Positive Deviation from Raoult's Law in a Two-Liquid Mixture
Two volatile, mutually miscible liquids, X (molar mass 58 g/mol) and Y (molar mass 74 g/mol), are studied at 25 °C. The vapor pressures of the pure liquids at this temperature are P°(X) = 120 torr and P°(Y) = 40 torr. For an equimolar liquid mixture (x_X = 0.50) the measured partial pressures in the vapor are p_X = 66 torr and p_Y = 30 torr, and the mixture remains a single liquid phase. A student predicts that the total vapor pressure will be 80 torr, and that the vapor will be richer in Y than the liquid is, because Y has the higher molar mass and therefore the stronger intermolecular forces.
1. Calculate the measured total vapor pressure and the vapor-phase mole fractions. Compare with Raoult's law, identify the sign of the deviation, and explain what it implies about X–X, Y–Y, and X–Y interactions.
Read the worked answer
Measured total: P_total = 66 + 30 = 96 torr. Vapor mole fractions: y_X = 66/96 = 0.688, y_Y = 30/96 = 0.312.
Raoult's law prediction: P_ideal = 0.50(120) + 0.50(40) = 60 + 20 = 80 torr, with y_X = 60/80 = 0.75 and y_Y = 0.25.
The measured pressure (96 torr) exceeds the ideal value, so this is a positive deviation: each component's partial pressure is greater than Raoult's law predicts (66 > 60 for X, 30 > 20 for Y). Molecules escape into the vapor more easily than from the pure liquids, which means X–Y attractions in solution are weaker than the average of X–X and Y–Y attractions. Forming the mixture sacrifices favorable contacts without replacing them fully, so the components behave as if they were "pushed out" of solution. The student's molar-mass argument fails: vapor pressure, not molar mass, is the direct evidence about intermolecular force strength.
2. Predict the sign of ΔH_mix for this pair of liquids and state what would be observed if they were mixed adiabatically. Then decide, from the data given, which pure liquid has the larger enthalpy of vaporization.
Read the worked answer
ΔH_mix is positive (endothermic). Breaking the stronger X–X and Y–Y contacts costs energy that is not fully recovered when the weaker X–Y contacts form, so heat is absorbed from the surroundings. In an adiabatic (insulated) container the temperature of the mixture would therefore fall; in a beaker open to the room the mixture would draw heat from the surroundings and feel cool. This is the thermodynamic signature that accompanies a positive deviation from Raoult's law - the two observations are two faces of the same molecular picture.
Liquid Y has the larger ΔH_vap. At the same temperature, Y's vapor pressure is much lower (40 vs 120 torr), so Y molecules are held more tightly in the liquid and require more energy per mole to escape. Equivalently, Y has the higher normal boiling point. Note that this conclusion comes from the vapor pressures, not from Y's larger molar mass; molar mass is only a rough proxy for dispersion forces and can mislead.
3. A simple distillation of this mixture is performed, and the first few drops of distillate are collected. Predict which component dominates the distillate and explain how the non-ideal behavior modifies a Raoult's-law-only prediction. Then note one practical consequence of the positive deviation for separating X and Y completely.
Read the worked answer
The first distillate is dominated by X. The vapor in equilibrium with the equimolar liquid contains y_X = 0.688 (or 0.75 if Raoult's law were obeyed), so the condensate is enriched in the more volatile component, X, and the boiling flask becomes enriched in Y. Each successive vaporization–condensation cycle pushes the distillate further toward pure X.
Because the deviation is positive, the measured vapor is actually less X-rich (y_X = 0.688) than Raoult's law predicts (0.75); each component evaporates more than ideal, so Y's presence in the vapor is higher than expected. The practical consequence: strong positive deviations often produce a minimum-boiling azeotrope, a composition at which vapor and liquid have the same composition. Simple or fractional distillation can purify X and Y only up to that azeotropic composition, and complete separation would require an alternative method.