I am a mathematician and I want to understand the molar free energy model given in Kim and Sanders (2020) Equation (1), that is, \begin{equation} f(c,T) = RT \big(c \log(c) + (1-c) \log(1-c)\big) + \big(A_0(T) + A_1(T) (1-2c)\big) \, c(1-c), \end{equation} where $T$ is the temperature, $c$ is the concentration (of $\text{SiO}_2$), $A_0$ and $A_1$ are temperature dependent coefficients of the Redlich-Kister interaction model which are given in $\text{kJ}\,\text{mol}^{-1}$ by \begin{equation} A_0(T) = 186.0575 - 0.3654\, T\quad \text{and}\quad A_1(T) = 43.7207 - 0.1401\, T \end{equation} (see also Kim and Sanders (1991)) and, as usual, $R = 0.0083144\ \text{kJ}\,(\text{mol K})^{-1}$ is the gas constant.

Problem: According to Kim and Sanders (2020), Figure 1, I would expect to get a double well potential for a temperature T = 900 K with two minima at approximately $c_1=0.82$ and $c_2=0.97$. However, plotting $f$ at $T = 900\ \text{K}$ yields:

Plot of molar free energy <span class=$f(\cdot,900)$" />

Questions: Did I misunderstand something concerning the energy model? Is there a transformation missing, especially concerning the composition $c_0 = 0.88$ mol of $\text{SiO}_2$ (and $0.12$ mol of $\text{Na}_2\text{O}$). Can someone give me good references to basics of similar problems (scaling etc., discussions of energy models ...)?


2 Answers 2


In an Ideal solution, based on how many ways one can randomly arrange the constituents, one gets:

$\Delta S_{mix} = -R(X_{A} \ln{X_{A}} + X_{B} \ln{X_{B}})$ with $X_{A}+X_{B} = 1$ being the relative molar concentrations.


$\Delta G_{mix} = RT(X_{A} \ln{X_{A}} + X_{B} \ln{X_{B}})$

that is your first bit.

For a Regular solution,

$\Delta H_{mix} = \Omega X_{A} X_{B}$

for some $\Omega$ that may well depend on temperature and composition. So in your equation, $A_{0}(T)$ is the part of $\Omega$ that depends only on $T$. Composition dependence is represented by a power series in $(X_{A}-X_{B})$, or equivalently $(2X_{A}-1)$ or $(1-2X_{B})$. The subscripts on the $A$ terms are the power of $(X_{A}-X_{B})$ for that term.

This representation is quite commonly used in the Calphad method for computing binary (and higher) phase diagrams.

So, in your particular case, what does this look like? Plugging those values into my homebrew Calphad program I get a miscibility gap that looks like:

Miscibility gap

And, stopping at 600K, a free energy curve that looks like:

Free energy at 600K

so I'm not quite sure what went wrong for your calculation. One thing to double check would be that your units (J/mol vs kJ/mol) for the entropy and enthalpies match.

Note also that you are not necessarily looking for minima in the curve - you are looking for common tangents to the curve which indicate that the Gibbs free energies of the two components are the same at the two compositions. The red line on the second plot shows the common tangent at that temperature (600K in this case). (That said, you can freely choose a component reference free energy that would make the curve at a particular temperature show those points as minima, but in general you will just get a common tangent.)

  • $\begingroup$ Thanks for your explanations regarding the model building. My problem is that the concrete energy $f$ with values $A_0$ and $A_1$ etc. as given above does not yield a double well potential (cp. figure in the question). However, I would expect a double well potential, that is, I would expect that $f$ has two minima. (Setting e.g. $\Omega = 20$ in your notation (example) would yield a double well potential at $T = 900\ K$). That's why I cannot accept your answer as an answer. $\endgroup$
    – Henning
    May 16, 2022 at 17:20
  • $\begingroup$ @Henning - I added my calculations based on those papers. $\endgroup$
    – Jon Custer
    May 16, 2022 at 18:35
  • $\begingroup$ Actually, your free energy curve does not look very different from my one in the question (up to a scaling factor). I am wondering if this is reasonable. I think that due to the miscibility there should be two minima, not just one minimum. $\endgroup$
    – Henning
    May 16, 2022 at 20:29
  • 1
    $\begingroup$ I missed somehow your last paragraph of your extended answer, sorry and thank you! In fact, that could be the confusion. I was obsessed with the fact that the system moves in two minima. OK, I have to think about it. Many thanks! $\endgroup$
    – Henning
    May 16, 2022 at 20:59
  • 1
    $\begingroup$ @Henning - I would highly recommend Porter and Easterling's Phase Transformations in Metals and Alloys. 2nd edition is best in my opinion, but see if you library has it. The first 40 pages or so cover a lot of ground on binary phase diagrams. $\endgroup$
    – Jon Custer
    May 16, 2022 at 21:03

As an extension of the above answer of Jon Custer, I would like to draw attention to the paper of Charles (1967). On page 634 Charles is concerned with a similar unfortunate fact that "the changes in mixing free energies, which might denote immiscibility, would be too slight to be observed on the scale necessary to present all the calculated free energy values." He therefore considered an additional linear term -- with arbitrary coefficient -- which enables a detection of a miscibility by a visual inspection of the corresponding (double well) potential.

Correspondingly, for the above system (cf. question) we may consider \begin{equation} f(c,T,B) = RT \big(c \log(c) + Bc + (1-c) \log(1-c)\big) + \big(A_0(T) + A_1(T) (1-2c)\big) \, c(1-c). \end{equation} This additional linear term $Bc$ does obviously not affect the conditions for the common tangent $$ \begin{gather} 0 =& f'(x,T) - f'(y,T)\\ 0 =& f(x,T) + f'(x,T) (y-x) - f(y,T). \end{gather} $$ Hence, the miscibility gap is not affected. However, the visual presentation of the common tangent seems to be more appealing. For the above Gibbs free energy at $T = 923.15 \text{K}$ I made the following graphs in MATLAB.

enter image description here


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