Phase Equilibrium II
- Two Component System
glucose
fermentation
filtration
Mixture of
ethanol & water
• How many components and phases in
this system?
• 2 components and 1 liquid phase
• Method to separate ethanol from
water??
• Fractional distillation?
• How?
• obtained only about 95% of ethanol
• Why not 100%???
21.1 Two Component System
- Mixture of 2 Miscible Liquids
• vapour pressure of a liquid (Pv)
- is a measure of the escape tendency of the liquid molecules
- influenced by its composition and the intermolecular
interaction between the 2 components
• composition of a solution
- shown by the mole fraction
• Ideal solutions
- mixtures of liquids of
similar chemical structures
- linear relationship of boiling
point and composition
(
nA
A 
n A  nB
)
• non-ideal solutions??
• -
mixtures of liquids having dissimilar
structures
have marked deviations from the linear
boiling point - composition relationship
21.2 Raoult's Law for Ideal Solutions
• states that the vapour pressure of a component in a
mixture at a given temperature is directly proportional
to its mole fraction, and is equal to the product of its
mole fraction and the vapour pressure of the pure
component at that temp.
• partial pressure of A:
partial pressure of B:
PA = XAPAo
PB = XBPBo
(Po is vapour pressure; X is mole fraction)
• total pressure of the mixture Pt = PA + PB
(Pt = XAPAo + XBPBo)
Example 21.1
• nA = 0.2mol ; nB = 0.5 mol ;
Pt = 40mmHg ; PAo = 20mmHg ; PBo = ?
• XA = 0.2 / (0.2+0.5) = 0.286
XB = 0.5 / (0.2+0.5) = 0.714
XA+ XB=1
• Pt = XAPAo + XBPBo
40 = (0.286)(20)+(0.714) PBo
• PBo= 48mmHg
Ex 21.1
• Pmethanolo = 90mmHg : Pethanolo = 45mmHg ; Pt = ?
• nmethanol = 64/32 = 2mol
nethanol = 92/46 = 2mol
• Xmethanol = 2 / (2+2) = 0.5
Xethanol = 2 / (2+2) = 0.5
• Pt = XmethanolPmethanolo + XethanolPethanolo
Pt = (0.5)(90)+(0.5)(45)
• Pt= 67.5mmHg
Ex 21.1
• PPo = 20kPa : PQo = 8kPa ; Pt = ?
nP = 6mol
nQ = 2mol
• XP = 6 / (6+2) = 0.75
XQ = 2 / (6+2) = 0.25
• Pt = XPPPo + XQPQo
Pt = (0.75)(20)+(0.25)(8)
• Pt= 17 kPa
• Mixtures of liquid that obey Raoult's Law are known
as ideal solutions
+
Strength of
 Strength of
Intermolecular
Intermolecular
attraction in pure A
attraction in pure B

Strength of
Intermolecular
attraction between
particle A and
particle B in mixture
• the escape tendency of molecule A or B in
mixture equals to their respective escape
tendency in pure A and pure B
• ===> no enthalpy change upon mixing
• ===> no volume change upon mixing
• Reasons: interaction doesn’t change upon
mixing
• more nearly alike the two substances are
chemically, the more nearly they exhibit ideal
behaviour
Actually, ideal solutions are seldom found and their existence only
holds for very dilute solutions.
Ex. 21.3
• Which of the following solution mixtures
are ideal solutions?
• a) benzene and methylbenzene
• b) cyclohexane and ethanol
• c) bromomethane and iodomethane
• d) propanone and trichloromethane
• e) propan-1-ol and propan-2-ol
• f) n-hexane and n-heptane
it’s a rough prediction only!!!
Vapour pressure-composition curve
• for ideal solution
1. PA a XA &
PB a XB
2. Pt = PA + PB
• ===>
gives a straight line in
the vapour pressurecomposition curve
Information from graph:
pure A is more volatile than pure B
boiling point-composition curve
• For ideal solution:
a linear variation of the b.p.
with the composition of the
solution
• slope of the graph is opposite to
the vapour pressurecomposition curve
• more volatile liquid, the vapour
pressure is higher at a fixed
temperature
==> a lower temperature is
required to set up the Pv
==> the liquid has a lower b.p.
Ex 21.4
• a)
• b) P(octane) = (0.3)(19.0) =5.7mmHg
P(2-methylheptane) =(1-0.3)(27.0)=18.9mmHg
Pt = 5.7 + 18.9 = 24.6mmHg
• c) It is assumed that the intermolecular interaction
between octane and 2-methylheptane molecules in the
mixture is the same as the that among octane and 2methylheptane molecules in their pure solutions.
(it is assumed that the solution mixture is ideal
solution.)
• d) b.p. = (126)(0.3) +(112)(0.7) = 116.2oC
Ex 21.5
• a) nmethanol = 20/32 = 0.625mol
nethanol = 100/46 = 2.174mol
•
Xmethanol = 0.625 / (0.625+2.174) = 0.223
Xethanol = 2.174 / (0.625+2.174) = 0.777
•
Pmethanol=(12.530)(0.223) = 2.79kPa
Pethanol = (5.866)(0.777) = 4.56kPa
• b) Pt = 2.79 + 4.56 = 7.35kPa
• c) compositon of methanol in vapour
= (2.79)/(7.35) = 0.380 (38%)
composition of ethanol in vapour
= (4.56)/(7.35) = 0.620 (62%)
Liquid contains
1 mol A ( )and
1 mol B (
)
PAo =10kPa ;PBo = 30kPa
Composition of liquid:
XA = 0.5 ; XB = 0.5
Composition of vapour?
Partial pressure of A = 10 x 0.5 = 5kPa
Partial pressure of B = 30 x 0.5 = 15kPa
Total vapour pressure = 5 + 15 = 20 kPa
mole fraction of A in vapour = 5/20 = 0.25
mole fraction of B in vapour = 15/20 =0.75
21.3 Deviations from Raoult's Law
- Non-ideal Solution
• How about those liquid mixtures do not obey
Raoult's law??
• ===> Non-ideal Solution
• 1) Solution with positive deviation from
Raoult’s Law
• 2) Solution with negative deviation from
Raoult’s Law
Positive Deviation from Raoult's Law
• Occurs when Pt > that predicted by Raoult's law
i.e. PA > XAPAo & PB > XBPBo
===> Pt > XAPAo + XBPBo
• Molecules A & B in the mixture escape from the liquid surface
more
than that expected for an ideal mixture.
+
Average
Strength of
+ Strength of
Intermolecular
Intermolecular
attraction in pure A
attraction in pure B
>
Strength of
Intermolecular
attraction between
particle A and
particle B in mixture
• The vapour pressure would be higher and the boiling
point would be lower compared with ideal behaviour
Example: Binary mixture of cyclohexane and ethanol
CH3CH2
O
H
CH3CH2
O
O
H
CH3CH2
H
Weak Van der Waal’s force
in pure cyclohexane
CH3CH2
Strong hydrogen bonding
in pure ethanol
O
H
CH3CH2
O
H
Weak dipole-induced dipole
interaction in the mixture
•
•
•
•
the interaction is now weaker after mixing the solution
===> escape tendency of molecule is higher
===> Pt is greater than expected
===> enthalpy change is endothermic
Energy(breaking interaction between molecules in pure solution) >
Energy(formation of new interaction between 2 kinds of molecules in mixture)
• ===>Volume expansion
weaker attraction between molecules in the mixture
=> intermolecular distance increase
Negative Deviation from Raoult's Law
• Occurs when Pt < that predicted by Raoult's law
i.e. PA < XAPAo & PB < XBPBo
===> Pt < XAPAo + XBPBo
• Molecules A & B in the mixture has a lower escape tendency
from the liquid surface compared with that of ideal mixture.
+
Average
Strength of
+ Strength of
Intermolecular
Intermolecular
attraction in pure A
attraction in pure B
<
Strength of
Intermolecular
attraction between
particle A and
particle B in mixture
• The vapour pressure would be lower and the boiling
point would be higher compared with ideal behaviour
Example: mixture of ethyl ethanoate & trichloromethane
O C
CH3
OC2H5
O C
CH3
Cl
H C Cl
OC2H5
Cl
Weak Van der Waal’s force
in pure ethyl ethanoate
Cl
Weak Van der Waal’s force
in pure trichloromethane
Cl
Cl C H
Cl
Cl
H C Cl
O C
CH3
OC2H5
Strong hydrogen bonding in the mixture
•
•
•
•
the interaction is now stronger after mixing the solution
===> escape tendency of molecule is lower
===> Pt is lower than expected
===> enthalpy change is exothermic
Energy(breaking interaction between molecules in pure solution) <
Energy(formation of new interaction between 2 kinds of molecules in mixture)
• ===>Volume contraction
stronger attraction between molecules in the mixture
=> intermolecular distance decrease
Ex. 21.6
•
•
•
•
•
•
•
•
•
•
a) Bromopropane and 2-bromobutane
obey
b) Tetrachloromethane and cyclohexane
obey
c) Benzene and methylbenzene
obey
d) Water and benzene
positive deviation from Raoult’s Law
e) Propanone and trichloromethane
negative deviation from Raoult’s Law
Explanation
1. The intermolecular force in pure water is strong
hydrogen bond.
2. The attraction between benzene molecules is weak
Van der Waals’ force in the pure solution.
3. However, the intermolecular force between water
and benzene is weak Van der Waals’ force in the
mixture.
4. As the average strength of intermolecular force in
the two pure solutions is greater than that in the
mixture,
5. - the escape tendency of molecules in the mixture is
higher than expected from the Raoult’s Law.
Therefore, this mixture has a positive deviation
compared with ideal solution.
- the interaction between molecules in the mixture
is weaker and the space between molecules is
larger. Thus, there will be an expansion in volume
after mixing.
- the energy absorbed for breaking the
intermolecular force between molecules in pure
solutions is greater than the energy released for
forming new intermolecular force between water
and benzene molecules in the mixture. Thus, the
process is endothermic and heat absorbed from
the surroundings.
Ex. 21.7
Set
Composition of liquid
mixture (XA)
1
2
3
4
5
6
7
8
1
0.9
0.7
0.6
0.5
0.3
0.1
0
Composition of liquid
mixture (XB)
0
0.1
0.3
0.4
0.5
0.7
0.9
1
Partial pressure of A
(PA)
30
27
21
18
15
9
3
0
Partial pressure of B (PB)
0
5
15
20
25
35
45
50
30
32
36
38
40
44
48
50
1
0.844
0.583
0.474
0.375
0.205
0.0625
0
0
0.156
0.417
0.526
0.625
0.795
0.9375
1
Vapour pressure of the
mixture (Pt)
Composition of vapour
(XA')
Composition of vapour
(XB')
Richer with B comparing with than in liquid
PAo
B.p.(B)
PBo
B.p.(A)
• What is the mole fraction of A in liquid and in vapour when
the vapour pressure is 1atm at ToC?
• Mole fraction of A in liquid is about 0.45
Mole fraction of A in vapour is about 0.55
XB
• (1) X boils at Tx
• (2) vapour y is
obtained
• (3) vapour y condense
to liquid y
• (4) liquid y boils at Ty
and vapour z is
obtained
• (5) vapour z condense
to liquid z
• (6) liquid z boils at Tz
• (7) ...... Pure A is
obtained as final
distillate
Tx
Ty
y(g)
z(g)
y(l)
x(l)
y(l)
Repeating boiling and condensation
Tedious!!!
Tz
z(l)
z(l)
Fractional distillation
fractionating tower for petroleum refining
21.5 Azeotropic Mixtures
Liquid mixtures which deviate negatively from Raoult's law
show a maximum in the boiling point-composition curve
Start from x
==> distillate = pure B
Start from w
==> distillate = pure A
Start from M
==> distillate = remain M
M is azeotropic mixture/
constant boiling mixture
(vapour composition =
liquid composition upon
boiling)
•
•
•
•
•
•
•
Start from x, residue?
Azeotropic mixture M
Start from w, residue?
Azeotropic mixture M
start from M, residue?
Azeotropic mixture M
What to do for M?
Discard?? Think about
it !!!
Positive Deviation from Raoult’s Law
Liquid mixtures which deviate positively from Raoult's law
show a minimum in the boiling point-composition curve
• Start from X, distillate and
residue?
• Distillate =
Azeotropic mixture M
Residue = pure B
• Start from Y, distillate and
residue?
• Distillate
=
Azeotropic
mixture
M
Residue = pure A
• start from M, distillate and
residue?
• Azeotropic mixture M
X
Y
• impossible to separate a non-ideal solution into
pure components by fractional distillation!
What should we do with azeotropic mixture??
Discard???
Solve the problem???
• Add a little amount of one of the component
and fractional distillation can be continued!!!
Ex 21.8
a) A constant boiling mixture (azeotropic mixture) is a
mixture of liquids with a fixed composition and it cannot
be separated by fractional distillation since the vapour
composition is the same as the liquid composition.
b) b.p. - composition diagram for HNO3 - water mixture
c) According to the diagram, it is known that the mixture
deviates negatively from Raoult’s Law. There will be a
stronger attraction between molecules in the mixture.
When nitric acid is added to water, there is an evolution of
heat. That means the reaction is exothermic. Moreover,
there is a reduction is volume when nitric acid and water
are mixed.
d) Raoult’s Law states that the partial vapour pressure of a
component in a mixture is directly proportional to its mole
fraction, and is equal to the product of its mole fraction and
the vapour pressure of the pure component at that
temperature.
e) The total vapour pressure above the nitric acid - water
mixture is less than the predicted value based on the
ideal behaviour. It shows a negative deviation from
Raoult’s law meaning that there is less tendency for the
molecules to escape from the solution than from the
pure
liquids.
f) The temperature of the mixture would rise gradually
until it reaches 122oC. The residual mixture is more
concentrated in nitric acid as water is distilled off until
its composition is 65% by mass of nitric acid.