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?
Spontaneity in regards to enthalpy and entropy
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Re: Spontaneity in regards to enthalpy and entropy
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.
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Re: Spontaneity in regards to enthalpy and entropy
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.
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Re: Spontaneity in regards to enthalpy and entropy
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.
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