I am calculating the Gibbs energy of dissolving one mole of solid glucose in pure water such that the final solution has a volume of one liter and a concentration of one mole per liter. I have three different ways of calculating it, and I get two different answers.
The Gibbs energy is a state function, so it should not matter which path I take. At the beginning of the reaction, there is one mole of solid glucose and about one liter of pure water, and at the end of the reaction, there is one mole of glucose dissolved in one liter of solution.
I will use $\Delta_r G$ for the molar Gibbs energy of reaction (dimensions: energy/amount) and $\Delta G$ for the change in Gibbs energy from start to finish (extensive quantity, dimensions: energy).
1. Calculation using Gibbs energies of formation
This corresponds to two processes, turning the reactants into elements and then turning the elements into products. In the problem, all species are at standard state, so there are no correction terms for concentration.
$$\Delta G_{\text{total}} = \pu{1 mol} \Delta G_f(\text{Glucose(s)}) - \pu{1 mol } \Delta G_f(\text{(Glucose(aq)})$$
$$= \pu{1 mol } \Delta_r G^\circ\text{(dissolution)}$$
2. Integrating over Gibbs energy
The reaction starts with no glucose in solution, and then the glucose concentration increases gradually until it reaches the final concentration. During this process, the Gibbs energy of reaction changes because it is concentration-dependent:
$$ \Delta_r G\text{(dissolution)} = \Delta_r G^\circ\text{(dissolution)} + R T \ln(Q)$$
I will use the concentration of glucose divided by 1 mol/L as the integration variable x. Q is equal to x. The amount of glucose dissolved is $c\ V = (x\ \pu{mol/L) }V = x\ \pu{ mol}$. We have to integrate $ \Delta_r G\text{(dissolution)}$ from zero to one:
$$\Delta G_{\text{total}} = \pu{1 mol }\int_0^1 \left[ \Delta_r G^\circ\text{(dissolution)} + R T \ln(x) dx \right] $$
Taking constants and constant factors out of the integral, we get:
$$\Delta G_{\text{total}} = \pu{1 mol } \Delta_r G^\circ\text{(dissolution)} + \pu{1 mol } R T \int_0^1 \ln(x) dx$$
The value of the integral is negative one, so overall we have:
$$\Delta G_{\text{total}} = \pu{1 mol } (\Delta_r G^\circ\text{(dissolution)} - R T) $$
3. Running the reaction at a constant concentration of 1 M
Here, we will use a process that keeps the glucose concentration constant. We place a semi-permeable membrane into the pure water, separating it into two compartments. At the beginning, one compartment (the one in contact with the solid glucose) has a volume of zero. As glucose dissolves, we move the membrane, increasing the volume of the compartment so that the glucose concentration remains at 1 mol/L. We keep doing that over the course of the dissolution reaction until the volume of solution is one liter at the end (and the volume of pure water is zero).
Because all species are at standard state at all time, we can use the standard Gibbs energy of reaction without a term correcting for concentration. This is one component of the total change in Gibbs energy. The other one is work against the osmotic pressure difference between pure water and 1 M glucose:
$$ w = \Pi \times V = \Delta c R T \times V = \pu{1 mol } R T$$
This work represents the difference between dissolving 1 mol glucose into pure water and dissolving 1 mol glucose into a 1 M glucose solution, so we have to add (or subtract?) it to get the Gibbs energy of our original process.
Which calculation is correct, and where are the problems with the other ones?
Calculation 2 and 3 are off by a difference of $\pu{1 mol }R T$ compared to calculation 1.
I have two hunches where the problem might be. First, the concentration of water changes during the reaction (in an ideal solution, it would change by 1 mol/L, I think). I did not include the water in the first calculation, and I wonder if this is related to the discrepancy. Second, I realize that 1 M glucose in water is not an ideal solution, and that I should use activities rather than concentrations. I don't know what happens in an example with much lower concentrations.