Advanced Phase Equilibrium Problems

Introduction

Real phase-equilibrium questions rarely test one idea at a time. A typical problem asks you to read a diagram, count phases and degrees of freedom, find the coexisting compositions, apply a material balance and then explain the result in terms of chemical potential. This page gathers the tools from the unit into one problem-solving strategy and applies it to several demanding cases, so that each step is justified by thermodynamics rather than by memorised diagram shapes.

Core explanation

One governing condition. Every phase-equilibrium problem rests on a single statement: at equilibrium, temperature and pressure are uniform, and the chemical potential of each component is the same in every phase in which it is present. For a component i distributed between phases α and β, μᵢ(α) = μᵢ(β). If μᵢ is lower in one phase, that component moves into it until the difference disappears, because transfer then lowers the total Gibbs energy.

Counting constraints: the phase rule. Each phase has C − 1 independent mole fractions, plus T and p are shared, giving P(C − 1) + 2 variables. Equality of chemical potential supplies C(P − 1) equations. The difference is the Gibbs phase rule, F = C − P + 2. When independent reactions or fixed stoichiometric constraints are present, C is replaced by the number of independent components, C = N − R − (special constraints). At fixed pressure one degree of freedom is used, so F′ = C − P + 1 on an isobaric binary diagram.

What a diagram encodes. A single-phase region is a patch where F is largest. A two-phase region in a binary diagram is crossed by horizontal tie lines; any overall composition on a tie line splits into the two end-point phases. An invariant point such as a eutectic or peritectic marks where three phases coexist in a binary isobaric diagram, so F′ = 0 and temperature and all three compositions are fixed. An azeotrope is different: only two phases coexist, but they share the same composition, which adds an extra constraint.

Quantities from geometry. The lever rule is a material balance: if the overall mole fraction of B is z and the coexisting phases have x α and x β, the fraction of material in phase β is (z − x α)/(x β − x α). This works on any tie line, whether the phases are liquid and vapour, two liquids, or a liquid and a solid.

Boundaries from calorimetry. Along any two-phase line of a pure substance, dμ(α) = dμ(β) gives the Clapeyron equation dp/dT = ΔH/(TΔV). For vaporisation, neglecting the liquid volume and treating the vapour as ideal leads to ln(p₂/p₁) = −(ΔH vap/R)(1/T₂ − 1/T₁). Many advanced questions combine these: calculate a boundary shift, then decide which phase is stable at the new conditions.

Non-ideal mixtures. For solutions, μᵢ = μᵢ° + RT ln(γᵢxᵢ). Activity coefficients above 1 indicate unfavourable mixing and can produce positive azeotropes or liquid–liquid splitting; values below 1 indicate favourable interactions and can produce negative azeotropes.

Formulae

F = C − P + 2 (F′ = C − P + 1 at fixed pressure). Lever rule: n β/(n α + n β) = (z − x α)/(x β − x α). Clapeyron: dp/dT = ΔH/(TΔV). Clausius–Clapeyron: ln(p₂/p₁) = −(ΔH/R)(1/T₂ − 1/T₁). Raoult's law: pᵢ = xᵢpᵢ . Modified Raoult's law: pᵢ = γᵢxᵢpᵢ .

Step-by-step reasoning

A reliable strategy for any multi-part problem:

1. Identify the components and any reactions, and fix C. 2. Locate the state point on the diagram and name the phases present, giving P. 3. Apply the phase rule to see what is fixed and what can vary. 4. If two phases coexist, draw the tie line and read both compositions. 5. Use the lever rule for amounts, and check the material balance adds up. 6. Explain the outcome using equality of chemical potentials.

Visual explanation

Imagine a binary temperature–composition diagram. Drop a vertical line at the overall composition as the system cools. Each time the line crosses a boundary, the number of phases changes. Inside a two-phase lens, a horizontal tie line runs through the state point, and its two ends slide along the boundaries as temperature changes, while the lever arms show how much of each phase is present.

Real-world analogy

A phase diagram is like a detailed map with contour lines. The phase rule tells you whether you are standing on open ground, on a path, or at a single crossroads. The tie line is the bridge connecting two towns, and the lever rule tells you how the population is divided between them.

Real-world example

In refrigeration, engineers choose working fluids whose vapour pressure curves place boiling near the evaporator temperature and condensation near room temperature. The Clausius–Clapeyron equation estimates the pressures needed on each side of the cycle, and the phase diagram confirms that the fluid never freezes or becomes supercritical under operating conditions.

Why?

Why do all these rules fit together? Because each is a consequence of minimising Gibbs energy at fixed T and p. Equal chemical potentials are the minimum condition, the phase rule counts how many such conditions constrain the variables, and the lever rule is conservation of mass applied to the phases that minimum selects.

Common misconception

"On a two-phase tie line, the phase compositions depend on the overall composition." They do not. At fixed T and p in a binary, the coexisting compositions are fixed by the tie-line ends; changing the overall composition only changes the relative amounts of the two phases.

Worked example

Question: A binary liquid of overall mole fraction z B = 0.40 is heated to a temperature at which liquid of x B = 0.25 coexists with vapour of y B = 0.65. What fraction of the moles is vapour, and how many degrees of freedom remain at fixed pressure?

Reasoning: Fraction of vapour = (z − x)/(y − x) = (0.40 − 0.25)/(0.65 − 0.25) = 0.15/0.40 = 0.375. With C = 2, P = 2 and fixed pressure, F′ = 2 − 2 + 1 = 1, so temperature alone fixes both compositions.

Answer: About 0.38 of the moles are vapour; F′ = 1.

Quick check

1. How many degrees of freedom exist at the triple point of a pure substance, and what does this mean? Answer: F = 1 − 3 + 2 = 0, so temperature and pressure are both fixed and the triple point is a single invariant state.

Exam focus

Always state C and P explicitly before using the phase rule, and say whether pressure is fixed. In lever-rule answers, show the tie-line compositions and the arm lengths. When using Clausius–Clapeyron, convert enthalpies to J mol⁻¹ and temperatures to kelvin.

Advanced insight

Near a critical point or a consolute point, the two coexisting compositions converge and the tie line shrinks to zero length. Here the curvature of the Gibbs energy with respect to composition vanishes, fluctuations become very large, and simple mean-field predictions of boundary shapes fail; measured coexistence curves follow universal power laws with exponents close to 0.33 rather than the mean-field value of 0.5.

Summary

Advanced phase-equilibrium problems are solved by combining a small set of tools: equality of chemical potentials, the phase rule F = C − P + 2, tie lines and the lever rule, and the Clapeyron family of equations for boundary slopes. Count components and phases first, read compositions from tie lines, calculate amounts by material balance and justify every conclusion in terms of Gibbs energy minimisation.

Practice questions

1. Calculate F for a binary system at fixed pressure in which a solid, a liquid and a vapour coexist. Answer: F′ = 2 − 3 + 1 = 0, so the state is invariant: temperature and all three phase compositions are fixed. 2. Water has ΔH vap ≈ 40.7 kJ mol⁻¹ and boils at 373 K at 101 kPa. Estimate its boiling point at 70 kPa. Answer: 1/T₂ = 1/373 − (8.314/40 700) ln(70/101) ≈ 0.002681 + 0.0000749 ≈ 0.002756 K⁻¹, so T₂ ≈ 363 K, about 90 °C. 3. In a liquid–liquid split, phase α has x B = 0.10 and phase β has x B = 0.80. For an overall z B = 0.30, what fraction of moles is in phase β? Answer: (0.30 − 0.10)/(0.80 − 0.10) = 0.20/0.70 ≈ 0.29. 4. Explain, using chemical potentials, why the melting point of ice falls when pressure increases. Answer: Liquid water is denser than ice, so ΔV of fusion is negative; raising pressure lowers the chemical potential of the liquid relative to ice, so the Clapeyron slope dp/dT = ΔH/(TΔV) is negative and melting occurs at a lower temperature.