Mon 09/28/2026 | Meeting 10
Ternary phase diagrams, the ternary lever rule, and phase-diagram construction from free energy
Reading: Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry
Ch. 4 §§4.2–4.3
Learning objectives — by the end of this lecture, you should be able to:
A binary system with a congruently melting binary compound: AB₂ melts to a liquid of its own composition, so the diagram reads as two eutectic diagrams side by side.
A real system of the same shape: the spinel MnAl₂O₄, a 1:1 ratio of Al₂O₃ to MnO, melts congruently between two eutectics.
MIT OCW 3.012, Fall 2005, lec19t.
The AB₂ line stops short of the liquidus, so at T₂ the solid melts to B plus a liquid richer in A than AB₂ itself.
At the peritectic point p, AB₂, B and a liquid richer in A than either solid coexist, which kinks the liquidus instead of dipping it to a minimum.
Compound and end members alike take up solid solution, so the vertical line phases widen into fields.
Five intermediate compounds and a eutectic at 396 °C, with the interpretation questions on the slides that follow.
An intermediate compound splits a binary diagram into two eutectic diagrams side by side, and how that compound melts decides everything that follows.
| Feature | On the diagram | Consequence for growth |
|---|---|---|
| Line phase (AB₂) | A vertical line at fixed stoichiometry | No composition range to work with |
| Congruent melting | That line runs all the way up to the liquidus | Cool a melt of the same composition slowly |
| Incongruent melting | The line stops short; a peritectic point appears | Cool from between the eutectic and the peritectic, then quench |
| Solid solution | A field of finite width instead of a line | Real systems show one around many compositions |
Melt to a liquid of the same composition and growth is easy; melt to a different liquid plus another solid and it is not.
Learning objectives — by the end of this lecture, you should be able to:
For a phase diagram with three components — a ternary phase diagram — we need four axes to plot the diagram at constant pressure.
An equilateral triangle for the base, and a vertical axis for temperature.
Each vertical face of the prism is one of the three binary diagrams; the draped surface across the top is the liquidus.
Ternary phase diagrams at constant pressure.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.9.
A 2D cut from the five-dimensional phase diagram at constant temperature and pressure, taken below the solidus so no liquid phase appears.
Each family of dashed lines is parallel to the edge opposite the component it measures; the three fractions read off them sum to 100%.
A(0.6)B(0.3)C(0.1)
A, B and AC₂ in equilibrium
B, C and AC₂ in equilibrium
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.10c.
What phases are present at each point?
a — ZrO₂ (60%) + Al₂O₃ (40%)
b — ZrTiO₄ alone
c — ZrO₂ + ZrTiO₄ + Al₂TiO₅
d — ZrO₂ + Al₂O₃ + Al₂TiO₅
e — TiO₂ + Al₂TiO₅ + ZrTiO₄
f — TiO₂ + Al₂TiO₅ + ZrTiO₄
Which phase(s) would be present if the overall composition was 50 mol% ZrO₂, 30 mol% TiO₂ and 20 mol% Al₂O₃?
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.11a.
Which phase(s) would be present if the overall composition was 50 mol% ZrO₂, 30 mol% TiO₂ and 20 mol% Al₂O₃?
P — 50% ZrO₂: halfway up, parallel to the TiO₂–Al₂O₃ base.
Q — 30% TiO₂: parallel to the ZrO₂–Al₂O₃ edge opposite TiO₂.
R — 20% Al₂O₃: parallel to the TiO₂–ZrO₂ edge opposite Al₂O₃.
All three lines meet at one point, which is the 50/30/20 composition.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.11a.
The 50/30/20 point falls inside the sub-triangle whose vertices are ZrTiO₄, ZrO₂ and Al₂TiO₅.
Three phases in equilibrium: ZrTiO₄, ZrO₂ and Al₂TiO₅.
Every region of a correctly drawn sub-solidus section is a triangle, and the three phases present are the three compounds at its corners.
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.11a.
123 = YBa2Cu3O7
211 = Y2BaCuO5
143 = YBa4Cu3O8.5
163 = YBa6Cu3O10.5
Which phase(s) would be present if the overall composition was Y₂Ba₂Cu₄O₉?
Woodward, Karen, Evans, and Vogt, Solid State Materials Chemistry, §4.3, Fig. 4.11b.
Phase diagrams represent the thermodynamic stability of phases as a function of composition, T, and P.
A full ternary system needs five dimensions — three compositions, T and P — so we fix the pressure and slice at one temperature, and what is left is a triangle of compositions below the liquidus.
| Where the composition lands | Phases | How to read it |
|---|---|---|
| On a vertex or a compound | 1 | That phase alone |
| On a tie-line | 2 | The two phases the line joins; lever rule along the line |
| Inside a tie-triangle | 3 | The three phases at its vertices; triangle (lever) rule |
Every point in the triangle is one composition, so the phase count follows from the geometry alone.
$F = U - TS$
Matter seeks to minimize the Helmholtz free energy at constant temperature and volume.
$$S_{\text{mix}} = -Nk\left[\ln(1-x) + x\ln\!\left(\frac{x}{1-x}\right)\right]$$
$N = N_A + N_B$ is the total number of atoms, $x$ is the mole fraction of B, and $k$ is Boltzmann's constant.
Entropy term is positive because there are more ways of rearranging 2 atoms in space than 1.
Maximum at $x = 0.5$, where $S_{\text{mix}}/Nk = \ln 2 \approx 0.693$; it falls to zero at both pure ends.
$$U_{\text{tot}} = N\frac{p}{2}\left[(1-x)\,u_{AA} + x\,u_{BB} + 2x(1-x)\,\varepsilon\right]$$
Internal energy term is from bonding, which can be energetically favorable or unfavorable.
$p$ is the number of nearest neighbors around each atom, so the $N$ atoms share $Np/2$ bonds; $u_{AA}$ and $u_{BB}$ are the A–A and B–B bond energies, and the interaction parameter is ${\varepsilon = u_{AB} - \tfrac{1}{2}(u_{AA} + u_{BB})}$.
$\varepsilon < 0$ bows the curve below the straight mixing line; $\varepsilon > 0$ bows it above.
$$F(x,T) = \Delta U(x) - T\,S_{\text{mix}}(x)$$
At high $T$ the entropy term dominates and mixing is favorable.
At low $T$ local minima appear, leading to phase segregation. They first split apart at $T_c = p\varepsilon/2k$.
$p$ = number of nearest neighbors per atom
$\varepsilon$ = energy penalty per unlike bond
$k$ = Boltzmann constant
$$\begin{aligned} F(x,T) &= \frac{Np}{2}\left[(1-x)u_{AA} + x\,u_{BB} + 2x(1-x)\varepsilon\right]\\ &\quad + NkT\left[(1-x)\ln(1-x) + x\ln x\right] \end{aligned}$$
Method of tangents can be used to find the compositions of each phase that are in equilibrium.
The atom count is fixed, so any split must average back to the overall composition $x_0$.
A two-phase mixture's free energy lies on the line joining the two phases, read at $x_0$.
The lowest such line is the common tangent: it touches the curve at $C_1 = 0.132$ and $C_2 = 0.859$, 13.2 % for alpha and 85.9 % for beta.
One shared line: $\mu_A$ and $\mu_B$ are the same in both phases.
Applying the method of tangents to free energy curves at different temperatures can be used to create a phase diagram.
Drop each pair of tangent points onto the temperature–composition plane and the pairs trace the miscibility dome, from $0.6\,T_c$ up to $T_c$ at its peak.
Above $T_c$ there is one minimum and one phase; below it, two minima and two coexisting compositions.
$$F_S(x,T) = RT\left(x\ln x + (1-x)\ln(1-x)\right)$$
$$\begin{aligned} F_L(x,T) &= RT\left(x\ln x + (1-x)\ln(1-x)\right)\\ &\quad + \Delta S\left[x(T_B - T) + (1-x)(T_A - T)\right] \end{aligned}$$
Nauman, Figure 12.7.
Example of system with heat of mixing of −30 kJ/mol.
Unlike neighbours are preferred, so the solid free-energy curve bows well below the straight mixing line and the alloy melts above either pure component: a continuous solid solution with a congruent maximum.
Positive heat of mixing means the melting point of the alloy is not higher than the melting point of the pure phases.
Example of system with positive heat of mixing.
The point where the liquid line minimum meets the tangents of the two solid phase minima leads to a lower melting point eutectic. The inset picks out $C_1$ and $C_2$ either side of the eutectic composition.
The free-energy curve of γ has a rounded nose, so the tangents from either side meet it at two compositions and γ occupies a band rather than a line.
In Al–Ni that phase is AlNi, stable from about 45 to 58 at% Ni and melting above both pure metals, while Al₃Ni and Al₃Ni₅ stay narrow.
A line compound exists at one exact stoichiometry only, so it is drawn as a vertical line and its free-energy well is a single point.
In Cd–Te that compound is αCdTe at 50 at% Te, melting congruently at 1098 °C with a eutectic on each side.
Work the Lecture 10 practice questions before the next class.