# Effect of temperature on equilibrium constant in terms of entropy change

This question was asked in the JEE examination of 2017 :

[A] option is incorrect since the system is losing heat and hence $$\Delta S = -ve$$

[B] Lets assume we have an endothermic reaction $$\ce { A <=> B - Heat}$$ Now if we increase the temperature , the reaction will shift to the product side and hence the entropy of the surroundings will decrease more as compared to the initial conditions. Therefore the Unfavourable change in entropy(decrease in entropy) should increase.

[C] Though the reaction is endothermic , since we are supplying heat to the system , the net change in entropy of the system is positive.

[D] Lets assume another exothermic reaction $$\ce{A<=>B + Heat}$$. When we increase the temperature , the reaction will shift to the left side and hence less heat will be given off to the surroundings. Therefore , favourable change in entropy(increase in entropy) decreases.

Hence, according to me , the correct answer should only be [D] , but [B] option is also given correct. What is the error in my explanation?

The key point is that the entropy of the surroundings changes, not that of the system (at least for a small change in T), so only [B] and [D] are relevant, as you correctly determined.

You can write

$$log(K) = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$$

This of course is one form of van't Hoff's equation. It tells you, assuming $$\Delta H$$ and $$\Delta S$$ are constant, that the temperature dependence is due to $$\Delta H$$ - the heat transferred to/from the surroundings (entropy increase/loss of the surroundings):

$$log(K) ∝ -\frac{\Delta H}{RT} = \frac{\Delta S_{surr}}{R}$$

Now increasing T makes the change in the entropy of the surroundings smaller. When $$\Delta H>0$$ (endothermic), ie $$S_{surr}<0$$, this makes the reaction more likely to happen. When $$\Delta H<0$$ (exothermic), ie $$S_{surr}>0$$, this makes the reaction less likely to happen.

So both [B] and [D] are correct.