Liquid-liquid and solid-liquid equilibrium

Why it matters

Liquid-liquid extraction, decanters after azeotropic distillation, and phase separation in reactors all depend on liquid-liquid equilibrium (LLE). Crystallisation, freeze concentration, wax deposition in pipelines and the design of solvents for solids depend on solid-liquid equilibrium (SLE). Both use the same rule as VLE, equal fugacity of each species in every phase, applied to condensed phases.

Key ideas

LLE criterion. Two liquid phases α and β at the same T and P share the same pure-liquid reference fugacity fᵢ, so equal fugacity reduces to equal activity: xᵢ^α·γᵢ^α = xᵢ^β·γᵢ^β for every species. Pressure has little effect because liquids are nearly incompressible.

When does a liquid split? Only if the Gibbs energy of mixing curve has a region of negative curvature (see the phase-equilibrium criteria topic). That needs strong positive deviations: for G^E/RT = A·x₁x₂ splitting requires A > 2. Activity-coefficient models that can describe LLE include Margules, van Laar (within limits), NRTL and UNIQUAC; Wilson's equation cannot.

Binary solubility diagrams. On a T-x diagram, the coexistence (binodal) curve encloses the two-phase region. Its top is an upper critical solution temperature (UCST): above it the liquids are fully miscible (phenol-water, near 66 °C). Some systems have a lower critical solution temperature (LCST) instead, becoming immiscible on heating (triethylamine-water), and a few show both.

Ternary LLE and extraction. On a triangular diagram, tie lines join coexisting phase compositions inside the binodal curve and shrink to a point, the plait point, where the two phases become identical. The distribution coefficient K_D = (solute fraction in extract)/(solute fraction in raffinate) and the selectivity β = K_D,solute/K_D,diluent measure how good a solvent is. Phase amounts follow from the lever rule along a tie line.

SLE criterion. For a pure solid i in equilibrium with a liquid solution, f̂ᵢ^l = fᵢ^s, giving xᵢγᵢ = fᵢ^s/fᵢ^l (ratio of pure solid to pure subcooled-liquid fugacity at T). Following a path through the melting point, neglecting the heat-capacity difference between liquid and solid, gives ln(xᵢγᵢ) = −(ΔH_fus/R)(1/T − 1/T_m).

Ideal solubility. With γᵢ = 1 the solubility depends only on the solute (its ΔH_fus and T_m), not on the solvent. It is a good estimate when solute and solvent are chemically similar (naphthalene in benzene) and poor for dissimilar pairs (naphthalene in water, where γ is huge). Higher melting point and larger heat of fusion mean lower solubility.

Simple eutectic systems. If the solids are mutually insoluble and the liquid is ideal, each component has its own liquidus (freezing-point) curve from the equation above. They intersect at the eutectic point, the lowest temperature at which liquid can exist, where both solids crystallise together. Below the eutectic temperature the mixture is entirely solid.

Formulas

xᵢ^α·γᵢ^α = xᵢ^β·γᵢ^β (LLE, each species) ln[x₁^α/(1 − x₁^α)] = A·(2x₁^α − 1) (symmetric two-suffix Margules, x₁^β = 1 − x₁^α) A > 2 (phase splitting for G^E/RT = A·x₁x₂) K_D = x_solute^E / x_solute^R, β = K_D,solute / K_D,diluent ln(xᵢ·γᵢ) = −(ΔH_fus,ᵢ/R)·(1/T − 1/T_m,ᵢ) (SLE, pure solid i, ΔCp neglected) ln xᵢ = −(ΔH_fus,ᵢ/R)·(1/T − 1/T_m,ᵢ) (ideal solubility) x₁ + x₂ = 1 on both liquidus curves (eutectic point)

Symbols: xᵢ^α, xᵢ^β mole fractions in the two liquid phases; γᵢ activity coefficient; A Margules parameter (dimensionless); K_D distribution coefficient; superscripts E and R extract and raffinate; ΔH_fus,ᵢ molar heat of fusion (J/mol); T_m,ᵢ melting point (K); T (K); R = 8.314 J/(mol·K). Take ΔH_fus and T_m from a data book.

Worked examples

Example 1 (standard): ideal solubility of a solid. Given: naphthalene, T_m = 353.4 K, ΔH_fus = 19.0 kJ/mol. Estimate its mole-fraction solubility in benzene at 25 °C assuming an ideal solution.

1. ln x = −(ΔH_fus/R)·(1/T − 1/T_m).
2. 1/T − 1/T_m = 1/298.15 − 1/353.4 = 0.0033540 − 0.0028297 = 5.243 × 10⁻⁴ K⁻¹.
3. ΔH_fus/R = 19 000/8.314 = 2285.3 K.
4. ln x = −2285.3 × 5.243 × 10⁻⁴ = −1.198, so x = 0.302. Answer: x_naphthalene ≈ 0.30 (close to the measured value because benzene and naphthalene are chemically similar).

Example 2 (GATE level): eutectic point and liquid-liquid split. (a) Benzene: T_m = 278.7 K, ΔH_fus = 9.87 kJ/mol. With naphthalene as above, find the eutectic temperature and composition (ideal liquid, immiscible solids).

1. Liquidus curves: x_N = exp[−2285.3(1/T − 1/353.4)] and x_B = exp[−(9870/8.314)(1/T − 1/278.7)] = exp[−1187.2(1/T − 1/278.7)].
2. The eutectic satisfies x_N + x_B = 1. Trial T = 270 K: x_N = 0.136, x_B = 0.872, sum 1.007. Trial T = 269 K: x_N = 0.131, x_B = 0.858, sum 0.989. Interpolating and refining gives T = 269.6 K.
3. At 269.6 K: x_N = 0.134, x_B = 0.866. Answer: eutectic at about 269.6 K (−3.6 °C), x_naphthalene ≈ 0.13

(b) A binary liquid follows G^E/RT = 2.5·x₁x₂. Find the compositions of the two coexisting liquids.

1. Because the model is symmetric, x₁^β = 1 − x₁^α and the LLE condition becomes ln[x/(1 − x)] = A(2x − 1) with x = x₁^α.
2. Solve by trial: at x = 0.145, left side = ln(0.1696) = −1.774; right side = 2.5 × (−0.710) = −1.775. Converged.
3. Check equal activities: x₁^αγ₁^α = 0.1448 × exp(2.5 × 0.8552²) = 0.1448 × 6.224 = 0.901; x₁^βγ₁^β = 0.8552 × exp(2.5 × 0.1448²) = 0.8552 × 1.054 = 0.901. Equal. Answer: the phases contain x₁ = 0.145 and x₁ = 0.855

Common mistakes

• Using VLE-style Pᵢ^sat in LLE; the vapour pressure cancels because both phases are liquids at the same T.
• Taking the trivial solution x₁^α = x₁^β (identical phases) as the answer.
• Using °C in the solubility equation or mixing kJ and J in ΔH_fus/R.
• Assuming ideal solubility holds in any solvent; it ignores γ, which can be very large for dissimilar pairs.
• Thinking the eutectic is a compound; it is a mixture of two solid phases that melt together at one temperature.
• Using Wilson's equation for a partially miscible system.

For GATE CH

• NAT problems on ideal solubility of a solid, freezing-point depression, and eutectic temperature from given ΔH_fus and T_m.
• LLE with a one-parameter Margules model: condition for splitting (A > 2), compositions for a symmetric system, or activities in two phases.
• Extraction-type questions using the lever rule and distribution coefficients on ternary data.
• Conceptual MCQs on UCST and LCST, plait points, and why Wilson cannot predict LLE.

Quick check

1. What is the equilibrium condition for species i between two liquid phases?
2. For G^E/RT = A·x₁x₂, what is the minimum A for phase splitting?
3. Does ideal solubility of a solid depend on the solvent?
4. Raising the melting point of a solute, at fixed ΔH_fus and T, makes its ideal solubility higher or lower?
5. What is the plait point on a ternary LLE diagram?

Answers: 1. xᵢ^αγᵢ^α = xᵢ^βγᵢ^β. 2. A = 2 (splitting for A > 2). 3. No, only on the solute's ΔH_fus, T_m and the temperature. 4. Lower. 5. The point where the tie lines shrink to zero length and the two liquid phases become identical.