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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.

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    $\begingroup$ For starters the molar mass of glucose is 180g/mol which means your solution is 15 wt% glucose which has a density of ~1.07 g/mL which translates to concentration of about 0.934 M. which is not trivially off $\endgroup$ – A.K. Apr 12 at 1:34
  • $\begingroup$ @A.K. - I changed the volume of water so that the final solution has a volume of exactly one liter and a concentration of exactly one molar. $\endgroup$ – Karsten Theis Apr 12 at 2:30
  • $\begingroup$ With some hindsight, I should have used something like benzene as solvent and naphthalene as solute to be a bit closer to an ideal solution. $\endgroup$ – Karsten Theis Apr 14 at 14:34
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I think the issue with your calculation 2 is that you are missing a term in the expression that you have integrated.

Let's start with the definition that $\Delta_r G = \frac{dG}{d\xi}$. For simplicity, I'm going to replace $\xi$ with $x$ from here on.

For each component of the reaction, the molar free energy $\overline{G}$ is given by $G^\circ + RT\ln a$, where $a$ is the activity, for which we will substitute the concentration in molarity, assuming an ideal solution, etc. Thus, the partial free energy contribution is the product of $\overline{G}$ and the number of moles.

For the aqueous glucose, the concentration is equal to $x$ and for the solid glucose the activity is $1$. The number of moles of the aqueous glucose is also equal to $x$, and the number of moles of solid glucose is $1-x$.

We can now write an expression for the total free energy of the reaction system as a function of $x$:

$$G(x)=xG_{gluc(aq)}^\circ + xRT\ln x + (1-x)G_{gluc(s)}^\circ + (1-x)RT\ln 1.$$

Taking the derivative with respect to $x$, we get

$$\frac{dG}{dx}=G_{gluc(aq)}^\circ + RT\ln x + xRT\left(\frac{1}{x}\right) - G_{gluc(s)}^\circ=\Delta_r G^\circ + RT\ln x + RT.$$

Integration of that expression from $x=0$ to $x=1$ gives the expected result of $\Delta G =\Delta_r G^\circ$ because of the extra $RT$ term that you are missing, which cancels out the extra $-RT$.

For calculation 3, I believe the osmotic work term should be left out, since there is no work done on the surroundings if you treat the entire container as the system. The expanding volume of the glucose solution and contracting volume of the water will move the (frictionless, massless) membrane without any energy input from outside the system.

UPDATE: For calculation 2, one can also deal explicitly with the water to reconcile the disagreement between the known $\Delta_r G$ at standard state and the derived $\frac{dG}{d\xi}$ expression

First, we need to change our reaction to:

$$\ce{Glucose(s) + water(l) -> Glucose (aq) + water (aq)}$$

By "water(aq)" I mean the water in the glucose solution. I'm not sure the correct notation for that.

First, we calculate the standard molar free energy of reaction as $G^\circ(products)-G^\circ(reactants)$:

For the water, I'm going to use a density of 55 mol/L. The exact value doesn't matter. Since we're assuming ideal solutions, one mol of aqueous glucose occupies the same space as 1 mol of water, so to make 1 L of a 1 M glucose solution (ie 1 mole of standard state aqueous glucose), we would add 1 mole of solid glucose to 54 moles of water. The volume of the "aqueous water" in the 1 M glucose solution is therefore 55/54 the volume of pure water. Therefore, the difference in free energy between the aqueous water in a 1 M solution and the same number of moles of pure water at constant temp is given by $-nRTln(V_f/V_i)=-54RTln(55/54)=-0.99RT$.

So our standard free energy change is

$$\Delta_r G^\circ = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99RT,$$

and "calculation 1" multiplies this by 1 mol to get $\Delta G = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s)) - 0.99RT$.

We can also see that the free energy of the water as a function of $x$ is $G_{wat}=54G^\circ_{pure}+54RT\ln\left(\frac{54}{54+x}\right)$, where the term in parentheses is the mole fraction of water in the aqueous mixture. We also need to modify the term for the activity of aqueous glucose since the volume is not constant. Instead of being equal to $x$, it is corrected for the volume being less than 1 liter by a factor of $\frac{54+x}{55}$.

Now we do calculation 2 over again. We start with

$$G(x)=x\left(G^\circ_{(glu(aq))}+RT\ln\left(x\cdot\frac{55}{54+x}\right)\right) + (1-x)(G^\circ_{(glu(s))}+RT\ln(1)) + 54G^\circ_{pure}+54RT\ln\left(\frac{54}{54+x}\right)$$

Taking the derivative with respect to $x$,

$$\frac{dG}{dx}=G^\circ_{(glu(aq))} + RT\ln\left(x\cdot\frac{55}{54+x}\right)+ xRT\left(\frac{1}{x}-\frac{1}{54+x}\right) - G^\circ_{(glu(s))} - 54RT\left(\frac{1}{54+x}\right).$$

$$=G^\circ_{(glu(aq))} - G^\circ_{(glu(s))} + RT\left(\ln\left(x\cdot\frac{55}{54+x}\right)+1-\frac{x+54}{x+54}\right)=G^\circ_{(glu(aq))} - G^\circ_{(glu(s))} + RT\ln\left(x\cdot\frac{55}{54+x}\right),$$

Now, we go back to our equation $\Delta_r G^\circ = \Delta_f G^\circ(Glucose(aq)) - \Delta_f G^\circ(Glucose(s))-0.99RT$, and make a substitution into the above expression to get

$$\Delta_r G = \frac{dG}{dx}=\Delta_r G^\circ + 0.99RT + RT\ln\left(x\cdot\frac{55}{54+x}\right).$$

To check this, we can integrate from $x=0$ to $x=1$, and we get $\Delta G = \Delta_r G^\circ$, so that matches up with calculation 1.

For the $\Delta_r G$ at standard state, things are a little trickier. There is no point in this reaction where the pure water and the aqueous glucose are both present, so if we explicitly include water as a reactant, we can't have all reactants and products at standard state at the same time for any value of $x$. That's the reason for the apparent discrepancy.

In order to determine $\Delta_r G^\circ$ using a function of $x$, we need to have a reaction path for which the standard state occurs at some value of $x$, such as your calculation 3. Another approach that I think will work is to have two beakers of water and add solid first to one and then the other. Halfway through the reaction, you have 1 L of 1 M glucose, 54/55 L of pure water and one mole of solid glucose, so all of the activities are 1. If you redo the above analysis for that reaction, you should get the correct result of $\Delta_r G$ at standard state is equal to $\Delta_r G^\circ$.

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  • $\begingroup$ Your comments about method 3 are very helpful: G is a state function, so if I find a path where I know how G changes, work and heat transfer does not matter. About method 2: If I understand correctly, you derive the expression $$\Delta_r G = \Delta_r G^\circ + RT\ln x + RT$$ I'm surprised that the RHS is not equal to the LHS for x = 1. $\endgroup$ – Karsten Theis Apr 13 at 18:03
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    $\begingroup$ I agree there's something problematic about that result, although $\xi =1$ and $Q=1$ (standard state) do not represent the same state, since solid is present when $Q=1$, but not when $\xi=1$. To get Q=1, we would need to start with more than 1 mole solid and have $\xi<1$, but I don't think that solves the problem. Dealing with the water explicitly also would only introduce a small correction, so definitely there's more to this. Certainly a really good question. $\endgroup$ – Andrew Apr 13 at 18:21
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    $\begingroup$ UPDATE: It looks like -RT is the free energy change of the water, so the +RT term is in the $\Delta_r G$ to correct for that. Explicitly dealing with the water is necessary to reconcile the two. I can add that to the answer or rewrite the answer if you like. $\endgroup$ – Andrew Apr 14 at 12:16
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    $\begingroup$ I added a section in the answer to explain where the -RT comes from and why we can't get to $\Delta_r G = \Delta_r G^\circ$ at standard state in this calculation. I think that clears everything up, but I may still be missing something. Thanks for a really thought-provoking question! $\endgroup$ – Andrew Apr 14 at 14:01
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    $\begingroup$ @KarstenTheis - the latter. The standard Gibbs energies of formation have units of J/mol, as does the 0.99RT term, which is the free energy change of the water per mole of 1 M aqueous glucose solution. So the entire expression is multiplied by 1 mol to get the $\Delta G$ in units of J. I tried to make that clear when I said "multiply this by 1 mol", but including units would have been better. Sorry for the confusion. $\endgroup$ – Andrew Apr 14 at 14:38

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