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](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
= Gibbs free energy
= enthalpy change
= entropy change
T = temperature in Kelvin
As we know that:
= +ve, reaction is non spontaneous
= -ve, reaction is spontaneous
= 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](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
![\Delta G=(+ve)-T(-ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28%2Bve%29-T%28-ve%29)
![\Delta G=(+ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28%2Bve%29)
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](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
![\Delta G=(+ve)-T(+ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28%2Bve%29-T%28%2Bve%29)
(at low temperature) (non-spontaneous)
(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](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
![\Delta G=(-ve)-T(+ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28-ve%29-T%28%2Bve%29)
![\Delta G=(-ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28-ve%29)
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](https://tex.z-dn.net/?f=%5CDelta%20G%3D%5CDelta%20H-T%5CDelta%20S)
![\Delta G=(-ve)-T(-ve)](https://tex.z-dn.net/?f=%5CDelta%20G%3D%28-ve%29-T%28-ve%29)
(at high temperature) (non-spontaneous)
(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