Question

Under what conditions does an increase in temperature turn a nonspontaneous process into a spontaneous process?

Answer 1

This, is an incomplete question, here is a complete question.

Under what conditions does an increase in temperature turn a nonspontaneous process into a spontaneous process?

Choose one or more:

ΔH < 0, ΔS > 0

ΔH > 0, ΔS > 0

ΔH > 0, ΔS < 0

ΔH < 0, ΔS < 0

(What I'm trying to figure out is if you need to consider T being negative OR positive to start)

Answer :  The correct option is, ΔH > 0, ΔS > 0

Explanation :

According to Gibb's equation:

$\Delta G=\Delta H-T\Delta S$

$\Delta G$ = Gibbs free energy  

$\Delta H$ = enthalpy change

$\Delta S$ = entropy change  

T = temperature in Kelvin

As we know that:

$\Delta G$= +ve, reaction is non spontaneous

$\Delta G$= -ve, reaction is spontaneous

$\Delta G$= 0, reaction is in equilibrium

As there are 4 conditions:

(A) ΔH is positive or (ΔH>0) and ΔS is negative or (ΔS<0).

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(+ve)-T(-ve)$

$\Delta G=(+ve)$

The reaction is non-spontaneous at all temperatures or spontaneous in reverse at all temperatures.

(B) ΔH is positive or (ΔH>0) and ΔS is positive or (ΔS>0).

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(+ve)-T(+ve)$

$\Delta G=(+ve)$  (at low temperature) (non-spontaneous)

$\Delta G=(-ve)$  (at high temperature) (spontaneous)

The reaction is spontaneous as written above a certain temperature.

(C) ΔH is negative or (ΔH<0) and ΔS is positive or (ΔS>0).

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(+ve)$

$\Delta G=(-ve)$

The reaction is spontaneous as written at all temperatures

(D) ΔH is negative or (ΔH<0) and ΔS is negative or (ΔS<0).

$\Delta G=\Delta H-T\Delta S$

$\Delta G=(-ve)-T(-ve)$

$\Delta G=(+ve)$  (at high temperature) (non-spontaneous)

$\Delta G=(-ve)$  (at low temperature) (spontaneous)

The reaction is spontaneous as written below a certain temperature.

From this we conclude that, the condition ΔH is positive or (ΔH>0) and ΔS is positive or (ΔS>0) at increase in temperature turn a nonspontaneous process into a spontaneous process.

Hence, the correct option is, ΔH > 0, ΔS > 0