Phase Equilibria
A( l )  A( g )
Evaporation-Condensation
A( s )  A(l )
Melting-Freezing
H 2O( l )  H 2O( g )
A( s )  A( g )
Sublimation-Condensation
A( solid I )  A( solid II )
Phase transition
dG   SdT  Vdp
d
G
S
V
  dT  dp
n
n
n
dGm  d   S m dT  Vm dp
  

  Sm
 T  p
Sg >> Sl > Ss
  solid 
solid

  Sm
 T  p
 liquid 

   S mliquid
 T  p
  gas 

   S mgas
 T  p
 Sm
slope in   T diagram
The most stable phase is that with lowest chemical potential.
dGm  d   S m dT  Vm dp
Pressure
Effect
  solid 

  Vmsolid
 p T
 liquid 

  Vmliquid
 p T
  gas 

  Vmgas
 p T
Vmgas  Vmliquid or Vmsolid
usually Vmliquid  Vmsolid
-T curve of
gases much
more largely
affected by
pressure change
than liquids or
solids
few subs tan ces : Vmsolid  Vmliquid
Pressure increase: - Boiling point elevation
- Freezing point elevation/depression
Phase Diagram
CO2
Clapeyron Equation
A( l )  A( g )
A( s )  A( g )
A( s )  A(l )
A( )  A(  )
   
At Equilibrium
d  d 
 S m dT  Vm dp   S m dT  Vm dp
V
V

m

m


 V dp  S


m
m
 S m
Vtr dp  S tr dT
Clapeyron Equation

dT
 Vm dp   S m  S m dT
dp S tr

dT Vtr
Solid-Liquid Equilibrium
A( s )  A(l )
dp Str S fus


dT Vtr V fus
S
l
m
S
dp 
s
m
  S
H fus
V fusTm
fus

H fus
Tm
dT
p2  p1 
Slope of pT-curve
H fus
dp

dT V fusTm
p2
H fus
p1
fus
 dp  V
H fus
V fus
Tm'
ln
Tm
Tm'  p2 
dT

T
Tm  p1  m
 Usually positive
 ~ 40 atm/K
 40 atm are needed to
change the melting
point by 1 K

V fus  Vml  Vms

usually , Vml  Vms
V fus  0
some systems, Vml  Vms
p2  p1 
Upon pressure
increase
p2  p1
H fus
V fus
V fus  0
Tm'
ln
Tm
 p2  p1   0
if V fus  0
Tm'
ln
0
Tm
Tm'  Tm
Melting point
elevation
if V fus  0
Tm'
ln
0
Tm
Tm'  Tm
Melting point
depression
water
Ice skating
p2  p1 
H fus
V fus
p2  p1 
Tm' H fus Tm  Tm'  Tm
ln

ln
Tm V fus
Tm
H fus
V fus
ln 1  x   x
 Tm'  Tm 

ln 1 
Tm 

if x is very small
H fus Tm'  Tm
p2  p1 
V fus Tm
H fus Tm
p 
V fus Tm
if pressure is changed by p, the
melting point will change by Tm
Liquid-Gas Equilibrium
A( l )  A( g )
S vap 
H vap
Tb
dp S vap

dT Vvap
dp H vap p


dT
T
RT
Slope of pT-curve
 always positive
 ~ 0.04 atm/K
 Boiling point
increases by 25 K
upon increasing the
pressure by 1 atm.
applies also to s-g
equilibrium

Vtr  Vvap  Vmg  Vml

Vmg  1000 Vml
V
g
m
V
l
m
V
g
m
RT

p
dp H vap dT

 2
p
R
T
T p 
dp H vap b 2 dT
p p  R  T p  T 2
1
b
1
p2
H vap  1
p2
1 


ln



p1
R  Tb  p2  Tb  p1  
H vap  1 1 
p T 
  
ln v 2  
pv T1 
R  T2 T1 
Clausius-Clapeyron
Equation
dp H vap dT
 p  R  T2
H vap 1
ln p  
 C
R T
p  eC e

H vap 1

R T
p  const  e
at T  Tb


H vap 1

R T
pv  patm
Determine the change in the freezing point of ice
upon pressure increase from 1 atm to 2 atm.
Vm(water)=18.02 cm3/mol and Vm(ice)=19.63 cm3/mol
at 273.15 K. Hfus=6.009 kJ/mol.
Benzene has a normal boiling point of 353.25 K. If
benzene is to be boiled at 30oC, to what value must
the pressure be lowered. Hvap=30.76 kJ/mol
Phase Rule
F: Number of degrees of freedom
Number of independent variables that can be changed without changing the number
of phases
C: number of independent components
P: number of coexisting phases
F=1
F=2
F=0
F=1
Liquid-Gas Equilibrium
of a binary mixture
Ideal solution:
H mix  0
Vmix  0
A  B   A  B mixture
H mix  0
Vmix  0
Energy of interaction AA,BB = A-B
Intramolecular forces AA,BB = A-B
100 mL A  100 mL B  200 mL mixture
Vm Apure   Vm B pure   Vm  Amixture   Vm Bmixture 
Ideal solutions obey Raoults Law
pi  xi  p
o
i
pi  xi  p
o
i
V
V
L
(pA)solvent >
L
(pA)solution
nA
xA 
1
n A  nB
psolution  p A  p B
psolution  x A p  xB p
o
A
p-x phase diagram
o
B
T=const.
psolution  1  xB  p Ao  xB p Bo
psolution  p  xB p  xB p
o
A
o
A

o
B
P Total  PA  PB  X APA  X BPB

psolution  p Ao  p Bo  p Ao xB
y
a 
b
psolution  pA  pB
x
PA 

X A PA
PB  X BPB
y B p Ao
xB  o
pB  p Ao  pBo y B

V



substitute in ptotal  p Ao  pBo  p Ao xB
L
pBo p Ao
ptotal  o
pB  p Ao  pBo y B

A+B
L
gas
ptotal
 p A  pB
p A  xVA  ptot
p B  xVB  ptot
p A  y A  ptot
p B  y B  ptot
pB
xBL p Bo
x A p Ao
yB 


ptot x AL p Ao  xBL p Bo x A p Ao  xB p Bo
xB p Bo
yB  o
p A   p Bo  p Ao xB
solve for xB
V

T const.
Ex. Benzene and Toluene
• Consider a mixture of benzene, C6H6, and
toluene, C7H8, containing 1.0 mol benzene and
2.0 mol toluene. At 20 °C, the vapor pressures
of the pure substances are:
P°benzene = 75 torr
P°toluene = 22 torr
• Assuming the mixture obeys Raoult’s law, what
is the total pressure above this solution?
23
T-x phase diagram
p=const.
Lever Rule



n L  xBa  xBc  nV  xBc '  xBa

n L  nV  ntot
n A  nB  ntot
Distillation
p=const.
Colligative Properties
Colligative Properties
Kf and Kb
31
Ex. Boiling Point Elevation
A 2.00 g sample of a large biomolecule was dissolved in 15.0 g of
CCl4. The boiling point of this solution was determined to be
77.85 °C. Calculate the molar mass of the biomolecule. For CCl4,
the Kb = 5.07 °C/m and BPCCl4 = 76.50 °C.
Tb  K b  msolute
nsolute
msolute 
wtsolvent / kg
nsolute
Tb  K b 
wtsolvent / kg
Tb  wtsolvent / kg 
nsolute 
Kb
Tb  T  T o  77.85o C  76.50o C  1.35o C
wtsolvent / kg  0.015 kg
K b  5.07 o C / m
nsolute  4.026 10 3 mol
Mwt solute 
msolute
2g

 497 g / mol
3
nsolute 4.026 10 mol
Ex. Freezing Point Depression
Estimate the freezing point of a permanent type of antifreeze
solution made up of 100.0 g ethylene glycol, C2H6O2, (MM = 62.07)
and 100.0 g H2O (MM = 18.02).
T f  K f  msolute
msolute  mEG
nEG

wt EG / kg
nEG 
mEG
100 g

 1.611 mol
Mwt EG 62.07 g / mol
nEG
1.611
T f  K f 
 1.86
 30o C
wt EG / kg
0.10
T f  T fo  T f
T f  T fo  T f  0o C  30o C  30o C
33
Membranes and Permeability
Membranes
– Separators
– Example: Cell walls
– Keep mixtures organized and
separated
Permeability
– Ability to pass substances
through membrane
Semipermeable Membrane
– Some substances pass, others don’t.
– Selective
Osmosis and Osmotic Pressure
A. Initially, Soln B separated from pure water, A, by osmotic
membrane (permeable to water). No osmosis occurred yet
B. After a while, volume of fluid in tube higher. Osmosis has
occurred.
35
Osmotic pressure (p): Pressure needed to stop the flow.
Flow of water
molecules
Net flow
Column rises
Pressure increases
Increase of flow from right to left
Finally:
Equilibrium established
Flow of water
molecules
Net flow = 0
Equation for Osmotic Pressure
• Assumes dilute solutions
p=iMRT
– p = osmotic pressure
– i = number of ions per formula unit
= 1 for molecules
– M = molarity of solution
• Molality, m, would be better, but M simplifies
• Especially for dilute solutions, where m  M
– T = Kelvin Temperature
– R = Ideal Gas constant
= 0.082057 L·atm·mol1K1
37
Eye drops must be at the same osmotic pressure as the human eye
to prevent water from moving into or out of the eye. A commercial
eye drop solution is 0.327 M in electrolyte particles. What is the
osmotic pressure in the human eye at 25°C?
p = MRT
T(K) = 25°C + 273.15
L  atm
p  0.327 M  0.08206
 298 K  8.00atm
K  mol
Using p to determine MM
The osmotic pressure of an aqueous solution of certain protein was
measured to determine its molar mass. The solution contained 3.50
mg of protein in sufficient H2O to form 5.00 mL of solution. The
measured osmotic pressure of this solution was 1.54 torr at 25 °C.
Calculate the molar mass of the protein.
 1atm 
1.54torr 

p
mol
 760torr 
M

 8.28 10 5
L  atm 
RT 
L
 0.08206
 298K
K  mol 



n  M V  8.28 105 M  5.00 103 L  4.14 107 mol
mass
3.50 10 3 g
3
Mwt 


8
.
45

10
g / mol
7
n
4.14 10 mol