CHEM 548: Materials Chemistry

Phase Diagrams: Binary, Ternary, and Free Energy

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

Binary Phase Diagrams Part 2

Learning Objectives

By the end of this lecture, you should be able to:

Interpret binary phase diagrams that include an intermediate compound.
Distinguish between congruent and incongruent melting behaviors.
Identify eutectic and peritectic points and explain their significance.
Predict which phases are in equilibrium at given regions of a diagram.
Assess strategies for growing single crystals of intermediate compounds.

Instructor source visual for this part of the phase-diagrams lesson.

This phase diagram shows one intermediate compound forming between the two end members.

Functionally two eutectic phase diagrams side by side: first eutectic between A and AB₂, second between AB₂ and B (last lecture's ran A to B)
White regions = two-phase fields; labels say which phases coexist
AB₂ is a line phase: stoichiometry fixed at a 1:2 ratio of A to B
Its vertical line runs up to the liquidus; melts directly to a liquid of the same composition → congruently melting

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Example: the MnO–Al₂O₃ system — one intermediate compound, MnAl₂O₄, a 1:1 molar ratio of Al₂O₃ to MnO
On the diagram it does not appear exactly at 50% composition, but we can set that detail aside

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Some intermediate compounds melt incongruently: the liquid composition differs from the solid; the vertical line does not reach the liquidus
If AB₂ melts incongruently: at T₂ the solid begins to melt; above T₂, phase B plus a liquid richer in A than AB₂
More liquid forms until the liquidus is crossed → homogeneous liquid of overall AB₂ composition

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Peritectic point (P): like a eutectic, a three-phase equilibrium of two solids and a liquid — here AB₂ and B with a liquid richer in A than either solid
Produces a kink in the liquidus (not a minimum) — complicates crystal growth: for a congruently melting compound, we can simply cool the liquid of the correct composition slowly to grow single crystals; for AB₂, cooling at its exact composition instead yields crystals of B
Crossing point U: the equilibrium solid suddenly switches from B to AB₂; unless cooled extremely slowly, B cannot fully transform — the microstructure may not reach equilibrium
To grow AB₂ crystals: start between E and P, cool very slowly through the AB₂ + liquid region, then quench rapidly below the solidus

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Realistic diagrams: compounds and end members can form solid solutions — not every intermediate composition is a line phase; often some solid solution forms around those compositions

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Finish the section with a real phase diagram and interpretation questions — if you’ve got a good grasp on things, I think these questions should be pretty easy

Now that the binary construction is complete, we can use the real chloride system to test its interpretation.

So my first question for you is in this phase diagram between lead chloride and rubidium chloride, what is the solidest temperature?

Solidest temperature = the temperature below which there are no liquid phases present
The marked lines are the points below which a liquid phase is not present; the lowest is 396 degrees C → the solidest temperature

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Five ternary phases on this diagram; congruently melting = a vertical line extending fully to the liquidus
Only two meet the criterion: RbPb₂Cl₅ and RbPbCl₃; the other three melt incongruently

Having found the solidest temperature, we can identify the equilibria and growth consequences at the marked points.

At the circled peritectic point, three phases are in equilibrium.

Above the liquidus is liquid → liquid is one phase — always true at a peritectic (or eutectic)
Region of liquid + RbCl → RbCl is present; region of liquid + Rb₂PbCl₄ → that compound is also in equilibrium
The three coexisting phases: liquid, RbCl, and Rb₂PbCl₄

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Easiest crystals: congruently melting compounds — RbPb₂Cl₅ and RbPbCl₃
But RbPbCl₃ is stable only down to about 320 °C; below that it decomposes → straightforward growth impractical
Routine growth: only RbPb₂Cl₅; growing RbPbCl₃ is not impossible — one would need to cool slowly at the 50–50 composition, then quench rapidly from high temperature to preserve the phase

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Converse question: which phases are the most difficult, practically impossible to grow?
Three ternary phases left: I think one of them it would be possible to grow crystals, and the other two would be very, very difficult
Reason: very difficult to grow any phase with no equilibrium with the liquid

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Solid + liquid regions highlighted in green: Rb₆Pb₅Cl₁₆ and Rb₃PbCl₅ never enter a liquid + solid field
No melt growth → would need a solid–solid transformation, generally ineffective for quality crystals

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Slow cooling RbPbCl₃ from the melt: stable only ~320–440 °C; below 320 °C it disappears from the diagram
Cooling through that range enters a two-phase region; the lever rule gives mainly Rb₆Pb₅Cl₁₆ with some RbPb₂Cl₅
A slow-cooled 1:1 RbCl + PbCl₂ mixture ends as that two-phase mixture, not pure RbPbCl₃; XRD at room temperature would confirm decomposition
Variable-temperature diffraction could reveal the stability window, and quenching might preserve it; but in practice, even rapid cooling from ~350 °C allows enough diffusion for RbPbCl₃ to decompose into the two ternary phases

These binary examples finish the phase-diagram construction, so we can turn to the assigned practice.

Summary

Binary diagrams with intermediate compounds can be treated as two eutectic diagrams side-by-side.
Intermediate compounds may be line phases with fixed stoichiometry (e.g., AB₂).
Congruently melting compounds: melt to a liquid of the same composition.
Incongruently melting compounds: melt to a liquid of different composition + another solid → introduces the peritectic point.
Crystal growth:
- Easier for congruently melting compounds.
- Incongruent compounds require controlled growth between eutectic and peritectic compositions, often followed by quenching.
Real systems often include solid solutions as well as intermediate line phases.

Homework

4.2 Refer to the phase diagram depicted below. (a) State which four phases are stable at 100 °C. (b) What is the name given to the horizontal line separating region 2 from 1 and 3? (c) What are the approximate melting points of A, AB, and B? (d) What happens if you try and melt

4.3 Using the phase diagram of Figure 4.8: (a) State how you would attempt to prepare a solid polycrystalline sample of ZrW₂O₈. (b) State how you would attempt to grow single crystals of ZrW₂O₈.

4.4 In the system Al₂O₃–BaO, five phases stable above 1300 °C were identified: Al₂O₃, Al₁₂BaO₁₉, Al₂BaO₄, Al₂Ba₃O₆, and BaO. Each was found to melt congruently at 2072 °C, 1900 °C, 1811 °C, 1616 °C, and 1918 °C, respectively. Eutectics form at x_BaO = 0.11, 0.32,

4.5 Perovskite chemists searching in the CaO–TiO₂ system initially found four phases stable above 1300 °C: CaO, Ca₃Ti₂O₇, CaTiO₃, and TiO₂. CaO, CaTiO₃, and TiO₂ were reported to melt congruently at 2600 °C, 1970 °C, and 1830 °C and Ca₃Ti₂O₇ to melt incongruently at 1750 °C. Eutectics were reported at x_TiO₂ = 0.29 and 0.76 with melting points of 1695 °C and 1460 °C. Sketch and fully label a phase diagram for this system.

Ternary Phase Diagrams

Learning Objectives

By the end of this lecture, you should be able to:

Explain the difference between binary and ternary phase diagrams.
Describe how ternary phase diagrams are represented and why simplifications (isothermal cuts) are used.
Locate a composition within a ternary diagram using the line/triangle method.
Identify which phases are present at a given point, line, or region of a ternary diagram.
Apply the triangle (lever) rule conceptually to determine phase fractions.
Interpret real ternary phase diagrams (e.g., TiO₂–ZrO₂–Al₂O₃, Y₂O₃–BaO–CuO systems).

A complete ternary phase diagram requires three axes for composition, one for temperature, and one for pressure—a five-dimensional plot that is impractical to visualize.

If we fix the pressure, we can reduce it to a three-dimensional diagram: equilateral triangular base for the three compositions, temperature as the vertical axis
Each vertical face of the prism = a binary phase diagram (eutectic diagrams on the front, left, and back faces)
Such diagrams can capture complex behavior, including intermediate phase formation, but quickly become difficult to interpret
Most applications, including this class: isothermal sections — constant temperature slices

With the triangular representation defined, we can learn how compositions are plotted on it.

We will take a constant temperature to obtain a 2D slice of the phase diagram, focusing on isothermal cuts below the solidus so no liquid phases appear.

Example: the ternary diagram of TiO₂–CaO–Cu₂O as an equilateral triangle; today's goal is to read and interpret such diagrams

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Locating a composition of A, B, and C: in a binary, drop a vertical line to the x-axis for the A:B ratio; in a ternary we work within a triangle
Several approaches in the textbook; I describe one of those, which I think is the simplest way

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Start at the upper vertex, where only A is present: 100% A, no B or C

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Take the edge opposite that vertex, at the bottom, and draw horizontal lines parallel to the opposite edge

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Each dashed red line = constant A; along the base from C to B, A = 0 — the binary C–B diagram
Lines above mark 20%, then 40%, 60%, 80% A; the right scale confirms (0.6 = 60%, 0.8 = 80%); the left scale measures C in the A–C binary, so 20% C = 80% A

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We could do the same for B: draw lines parallel to the AC side — the lines of constant B

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Analogous for C: lines of constant C parallel to the side representing the mixture of A and B

Once composition lines are in place, we can read a point and identify its equilibrium phases.

Asking what is the composition of point d

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Draw lines parallel to each edge: A from a line parallel to BC, about 60%; B parallel to AC, about 30%; C parallel to AB, about 10%
They add to 100% → 60% A, 30% B, 10% C

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Next: which phases are in equilibrium at a located point — three cases
First, the point may fall directly on a phase: at AC₂ — just like at pure A, B, or C — the only phase present is that compound

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If the composition lies on a line, the system is a two-phase mixture — between AC₂ and B the phases are AC₂ and B; bottom edge B + C; right edge A + B
Left edge depends on position: above the junction A + AC₂, below it AC₂ + C

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The rest of the phase diagram is divided into triangular regions: each region = equilibrium between the three phases at its vertices
Green triangle (B, C, AC₂) and blue triangle (A, B, AC₂) — mixtures of those three phases

The same triangular logic lets us determine phase fractions in a real ternary diagram.

Now consider a simple real ternary diagram: TiO₂–ZrO₂–Al₂O₃.

Two binary compounds — ZrTiO₄ and Al₂TiO₅ — and no ternary compounds involving all three components
Points A–F are marked: identify the phase(s) present at each point — write down the phases

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A: on the ZrO₂–Al₂O₃ binary, no intermediates → two-phase mixture, ~60% ZrO₂ / 40% Al₂O₃
B: midway on TiO₂–ZrO₂ → the binary compound ZrTiO₄
C: inside triangle {ZrO₂, ZrTiO₄, Al₂TiO₅}; D: inside triangle {ZrO₂, Al₂O₃, Al₂TiO₅} — three-phase equilibria of those vertices
E, F: both inside triangle {TiO₂, Al₂TiO₅, ZrTiO₄} — same three phases at equilibrium; only phase proportions differ

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Phase ratios at E: the triangle rule, analogous to the lever rule — a line from each vertex through E to the opposite side
For each component the fraction = opposite segment / total line length; repeat for all three vertices

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Suppose 50 mol% ZrO₂, 30 mol% TiO₂, 20 mol% Al₂O₃: mix, grind, heat to the isothermal condition (say 1000 °C), measure the equilibrium phases
Where does the 50/30/20 point plot, and which phases are present there?

Having placed the composition, we can check it directly on the diagram.

Now let’s examine a more complex ternary diagram: Y₂O₃–BaO–CuO.

Historically significant: high-temperature superconductors were discovered from systems like this
Two binary compounds between Y₂O₃ and BaO, two between BaO and CuO, one between Y₂O₃ and CuO; plus four ternary compounds, shown by abbreviations
The diagram breaks into triangular regions; a non-triangular region means the diagram is drawn incorrectly — here every region is triangular

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Question: for Y₂Ba₂Cu₄O₉, is the equilibrated sample single-phase, two-phase, or three-phase, and which phases?
Key step: convert the formula to oxide components — 1 Y₂O₃ + 2 BaO + 4 CuO (O totals check: 3+2+4 = 9)
Mole fractions: Y₂O₃ = 1/7 ≈ 14.3%, BaO = 2/7 ≈ 28.6%, CuO = 4/7 ≈ 57.1%; plot (14.3, 28.6, 57.1) and read the triangular region

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50% copper oxide: the horizontal red line; 25% yttrium oxide: the line parallel to the barium oxide, copper oxide, face; 25% barium oxide: along the other red line
All three intersect at the point drawn with the white circle

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That point sits in the triangle shaded green

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That point sits in the triangle shaded green

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The three-phase region has vertices CuO (apex), YBa₂Cu₃O₇ (“123”), and Y₂BaCuO₅ (“211”)
Start with Y₂Ba₂Cu₄O₉, grind, heat, quench: XRD reveals a mixture of those three → the target is not a stable single phase — it decomposes into CuO + 123 + 211 at the stated conditions

This complex diagram leads naturally to the associated practice questions.

Summary: Ternary Phase Diagrams

Phase diagrams represent thermodynamic stability of phases as a function of composition, T, and P.
A full ternary system requires 5D (3 compositions + T + P) → simplified by fixing pressure, often slicing at constant T.
2D triangular diagrams represent isothermal sections below the liquidus.
Each point in the triangle = a unique composition (relative fractions of A–B–C).
Tie-lines connect coexisting phases; tie-triangles define three-phase equilibria.
Lever rule applies inside two-phase and three-phase regions.

$F(x,T)=\Delta U(x)-T S_{mix}(x)$

At high temperature entropy dominates and the free-energy curve is smooth with one minimum so the mixture is completely miscible, but at lower temperature if unlike bonds cost energy (ε > 0) the curve develops two dips near the pure ends with a barrier in the middle so the alloy separates into two phases, and the special point where the single dip just splits into two is called the consolute temperature given by T_c = (p × ε) / (2 × k_B), for example with p = 12 and ε = 0.010 eV (10 meV) the consolute temperature is about 696 K while with ε = 0.015 eV (15 meV) it is about 1040 K, which means that the larger the energy penalty the higher the consolute temperature because more thermal disorder is needed to keep the system mixed

High T: entropy dominates — smooth curve, one minimum, completely miscible
Lower T with ε > 0: two dips near the pure ends, barrier in the middle → the alloy separates into two phases
Single dip just splits into two: the consolute temperature, T_c = (p × ε) / (2 × k_B)
p = 12: ε = 0.010 eV → about 696 K; ε = 0.015 eV → about 1040 K — larger penalty, higher T_c

Once the free-energy curve is known, a common tangent gives the coexisting compositions.

Method of tangents lets us find the compositions of the two phases that can coexist at a given temperature by drawing a straight line that just touches the free-energy curve at two points, and these tangent points give the equilibrium compositions directly, so in this case the line touches at x = 0.132 (13.2%) for the alpha phase and at x = 0.859 (85.9%) for the beta phase, meaning if the overall composition lies anywhere between these values the system will split into a mixture of alpha and beta with those compositions.

Method of tangents: a straight line just touching the free-energy curve at two points
The tangent points are the equilibrium compositions: x = 0.132 (13.2%) alpha, x = 0.859 (85.9%) beta
Overall composition between these values → the system splits into alpha + beta at those compositions

Repeating that tangent construction at different temperatures builds the phase diagram.

By applying the method of tangents to the free-energy curves at different temperatures we can trace out the pairs of compositions that coexist, and when we project those tangent points down onto the temperature–composition plane we generate the phase diagram, so at high temperature the curve has only one minimum and the system is fully mixed, while at lower temperatures two minima appear and the tangent gives two coexisting compositions, and by repeating this at many temperatures we build the binodal curve that outlines the miscibility gap with the consolute temperature at its peak.

Tangent points at each temperature, projected down onto the temperature–composition plane → the phase diagram
High T: one minimum, fully mixed; lower T: two minima, two coexisting compositions
Repeating at many temperatures builds the binodal curve — the miscibility gap, consolute temperature at its peak

The same method can be applied to a solid-liquid system such as silicon-germanium.

In the Si–Ge system there is no heat of mixing, so the solid free energy is F_s(x,T) = RT[x ln(x) + (1–x) ln(1–x)] and the liquid free energy is F_l(x,T) = RT[x ln(x) + (1–x) ln(1–x)] + ΔS[x(T_B – T) + (1–x)(T_A – T)], where ΔS is the entropy of fusion per mole and T_A and T_B are the melting points of pure Si and pure Ge, so the entropy of mixing is the same in both phases but the liquid curve is shifted by how far the temperature is below each melting point, which means at low temperature the solid has lower free energy and is stable, at high temperature the liquid has lower free energy and is stable, and in between the curves cross so a common tangent defines the liquidus and solidus, giving the smooth phase diagram with complete miscibility that we see for Si–Ge.

No heat of mixing: solid F_s(x,T) = RT[x ln(x) + (1–x) ln(1–x)]
Liquid adds ΔS[x(T_B – T) + (1–x)(T_A – T)] — ΔS = entropy of fusion per mole; T_A, T_B = melting points of pure Si and pure Ge
Same mixing entropy in both phases; the liquid curve is shifted by how far T sits below each melting point
Low T: solid stable; high T: liquid stable; between, the curves cross and a common tangent defines the liquidus and solidus
Result: the smooth diagram with complete miscibility we see for Si–Ge

Changing the heat of mixing shows how a congruent melting point can emerge.

From high T (1500 K, single-phase liquid L), through 1100–1000 K where the liquid free energy rises and mutual tangents to the solid minima define α+L and β+L fields with tie-lines giving equilibrium compositions (lever rule), down to the eutectic at ~940 K where the liquid minimum just touches the common tangent between α and β (an invariant three-phase point, L ⇌ α+β, depressed below the pure melting points by positive solid heat of mixing), and finally at lower T (e.g. 700 K) the liquid lies above the α–β tangent leaving only two solids (α+β), with phase compositions from tangent points and fractions from the lever rule—illustrating how positive enthalpy of mixing produces the V-shaped liquidus, eutectic reaction, and α+β solid region [oai_citation:0‡Nauman_ch12_phase_equilibria.pdf](file-service://file-Q1ewSHDHZyMzxiqeDyNy2m).

1500 K: single-phase liquid L
1100–1000 K: liquid free energy rises; mutual tangents to the solid minima define α+L and β+L fields, tie-lines give equilibrium compositions (lever rule)
Eutectic at ~940 K: the liquid minimum just touches the common tangent between α and β — an invariant three-phase point, L ⇌ α+β, depressed below the pure melting points by positive solid heat of mixing
Lower T (e.g. 700 K): the liquid lies above the α–β tangent → only two solids, α+β; compositions from tangent points, fractions from the lever rule
Positive enthalpy of mixing → the V-shaped liquidus, eutectic reaction, and α+β solid region

The next example shows how a rounded free-energy feature produces an intermediate phase.

What is a “line compound,” and why is it a vertical line?Some compounds only form at one exact composition.

Cd–Te: that's CdTe (50 at% Te) — any off-stoichiometric alloy splits into Cd + CdTe (left) or Te + CdTe (right)
Why vertical: the solid-CdTe free-energy curve has a sharp, wedge-like minimum at stoichiometry, so the common tangent from the liquid touches it at a single composition
Mapped to the T–x diagram, that single tangent point projects to a vertical CdTe line — no composition range
A rounded minimum would give tangents at two compositions → a finite γ + L two-phase band, a solid-solution instead of a line compound
Right panel, real Cd–Te data: congruent melting of CdTe at ≈ 1098 °C, far above Cd ≈ 321 °C and Te ≈ 450 °C, with the vertical α-CdTe field; left sketch = the generic A–γ–B topology

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Homework Questions

Practice questions

Work the Lecture 10 practice questions before the next class. They cover binary diagrams with intermediate compounds, congruent and incongruent melting, reading a ternary triangle, the triangle (lever) rule, and building a phase diagram from free-energy curves and common tangents.

Open the Lecture 10 practice questions

Every question carries a worked explanation, so you can check your reasoning as you go.