1. Phase Equilibria H20 (g) H20 (l) X(phase 1) X(phase 2)
At equilibrium, DG = 0 so
µ(phase 1) = µ(phase 2)
For boiling: K = a(g)/a(l)
Gases fairly ideal so a(g) = P(atm)
Activity of liquid set to 1: a(l) = 1
Equilibrium constant K = P P1, P2 are equilibium vapor pressures at T1 and T2
van’t Hoff equation Clausius-Clapeyron equation
H20 (g)
H20 (l)

2. Vapor Pressure DHvap ln P(atm) DHsub 1/T
DH can be determined from measurements of vapor pressure at multiple temperatures.
From Clausius-Clapeyron equation we see that -DH/R is the slope lnP vs 1/T.
DHvap
DHsub
ln P(atm)
1/T
liquid
water
solid
ice
vapor
Example Problem:
At what pressure (or altitude) does water boil at body temperature?
Get the pressure first, then estimate the altitude.
DH not completely independent of T

3. Pressure Dependence of Melting Temperature
For reversible processes:
Heat of fusion is always positive.
Volume of fusion is usually positive, but it is negative for water.

4. Mixtures: vapor pressure
For each component, i, in the mixture:
µi(phase 1) = µi (phase 2)
Vapor pressure
µA(solution) = µA(g, PA)
Letting µ*A and P*A refer to the pure solvent,
µA(g, PA) = µ*A(g, P*A) + RTln(PA/ P*A)
At equilibrium, µA(g, PA) = µA(solution) so,
µA(solution) = µ*A(pure solvent) = RTln(PA/ P*A)
= RTln(aA)
Solvent activity can be determined by measuring the vapor pressure.
aA = PA/ P*A
H2O (g)
O2(g)
H2O (l)
O2(aq)

5. Mixtures: Raoult’s law
Solvent activity is related to the vapor pressure
aA = PA/ P*A
(PA = vapor pressure solution)
(P*A = vapor pressure pure solvent)
Activity of the solvent can be expressed as
aA = gAXA
(where XA is the mole fraction)
Then for ideal solutions, gA = 1 and
PA = XA P*A
This is Raoult’s law for ideal solutions. Applies to the solvent.
H2O (g)
O2(g)
H2O (l)
O2(aq)

6. Mixtures: Henry’s law H2O (g) O2(g) H2O (l) O2(aq)
Focus on the solute (e.g. O2)
O2(g) O2(aq)
Keq = PO2 = kO2XO2
PO2 = vapor pressure of O2
XO2 = mole fraction of O2 in liquid
kO2 = Henry’s law constant
(measured experimentally; tabulated TSWP p. 194)
Gas solubility is directly proportional to the gas partial pressure.
Consider the relation between a faming bottle of champagne and the Bends.
H2O (g)
O2(g)
H2O (l)
O2(aq)
Facts about O2:
XO2(air) = 0.2
XO2(water) = 4.7 x 10-6
[O2](water) = 0.03[O2](air)
(mol/liter)

7. Example
H2O (g)
O2(g),N2(g)
Ar(g)
Calculate the vapor pressure of water in equilibrium with air at 1atm, 25 °C.
Calculate the total mole fraction of air dissolved in the water first.
From Raoult’s law we have,
The vapor-pressure lowering of water due to air solubility is negligible. The 3% of water in air (humidity at equilibrium) is not negligible, however.
H2O (l)
O2(aq), N2(aq)
Ar(aq)
Air data
O2(g) 20.95%
N2(g) 78.08%
Ar(g) 0.93%
Henry’s constants
kO2 = 4.3 x 102
kN2 = 8.5 x 102
kAr = 3.9 x 102
Water data
P*H2O = atm (25 °C)

8. Multiple Phases
At equilibrium, chemical potential for each component is the same in all three phases. Direct contact is not needed.
Immiscible liquids, such as decane and water, can form phases.
Semipermeable barriers can also be used to create effectively different phases
- cell membranes (lipid bilayers)
- dialysis membranes (nanoporous materials)
Even molecular binding can be thought of in this way
- bound vs free ligand (O2, hemoglobin)
H2O (g)
O2(g), N2(g)
Ar(g), C7H16(g)
H2O (l)
O2(aq), N2(aq)
Ar(aq), C7H16(aq)
H2O (C7H16)
O2(C7H16), N2(C7H16)
Ar(C7H16), C7H16(l)

9. Multiple Phases: equilibrium without contact
µA(s) = µA(aq) = µA(C7H16)
The solute in water is in equilibrium with its counterpart in heptane even though they are not in contact.
The activity of the solute in each saturated (equilibrium) solution can be used to calculate the standard chemical potential for the solute in that solvent relative to the solid.
This provides a basis for comparison and we can determine
Dµ0A = µ0A (C7H16) - µ0A (aq)
the change in chemical potential when the solute is transferred from water to heptane.
H2O (l)
Solute (aq)
Solid
C10H22(l)
Solute (C7H16)
Solid

10. Example: Hydrophobicity
For the series of fatty acids CnH2n+1COOH
With n ≥ 5, generally more soluble in organic solvent than water
For palmitic acid (C15H31COOH)
Dµ0A = µ0A (C7H16) - µ0A (aq) = -38 kJ/mol
For the series 8 ≤ n ≤ 22
Dµ0A = µ0A (C7H16) - µ0A (aq) = n kJ/mol
The hydrophobic core of a lipid membrane is a similar environment to organic solvents such as heptane.
Measurements of Dµ0A such as above relate to how molecules partition into cell membranes.
See TSWP p. 208 for experimental data
COOH
(Hydrophilic)
CH2 groups
Hydrophobic

11. Hydropathy Index
Hydropathy index is the relative hydrophobicity-hydrophilicity of the side-chain groups in proteins at neutral pH.
Various methods of ranking exist
Negative values characterize hydrophilic groups
Positive values characterize hydrophobic groups
This has been used as a starting point to understand basic thermodynamics of protein folding.
Isoleucine 4.5
Valine 4.2
Leucine 3.8
Phenylalanine 2.8
Cystein 2.5
Methionine 1.9
Alanine 1.8
Glycine -0.4
Threonine -0.7
Serine -0.8
Tryptophan -0.9
Tyrosine -1.3
Proline -1.6
Histidine -3.2
Glutamic Acid -3.5
Glutamine -3.5
Aspartic Acid -3.5
Asparigine -3.5
Lysine -3.9
Arginine -4.5
Side chain Index

12. Transfer between same solvents
For transfer of a solute molecule between different solutions in the same solvent:
DµA = µA(solution 2) - µA(solution 1)
= RT ln(aA(solution 2)/aA(solution 1))
If the activity coefficients are the same in the two solutions
DµA = RT ln(cA(solution 2)/cA(solution 1))
This is the free energy that drives diffusion and is the free energy of mixing.

13. Transfer of charged molecules
Electric field does work on ion as it passes between solutions.
DµA = µA(solution 2) - µA(solution 1) + ZFV
V = (f2 - f1) potential difference in volts between the two solutions.
F = Faraday = 96,485 (eV-1) = charge of mole of electrons.
Z = charge of the ion (e.g. ±1)
Note: this is true for any f(x). Convenient to write:
µA,tot = µA + ZFf
[Na+](1)
[Cl-](1) f1
[Na+](2)
[Cl-](2) f2 f1
Electrical Potential (f) f2
Position (x)

14. Equilibrium Dialysis Example
At equilibrium:
O2(out) = O2(in, aq)
[Mb·O2]/[Mb][O2(aq)] = Keq
If we are able to asses the total ligand concentration in the dialysis bag:
[O2(aq)] + [Mb·O2] = [O2 (in, total)]
Then [Mb·O2] = [O2 (in, total)] - [O2(aq)]
(these are measurable)
Can compute Keq.
If we have direct a probe for Mb·O2, then we don’t need the dialysis, can read of concentrations and compute Keq.
Dialysis can also be used to exchange solution (eg. change [salt])
H2O (l)
O2(aq), N2(aq)
etc.
H2O (l)
O2(aq), N2(aq)
Mb(aq), Mb·O2(aq)
Semipermeable membrane (cellulose): allows water and dissolved small solutes to pass, blocks passage of large proteins such as myoglobin (Mb)

15. Scatchard Equation General version: M + A M·A Keq = [M·A]/([M][A])
Simplify by introducing n, the average number of ligand molecules (A) bound to the macromolecule (M) at equilibrium:
Scatchard plot
NKeq
Slope = -Keq
n/[A]
For one ligand binding site per macromolecule
Scatchard equation
N independent binding sites per macromolecule. n N

16. Cooperative Binding
For a macromolecule with multiple binding sites, binding to one site can influence binding properties of other sites.
Failure of data plotted in a Scatchard plot to give a straight line indicates cooperative or anticooperative binding among binding sites.
Cooperative = binding of second ligand is made easier
Anticooperative = binding of second ligand is made more difficult
Hemoglobin is a favorite example of a protein with cooperative binding behavior.
Binds up to 4 O2
Cooperative: most O2 released in tissue while binding O2 maximally in the lungs
Binding curve shows characteristic sigmoidal shape 1
Myoglobin
Hemoglobin f
p50 = 1.5 Torr
p50 = 16.6 Torr
f = fraction of sites bound
PO2(Torr) 40

17. Hill Plot Scatchard equation (non-cooperative binding)
For cooperative binding.
n = Hill coefficient
K = a constant, not the Keq for a single ligand
Slope of each line give the Hill cooperativity coefficient.
Slope = 1 no cooperativity
Slope = N maximum (all-or-nothing)
cooperativity
See Example 5.4 (TSWP p ) for a detailed study of Hemoglobin
1.5
Myoglobin
n = 1.0
log[f /(1-f)]
Hemoglobin
n = 2.8
-1.5
-0.5 +2
log[P02] (Torr)