Quantum calculation of free energy change in neat conditions
One needs to distinguish between the standard free energy (of reaction) for a process, $\Delta _r G^\circ$, and the free energy in general, $\Delta _r G$. You show how to use quantum methods to compute $\Delta _r G^\circ$. This tells you how much the free energy changes if you take 1 mole of each reagent at 1 M concentration and generate one mole of complex at 1 M concentration. At other concentrations one uses
$$\Delta_r G = \Delta_r G^0 + RT \ln\left(\frac{C_{AB}/C^\circ}{(C_A/C^\circ )(C_B/C^\circ)}\right) \tag{Eq. 1}$$
for the reaction $\ce{A +B&lt;=&gt;AB}$ (assuming all components behave ideally; $\pu{C^\circ= 1 M}$ is the standard concentration). In fact, you use an analogous expression to compute the free energy difference between 1 atm (at $\pu{0^\circ C}$) and 1 M concentrations for each reagent or product:
$$\begin{align}\Delta_r G(\pu{1 atm}\rightarrow\pu{1 M}) &amp;= RT \ln\left(\frac{C_X(\pu{1 M})}{C_X(\pu{1 atm})}\right)\\ &amp;= RT \ln\left(\frac{\pu{1 M}}{(\pu{1e-3 \pu{L/m^3}})\cdot(1/(RT/\pu{1 atm})}\right)\end{align} $$
You use analogous relations (usually in the form of Eq. (1)) if you intend to compute the free energy change for one mole of complex formation when reagents and products are at particular concentrations (information which you then have to provide).
If you are interested in computing the equilibrium concentration of complex given specified concentrations of A and B (assumed constant) then you solve for $C_{AB}$ in Eq. (1) after plugging in values for $C_A$ and $C_B$, remembering again that you are assuming your solution behaves ideally.