Symmetric regular-solution binodal ($\log (x_{\alpha }/(1-x_{\alpha }))+\chi (1-2x_{\alpha })=0(0<x_{\alpha }<1/2,\chi >2)$)

For a symmetric binary regular solution, define the dimensionless molar mixing Gibbs free energywhere all logarithms are natural, $x$ is one component's mole fraction and $\chi$ the dimensionless interaction parameter. The 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.^[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 requiresThis 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:For $\chi \le 2$, this is nonnegative throughout the interval, so the mixing free energy is convex. For $\chi >2$, let $f=g^{\prime }$. On $(0,1/2)$, $f^{\prime }$ 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, $\chi =3$ yieldsThe spinodal endpoints instead solve $g^{\prime \prime }=0$: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.