Chapter 11: Other Types of Phase Equilibria in Fluid Mixtures (selected topics)

Partitioning a solute among two coexisting liquid phases Two partially miscible or completely immiscible liquids How the solute distributes between the two phases Purification – LL extraction, Partition chromatography Drug distribution – Lipids, body fluids Pollutant distribution – air.,water,soil

Partition of a solute Distribution coefficient The LLE equilibrium condition is: Then:

Case I: solute does not affect solubility of the solvents Little amount of solute or totally immiscible liquids I-a) N i moles of solute completely dissolved and distributed between immiscible solvents

Case I-b: some undissolved solute (solid or gas) in equilibrium with two immiscible solvents

Case Ic: partially miscible liquids

Case II: solute affects the LLE (partially miscible solvents)

Liquid-liquid equilibrium (LLE) Extraction problems involve at least three components: the solute and two solvents. It is usual to represent their phase behavior in triangular diagrams.

Liquid-liquid equilibrium (LLE) Extraction problems involve at least three components: the solute and two solvents. It is usual to represent their phase behavior in triangular diagrams. Binodal curve Tie line Plait point

Liquid-liquid equilibrium (LLE) Reading the scale in a triangular diagram

Example 4 It is desired to remove some acetone from a mixture that contains 60 wt% acetone and 40 wt% water by extraction with methyl isobutyl ketone (MIK). If 3 kg of MIK are contacted with 1 kg of this acetone+water solution, what will be the amounts and compositions of the phases in equilibrium? Solution

Example 4 60 wt% acetone and 40 wt% water (1 kg) Pure MIK (3 kg)

Example 4 Next… Find the point that represents the global composition of the system. Based on the information given, the total amounts of acetone, water, and MIK are equal to 0.6 kg, 0.4 kg, and 3 kg. The corresponding weight fractions are: 0.15, 0.10, and Locate this point in the diagram.

Example 4 Global composition

Example 4 Approximated tie line

Example 4 Approximated tie line MIK-rich phase 80.5% MIK 15.5% Acetone 4.0% Water Water-rich phase 2.0% MIK 8.0% Acetone 90.0% Water

Example 4 Calculation of the phase amounts (L I and L II ) Global mass balance One component mass balance (water for example)

Liquid-liquid equilibrium (LLE)

Osmotic equilibrium Consider two cells at the same temperature, separated by a membrane permeable to some of the species present, but impermeable to others. For simplicity, assume a binary solute+solvent system and that the membrane is permeable to the solvent only. Cell I contains the pure solvent and cell II contains the mixture. At equilibrium, the following equation is valid:

Osmotic equilibrium

Assuming the liquid is incompressible:

Osmotic equilibrium Applying logarithm:

Example 5 Compute the osmotic pressure at K between an ideal aqueous solution 98 mol% water and pure water. For an ideal solution, the activity coefficient is equal to 1 and the molar volume of water is approximate equal to 18x10 -6 m 3 /mol. Solution

Example 5 Compute the osmotic pressure at K between an ideal aqueous solution 98 mol% water and pure water. For an ideal solution, the activity coefficient is equal to 1 and the molar volume of water is approximate equal to 18x10 -6 m 3 /mol. Solution

Osmotic equilibrium For ideal solutions: For a dilute ideal solution, by using a Taylor series expansion of the logarithm of the solvent mole fraction, the following approximated expression can be derived:

Osmotic equilibrium Assuming: and

Osmotic equilibrium For simplicity, let us drop superscript II, then: where m solute is the solute’s molar mass. A practical application of this equation is to use it to find the molar mass of polymers and proteins.

Osmotic equilibrium Schematics of an osmometer

Example 6 Polyvinyl chloride (PVC) is soluble in cyclohexanone. At 25 o C, if a solution of PVC batch with 2 g/L of solvent is placed in an osmometer, the height h in the osmometer is 0.85 cm. Knowing that the density of pure cyclohexanone is 0.98 g/cm 3, estimate the molar mass of this PVC batch.

Example 6 Solution At the membrane, in the mixture side: At the membrane, in the pure solvent side: H Assuming the density of the solution and of the solvent are equal:

Example 6 Solution H

Example 6 Solution H

Recommendation Read the sections of chapter 11 covered in these notes and review the corresponding examples.