Ideal solutions, Lewis-Randall rule and excess properties

Why it matters

Real liquid mixtures are described as "ideal solution plus correction". The ideal solution gives a simple, exact-looking baseline (no heat or volume change on mixing, fugacity proportional to mole fraction), and excess properties measure the departure from it. Activity coefficients, azeotropes, liquid-liquid splitting and the heat released on mixing all come out of this framework.

Key ideas

Ideal solution. A mixture in which every species' chemical potential has the form μᵢ = Gᵢ + RT ln xᵢ, where Gᵢ is the molar Gibbs energy of pure i at the same T, P and physical state. Physically, the molecules are similar in size and the unlike-pair interactions equal the average of like-pair interactions (benzene-toluene, isomers, neighbouring members of a homologous series). Consequences:

• V̄ᵢ = Vᵢ and H̄ᵢ = Hᵢ: no volume change and no heat effect on mixing.
• S̄ᵢ = Sᵢ − R ln xᵢ: mixing still creates entropy, so ΔS_mix > 0 and ΔG_mix < 0. An ideal solution is not a "no-change" mixture. An ideal-gas mixture is a special case of an ideal solution.

Lewis-Randall rule. In an ideal solution the fugacity of each species is f̂ᵢ^id = xᵢfᵢ, with fᵢ the pure-species fugacity at the mixture T and P in the same phase. Combined with ideal-gas vapour and low pressure, it gives Raoult's law, yᵢP = xᵢPᵢ^sat. For a gas mixture the same idea gives φ̂ᵢ = φᵢ, a useful approximation when the gas components are similar or one is dilute.

Henry's law. For a dilute solute (often a supercritical gas dissolved in a liquid), f̂ᵢ = xᵢℋᵢ as xᵢ → 0, where ℋᵢ is Henry's constant (an empirical value from data tables). Lewis-Randall is the limiting law as xᵢ → 1; Henry's law is the limiting law as xᵢ → 0. Gibbs-Duhem guarantees that in a binary, if the solute obeys Henry's law the solvent obeys Lewis-Randall over the same dilute range.

Excess properties. M^E = M − M^id at the same T, P and composition. Because ΔV and ΔH are zero for an ideal solution, V^E = ΔV_mix and H^E = ΔH_mix. For G and S they differ: G^E = ΔG_mix − RT Σxᵢ ln xᵢ. Excess properties behave like ordinary properties: G^E = H^E − TS^E, and partial excess properties obey summability and Gibbs-Duhem.

Activity coefficient. γᵢ = f̂ᵢ/(xᵢfᵢ). It is the ratio of the actual fugacity to the ideal-solution value. ln γᵢ is the partial molar property of G^E/RT, so G^E/RT = Σxᵢ ln γᵢ and, at constant T and P, Σxᵢ d ln γᵢ = 0. γᵢ > 1 means positive deviation (unlike molecules repel relative to like ones, higher vapour pressure than Raoult); γᵢ < 1 means negative deviation.

Formulas

μᵢ^id = Gᵢ + R·T·ln xᵢ f̂ᵢ^id = xᵢ·fᵢ (Lewis-Randall) f̂ᵢ = xᵢ·ℋᵢ (Henry's law, dilute solute) ΔV_mix^id = 0, ΔH_mix^id = 0 ΔS_mix^id = −R·Σ xᵢ·ln xᵢ ΔG_mix^id = R·T·Σ xᵢ·ln xᵢ M^E = M − M^id G^E = H^E − T·S^E γᵢ = f̂ᵢ / (xᵢ·fᵢ) G^E/(R·T) = Σ xᵢ·ln γᵢ ln γᵢ = [∂(n·G^E/RT)/∂nᵢ]_{T,P,nⱼ} ΔG_mix = G^E + R·T·Σ xᵢ·ln xᵢ

Symbols: xᵢ liquid mole fraction; Gᵢ, fᵢ pure-species molar Gibbs energy (J/mol) and fugacity (Pa); f̂ᵢ fugacity in solution (Pa); ℋᵢ Henry's constant (Pa); M^E excess molar property; G^E, H^E (J/mol), S^E (J/mol·K); γᵢ activity coefficient (dimensionless); R = 8.314 J/(mol·K); T (K).

Worked examples

Example 1 (standard): mixing functions of an ideal solution. Given: an ideal liquid solution with x₁ = 0.4 at 350 K. Find ΔS_mix, ΔG_mix, ΔH_mix and ΔV_mix.

1. ΔS_mix = −R·(x₁ ln x₁ + x₂ ln x₂) = −8.314 × (0.4 × (−0.9163) + 0.6 × (−0.5108)) = −8.314 × (−0.6730) = 5.595 J/(mol·K).
2. ΔG_mix = −T·ΔS_mix (since ΔH_mix = 0) = −350 × 5.595 = −1958 J/mol.
3. ΔH_mix = 0 and ΔV_mix = 0 by definition of an ideal solution. Answer: ΔS_mix = 5.60 J/(mol·K), ΔG_mix = −1958 J/mol, ΔH_mix = ΔV_mix = 0

Example 2 (GATE level): excess functions of a non-ideal binary. Given: at 350 K a binary liquid follows G^E/RT = A·x₁x₂ with A = 1.2, and the measured H^E at x₁ = 0.4 is 1200 J/mol. Pure-1 fugacity is 95 kPa. At x₁ = 0.4, find γ₁, γ₂, G^E, S^E, ΔG_mix and f̂₁.

1. For this model ln γ₁ = A·x₂² and ln γ₂ = A·x₁² (from ln γᵢ as the partial property of nG^E/RT).
2. ln γ₁ = 1.2 × 0.36 = 0.432, so γ₁ = 1.540. ln γ₂ = 1.2 × 0.16 = 0.192, so γ₂ = 1.212.
3. Check summability: x₁ ln γ₁ + x₂ ln γ₂ = 0.1728 + 0.1152 = 0.288 = A·x₁x₂. Consistent.
4. G^E = 0.288 × 8.314 × 350 = 838 J/mol.
5. S^E = (H^E − G^E)/T = (1200 − 838)/350 = 1.03 J/(mol·K).
6. ΔG_mix = G^E + RT Σxᵢ ln xᵢ = 838 − 1958 = −1120 J/mol (still negative, so mixing is favourable at this composition).
7. f̂₁ = γ₁·x₁·f₁ = 1.540 × 0.4 × 95 = 58.5 kPa, versus 38.0 kPa for an ideal solution. Answer: γ₁ = 1.54, γ₂ = 1.21, G^E = 838 J/mol, S^E = 1.03 J/(mol·K), ΔG_mix = −1120 J/mol, f̂₁ = 58.5 kPa

Common mistakes

• Saying ΔG_mix = 0 or ΔS_mix = 0 for an ideal solution; only ΔH and ΔV are zero.
• Confusing ΔG_mix with G^E; they differ by RT Σxᵢ ln xᵢ.
• Using Raoult's law (yᵢP = xᵢPᵢ^sat) at high pressure, where neither the vapour nor the Lewis-Randall reference is adequate without φ corrections.
• Applying Lewis-Randall to a dilute supercritical gas, whose pure-liquid fugacity does not exist; use Henry's law.
• Writing G^E/RT = Σ ln γᵢ without the mole-fraction weights.

For GATE CH

• NAT problems on ΔS_mix and ΔG_mix of ideal solutions and on G^E, H^E, S^E from given data.
• Given G^E/RT as a function of composition, find ln γ₁, ln γ₂ or γ at infinite dilution.
• Conceptual MCQs on which mixing properties vanish for an ideal solution, the Lewis-Randall rule versus Henry's law, and the sign of deviations.
• Practise the relation G^E/RT = Σxᵢ ln γᵢ in both directions.

Quick check

1. Which two mixing properties are zero for an ideal solution?
2. Write the Lewis-Randall rule.
3. What is ΔS_mix for an equimolar ideal binary, in terms of R?
4. If G^E/RT = 0.8x₁x₂, what is γ₁ at infinite dilution?
5. Is ln γᵢ an ordinary molar or a partial molar property, and of what?

Answers: 1. ΔH_mix and ΔV_mix. 2. f̂ᵢ = xᵢfᵢ. 3. R ln 2 = 0.693R. 4. e^0.8 = 2.23. 5. Partial molar property of G^E/RT.