# reconciling free energy equations

I am struggling with a very basic concept and I think the issue lies in the definition of standard states.

When calculating standard state free energy change via Gibbs equation $$\Delta G^ \circ = \Delta H^\circ – T \Delta S ^\circ$$ one simply plugs in the standard state entropy and enthalpy, and the corresponding temperature, assuming that $\Delta H^\circ$ and $\Delta S^\circ$ are relatively temperature independent. Since these are standard state values, all species are at $1~ \mathrm{M}$ (in solution). But if all species are at $1~ \mathrm{M}$, then using the equation $\Delta G^ \circ = -RT ~ \ln K$ yields $\Delta G^ \circ = 0$. What am I missing?

• K is the equilibrium constant; the concentrations in it are the equilibrium concentrations, and those are not 1. – Ivan Neretin Mar 15 '16 at 7:34
• Any assumption that standard state entropy and enthalpy are relatively constant is doomed to failure. There is a reason that there are publications on, say, the accepted Gibbs free energy functions vs temperature for the elements (A.T. Dinsdale, 'SGTE Data for Pure Elements', CALPHAD 15(4) 317-425 (1991)). – Jon Custer Mar 15 '16 at 14:03
• Ivan, I agree that the equilibrium concentrations will not be 1 M. But if dGo is the standard state free energy of formation, and the standard state is the most stable configuration with concentrations at 1 M, then how do I reconcile this definition with the equilibrium constant not simply being K = 1 (because dGo = -RT ln K, so for dGo we have concentrations at 1 M, thus K = 1 by this reasoning)? – Andrew U Mar 15 '16 at 18:00
• 'if dGo is the standard state free energy of formation' - I believe that dG0 is not quite that. There is a difference between dG0 for formation and dG0 for reaction. dG0(reaction) = dGo(formation products) - dG0(formation reactants). Also I don't believe it's accurate to say that the standard state is the most stable one. It's just a convenient comparison point. Most of the time, dG0(rxn) != 0 as the equilibrium concentration will not be 1, so at 1 concentration the reaction will want to go forward or backward. – Daniel Jun 21 '16 at 23:41