So, I'm trying to derive the equation: $$\Delta G = \Delta G^o + RT \ln Q\tag{1}$$ And trying to follow the steps detailed on this website here. I am having trouble deriving equation 7, which tries to sum chemical potentials. Previously, it derives the change in chemical potential, $\Delta \mu _i$ when the reaction is poised away from equilibrium, and for the reaction $A + B \rightarrow C + D$, four equations, each for $A, B, C$ and $D$ respectively. $$\Delta \mu _A = \mu_A' -\mu_A = RT \ln([A']/[A])\tag{2}$$ Where $[A']$ is the activity away from equilibrium. Afterwards, it proceeds by: $$\Sigma\Delta\mu_i = RT \ln\left(\frac{[C'][D'][A][B]}{[C][D][A'][B']}\right)\tag{3}$$ My confusion is regarding the arrangement of the activities in the $\ln()$ function. I assumed that: $$\Sigma\Delta\mu_i = \Delta\mu_A + \Delta\mu_B + \Delta\mu_C + \Delta\mu_D = RT \ln\left(\frac{[A'][B'][C'][D']}{[A][B][C][D]}\right)\tag{4}$$

It says that we assume the concentration of the products increases by convention, but I don't get it. Even if the concentration of the reactants decrease and the products increase, the change in sign will be due to the $\ln()$ function. If you flip the activities of the reactants around, it doesn't give the net change in Gibb's free energy anymore.

If anyone can help explain this, I will be very grateful.

P.S. Sorry if the question seemed low effort, but the context is mostly given in the website, and I can't find an explanation anywhere :(

  • 2
    $\begingroup$ Equation (4) lacks the stoichiometric coefficients, which are taken as negative for reactants (because they are used up, not made). Equation (9) of the source cited shows the correct relationship between Gibbs energy of reaction and chemical potentials. $\endgroup$
    – Karsten
    Commented Jan 8, 2021 at 12:38

1 Answer 1


It seems to me, it goes like this:

$$\mu_A=\mu_A^0+RT\ln{[A]}$$ $$\mu'_{A}=\mu_{A}^0+RT\ln{[A']}$$

So, $$\Delta G=\mu'_C+\mu'_D-\mu'_A-\mu'_B=\mu_{C}^0+RT\ln{[C']}+\mu_{D}^0+RT\ln{[D']}-\mu_{A}^0-RT\ln{[A']}-\mu_{B}^0-RT\ln{[B']} =\Delta G^0+RT\ln{Q}$$where $$Q=\frac{[C'][D']}{[A'][B']}$$ But, $$\Delta G^0=-RT\ln{K}$$where $$K=\frac{[C][D]}{[A][B]}$$


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