# Connection between Gibbs Energy of Formation and Gibbs Energy of Mixing

Recently I found the following formula, which I cannot verify to be correct. Since I found this formula in several publications I assume that it is correct but I don't know why.

First there is the Gibbs Energy of Formation $$\Delta_fG$$ and second the Gibbs Energy of Mixing $$\Delta_mG$$ The formula I found says, that $$\Delta_mG = \frac{\Delta_fG^{\circ}}{N}$$ where N is the number of particles. I am not sure if this is really N or n (amount of substance). Have anyone seen this formula and/or can explain why it is valid/how it can be derived?

Thanks!

Edit:

Indeed, my problem is also in the frame of binary alloys. Consider the reaction $$2A + B = A_2B$$ which is performed in an electrochemical cell (emf measurement). Then $$\Delta_fG^\circ = -zFE$$ because the reactants are elements and A is the active species. So far this should be correct. In this context the formula in question reads as follows $$\Delta_mG = \frac{\Delta_fG^\circ}{3}$$

The full definition of the Gibbs Energy of Mixing is $$\Delta_mG = x_A RT\ln{a_A}+x_B RT\ln{a_B}$$

Unfortunately, I cannot give more detail, because I don't understand the formula. :)

• Perhaps you could edit in more context. Where I play most with thermodynamics (binary alloy phase diagrams), that relationship would definitely not hold. – Jon Custer Aug 29 '18 at 15:36
• Nobody? What information should I add to make answering easier? – user67367 Sep 1 '18 at 10:04
• Can you give some links to the places where you have seen this formula? You mention several publications. It might help give more background to show where it came from. – user64968 Sep 2 '18 at 8:06
• here are two documents (they are not open access): DoIs: 10.1149/1.2127590 and 10.1016/j.jct.2004.07.033 – user67367 Sep 2 '18 at 9:24

Thanks for the links in the comments above. The second one J Chem Thermo, 37, 97 (2005) seemed to be clearer. Eqn (12) reads $$\Delta_M G=\Delta_f G^\circ /4$$ but is preceded by the sentence

According to the definition of the molar free energy of mixing, $\Delta_M G$ for Ni$_3$Al would refer to the formula Ni$_{3/4}$Al$_{1/4}$, and thus the equation ... was obtained.

In the first link J Electrochem Soc, 128, 1181 (1981) it is harder to spot your equation. However, I think I see it, in comparing eqns  and  which relate $\Delta G_m$ and $\Delta G_f^\circ$ to a Coulombic titration expression involving an integral over the compositional variable $y$. This is for the alloy Li$_y$Sn, and the expressions look identical except for a factor $1/(1+y)$. So I guess that's it. However there's no helpful explanation.

Bearing in mind the sentence I quoted from the (2005) paper, I suspect that this factor is simply a question of how these quantities are defined, in relation to the formula of the alloy under consideration (i.e. per atom or per formula unit), and not intimately connected with the expression for the free energy of mixing in terms of mole fractions and activities (except in that it is written per atom, or more accurately, per mole of atoms).

[Edit following OP comment]

I think that this is less complicated than it may have looked. $\Delta G$ is just the difference in Gibbs energy between two well defined states. I see, for instance, on p43 of Statistical Thermodynamics of Alloys by NA Gokcen (1986) the definition

The molar Gibbs energy of mixing refers to the process prescribed by

$x_1$ (pure component 1) $+$ $x_2$ (pure component 2) $\rightarrow$ $1$ gram atom of alloy

where, of course, "gram atom" is old terminology; but the meaning is clear. If the original metals are in their standard states, this is essentially the same as the definition of the standard Gibbs free energy of formation of the alloy, except that the latter is usually expressed in terms of one mole of the compound. I think this difference is what is intended by the numerical factor, irrespective of any particular ways of writing the Gibbs energy of mixing in terms of other variables (for example, one can sometimes use regular solution theory to estimate $\Delta S$ and $\Delta H$ of mixing, but I don't think that's relevant here).

• You are looking at the right equations. But how can this be a matter of definition, when therms like $\Delta_fG$ and $\Delta_{mix}G$ have their own fixed definitions. There is even more: $\Delta_fG$ is a differential value $(\partial G/\partial \xi)$ and $\Delta_{mix}G$ an integral value. This is really confusing... – user67367 Sep 3 '18 at 8:56
• I think the two quantities are defined to be essentially the same (apart from this numerical factor), and I've edited my answer accordingly. – user64968 Sep 3 '18 at 15:20
• Sorry for the late response, I start thinking that you might be right. Therefore, I will accept your answer until I found a better explanation (if there is one :) ). Thank you! – user67367 Sep 17 '18 at 10:02