Spontaneity in regards to enthalpy and entropy

[Nane Onanyan 1G]

When deltaH is positive and deltaS is positive, the reaction is spontaneous above a certain temperature.

Does this mean that for the reverse reaction, the reaction is spontaneous below a certain temperature? If yes, is that always the case?

[Marc Lubman 3B]

I think yes, both are correct, because the reverse reaction will have a negative enthalpy change and a negative entropy change, so whether the enthalpy change is negative enough to overcome the positive value of -TdeltaS will depend on whether the T value is low enough.

[edward_brodell_2I]

Since deltaG = deltaH - TdeltaS if we see that the reverse reaction will result in an opposite -deltaH and since TdeltaS is negative at a temperature it will be favorable.

[Yichen Fan 3A]

Yeah, for the reverse reaction you will just have negative deltaH and deltaS which if you put into deltaG = deltaH - TdeltaS, it will be a spantaneous reaction below certain temperature.