Binary Solid–Liquid Equilibrium

Introduction

Cooling a mixture of two substances can produce a pure or nearly pure solid of one component before the remaining liquid reaches a special eutectic composition. At the eutectic temperature, liquid transforms into two solid phases together. A temperature–composition diagram shows when each phase appears and how much is present. Reading it requires separating phase composition from overall alloy composition and separating an equilibrium phase from a microstructural constituent.

Core explanation

In a simple eutectic binary system at fixed pressure, two liquidus branches descend from the pure-component melting points and meet at the eutectic point (x E,T E). Above the liquidus, one liquid is stable. On the A-rich side between liquidus and eutectic temperature, liquid coexists with A-rich solid α; on the B-rich side, liquid coexists with B-rich solid β. At the eutectic point, the reaction on cooling is L(x E) → α + β. Below T E, a suitable intermediate overall composition contains the two solids.

The eutectic point is invariant at fixed pressure in the simple binary system. The general phase rule gives F = C − P + 2 = 2 − 3 + 2 = 1 when L, α and β coexist; fixing p removes that remaining freedom, leaving zero adjustable T or composition of the coexisting phases at the eutectic. The overall sample composition can differ from x E, but the liquid just before the eutectic reaction approaches x E along its liquidus branch.

Inside any two-phase region, an isothermal tie line connects the equilibrium compositions of the two phases. If α has component-B fraction x α, liquid has x L and the overall fraction is z B between them, material balance gives f L = (z B − x α)/(x L − x α) and f α = 1 − f L on a mole basis. The same idea applies to the α+β region with solid endpoints. If a diagram's horizontal axis is mass fraction, the lever result gives mass fractions, not mole fractions; basis matters.

Cooling history affects microstructure. For an A-rich overall alloy, primary α can crystallise above T E, then the remaining liquid freezes as a fine α+β eutectic mixture. “Eutectic microconstituent” includes both α and β phases, so one should not count it as a third equilibrium phase below T E. Slow equilibrium cooling and rapid quenching may produce different observed textures even when the diagram describes equilibrium boundaries.

Step-by-step reasoning

Check whether pressure is fixed and whether horizontal composition is mole or mass fraction. Mark the overall z and follow a vertical cooling path. At each T, identify its phase region; within two-phase regions draw a horizontal tie line to read endpoint compositions. Apply material balance for phase fractions, then describe what happens as liquid reaches the eutectic point.

Visual explanation

Draw temperature vertically and fraction B horizontally. Two descending liquidus curves meet in a V at the eutectic. Label L above, L+α left, L+β right and α+β below. Draw an A-rich cooling line that first crosses the left liquidus, then reaches T E. A horizontal tie line in L+α shows endpoint compositions and opposite lever arms.

Real-world analogy

As a mixed liquid cools, one ingredient may leave first, changing the composition of what remains. Eventually the residual mixture reaches a special ratio where two solids form together. The analogy resembles selective removal during separation, though phase equilibrium is fixed by chemical potentials and not a conscious sorting process.

Real-world example

Solder alloys are chosen partly for their melting behaviour. Near a eutectic composition, an alloy can melt or freeze over a narrow temperature interval under equilibrium conditions, useful for joining components. Actual solder formulations may have additional components or nonideal solid solubility, so a simple binary diagram is a model rather than a universal specification.

Why?

The liquid's chemical potentials must match those of the coexisting solid at each point. As one solid crystallises, material balance changes the liquid composition, moving it along a liquidus branch. At the eutectic, both solid phases can match the same liquid simultaneously at one fixed T for a fixed p, producing an invariant three-phase transformation.

Common misconception

The eutectic point is not necessarily the composition of every solid formed from an off-eutectic sample. Primary solid may form earlier and has its own composition. Another error is to call the fine eutectic mixture a single phase; it consists of two solid phases. A lever-rule result depends on whether the diagram uses moles or mass.

Worked example

At a temperature above T E in an A-rich L+α region, suppose solid α has x B = 0.10, liquid has x B = 0.50 and overall alloy has z B = 0.30 on a mole basis. Then f L = (0.30 − 0.10)/(0.50 − 0.10) = 0.50, and f α = 0.50. These are phase amounts at that temperature, not the final low-temperature α and β fractions after all remaining liquid solidifies.

Quick check

1. In a simple binary eutectic at fixed pressure, how many phases coexist at the eutectic point? Answer: Three: liquid L and two solids α and β. Their equilibrium compositions and temperature are fixed for that pressure.

Exam focus

Track the vertical cooling path and draw tie lines at the requested T, not automatically at T E. Read endpoint phase compositions before using the lever rule. State the composition basis. Distinguish primary solid, eutectic microconstituent and thermodynamic phase count.

Advanced insight

Limited solid solubility shifts α and β endpoint compositions away from pure A and B, while compound formation or peritectic reactions create different diagram topologies. The common-tangent construction on free-energy curves supplies the underlying equilibrium condition for each tie line.

Summary

A simple eutectic diagram has liquidus branches leading to an invariant L → α + β point at fixed pressure. Cooling off-eutectic mixtures produces primary solid before residual liquid undergoes the eutectic reaction. Tie-line endpoints give phase compositions, and the lever rule gives amounts on the diagram's chosen composition basis.

Practice questions

1. In L+α, x α = 0.05, x L = 0.45 and overall z = 0.25. Find liquid fraction. Answer: f L = (0.25 − 0.05)/(0.45 − 0.05) = 0.50 on the same mole or mass basis as the composition axis. 2. At fixed pressure, why is the binary eutectic temperature invariant while L, α and β coexist? Answer: C = 2 and P = 3 give F = 1 generally; fixing p uses that freedom, leaving no independently variable T or phase composition during three-phase equilibrium. 3. Is a eutectic microstructure composed of one solid phase because it formed at one temperature? Answer: No. The eutectic microconstituent contains two distinct solids α and β, even though they formed together at the eutectic temperature.