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!

  • 1
    $\begingroup$ en.wikipedia.org/wiki/Chemical_equilibrium this contains a pretty rigorous treatment of gibbs-helmholtz equation. $\endgroup$ Commented Nov 6, 2013 at 7:43
  • $\begingroup$ 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 $\mu_A=\mu_A^o+RT\ln [A]$? $\endgroup$
    – Greg
    Commented Nov 6, 2013 at 7:51
  • $\begingroup$ see the thermodynamic derivation part here.en.wikipedia.org/wiki/… $\endgroup$ Commented Nov 6, 2013 at 8:11
  • $\begingroup$ 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. $\endgroup$
    – Ben Norris
    Commented Nov 6, 2013 at 11:54

1 Answer 1


Using the fundamental equations for the state function (and its natural variables): \begin{align} \mathrm{d}G &= -S\mathrm{d}T + V\mathrm{d}P\\ V &= \left(\frac{\partial G}{\partial P}\right)_T\\ \bar{G}(T,P_2) &= \bar{G}(T,P_1) + \int_{P_1}^{P_2}\bar{V} \mathrm{d}p \end{align} Here $\bar{x}$ represents molar $x$, i.e. $x$ per mole \begin{align} \bar{V} &= \frac{RT}{P}\\ \bar{G}(T,P_2) &= \bar{G}(T,P_1) + RT \ln\frac{P_2}{P_1} \end{align} Defining standard state as $P = \pu{1 bar}$ and $\bar{G}=\mu$ $$\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$$ consider the general gaseous reaction $\ce{a A + b B -> c C + d D}$ $$\Delta G=(c\mu_\ce{C} + d\mu_\ce{D} - a\mu_\ce{A} - b\mu_\ce{B})$$ for "unit progress" in reaction. Using $\mu_i = \mu^\circ_i + RT\ln \frac{P_i}{\pu{1 bar}}$ \begin{align} \Delta G &= (c\mu^\circ_\ce{C} + d\mu^\circ_\ce{D} - a\mu^\circ_\ce{A} - b\mu^\circ_\ce{B}) + RT \ln\frac{P_\ce{C}^c P_\ce{D}^d}{P_\ce{A}^a P_\ce{B}^b}\\ \Delta G &= \Delta G^\circ + RT\ln Q \end{align}

  • $\begingroup$ Gibbs free energy is the free energy at constant pressure and temperature, so why did you use dT and dP in your equation? It would seem that this violates the definition of the Gibb's free energy. Also, why did you take the path where entropy and volume remains constant, it can take any path. $\endgroup$
    – user7304
    Commented Aug 1, 2014 at 12:12
  • $\begingroup$ This discussion is further continued in a new post [chemistry.stackexchange.com/questions/104950/… $\endgroup$ Commented Nov 30, 2018 at 10:29
  • $\begingroup$ In this equation $\mu(T,P)=\mu^\circ (T) + RT\ln \frac{P}{P_o}$ is $P$ the total pressure or the partial pressure of the gas in the system? $\endgroup$
    – ado sar
    Commented Nov 25, 2020 at 14:46
  • 1
    $\begingroup$ @ado sat it's partial pressure $\endgroup$
    – C.X.F.
    Commented Feb 10, 2021 at 0:01

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