Gibbs energy derivative vs equilibrium for aqueous reaction

Question

I am trying to find the Gibbs-energy equilibrium for the reaction $$\mathrm{A}+\mathrm{B}\rightleftharpoons\mathrm{C}.$$ According to most sources the equilibrium concentrations in an ideal solution should form a ratio $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}.$$

I am having trouble seeing this when the stoichiometric coefficients on either side of the reaction are unequal, as in this example above.

The Gibbs energy for the solution should be $$G=n_A(\mu_A^0+RT\ln[\mathrm{A}])+n_B(\mu_B^0+RT\ln[\mathrm{B}])+n_C(\mu_C^0+RT\ln[\mathrm{C}]).$$

Assuming we started with a 1 L solution containing 1 mol of A and 1 mol of B, $\xi$ mol of each react to form $\xi$ mol of C, this equation should then be $$G=(1-\xi)(\mu_A^0+RT\ln(1-\xi))+(1-\xi)(\mu_B^0+RT\ln(1-\xi))+\xi(\mu_C^0+RT\ln\xi).$$

My understanding is that if the Gibbs free energy is at its minimum the reaction above will not be spontaneous in either direction and so the solution will be at equilibrium. The derivative of the above equation gives $$0=-(\mu_A^0+RT\ln(1-\xi))-RT-(\mu_B^0+RT\ln(1-\xi))-RT+(\mu_C^0+RT\ln\xi)+RT.$$

Some rearrangements result in $$\Delta G^0=-RT\ln\frac{[\mathrm{C}]}{[\mathrm{A}][\mathrm{B}]}+RT.$$

In other words, my expression is off by $\Delta nRT$. Having been unable to find any help on this matter on the internet, I feel I have either made some silly logical/mathematical mistake or have some fundamental misunderstanding. Please point out the error.

Answer 1

To derive the Gibbs' free energy of an ideal solution you would invoke Raoult's law, which says that $$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{p_i}{p*}\right)= \mu_i^*(l)+RT\ln\chi_i$$ Here the asterisk refers to the pure substance. Using the notation in the OP, the total number of moles of reagent and product in solution is

$$n_{tot}=2(n_{ini}-\xi)+\xi+n_{solv}=2n_{ini}-\xi+n_{solv}$$

where $n_{ini}$ is the initial amount of A or B (assumed equal), so that one can write for reagents A and B: $$\mu_i(l) = \mu_i^*(l)+RT\ln\left(\frac{n_{ini}-\xi}{2n_{ini}-\xi+n_{solv}}\right)$$ and for product C: $$\mu_C(l) = \mu_C^*(l)+RT\ln\left(\frac{\xi}{2n_{ini}-\xi+n_{solv}}\right)$$ There is also a term for the solvent: $$\mu_{solv}(l) = \mu_{solv}^*(l)+RT\ln\left(\frac{n_{solv}}{2n_{ini}-\xi+n_{solv}}\right)$$ which I ignored in earlier edits. Adding terms to compute the total Gibbs energy and taking the derivative wrt the extent of reaction $\xi$ then results in

$$\begin{align} \Delta G' &= \Delta \mu^* + RT \ln\left(\frac{\xi(2n_{ini}-\xi+n_{solv})}{(n_{ini}-\xi)^2}\right) \\ &= \Delta \mu^* + RT \ln\left(\frac{n_Cn_{tot}}{n_A n_B}\right) \\&= \Delta \mu^* + RT \ln \left(\frac{\chi_C}{\chi_A \chi_B}\right) \end{align} $$

($^\dagger$)

Note that here the standard state for each substance refers to the pure liquid in equilibrium with its vapor at the same temperature. To obtain an expression relating a change in standard free energy to the concentration equilibrium constant requires expressing mole fractions in terms of concentrations. This is generally done for dilute solutions where $n_{solv} \approx n_{tot}$ by writing $\chi_i\approx c_iM_{solv}/\rho_{solv}$, which leads to

$$\Delta G' = \Delta \mu^* - RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) + RT \ln \left(\frac{c_C}{c_A c_B}\right) $$ Then by redefining the standard states as $$\mu_i^\circ = \mu_i^* + RT \ln \left( \frac{M_{solv}}{\rho_{solv}} \right) $$ leads to $$\Delta G' = \Delta \mu^\circ + RT \ln \left(\frac{c_C}{c_A c_B}\right) $$ If you think the standard state looks strange, consider that if $\chi_i = c_iM_{solv}/\rho_{solv}$ then $$\chi_i(\pu{c^\circ_i= 1 M}) = c_i^\circ \frac{M_{solv}} {\rho_{solv}}$$ and so we can write

$$\mu_i^\circ = \mu_i^* + RT \ln \left( \chi_i(\pu{c^\circ_i= 1 M}) \right) $$ which is more in line with canonical definitions of standard states. The new standard state is defined as a 1 M ideal solution with chemical potential related to that of the pure solute by the above equation.

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$^\dagger$/TL;DR : The mysterious extra $RT$ term in the OP is gone. It's important to keep a term for the solvent. And a lesson from these repeated edits: careful doing math when tired!