## Standard Free Energy of Formation Units

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Dayna Pham 1I
Posts: 98
Joined: Fri Sep 28, 2018 12:16 am
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### Standard Free Energy of Formation Units

In class today, Dr. Lavelle calculated the change in standard free energy of formation by using the summation method, and his deltaG standard of formation was in kJ/mol. He said that the deltaG standard of the reaction is a per mole equivalent, that is, per mole of reaction.

Now I'm confused, because in the textbook, in the picture I attached below (from page 354 of the sixth edition), they cancelled out moles using the stoichiometric coefficients and ended up with a deltaG standard value for the reaction in kJ, not kJ per mole of reaction. Can anyone offer any insight as to when it's kJ/mol versus kJ?

My same inquiry applies for deltaS standard of reaction, and deltaH standard of reaction.

Eshwar Venkat 1F
Posts: 32
Joined: Fri Sep 28, 2018 12:22 am

### Re: Standard Free Energy of Formation Units

I can't offer any insight as to whether it is kJ/mol or just kJ, but for what it's worth, the molecules in the problem we did in class yesterday all had stoichiometric coefficients of 1, so the moles could cancel out to leave us with units of kJ, and the answer would be the same.

Shutong Hou_1F
Posts: 117
Joined: Sat Sep 14, 2019 12:17 am

### Re: Standard Free Energy of Formation Units

I think that the unit of deltaS standard of reaction is J/K, because delta S = q/T; the unit of deltaH standard of reaction is kJ/mol. Since G = H -TS, G should have the same unit as H, so I think that the unit of standard free energy of formation is also kJ/mol.