Why does ΔG = ΔG° + RT ln(Q) work with both Qp and Qc?

[breez]

I'm not sure whether this belongs in the chemistry section, but since it concerns free energy, I decided to post it here.

My chemistry text, Zumdhal, says (without proof, I am self-studying a first year college-level chem course) that $$\Delta G = \Delta G^{\circ} + RTln(Q)$$, where Q is the reaction quotient in terms of partial pressures.

Later, when deriving the Nernst Equation, the text uses this previous identity, but in further example problems, partial pressures are not used for the reaction quotient, but rather concentrations.

I am extremely confused on this respect since $$K_p = K_c(RT)^{\Delta n}$$, where n is the difference in coefficients of reactants from products in the balanced reaction.

So how can

$$\Delta G^{\circ} + RTln(Q_p)$$ = $$\Delta G^{\circ} + RTln(Q_c)$$ ??

 

[breez]

I guess I answered my own question: http://www.nyu.edu/classes/tuckerman/honors.chem/lectures/lecture_19/lecture_19.html

That took me a while to dig up, and its explanations mathematically show, using reference points in the mass-action expression, that this same general relationship holds for the free energy identity posted above.

Also, this means that one can use either partial pressures or concentrations in the Nernst equation.