculating $\Delta S^{o}_{surr}$, we have to divide $q_{rev, surr} = q_{surr}$ by T, and T is now higher. Hence, for the same heat flow from the surroundings (which is given by $-\Delta H^{o}_{rxn}$), the resulting unfavorable entropy change in the surroundings is less.

[I've used standard state here, which means it's carried out at a constant external pressure of 1 bar, but what I've written applies at any constant external pressure.]

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Since the process is endothermic and the temperature is rising, the value of K increases, so heat must flow from surroundings to the system to favor the reaction which implies increase in an unfavorable change in entropy of surroundings rather than decrease as given in the option.

This describes what happens if you let a reaction go to equilibrium, and then increase the temperature. K increases, the reaction goes forward to reach equilibrium, and heat is transferred from surrounding to the system (because it is endothermic), decreasing the entropy of the surrounding.

This is no surprise. Whenever an endothermic reaction approaches equilibrium, the entropy of the environment decreases and the entropy of the system increases (by a larger amount). At equilibrium, further forward reaction would go against the 2nd law of thermodynamics (as the reaction proceeds, the change in entropy of the system gets smaller and smaller: $\Delta_r S = \Delta_r S^\circ - R \ln(Q)$).

Now, I am having trouble understanding option (B) (which in fact, is one of the correct options)

The exercise does not specify any specific scenario. It is just asking about the change in equilibrium constant. You can solve this by looking at the relationship between standard Gibbs energy of reaction and K, as well as the definition of Gibbs energy via enthalpy and entropy:

$$ \ln(K) = - \frac{\Delta_r G^\circ }{R T} = - \frac{\Delta_r H^\circ }{R T} + \frac{\Delta_r S^\circ }{R}$$

If reaction enthalpy and entropy are independent of temperature in the range of interest, raising the temperature will increase ln(K) if the reaction enthalpy is positive. In a more advanced treatment, you will learn that $\frac{d \ln(K)}{dt} = - \frac{\Delta_r H^\circ }{R T^2}$ exactly.

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    $\begingroup$ "At equilibrium, further forward reaction would go against the 2nd law of thermodynamics" beautifully sums it up since it incorporates the very definition of Equilibrium. And shouldn't it be $\frac{d \ln(k)}{dt}=-\frac{\Delta_r H^\circ }{R T^2}$? $\endgroup$ – Samarth May 10 '20 at 7:06

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