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### ∆G, ∆H, ∆S

Posted: Tue Feb 11, 2020 1:13 am
What's the relationship between ∆G, ∆H, and ∆S? I know that the Gibbs Free Energy equation is ∆G = ∆H - T∆S, but I'm not sure how exactly it all fits together.

### Re: ∆G, ∆H, ∆S

Posted: Tue Feb 11, 2020 8:49 am
Based on where we are in class, this has been the only relationship established. You can also think of how relative the values of ∆H and ∆S make ∆G negative at which temperatures to determine whether a reaction is spontaneous.

### Re: ∆G, ∆H, ∆S

Posted: Tue Feb 11, 2020 4:39 pm
^^ as they said above, in order for a spontaneous reaction to occur, deltaG must be negative. This means that the value for T*deltaS must be greater than the change in enthalpy. For an endothermic reaction where enthalpy is a positive number this is often not true except at very high temperatures.

### Re: ∆G, ∆H, ∆S

Posted: Tue Feb 11, 2020 6:38 pm
They relate because they can tell you lots of things about if a reaction will occur spontaneously and if it's favorable. It is expected that how this could be presented on a test would be if you were given all or some of these terms and asked to deduce overall conclusions about the system.

### Re: ∆G, ∆H, ∆S

Posted: Tue Feb 11, 2020 7:03 pm
basically, deltaG depends on deltaH and deltaS to be a spontaneous(-) or nonspontaneous(+) rxn. Since it is a new topic i dont think we need to know much more than that

### Re: ∆G, ∆H, ∆S

Posted: Wed Feb 19, 2020 6:42 pm
Here's a table that relates the three fairly well, with respect to T in order to determine spontaneity:

### Re: ∆G, ∆H, ∆S

Posted: Wed Feb 19, 2020 6:49 pm
To add on, ∆G = ∆H - T∆S is very useful for predicting if a given reaction is spontaneous or non-spontaneous. When ∆G=0 the system is at equilibrium. When ∆G is negative, the forward reaction is spontaneous. When ∆G is positive, the reverse reaction is spontaneous. The sign of ∆G depends on the enthalpy, entropy, and temperature; Its relationship is modeled by the equation ∆G = ∆H - T∆S.