Solid–Liquid Eutectic Diagrams

Consider two substances A and B that are completely miscible as liquids but essentially immiscible as solids, each crystallizing in its own pure solid phase. Start with pure liquid A at high temperature. As you cool it, nothing happens until you reach the melting point of pure A, $T_A^*$, where solid A begins to appear. Now do the same experiment with a mixture. Adding B to liquid A lowers the freezing point of A, exactly as in colligative freezing-point depression. As more B is dissolved, the temperature at which solid A first appears keeps dropping. This traces out a curve, the A-liquidus, descending from $T_A^*$ as composition shifts toward more B. Symmetrically, starting from pure B and adding A traces the B-liquidus, descending from $T_B^*$ as composition shifts toward more A. These two liquidus curves must intersect somewhere. At that intersection, the liquid is simultaneously saturated with respect to both solid A and solid B. This is the eutectic point ($T_E$, $x_E$), the single lowest-melting composition, the invariant point where three phases, liquid, solid A, and solid B, coexist. Below the eutectic temperature, no liquid phase can exist in this simple system; everything is solid, present either as regions of pure crystalline A, pure crystalline B, or a fine intimate mixture of both called the eutectic solid, formed when liquid of eutectic composition solidifies simultaneously into both crystal types. The full phase diagram divides into regions: a single liquid region above both liquidus curves, two two-phase regions (liquid + solid A, and liquid + solid B) bounded by liquidus above and the eutectic horizontal (the solidus, here degenerate to a single line since the solids are pure) below, and below that horizontal line, a two-phase solid region everywhere except at pure A or pure B.

A second essential picture is the cooling curve: temperature versus time as a fixed-composition melt loses heat at constant rate. For a composition off the eutectic, the curve is smooth until it hits the liquidus, where a slope change (a "kink") appears as solid begins to form and releases latent heat. The slope stays shallower through the two-phase region, then the curve flattens into a plateau at $T_E$ while the remaining liquid freezes at constant temperature (a three-phase invariant), before dropping again once all liquid is gone. A melt of exactly the eutectic composition shows no kink at all, only a single plateau at $T_E$, behaving thermally like a pure substance even though it is a mixture.

where: $F$ = degrees of freedom, $C = 2$ components, $P$ = number of phases present; pressure is fixed so the "+2" phase rule loses one degree of freedom.

where: $x_A$ = mole fraction of A in the liquid on the A-liquidus, $\Delta_{\text{fus}}H_A$ = molar enthalpy of fusion of pure A, $T_A^*$ = melting point of pure A, $T$ = liquidus temperature.

where: lever rule giving moles of phase $\alpha$ at overall composition $x$, with the tie-line endpoints at $x_\alpha$ and $x_\beta$.

where: the eutectic point is defined as the unique intersection of the two liquidus curves computed from each component's freezing-point-depression law.

Example 1 — Foundational: locating a eutectic with ideal liquidus curves. Consider a simplified binary system of two organic solids, A and B, forming an ideal liquid solution with no solid solubility. Data: $T_A^* = 400\ \text{K}$, $\Delta_{\text{fus}}H_A = 18.0\ \text{kJ mol}^{-1}$; $T_B^* = 350\ \text{K}$, $\Delta_{\text{fus}}H_B = 12.0\ \text{kJ mol}^{-1}$. Find the eutectic temperature and composition. Step 1: write both liquidus equations.

Step 2: scan temperature to find where both give consistent, complementary mole fractions. Try $T = 320\ \text{K}$: A-liquidus:

B-liquidus:

The two values (0.259 vs 0.321) don't yet agree; at $T=320$ K the A-liquidus $x_A$ is below the B-liquidus $x_A$, meaning we've gone slightly too cold. Try $T = 324\ \text{K}$: A-liquidus: $\ln x_A = -2164.5(0.003086-0.002500) = -1.269$, so $x_A = 0.281$. B-liquidus: $\ln(1-x_A) = -1443.4(0.003086-0.002857) = -0.330$, so $1-x_A = 0.719$, $x_A = 0.281$. The two agree to three significant figures at $T \approx 324\ \text{K}$. Result: $T_E \approx 324\ \text{K}\ (51^{\circ}\text{C})$, $x_E(\text{A}) \approx 0.28$, $x_E(\text{B}) \approx 0.72$. Sanity check: $T_E$ lies below both $T_B^*$ (350 K) and $T_A^*$ (400 K), as required — the eutectic must be the lowest-melting point in the system. The eutectic composition sits closer to pure B, consistent with B's smaller $\Delta_{\text{fus}}H_B$, which makes B's freezing point depress less steeply, pulling the crossing point toward higher $x_B$.

Example 2 — Applied: cooling curve, phase amounts, and the lever rule. A 100 g melt of composition $x_B = 0.55$ (using the same system as Example 1, with eutectic at $x_E = 0.72$ in B, $T_E = 324\ \text{K}$) is cooled slowly from 380 K to 300 K. Suppose at $T = 340\ \text{K}$ the B-liquidus gives the liquid composition as $x_B^{\text{liq}} = 0.60$ (solid B is the phase precipitating, since overall composition $x_B=0.55$ is less rich in B than the liquid, meaning we're on the B-liquidus side — reread: overall $x_B = 0.55 < x_E = 0.72$ places this melt in the "A-rich" partition, so solid A crystallizes first as the melt cools below the A-liquidus). Correcting the setup for internal consistency: overall composition $x_B = 0.55$ lies between pure A ($x_B=0$) and the eutectic ($x_B = 0.72$), so this is on the A-liquidus side. As the melt cools, solid A begins to separate once $T$ drops to the A-liquidus temperature for $x_B = 0.55$ (i.e., $x_A = 0.45$). Step 1: find the temperature at which solid A first appears. Using the A-liquidus equation from Example 1 with $x_A = 0.45$:

So the cooling curve shows a kink at $T = 348.6\ \text{K}$; above this, pure liquid; below, liquid + solid A. Step 2: find phase compositions and amounts at $T = 330\ \text{K}$, using the lever rule. At 330 K, evaluate the A-liquidus to find the liquid composition:

The tie-line at 330 K runs from pure solid A ($x_B = 0$) to liquid at $x_B^{\text{liq}} = 0.683$. Overall composition is $x_B = 0.55$. By the lever rule, the mole fraction of material that is liquid is:

So 80.5% of the total moles remain liquid, and 19.5% have solidified as pure solid A. Step 3: convert to mass. Take molar masses $M_A = 120\ \text{g mol}^{-1}$, $M_B = 140\ \text{g mol}^{-1}$. Overall moles in 100 g at $x_B=0.55$: average molar mass $= 0.45(120)+0.55(140) = 54+77=131\ \text{g mol}^{-1}$, so $n_{\text{total}} = 100/131 = 0.763\ \text{mol}$. Moles solid A $= 0.195 \times 0.763 = 0.1489\ \text{mol}$, mass $= 0.1489\times120 = 17.9\ \text{g}$. Moles liquid $= 0.805\times0.763 = 0.614\ \text{mol}$, mass $= 100 - 17.9 = 82.1\ \text{g}$ (by mass balance). Sanity check: As cooling continues toward $T_E = 324\ \text{K}$, more solid A must form and the liquid composition slides down the A-liquidus toward $x_E$; at exactly $T_E$ the remaining liquid (whatever fraction is left) has composition $x_E = 0.72$ and then freezes isothermally into eutectic solid, producing the plateau on the cooling curve. The amount of eutectic solid finally formed can be found the same way, by the lever rule at $T_E^-$.

The term "eutectic," from the Greek eutektos meaning "easily melted," was coined in the 1880s by the British chemist Frederick Guthrie, who systematically studied freezing-point curves of salt mixtures. The theoretical grounding of the whole diagram, however, rests on the chemical potential and phase-rule framework developed by Josiah Willard Gibbs a decade earlier, later popularized and applied to metallurgy by scientists such as Gustav Tammann, whose triangulation method for constructing binary phase diagrams from cooling-curve data became standard practice in twentieth-century metallurgy.

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