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!

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    $\begingroup$ en.wikipedia.org/wiki/Chemical_equilibrium this contains a pretty rigorous treatment of gibbs-helmholtz equation. $\endgroup$ Nov 6 '13 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
    Nov 6 '13 at 7:51
  • $\begingroup$ see the thermodynamic derivation part here.en.wikipedia.org/wiki/… $\endgroup$ Nov 6 '13 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
    Nov 6 '13 at 11:54

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
    Aug 1 '14 at 12:12
  • $\begingroup$ This discussion is further continued in a new post [chemistry.stackexchange.com/questions/104950/… $\endgroup$ Nov 30 '18 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
    Nov 25 '20 at 14:46
  • 1
    $\begingroup$ @ado sat it's partial pressure $\endgroup$
    – C.X.F.
    Feb 10 '21 at 0:01

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