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Two condensed phases in equilibrium: examples: – ice and water – oil and water

Phase equilibria. Two condensed phases in equilibrium: examples: – ice and water – oil and water – ocean water and moist air. Let's say the two phases are two solvents 1 and 2 , and B is a solute. Then at equilibrium:. μ B1 = μ B2. μ B1 = μ° B1 + RT ln a B1.

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Two condensed phases in equilibrium: examples: – ice and water – oil and water

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  1. Phase equilibria Two condensed phases in equilibrium: examples: – ice and water – oil and water – ocean water and moist air. Let's say the two phases are two solvents 1 and 2 , and B is a solute. Then at equilibrium: μB1 = μB2 μB1 = μ°B1 + RT ln aB1 μB2 = μ°B2 + RT ln aB2, μ°B1 + RT ln aB1 = μ°B2 + RT ln aB2 RT (ln aB2 – ln aB1) = –(μ°B2 – μ°B1) aB2/aB1 = exp[–(μ°B2 – μ°B1)/RT] = exp[–(Δμ°B)/RT] aB2/aB1 is the partition coefficientP CHEM 471: Physical Chemistry The hexane/water partition coefficient of phenol at 298.15 K is 0.20; calculate the free energy of transfer of phenol from water to hexane. aB2/aB1 = 0.20 ⇒ Δμ°B = –RT ln (aB2/aB1) = –8.3145 × 298.15 × ln(.2) = 4.0 kJ/mol

  2. Partition coefficients If the solutes in phase 1 and phase 2 are in equilibrium with a solid (saturated solution) they are in equilibrium with each other. The solubility of phenol in water is 83 g/L. What is its solubility in hexane? Assuming the partition coefficient is based on molarity, then the solubility of phenol in hexane must be 0.2 × the solubility in water, or in other words, 16.6 g/L Partition coefficients between octanol and water are often used as a measure of hydrophobicity. Example: benzoic acid has an octanol/water partition coefficient of 76. Solubility of benzoic acid in water is 3.5 g/kg water. M = 122.12. Solubility = 3.5/122.12 = 0.029 m. CHEM 471: Physical Chemistry 0.029 m benzoic acid in water 0.029 x 76 = 2.204 m benzoic acid in octanol Solid benzoic acid Solubility of benzoic acid in octanol = 2.204 m

  3. Amphiphilic molecules Hydrocarbon in hydrocarbon µH1 = µ°H1 + RT ln aH1 ~ µ°H1 Hydrocarbon in water µH2 = µ°H2 + RT ln aH2 ~ µ°H21 + RT ln xH2 hydrocarbon (Concentration is low, so aH2 ~ xH2 Free energy of transfer water ΔtransG = µH2 – µH1 = µ°H2 – µ°H1 + RT ln xH2 At equilibrium ΔtransG = 0 so 0 = µ°H2 – µ°H1 + RT ln xH2 =ΔtransG° + RT ln xH2 CHEM 471: Physical Chemistry Cratic Unitary ΔtransG° = –RT ln xH2 ln xH2 = –ΔtransG°/RT = –ΔtransH°/R(1/T) + ΔtransS°/R Plot the log of the solubility vs. 1/T; slope is –ΔtransH°/R; intercept is ΔtransS°/R

  4. Hydrophobic effect CHEM 471: Physical Chemistry Insolubility of hydrocarbons in water is entropic, not enthalpic

  5. Hydropathy index CHEM 471: Physical Chemistry

  6. Amphiphilic molecules Amphiphilic (Hydrophobic on one end, hydrophilic on the other CHEM 471: Physical Chemistry

  7. Amphiphilic structures Detergent micelle Monolayer Micelle Bilayer CHEM 471: Physical Chemistry Vesicle

  8. Surface tension Surface tension is the two-dimensional equivalent of pressure p = (∂G/∂V)T γ = (∂G/∂A)T = (∂F/∂l)T Free energy per unit area or force per unit length CHEM 471: Physical Chemistry

  9. Surface tension CHEM 471: Physical Chemistry

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