Binary Vapour–Liquid Equilibrium

Introduction

A binary liquid generally evaporates into a vapour with a different composition. This difference makes distillation possible. A phase diagram displays both the liquid composition at which boiling begins and the vapour composition at which condensation begins. Bubble and dew curves enclose a two-phase region, while tie lines connect compositions that coexist at one temperature and pressure.

Core explanation

For ideal liquid mixture A and B with ideal vapour at fixed T, Raoult's law gives p A = x Ap A and p B = x Bp B . Dalton's law gives total p = p A + p B = x Ap A + (1 − x A)p B . This is the bubble-pressure relation: it states the total pressure at which a liquid of composition x A can coexist with its first vapour at that T. The vapour mole fraction is y A = p A/p = x Ap A /p. If p A > p B , then y A > x A for an interior ideal composition: the vapour is enriched in more volatile A.

At fixed T, one can instead start with a vapour composition y A and ask the pressure at which the first liquid drop appears. Using x i = y ip/p i and Σx i = 1 gives 1/p dew = y A/p A + y B/p B . Bubble and dew curves are different functions of composition because one uses x and the other y. A horizontal tie line at one T and p connects the liquid point x A with vapour point y A. Between the curves, both phases are present; outside them, only liquid or only vapour is stable under the chosen conditions.

At fixed pressure, a temperature–composition diagram shows bubble and dew temperatures instead. The same equilibrium chemical-potential conditions hold, but pure vapour pressures now vary with T, so the ideal equations generally require iterative solution. The direction of a plotted curve depends on whether pressure or temperature is held fixed; memorising “upper curve means vapour” without checking axes can cause errors.

Overall composition z A lies between x A and y A inside a two-phase region. Material balance gives z A = Lx A + Vy A when L and V are liquid and vapour mole fractions with L + V = 1. The lever rule then determines phase amounts, but it does not alter the equilibrium endpoint compositions of a tie line at a specified T,p.

Real mixtures replace simple Raoult law with activity coefficients and may show azeotropes or miscibility gaps. The diagram must also state whether the phases are at equilibrium and whether chemical reaction occurs; a changing reaction composition adds constraints beyond the nonreactive binary picture.

Step-by-step reasoning

Read whether T or p is fixed and whether composition symbols are liquid x, vapour y or overall z. For an ideal fixed-T problem, calculate pure vapour pressures, form p bubble = Σx ip i , then y i = x ip i /p. For a dew calculation use 1/p = Σy i/p i . Draw or read a tie line before applying z = Lx + Vy to phase amounts.

Visual explanation

Draw pressure vertically and A composition horizontally at fixed T. Show a bubble curve and a dew curve enclosing a two-phase lens. A horizontal line crosses the lens at x A on one boundary and y A on the other. Place overall z A between them and draw lever arms illustrating that the farther endpoint has the smaller corresponding phase fraction.

Real-world analogy

Selecting raisins from a mixed bowl can give a handful richer in raisins than the bowl's average if one ingredient is more readily picked up. Vapour is similarly enriched in the more volatile component relative to the liquid. The analogy conveys composition separation, not the molecular basis of vapour pressure.

Real-world example

In a mixture of two volatile liquids with different pure vapour pressures, the first vapour produced on heating can be richer in the more volatile component than the liquid. A condenser collecting that vapour yields a distillate of altered composition. Repeated equilibrium stages can increase separation, subject to nonidealities and equipment performance.

Why?

At equilibrium, each component's μ matches across liquid and vapour. For the ideal model this leads to p i = x ip i . Components with larger p i contribute more strongly to vapour at the same liquid x i, so y i and x i differ. Two independent component equalities determine a paired liquid and vapour composition at one T,p.

Common misconception

The horizontal coordinate in a binary phase diagram may represent x, y or overall z depending on the curve and question. They are not interchangeable inside a two-phase region. Also, the bubble curve is not the amount of vapour present; it marks onset of vapour for a liquid composition. Tie-line endpoints give phase compositions, while the lever rule gives amounts.

Worked example

At a chosen T, let p A = 80 kPa and p B = 20 kPa. A liquid has x A = 0.40 and x B = 0.60. Its bubble pressure is p = 0.40(80) + 0.60(20) = 44 kPa. The first equilibrium vapour has y A = 32/44 ≈ 0.727 and y B ≈ 0.273. A is enriched in vapour because it has the larger pure vapour pressure.

Quick check

1. If x A = 0.40 but y A = 0.70 on one tie line, which phase is richer in A? Answer: The vapour is richer in A because y A > x A. At equilibrium the liquid and vapour still share one T and p despite differing compositions.

Exam focus

Label fixed T or fixed p, then distinguish x, y and z. Use partial pressures p i = x ip i and y i = p i/p only under stated ideal assumptions. For dew pressure use the reciprocal sum rather than reusing the bubble equation with y substituted for x. Apply material balance separately for phase fractions.

Advanced insight

Relative volatility measures the ratio of component enrichment factors and can vary with composition and temperature in real mixtures. When it approaches one, vapour and liquid compositions become similar and separation by ordinary distillation becomes difficult. Activity-coefficient models are used to predict such deviations from ideal curves.

Summary

Binary vapour–liquid equilibrium has paired liquid x and vapour y compositions. Ideal Raoult and Dalton laws give p bubble = Σx ip i and y i = x ip i /p; the dew relation starts from y. Bubble and dew curves bound a two-phase region, tie lines join coexisting compositions, and material balance determines phase amounts.

Practice questions

1. At fixed T, p A = 60 kPa and p B = 30 kPa. For x A = 0.50, find bubble pressure and y A. Answer: p = 0.5(60) + 0.5(30) = 45 kPa. y A = 30/45 = 2/3, so vapour is enriched in A. 2. For a fixed-T ideal vapour with y A = y B = 0.5 and p A = 60, p B = 30 kPa, find dew pressure. Answer: 1/p = 0.5/60 + 0.5/30 = 0.025 kPa⁻¹, so p dew = 40 kPa. 3. A two-phase sample has x A = 0.20, y A = 0.80 and overall z A = 0.50. Find vapour mole fraction. Answer: z A = (1−V)0.20 + V0.80 = 0.20 + 0.60V, so V = 0.50 and liquid fraction is also 0.50.