Gibbs free energy equation

[549]

If you are given data for ∆Gf˚, T, S˚, and ∆Hf˚ for a molecule and asked to find ∆H˚ for the reaction forming that molecule, why can't you use the Gibbs free energy equation? I understand now that ∆H˚= ∆Hf˚ (which was given) for a molecule being formed from its elements, but I don't get why the Gibbs equation isn't applicable here. I can never seem to get this concept straight :-/

Also, this is AAMC FL7 PS Question 3.

Thanks!

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[Rabolisk]

To use the Gibbs equation, you have to be given all the relevant variables. It seems that's not the case here.

 

[549]

The problem gives ∆Gf˚, T, S˚. Is that not adequate because the entropy term is not given as change in entropy? If it did, would that give the correct answer? Thanks for helping!

 

[Rabolisk]

Yes, you're right. Entropy is an absolute term. The entropy of elements in standard form is not zero.