Ternary Phase Diagrams and Tie Lines

Introduction

A binary composition needs one independent fraction; a ternary composition needs two. An equilateral triangle provides a compact map because every point can represent three fractions that add to one. Ternary phase diagrams are used to interpret solvent extraction, liquid–liquid separation and multicomponent materials. Their geometry is simple once vertex, edge and tie-line meanings are kept distinct.

Core explanation

Let x A+x B+x C = 1. Put pure A, B and C at the three vertices. An edge contains only its two endpoint components, with the opposite component at zero. An interior point contains all three. Lines parallel to the side opposite A have constant x A; analogous parallel lines read x B and x C. A point's three coordinate readings must sum to one, a useful check against using the wrong axis direction.

At fixed T and p, a ternary liquid system may have a one-phase region and a two-phase miscibility gap bounded by a binodal. Inside the two-phase region, an overall composition z splits into phases α and β with compositions x^α and x^β at the endpoints of a tie line passing through z. The endpoints satisfy equality of each component's chemical potential across phases. Tie lines are not necessarily parallel and can rotate across the diagram; they must be read from data or a model, not invented by drawing an arbitrary line to the binodal.

If f α and f β are phase mole fractions with f α+f β = 1, material balance is z i = f α x i^α + f β x i^β for i = A,B,C. The overall point lies on the straight segment between endpoint compositions. The lever rule applies along that tie line: the fraction of one phase is proportional to the distance from z to the opposite endpoint. Using mole fractions gives mole-based phase fractions; mass-fraction diagrams give mass-based balances.

A mixture path can be shown by connecting an initial composition with the pure component being added. For example, adding pure solvent C to an A/B mixture moves the overall point along a straight line toward vertex C before phase equilibrium is re-evaluated. As soon as the path enters or leaves a two-phase region, the number of phases changes. An actual extraction process also requires a distribution of solute between phases, which the tie-line endpoints encode.

Step-by-step reasoning

Read diagram conditions and component labels. Identify vertices, edges and parallel-coordinate directions, then verify x A+x B+x C = 1. Locate overall z relative to the binodal. If z is in a two-phase region, find the measured or supplied tie line through it and read endpoint compositions. Solve one component's balance for a phase fraction and check the other two balances for consistency.

Visual explanation

Draw an equilateral triangle with A, B, C vertices and a marked interior point z. Through z draw three coordinate-parallel guide lines showing fractions that sum to one. Add a lens-shaped binodal and one tie line whose endpoints α and β straddle z. Show distances from z to endpoints as lever arms.

Real-world analogy

Choosing a recipe with three ingredients can be represented by one point in a triangle; pure ingredients sit at corners and two-ingredient recipes lie on edges. If the recipe separates into two batches of different compositions, the average recipe point lies on the line between their compositions. The analogy matches material balance, while actual phase separation depends on thermodynamic stability.

Real-world example

In liquid–liquid extraction, water, an organic solvent and a solute may form two liquid phases. A ternary diagram shows whether a feed mixed with solvent will split and what compositions the two layers will have. Process design uses tie-line data to estimate how much solute remains in the raffinate and enters the extract.

Why?

Three fractions constrained to sum to one have two independent degrees of composition freedom, matching a two-dimensional triangle. Phase equilibrium selects paired endpoint compositions through chemical-potential equality. Overall material conservation makes any mixture of those endpoints lie on their connecting line, which is the geometric basis of the ternary lever rule.

Common misconception

An interior point's three fractions are not independent; their sum must be one. A randomly drawn chord across a binodal is not necessarily a true tie line. Another mistake is to read phase compositions as the overall feed composition after separation; each layer usually has a different composition.

Worked example

Suppose a tie line joins α with (x A,x B,x C) = (0.70,0.20,0.10) and β with (0.10,0.30,0.60). If equal mole amounts of the phases coexist, the overall composition is their average: z = (0.40,0.25,0.35). The fractions sum to one. Conversely, a feed with this z on that tie line gives f α = f β = 0.50 by the A balance, and B and C balances confirm it.

Quick check

1. What composition is represented by a point on the edge connecting A and B vertices? Answer: It contains A and B but x C = 0. The exact A/B fractions are read along that edge and sum to one.

Exam focus

Verify all three fractions sum to one and match the diagram's mass or mole basis. Read tie-line endpoints from supplied data; do not infer them from the overall point alone. Use z i = f αx i^α+f βx i^β and check at least one second component. State fixed T,p for a two-dimensional composition slice.

Advanced insight

The ternary two-phase region and tie-line field can change with temperature and pressure. At a plait point, the two endpoint compositions merge, analogous to a critical solution point. More complicated ternary diagrams can contain multiple two-phase regions or three-phase triangles, requiring additional phase-rule and material-balance analysis.

Summary

An equilateral triangle represents ternary compositions with fractions summing to one. Binodals delimit phase separation, and true tie lines join equilibrium phase compositions. The overall composition lies between endpoints, with phase amounts set by material balance or lever distances on the correct composition basis.

Practice questions

1. A ternary point has x A = 0.20 and x B = 0.35. Find x C. Answer: x C = 1 − 0.20 − 0.35 = 0.45. All fractions are nonnegative and sum to one. 2. On a tie line, α has x A = 0.80 and β has x A = 0.20. If overall z A = 0.50, find f α. Answer: 0.50 = f α(0.80)+(1−f α)(0.20), so f α = 0.50. This assumes the same mole-fraction basis for z and phase compositions. 3. Why is it unsafe to draw any line through an overall ternary composition and call it a tie line? Answer: Coexisting endpoint compositions must satisfy chemical-potential equality and come from measured or modelled equilibrium data. Geometry and overall balance alone do not determine the tie-line direction.