Phase Equilibria
)()( gl AA )(2)(2 gl OHOH
)()( ls AA )()( gs AA Melting-Freezing
Evaporation-Condensation
Sublimation-Condensation
)()( IIsolidIsolid AA Phase transition
mp
mmm
ST
dpVdTSddG
dpnVdT
nS
nGd
VdpSdTdG
gasm
p
gas
liquidm
p
liquid
solidm
p
solid
ST
ST
ST
diagramTinslope
Sm
Sg >> Sl > Ss
The most stable phase is that with lowest chemical potential.
Pressure Effect
liquidm
solidm
solidm
liquidm
solidm
liquidm
gasm
gasm
T
gasliquidm
T
liquidsolidm
T
solid
mmm
VVcessubsfewVVusually
VorVV
Vp
Vp
Vp
dpVdTSddG
:tan
-T curve of gases much more largely affected by
pressure change than liquids or
solids
Pressure increase: - Boiling point elevation - Freezing point elevation/depression
CO2
Clapeyron Equation)()( gl AA )()( ls AA )()( gs AA
At Equilibrium
)()( AA
tr
tr
trtr
mmmm
mmmm
mmmm
VS
dTdp
dTSdpV
dTSSdpVV
dTSSdpVV
dpVdTSdpVdTS
dd
Clapeyron Equation
Solid-Liquid Equilibrium
)()( ls AA
m
m
fus
fus
pT
pT mfus
fusp
pmfus
fus
mfus
fus
m
fusfus
sm
lm
fus
fus
tr
tr
TT
VH
pp
TdT
VH
dpdTTV
Hdp
TVH
dTdp
TH
SSS
VS
VS
dTdp
m
m
'
12 ln
2'
1
2
1
Slope of pT-curve Usually positive ~ 40 atm/K 40 atm are needed to
change the melting point by 1 K
0,
0,
fuss
ml
m
fuss
ml
ms
ml
mfus
VVVsystemssome
VVVusuallyVVV
m
m
fus
fus
TT
VH
pp'
12 ln
mmm
mfus
mmm
mfus
TTTTVif
TTTTVif
pppp
''
''
1212
0ln0
0ln0
0Upon pressure increase
Melting point elevation
Melting point depression
waterIce skating
m
m
fus
fus
m
mm
fus
fus
m
mm
fus
fus
m
mmm
fus
fus
m
m
fus
fus
TT
VH
p
TTT
VH
pp
smallveryisxifxx
TTT
VH
pp
TTTT
VH
TT
VH
pp
'
12
'
12
''
12
1ln
1ln
lnln
if pressure is changed by p, the
melting point will change by Tm
Liquid-Gas Equilibrium )()( gl AA
vap
vap
VS
dTdp
p
RTVVV
VV
VVVV
gm
lm
gm
lm
gm
lm
gmvaptr
1000b
vapvap T
HS
121
2
121
2
2
2
11ln
11ln
2
1
2
1
TTRH
TpTp
pTpTRH
pp
TdT
RH
pdp
TdT
RH
pdp
RTp
TH
dTdp
vap
v
v
bb
vap
pT
pT
vapp
p
vap
vap
b
b
Slope of pT-curve always positive ~ 0.04 atm/K Boiling point
increases by 25 K upon increasing the pressure by 1 atm.
Clausius-Clapeyron Equation
applies also to s-g equilibrium
CTR
Hp
TdT
RH
pdp
vap
vap
1ln
2
TRH
TRH
C
vap
vap
econstp
eep
1
1
atmvb ppTTat
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
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.
Phase RuleF: Number of degrees of freedomNumber of independent variables that can be changed without changing the number of phases
C: number of independent componentsP: number of coexisting phases
F=1
F=2
F=1
F=0
Liquid-Gas Equilibrium
of a binary mixtureIdeal solution: 00 mixmix VH
mixtureBABA
0 mixH Energy of interaction AA,BB = A-BIntramolecular forces AA,BB = A-B
0 mixV mixturemmixturempurempurem BVAVBVAV
mixturemLBmLAmL
200100100
Ideal solutions obey Raoults Law
oiii pxp
L
V
L
V
(pA)solvent > (pA)solution
oiii pxp
1
BA
AA nn
nx
BA pppsolution
BBAABATotal PXPXPPP
AAA PXP
BBB PXP
xbay
xpppp
pxpxpp
pxpxp
pxpxp
ppp
BoA
oB
oAsolution
oBB
oAB
oAsolution
oBB
oABsolution
oBB
oAAsolution
BAsolution
1
p-x phase diagramT=const.
A+BL
V
BoA
oB
oA
oBB
B
oBB
oAA
oAA
oB
LB
oA
LA
oB
LB
tot
BB
totBBtotAA
totVBBtot
VAA
BAgastotal
xppppxy
pxpxpx
pxpxpx
ppy
pyppyp
pxppxp
ppp
BoB
oA
oB
oA
oB
total
BoA
oB
oAtotal
BoB
oA
oB
oAB
B
ypppppp
xppppinsubstitute
yppppyx
solve for xB
L
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 torrP°toluene = 22 torr
• Assuming the mixture obeys Raoult’s law, what is the total pressure above this solution?
23
T-x phase diagramp=const.
aB
cB
VcB
aB
L xxnxxn 'Lever Rule
totBA
totVL
nnn
nnn
Distillationp=const.
Colligative Properties
Colligative Properties
Kf and Kb
31
Ex. Boiling Point ElevationA 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.
b
solventbsolute
solvent
solutebb
solvent
solutesolute
solutebb
KkgwtTn
kgwtnKT
kgwtnm
mKT
/
/
/
mCKkgkgwt
CCCTTT
obsolvent
oooob
/07.5015.0/
35.150.7685.77
molnsolute310026.4
molgmol
gnmMwt
solute
solutesolute /497
10026.42
3
Ex. Freezing Point DepressionEstimate 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).
33CCCTTT
TTT
Ckgwt
nKT
kgwtnmm
mKT
ooof
off
foff
o
EG
EGff
EG
EGEGsolute
soluteff
30300
3010.0
611.186.1/
/
molmolg
gMwt
mnEG
EGEG 611.1
/07.62100
Membranes and PermeabilityMembranes – 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
Flow of water molecules
Net flow
Column risesPressure increases
Increase of flow from right to leftFinally:
Equilibrium established
Flow of water molecules
Net flow = 0
Osmotic pressure (p): Pressure needed to stop the flow.
Equation for Osmotic Pressure
• Assumes dilute solutions
p = i M R T– 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·mol1K1
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?
atmKmolKatmLM 00.829808206.0327.0
p
p = MRT T(K) = 25°C + 273.15
Using p to determine MMThe 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.
Lmol
KmolKatmL
torratmtorr
RTM 51028.8
29808206.0
760154.1
p
molLMVMn 735 1014.41000.51028.8
molgmolg
nmassMwt /1045.8
1014.41050.3 3
7
3