= Symmetric regular-solution binodal
{title2=$\log(x_\alpha/(1-x_\alpha))+\chi(1-2x_\alpha)=0\quad(0<x_\alpha<1/2,\ \chi>2)$}
For a symmetric binary <regular solution>, define the dimensionless molar mixing <Gibbs free energy>
$$
g(x)=x\log x+(1-x)\log(1-x)+\chi x(1-x),\qquad0<x<1,
$$
where all logarithms are natural, $x$ is one component's <mole fraction> and $\chi$ the dimensionless interaction parameter. The \b[symmetric regular-solution binodal] gives the compositions of two coexisting phases, rather than the overall composition of their mixture. The <common-tangent construction> supplies the equilibrium condition.<Unstable Solutions: MIT 3.00 Lecture 32 by W. C. Carter>[Printed pp218–221, regular solution and symmetric common tangent]{ref}
Symmetry gives $g(x)=g(1-x)$. For distinct coexistence endpoints $x_\alpha<1/2$ and $x_\beta=1-x_\alpha$, their joining line is horizontal. Tangency at both ends therefore requires
$$
g'(x_\alpha)=\log\frac{x_\alpha}{1-x_\alpha}+\chi(1-2x_\alpha)=0.
$$
This derives the displayed binodal equation. The ever-present stationary solution $x=1/2$ is not a distinct two-phase pair.
To see when the nontrivial pair exists, differentiate:
$$
g''(x)=\frac1{x(1-x)}-2\chi.
$$
For $\chi\leq2$, this is nonnegative throughout the interval, so the mixing free energy is convex. For $\chi>2$, let $f=g'$. On $(0,1/2)$, $f'$ decreases from infinity to $4-2\chi<0$. Consequently $f$ first increases and then decreases. It starts at negative infinity, is positive just below1/2, and ends at0. There is exactly one zero in the open interval before its maximum. That zero has positive curvature and gives the lower coexistence composition; symmetry gives the other.
For an original numerical example, $\chi=3$ yields
$$
x_\alpha\simeq0.0707202,\qquad x_\beta\simeq0.9292798.
$$
The <spinodal> endpoints instead solve $g''=0$:
$$
x_{s,\pm}=\frac{1\pm\sqrt{1-2/\chi}}2,
$$
which give approximately 0.211325 and 0.788675 at $\chi=3$. The binodal marks equilibrium phase compositions; the spinodal marks loss of local curvature stability. Their intervening metastable ranges are therefore not interchangeable boundaries. At $\chi=2$ the endpoints meet at the symmetric critical composition 1/2.
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