[orthoformate]

Extensive (J) or intensive (J/mol)?

Thank you! then ΔH is calculated as you described (ΔHm) as an intensive unit when dealing with ΔH-TΔS

what I was trying to illustrate with my earlier post is that you could calculate this conversion for one mol, but then you would have to divide it in half for the actual ΔG if you wanted to calculate it via ΔG=ΔH-TΔS. Is this correct?

[mjc123]

Careful! What is the "ΔG" you are trying to calculate?

The diagram below shows the molar Gibbs energy of a mixture of A and B at a total concentration [A] + [B ] = 1M as a function of mole fraction of B in the mixture (that is, x_B = n_B/(n_A + n_B), not counting the solvent). I have used your example of K = 3, so ΔG_m° = -RTln3 (T = 298 K), and for reference G_m,A° = 0. For simplicity I have assumed ideal mixing, i.e. ΔG_mix = RT(x_Alnx_a + x_Blnx_B).
Not surprisingly, G is a minimum at the equilibrium composition x_B = 0.75. If you want the molar ΔG for the conversion you described, from x_b = 0.5 to 0.75, that is given by "ΔG_m" on the graph. Note that this "ΔG_m" is not equal to 0.25*ΔG_m°, and would also not be equal to ΔG_m for conversion from, say, x_B = 0 to 0.25.
What is "ΔG" in the equation ΔG = -RTlnK + RTlnQ? Well, at the equilibrium position, Q = K, so "ΔG" = 0. What is zero at the equilibrium point? Answer: the slope of the G-x curve, the partial molar Gibbs energy of reaction dG_m/dx_B, or ΔG_m(x_B) - the free energy per mole of converting a small amount of A to B at constant composition x_B.

Now in your scenario, you start with 1L of 1M of both A and B - total concentration 2M, total amount 2 moles. In this case, again assuming ideality, the value of G_m,A = G_m,A° + RTln[A], and the same for G_m,B. Both values are increased by RTln2, but the difference between them is unchanged (this would not be the case if there were unequal moles on either side of the equation). Then G = n*G_m, so all the ΔG values we have talked about are multiplied by 2. In other words, the curve is shifted along the y-axis to account for the change in concentration, and stretched in the y-direction to account for the number of moles.

I'm sorry if this seems unduly complicated, but it reflects the complexity of the situation, and shows why people often go wrong when they just ask "what is ΔG?"

[orthoformate]

Are you saying that since ΔG=-RTlnK+RTlnQ is not a linear function you cannot simply multiply the 0.25*ΔGm° to get the correct ΔG. I can see from your graph (thank you very much for the graph b.t.w.) that the first 0.25m converted would give a larger (or more negative because of the sign) ΔG than the last 0.25 mols.

Also, you said that ΔG at any specific point is equal to δG/δxb so as you said "the partial molar Gibbs energy of reaction dGm/dxB, or ΔGm(xB) - the free energy per mole of converting a small amount of A to B at constant composition xB"

meaning that that value of of ΔGm changes as you change composition.

If you wanted to calculate the ΔG change from beginning to equilibrium, could you simply take the integral of δG/δxb bounded from the starting point to equilibrium with respect to xb?

Thank you for taking the time to make this post. It was very eye opening for me.

[mjc123]

Are you saying that since ΔG=-RTlnK+RTlnQ is not a linear function you cannot simply multiply the 0.25*ΔGm° to get the correct ΔG.

Careful - what "correct ΔG" are you after? Is it the molar (i.e. per mole of mixture) free energy change on going from x_B = 0.5 to 0.75 (or generally x to x+0.25)? This is G_m(x+0.25) - G_m(x), and is generally not equal to 0.25*ΔG_m°. It is also not equal to -RTlnK + RTlnQ, which is equal to the slope of the G_m curve, and is defined for a particular value of x_B (and Q), not a range.
It would be more correct to phrase your statement as "since G_m is not a linear function of x_B, you cannot simply multiply the 0.25*ΔGm° to get the correct ΔG." Or more generally, "since G_m is not a linear function of x_B, you cannot simply say G_m(x_f) - G_m(x_i) = (x_f - x_i)*ΔG_m°."

If you wanted to calculate the ΔG change from beginning to equilibrium, could you simply take the integral of δG/δxb bounded from the starting point to equilibrium with respect to xb?

Yes, but is there a situation where you are likely to know dG_m/dx_B as a function of x_B, but not G_m as a function of x_B?

[orthoformate]

This is very interesting.

Why is it that when you use the table values of ΔH and ΔS you can use arithmetic to solve for ΔG? via ΔG=ΔH-TΔS

Are these table values only accurate for exactly one mol gone to complete conversion?

[mjc123]

Tables usually give you molar standard heats of formation and molar standard entropies. If you know these values for A and B, what value of ΔG do they give you? What does that correspond to on the diagram?

[orthoformate]

The standard molar entropy and enthalpy for A and B would allow you calculate a molar G for both A and B. On the diagram, the 100 mol% of A is on the far right, and the 100 mol% of B is on the left. you would use these values to get the ΔG, which would be the change on the y axis.

using my example ΔG=RTln(3): ΔG= 2.7 Kj/mol

but, we know thermodynamic equilibrium would allow the reaction to achieve a greater ΔG than this.

[mjc123]

On the diagram, the 100 mol% of A is on the far right, and the 100 mol% of B is on the left

Other way round on my diagram.

using my example ΔG=RTln(3): ΔG= 2.7 Kj/mol

ΔG° = -RTlnK

but, we know thermodynamic equilibrium would allow the reaction to achieve a greater ΔG than this.

In my calculations, assuming ideality, ΔG_m (starting from 100% A) = x_BΔG_m° + RT(x_Alnx_A + x_Blnx_B)