This Demonstration illustrates the behavior of a system of two pure substances (A and B) and a solid compound of the two (A₂B₃). Solid–solid equilibrium, solid–liquid equilibrium, and solid–solid–liquid equilibrium are represented in the phase diagram. Mixtures of the pure solids (A + A₂B₃ and B + A₂B₃) are immiscible.

At the selected temperature and mole fraction of B (represented by the black point on the T–mole fraction diagram), the relative amounts of the four possible phases are shown in the bar graph. The mole fraction of B in the liquid phase (mixture of A and B) is given above the liquid bar in the bar graph. Move the black point by adjusting the sliders for the mole fraction B and heat added.

When heat is added, the temperature increases, except when the point is on one of the two horizontal lines (at about 200 °C and 300 °C) or when pure A₂B₃ is in equilibrium with the liquid phase. On the lines at 200 °C and 300 °C, three phases can be in equilibrium: solid A₂B₃, liquid, and either solid A or B. A mole balance is used to find the relative amounts of each phase on these lines. The amount of heat added represents what phases are present in this system; it is not meant to represent a real system.

In the two-phase regions, the relative amounts of each phase are obtained using the lever rule, and the mole fraction of B in the liquid phase is shown by a vertical dotted line.

The lever rule is used to find the relative molar contents of each phase. An example in the solid A + liquid region is given by:

r_liquid = (z_B − 0) / (x_B − 0) = z_B / x_B ,

r_solid of A = (z_B − x_B) / (0 − x_B) = 1 − (z_B − 0)/(x_B − 0) = 1 − r_liquid ,

where z_B is the overall mole fraction of the mixture (the mole fraction of the point on the T–x–y diagram), x_B is the mole fraction of B in the liquid phase, the mole fraction of B in the solid A phase is zero, r_liquid is the relative amount of liquid, and r_solid is the relative amount of solid A.

When the system is in solid–solid–liquid equilibrium, the relative amounts of each phase are found from mass balances. For example, using a value of 10% melted and an initial mole fraction of B of 0.7 in the diagram:

1. Determine the initial mole fraction of B: z_B = 0.7.
2. Determine the mole fraction of component B in each phase: S_{solid, A2B3} = 0.6, S_{solid, B} = 1, x_B = 0.8.
3. Set a basis for the total moles in the system: S_total = 1.
4. Determine the percentage melted: μ = 0.1.
5. Determine the amount of liquid in the system: L = μ S_total = 0.1(1) = 0.1.
6. Mole balance on the whole system: 1 = S_{solid, A2B3} + S_{solid, B} + 0.1.
7. Mole balance on component B: 0.7(1) = 0.6 S_A2B3 + 1 S_B + 0.8(0.1), so 0.7 = 0.6 S_A2B3 + S_B + 0.08.
8. Simultaneously solve the two balance equations (number of equations equals number of unknowns).
9. Relative amounts of each phase: S_{solid, A2B3} = 0.7 (solid A₂B₃), S_B = 0.2 (solid B), V = 0.1 (liquid).

This simulation was created in the Department of Chemical and Biological Engineering at University of Colorado Boulder for LearnChemE.com by John L. Falconer using Claude AI. It is a JavaScript/HTML5 implementation of a Mathematica simulation by Megan E. Maguire, Neil Hendren, Rachael L. Baumann, and John L. Falconer. It was prepared with financial support from the National Science Foundation (DUE 2336987 and 2336988) in collaboration with Washington State University. Address any questions or comments to LearnChemE@gmail.com.