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

With increase in temperature, the value of K. for endothermic reaction increases because unfavorable change in entropy of the surroundings decreases, with increase in temperature, the value of K for exothermic reaction decreases because favorable change in entropy of the surroundings decreases

Thermodynamics PYQ — JEE Advanced 2017 · Shift 2 · Q14 with Solution

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Correct answer: B, D

Use the relation between equilibrium constant and temperature

For a reaction,

\Delta G^\circ = -RT\ln K

and

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

So the temperature dependence of $K$ is governed mainly by the sign of $\Delta H^\circ$ .

Also, from van't Hoff equation,

\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}

Hence:

If $\Delta H^\circ > 0$ (endothermic), then $K$ increases with $T$ .
If $\Delta H^\circ < 0$ (exothermic), then $K$ decreases with $T$ .

Interpret in terms of entropy of surroundings

For a process in equilibrium with surroundings,

\Delta S_{\text{surr}} = -\frac{\Delta H}{T}

For an exothermic reaction, $\Delta H < 0$ , so
$\Delta S_{\text{surr}} > 0$
This is a favorable entropy change of surroundings. But as $T$ increases, its magnitude decreases because of division by $T$ . Therefore, the favorable contribution from surroundings decreases, so $K$ decreases.
For an endothermic reaction, $\Delta H > 0$ , so
$\Delta S_{\text{surr}} < 0$
This is an unfavorable entropy change of surroundings. As $T$ increases, its negative magnitude decreases, i.e. it becomes less unfavorable. Therefore, $K$ increases.

Check each option

Option A

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

First part: $K$ decreases for exothermic reaction with increase in $T$ — true.
Reason given: because entropy change of the system is positive — not generally true / not the correct reason.

So, A is incorrect.

Option B

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

For endothermic reaction, $K$ increases with $T$ — true.
Since $\Delta S_{\text{surr}} = -\Delta H/T < 0$ , the surroundings entropy change is unfavorable.
As $T$ increases, this unfavorable effect decreases in magnitude — true.

So, B is correct.

Option C

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

First part is true.
But the reason is incorrect: for endothermic reaction, system entropy need not be negative; this is not the governing general reason.

So, C is incorrect.

Option D

"with increase in temperature, the value of $K$ for exothermic reaction decreases because favorable change in entropy of the surroundings decreases"

For exothermic reaction, $\Delta S_{\text{surr}} > 0$ is favorable.
As $T$ increases, this favorable change decreases in magnitude.
Hence $K$ decreases.

So, D is correct.

Final derived answer

The correct options are:

\boxed{B,\ D}

Comparison with stored answer

Stored correct answer: $B, D$

This matches the derived answer exactly.

More from Thermodynamics

View written solutionFree

Correct answer: B, D

Use the relation between equilibrium constant and temperature

For a reaction,

\Delta G^\circ = -RT\ln K

and

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

So the temperature dependence of $K$ is governed mainly by the sign of $\Delta H^\circ$ .

Also, from van't Hoff equation,

\frac{d\ln K}{dT} = \frac{\Delta H^\circ}{RT^2}

Hence:

If $\Delta H^\circ > 0$ (endothermic), then $K$ increases with $T$ .
If $\Delta H^\circ < 0$ (exothermic), then $K$ decreases with $T$ .

Interpret in terms of entropy of surroundings

For a process in equilibrium with surroundings,

\Delta S_{\text{surr}} = -\frac{\Delta H}{T}

For an exothermic reaction, $\Delta H < 0$ , so
$\Delta S_{\text{surr}} > 0$
This is a favorable entropy change of surroundings. But as $T$ increases, its magnitude decreases because of division by $T$ . Therefore, the favorable contribution from surroundings decreases, so $K$ decreases.
For an endothermic reaction, $\Delta H > 0$ , so
$\Delta S_{\text{surr}} < 0$
This is an unfavorable entropy change of surroundings. As $T$ increases, its negative magnitude decreases, i.e. it becomes less unfavorable. Therefore, $K$ increases.

Check each option

Option A

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

First part: $K$ decreases for exothermic reaction with increase in $T$ — true.
Reason given: because entropy change of the system is positive — not generally true / not the correct reason.

So, A is incorrect.

Option B

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

For endothermic reaction, $K$ increases with $T$ — true.
Since $\Delta S_{\text{surr}} = -\Delta H/T < 0$ , the surroundings entropy change is unfavorable.
As $T$ increases, this unfavorable effect decreases in magnitude — true.

So, B is correct.

Option C

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

First part is true.
But the reason is incorrect: for endothermic reaction, system entropy need not be negative; this is not the governing general reason.

So, C is incorrect.

Option D

"with increase in temperature, the value of $K$ for exothermic reaction decreases because favorable change in entropy of the surroundings decreases"

For exothermic reaction, $\Delta S_{\text{surr}} > 0$ is favorable.
As $T$ increases, this favorable change decreases in magnitude.
Hence $K$ decreases.

So, D is correct.

Final derived answer

The correct options are:

\boxed{B,\ D}

Comparison with stored answer

Stored correct answer: $B, D$

This matches the derived answer exactly.

Thermodynamics question

Thermodynamics question

Option A

Option B

Option C

Option D

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Thermodynamics question

Thermodynamics question

Option A

Option B

Option C

Option D

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