I have a question, if a process is nonspontaneous, it does not always need to follow the second law of thermodynamics, right?
Thank you ahead.
Thank you ahead.
If a process is non spontaneous then it could be reasonable to say the ∆S (change in entropy) is negative for just that reaction according to the Gibbs free energy equation:
G = ∆H - T∆S
Your question could also be asking about a spontaneous reaction. If the reaction is spontaneous (G<0) it is still possible to have ∆S < 0 if temperature is low and ∆H is also negative, or the rxn. is exothermic.
The 2nd law of thermo says that in any spontaneous rxn. there is always an increase in the entropy of the universe, and yet we can show using the Gibbs equation that entropy could decrease and the reaction can be spontaneous. However, this is just the entropy of the reaction, not the entropy of the universe. These are exclusive and the entropy of the universe obviously takes into consideration the surroundings, while the reactant/product entropy change only takes into consideration those 2 compounds. The entropy of the universe is always increasing.
Hi, I understand that, but second law of thermodynamics only seems to focus an increase in entropy for spontaneous reactions. However, it does not say that there will be an increase in entropy for non spontaneous reaction, right? Thank you for your explanation, though. I really like your books, as well.