Phase Equilibria

X(phase 1) X(phase 2)

At equilibrium, G = 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) = 1Equilibrium constant K = P P1, P2 are equilibium vapor pressures at T1 and T2

van’t Hoff equation Clausius-Clapeyron equation

H20 (l)

H20 (g)

lnK2

K1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

−ΔH0

R1T2

−1T1

⎛

⎝ ⎜

⎞

⎠ ⎟ ln

P2

P1

⎛

⎝ ⎜

⎞

⎠ ⎟ =

−ΔH vap

R1T2

−1T1

⎛

⎝ ⎜

⎞

⎠ ⎟

Vapor Pressure

H can be determined from measurements of vapor pressure at multiple temperatures.

From Clausius-Clapeyron equation we see that -H/R is the slope lnP vs 1/T.

Example Problem:

At what pressure (or altitude) does water boil at body temperature?

Get the pressure first, then estimate the altitude.

Hvap

Hsub

ln P

(atm

)

1/T

liquidwater

solidice

watervapor

H not completely independent of T

Pressure Dependence of Melting TemperatureG (s) =G (l)

dG (s) =dG (l) dG =−S dT+V dP

−S (s)dT+V (s)dP=−S (l)dT+V (l)dP

dTdP

=ΔV fus

ΔS fus ΔV fus=V (l) −V (s) and ΔS fus=S (l) −S (s)

For reversible processes:

Heat of fusion is always positive.Volume of fusion is usually positive, but it is negative for water.

ΔS=qrev

T so ΔS fus=

ΔH fus

T

dTdP

=TΔV fus

ΔH fus

T2 −T1

P2 −P1=

T1+T2

2⎛ ⎝

⎞ ⎠

ΔV fus

ΔH fus (approximation)

Mixtures: vapor pressure

H2O (l)O2(aq)

H2O (g)O2(g)

For each component, i, in the mixture:µi(phase 1) = µi (phase 2)

Vapor pressureµA(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

Mixtures: Raoult’s law

H2O (l)O2(aq)

H2O (g)O2(g)

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 = AXA

(where XA is the mole fraction)

Then for ideal solutions, A = 1 and

PA = XA P*A

This is Raoult’s law for ideal solutions. Applies to the solvent.

Mixtures: Henry’s law

H2O (l)O2(aq)

H2O (g)O2(g)

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 liquidkO2 = 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.

Facts about O2:XO2(air) = 0.2XO2(water) = 4.7 x 10-6

[O2](water) = 0.03[O2](air) (mol/liter)

ExampleCalculate 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)

H2O (g)O2(g),N2(g)

Ar(g)

Air dataO2(g) 20.95%N2(g) 78.08%Ar(g) 0.93%

Henry’s constantskO2 = 4.3 x 102

kN2 = 8.5 x 102

kAr = 3.9 x 102

Water dataP*H2O = 0.031 atm (25 °C)

Xair =XN2+XO2

+XAr Henry's law: PA =kAXA

=PN2

kN2

+PO2

kO2

+PAr

kAr=14.18×10−6

PH2O =XH2OPH2O*

=(1−Xair)PH2O* ≈PH2O

* since Xair is so small

Multiple Phases

H2O (l)O2(aq), N2(aq)

Ar(aq), C7H16(aq)

H2O (g)O2(g), N2(g)

Ar(g), C7H16(g)

H2O (C7H16)O2(C7H16), N2(C7H16)Ar(C7H16), C7H16(l)

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)

Multiple Phases: equilibrium without contact

H2O (l)Solute (aq)

Solid

C10H22(l)Solute (C7H16)

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

µ0A = µ0

A (C7H16) - µ0A (aq)

the change in chemical potential when the solute is transferred from water to heptane.

Solid

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)

µ0A = µ0

A (C7H16) - µ0A (aq) = -38 kJ/mol

– For the series 8 ≤ n ≤ 22

µ0A = µ0

A (C7H16) - µ0A (aq) = 17.82 - 3.45n kJ/mol

• The hydrophobic core of a lipid membrane is a similar environment to organic solvents such as heptane.

– Measurements of µ0A such as above relate to how molecules partition into cell

membranes.

COOH(Hydrophilic)

CH2 groupsHydrophobic

See TSWP p. 208 for experimental data

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 4.5Valine 4.2Leucine 3.8Phenylalanine 2.8Cystein 2.5Methionine 1.9Alanine 1.8Glycine -0.4Threonine -0.7Serine -0.8

Tryptophan -0.9Tyrosine -1.3Proline -1.6Histidine -3.2Glutamic Acid -3.5Glutamine -3.5Aspartic Acid -3.5Asparigine -3.5Lysine -3.9Arginine -4.5

Side chain Index Side chain Index

Transfer between same solvents

For transfer of a solute molecule between different solutions in the same solvent:

µA = 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

µA = RT ln(cA(solution 2)/cA(solution 1))

This is the free energy that drives diffusion and is the free energy of mixing.

Transfer of charged molecules

[Na+](2)[Cl-](2)

2

[Na+](1)[Cl-](1)

1

Ele

ctri

cal P

oten

tial

()

Position (x)

1

2

∂φ∂x

=E

Electric field does work on ion as it passes between solutions.

µA = µA(solution 2) - µA(solution 1) + ZFV

V = (2 - 1) 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 (x). Convenient to write:

µA,tot = µA + ZF

w= f1

2

∫ dx

f =ZFE=ZF∂φ∂x

w= ZF1

2

∫∂φ∂x

dx=ZF φ2 −φ1( )

Equilibrium Dialysis Example

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)

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])

Scatchard Equation

Keq=cA bound( )

cM −cA(bound)( )cA outside( )

General version: M + A M·A Keq = [M·A]/([M][A])

Simplify by introducing , the average number of ligand molecules (A) bound to the macromolecule (M) at equilibrium:

ν ≡cA bound( )

cM

Keq=ν

1−ν( ) A[ ]

νA[ ]

=Keq1−ν( )

For one ligand binding site per macromolecule

νA[ ]

=Keq N −ν( )

Scatchard equationN independent binding sites per macromolecule.

/[A

]

Slope = -Keq

N

NKeq

Scatchard plot

Cooperative BindingFor 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 easierAnticooperative = 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 lungsBinding curve shows characteristic sigmoidal shape f

0

1

PO2(Torr)0 40

p50 = 16.6 Torr

HemoglobinMyoglobin

p50 = 1.5 Torr

f = fraction of sites bound

Hill Plotν N

1−ν N( )=

f1−f( )

=Keq A[ ]

f1− f( )

=K A[ ]n

Scatchard equation (non-cooperative binding)

For cooperative binding.

n = Hill coefficientK = a constant, not the Keq for a single ligand

Slope of each line give the Hill cooperativity coefficient.

Slope = 1 no cooperativitySlope = N maximum (all-or-nothing)

cooperativity

See Example 5.4 (TSWP p. 204 - 207) for a detailed study of Hemoglobin

logf

1− f( )

⎛

⎝ ⎜

⎞

⎠ ⎟ =nlogA[ ]+logK

log[P02] (Torr)

log[

f /(

1-f)

]

1.5

-1.5-0.5 +2

Hemoglobinn = 2.8

Myoglobinn = 1.0