# Derivation of relationship between equilibrium constant and Gibbs free energy change

Why is $\Delta G=\Delta G^o+RT\ln Q?$

It feels like all online sources were written for introductory Chemistry students! Where do I find a rigorous proof of this identity? Greatly appreciate it!

• en.wikipedia.org/wiki/Chemical_equilibrium this contains a pretty rigorous treatment of gibbs-helmholtz equation. – Satwik Pasani Nov 6 '13 at 7:43
• In the following section: The chemical potential of a reagent A is a function of the activity, {A} of that reagent. How do we get $\mu_A=\mu_A^o+RT\ln [A]$? – Greg Nov 6 '13 at 7:51
• see the thermodynamic derivation part here.en.wikipedia.org/wiki/… – Satwik Pasani Nov 6 '13 at 8:11
• I edited the question to remove MathJax from the title. Copious Mathjax in titles makes questions hard to locate using searches (both internal and external). Let me know if the new title is inappropriate. – Ben Norris Nov 6 '13 at 11:54

Using the fundamental equations for the state function (and its natural variables):_ $$dG=-SdT+VdP$$ $$V=(\dfrac {\partial G}{\partial P})_T$$ $$\bar G(T,P_2)=\bar G(T,P_1)+\int_{P_1}^{P_2}\bar V dp$$ Here $\bar x$ represents molar $x$, i.e. $x$ per mole $$\bar V=\frac {RT}{P}$$ $$\bar G(T,P_2)=\bar G(T,P_1)+RT \ln\frac {P_2}{P_1}$$ Defining standard state as $P=1\text{bar}$ and $\bar G=\mu$ $$\mu(T,P)=\mu^o (T)+RT\ln \frac{P}{P_o}$$ consider the general gaseous reaction $aA+bB\rightarrow cC+dD$ $$\Delta G=(c\mu_C+d\mu_D-a\mu_A-b\mu_B)$$ for "unit progress" in reaction. Using $\mu_i=\mu^o_i+RT\ln \frac{P_i}{1bar}$ $$\Delta G=(c\mu^o_C+d\mu^o_D-a\mu^o_A-b\mu^o_B)+RT\ln \frac{P_C^cP_D^d}{P_A^aP_B^b}$$ $$\Delta G=\Delta G^o+RT\ln Q$$