Modified Raoult's law and azeotropes

Why it matters

Most industrial liquid mixtures (alcohol-water, acetone-chloroform, solvents with water) are not ideal. Modified Raoult's law adds activity coefficients so the same bubble and dew calculations still work, and it predicts azeotropes: compositions where vapour and liquid are identical and ordinary distillation stops separating. Spotting an azeotrope early decides whether a plant needs pressure-swing, extractive or azeotropic distillation.

Key ideas

Modified Raoult's law. yᵢP = xᵢγᵢPᵢ^sat. It keeps the ideal-gas vapour and negligible Poynting assumptions of Raoult's law (so it is for low to moderate pressure) but allows a non-ideal liquid through γᵢ(T, x), from a model such as Margules, van Laar, Wilson or NRTL. Bubble and dew calculations use the same sums as before with xᵢγᵢPᵢ^sat in place of xᵢPᵢ^sat; DEW P and DEW T now need iteration because γ depends on the unknown x.

Positive and negative deviations.

• γᵢ > 1: positive deviation. Unlike molecules attract less than like ones; the total pressure lies above the Raoult straight line on a P-x diagram. Large positive deviation gives a maximum on the P-x-y curve, i.e. a maximum-pressure, minimum-boiling azeotrope (ethanol-water, about 89 mol% ethanol at 78.2 °C and 1 atm).
• γᵢ < 1: negative deviation. Unlike molecules attract more strongly (for example hydrogen bonding between acetone and chloroform); large negative deviation gives a minimum-pressure, maximum-boiling azeotrope. Positive deviation does not guarantee an azeotrope; it needs to be large compared with the difference in vapour pressures.

Azeotrope condition. At an azeotrope xᵢ = yᵢ, so the relative volatility α₁₂ = 1. With modified Raoult's law this gives γ₁P₁^sat = γ₂P₂^sat, and at that point P = γᵢPᵢ^sat for each component. This last relation lets you get activity coefficients directly from a measured azeotrope.

Testing for an azeotrope. α₁₂ = γ₁P₁^sat/(γ₂P₂^sat) changes continuously with composition. Evaluate it at both ends:

• at x₁ → 0: α₁₂ = γ₁^∞P₁^sat/P₂^sat;
• at x₁ → 1: α₁₂ = P₁^sat/(γ₂^∞P₂^sat). If one end is above 1 and the other below 1, α₁₂ passes through 1 and an azeotrope exists at that temperature.

Breaking azeotropes.

• Pressure-swing distillation: the azeotropic composition shifts with pressure, so two columns at different pressures can cross it.
• Extractive distillation: a heavy, high-boiling solvent changes the activity coefficients and leaves the column with one component.
• Azeotropic (heteroazeotropic) distillation: an entrainer forms a new low-boiling azeotrope that splits into two liquid phases in the decanter.
• Non-distillation routes: membranes (pervaporation), adsorption (molecular sieves for ethanol drying).

Formulas

yᵢ·P = xᵢ·γᵢ·Pᵢ^sat P = Σ xᵢ·γᵢ·Pᵢ^sat (bubble pressure) 1/P = Σ yᵢ/(γᵢ·Pᵢ^sat) (dew pressure) α₁₂ = (y₁/x₁)/(y₂/x₂) = γ₁·P₁^sat / (γ₂·P₂^sat) γ₁·P₁^sat = γ₂·P₂^sat, P_az = γᵢ·Pᵢ^sat (azeotrope) α₁₂(x₁→0) = γ₁^∞·P₁^sat/P₂^sat, α₁₂(x₁→1) = P₁^sat/(γ₂^∞·P₂^sat) ln γ₁ = A·x₂², ln γ₂ = A·x₁² (two-suffix Margules, used below) x₁,az = [1 − ln(P₂^sat/P₁^sat)/A] / 2 (two-suffix Margules azeotrope)

Symbols: xᵢ, yᵢ liquid and vapour mole fractions; P total pressure (kPa); Pᵢ^sat vapour pressure (kPa) at system T; γᵢ activity coefficient; γᵢ^∞ infinite-dilution value; α₁₂ relative volatility; A Margules parameter (dimensionless).

Worked examples

For both examples: a binary at fixed T with P₁^sat = 80 kPa, P₂^sat = 50 kPa, and G^E/RT = 1.2·x₁x₂.

Example 1 (standard): BUBL P with activity coefficients. Given: liquid with x₁ = 0.3. Find P and y₁.

1. ln γ₁ = 1.2 × 0.7² = 0.588, γ₁ = 1.800; ln γ₂ = 1.2 × 0.3² = 0.108, γ₂ = 1.114.
2. Partial pressures: x₁γ₁P₁^sat = 0.3 × 1.800 × 80 = 43.21 kPa; x₂γ₂P₂^sat = 0.7 × 1.114 × 50 = 38.99 kPa.
3. P = Σ xᵢγᵢPᵢ^sat = 43.21 + 38.99 = 82.20 kPa.
4. y₁ = 43.21/82.20 = 0.526.
5. Raoult's law would give P = 0.3 × 80 + 0.7 × 50 = 59 kPa: a strong positive deviation. Answer: P = 82.2 kPa, y₁ = 0.526

Example 2 (GATE level): does an azeotrope exist, and where?

1. γ₁^∞ = γ₂^∞ = e^1.2 = 3.320.
2. α₁₂(x₁→0) = 3.320 × 80/50 = 5.31 > 1; α₁₂(x₁→1) = 80/(3.320 × 50) = 0.482 < 1. α₁₂ crosses 1, so an azeotrope exists.
3. Condition γ₁P₁^sat = γ₂P₂^sat gives ln(γ₁/γ₂) = A(x₂² − x₁²) = A(1 − 2x₁) = ln(P₂^sat/P₁^sat).
4. 1 − 2x₁ = ln(50/80)/1.2 = −0.4700/1.2 = −0.3917, so x₁ = 0.696.
5. P_az = γ₁P₁^sat = 80 × exp(1.2 × 0.3042²) = 80 × 1.1174 = 89.39 kPa. Check with component 2: 50 × exp(1.2 × 0.6958²) = 50 × 1.7878 = 89.39 kPa.
6. P_az exceeds both vapour pressures, so this is a maximum-pressure (minimum-boiling) azeotrope. Answer: azeotrope at x₁ = y₁ = 0.696, P = 89.4 kPa

Common mistakes

• Using Raoult's law for alcohol-water or other polar mixtures and missing the azeotrope.
• Equating "maximum-pressure" and "maximum-boiling"; a maximum on the P-x-y diagram is a minimum on the T-x-y diagram.
• Taking γᵢ = 1 for the dilute component; it is the dilute component whose γ departs most from 1.
• Forgetting to iterate γ in DEW P or DEW T calculations.
• Assuming the azeotropic composition is fixed; it shifts with pressure (the basis of pressure-swing distillation).

For GATE CH

• NAT problems: bubble pressure and vapour composition with given γ or Margules/van Laar parameters; azeotrope composition and pressure from a one-parameter model.
• Activity coefficients or model parameters from azeotropic data (γᵢ = P/Pᵢ^sat).
• Conceptual MCQs linking sign of deviation, type of azeotrope and molecular interactions, and methods of breaking azeotropes.
• Practise the end-point α₁₂ test for azeotrope existence.

Quick check

1. At an azeotrope, what is the relative volatility?
2. A maximum-pressure azeotrope corresponds to what kind of boiling-point behaviour?
3. At a low-pressure azeotrope with P = 100 kPa and P₁^sat = 80 kPa, what is γ₁?
4. Do negative deviations give γ greater or less than 1?
5. Name one method that exploits the pressure dependence of the azeotrope.

Answers: 1. 1. 2. Minimum-boiling. 3. 1.25. 4. Less than 1. 5. Pressure-swing distillation.