Standard form

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

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

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

KarenaKaing_1D
Posts: 37
Joined: Fri Sep 25, 2015 3:00 am

Standard form

When calculating gibbs free energy, do you take into account the stable form, which would made it equal to 0? Or is that only for enthalpy?

Chem_Mod
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Joined: Thu Aug 04, 2011 1:53 pm
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Re: Standard form

For pure elements at standard state, the deltaG of formation will also be zero, just like the enthalpy