Spontaneous with Temperature Increase?

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Dhwani Krishnan 1G
Posts: 63
Joined: Fri Sep 28, 2018 12:17 am

Spontaneous with Temperature Increase?

Can a nonspontaneous process with a positive deltaS become spontaneous if the temperature is increased (assuming that deltaH and deltaS are both independent of temperature)?

Kevin Tang 4L
Posts: 83
Joined: Fri Sep 28, 2018 12:28 am

Re: Spontaneous with Temperature Increase?

Yes, when delta H is positive and delta S is positive, it will be spontaneous at high temperatures and nonspontaneous at low temperatures.
This is because when delta S is positive, -T Delta S will be negative. If the T in the equation is large enough, then the whole combined -T Delta S can be greater than the delta H, making the Delta G negative, making it spontaneous.

Hope this helps correct me if I'm wrong.

Michael Novelo 4G
Posts: 64
Joined: Fri Sep 28, 2018 12:28 am

Re: Spontaneous with Temperature Increase?

The previous statement is correct. For Gibbs free energy to be spontaneous it has to be negative so with a high temperature and positive value of delta S and a positive value of delta H the answer will always be negative of delta G which indicates it's spontaneous. Higher temperature increase does not mean it's always spontaneous however if delta S is a negative value and delta H is a positive value then it will give you a positive value of delta G since it's H - TS. The signs of H and S are important when figuring out whether or not temperature increase indicates if it is spontaneous.