PHASE EQUILIBRIUM
(GROUP SECOND)
1) Muhammad Shafique
2) Noor Yussuf
3) Noor Nadia Syahira
4) Norfaizah
5)Nur Alya
6) Nur Azahariah
7) Wan Muhammad Hakimi
TWO COMPONENT :
(LIQUID – LIQUID SYSTEM)
Definition & Application of Roult Law
Raoult’s law
-vapour pressure of the solute-containing solution
(P A) is equal to the mole fraction of the solvent
(XA) times the vapour pressure of the pure
solvent (P*A).
P A=XAP*A
PSOLUTION=XSOLVENTP*SOLVENT
Pressure/atm
Phase Diagram for Water
Solvent
Solution
Temp. /K
Phase Diagram for Water
Pressure/atm
Vapour pressure lowering
ΔP
Boiling point
of the solution
Freezing point
of the solution
ΔTb
ΔTf
Temp. /K
(a) Vapour-pressure lowering
If a nonvolatile solute (one that has no
tendency to escape from a solution) is
dissolved in a liquid solvent, the vapour
pressure of the solvent is lowered.
Psolution = XsolventPosolvent
(Raoult’s law )
If the solution contains only one solute:
XA + XB =1
PA = (1-XB)PoA
PA = PoA - XBPoA
(PoA – PA) = XBPoA
∆P = XBPoA
∆PA = vapour-pressure lowering
XB = the mole fraction of solute
PoA = vapour-pressure of the pure solvent
• Example :
• The vapour pressure of pure water at 20○C is
25.21 mmHg. If a nonvolatile solute,sucrose is
added to a mole fraction of 20%,what is the
resulting vapour pressure of the solution at 20
○C?
• Solution
• P A=XAP*A
•
•
0.80 x 17.5mmHg
14 mmHg
(b) Boiling point elevation
• The boiling point of the solution is higher than the
boiling point of the pure solvent.
• ∆Tb = boiling point elevation.
• For dilute solutions, the increase in boiling point
depends on the molality of the solute in the solution.
• ∆Tb = kb m
• ∆Tb = boiling point of solution – boiling point of
solvent
• kb = molal boiling point elevation constant
(b) Boiling point elevation
DTb = Tb – T ob
Tb - boiling point of
the pure solvent
T b - boiling point of
the solution
o
Tb > T b
DTb > 0
DTb = Kb m
m - molality of the solution
Kb - molal boiling-point
elevation constant (oC/m)
(c) Freezing point depression
• The freezing point of the solution is lower
than the freezing point of the pure solvent.
• The freezing point depression = ∆Tf.
• For dilute solutions, the decrease in freezing
point depends on the molality of the solute in
the solution.
Freezing-Point Depression
DTf = T f – Tf
Tf - freezing point of
the pure solvent
Tf - freezing point of
the solution
o
Tf > Tf
DTf > 0
DTf = Kf m
m - molality of the solution
Kf - molal freezing-point
depression constant (0C/m)
• Example :
• What are the freezing point and boiling point
of a solution containing 6.50 g of ethylene
glycol (C2H6O2), commonly used as an
automotive antifreeze, in 200 g water?
Solution
• To determine ∆Tf and ∆Tb. The molality of the solution has
to be calculated.
• The number of moles of C2H6O2 is,
•
•
6.50g = 0.105 mol
62.1gmol-1
• The mass of solvent in kg is,
•
•
200 kg
1000
= 0.200 kg H2O
• The molality is therefore,
• = 0.525 m C2H6O2
• For H2O , Kf = 1.86 ○C m-1
• Kb = 0.51 ○C m-1
The changes in the freezing point and boiling point:
∆Tf = (1.86○C/m)(0.525m) = 0.967○C
∆Tb = (0.51○C/m)(0.525m) = 0.270○C
DTf = T f – Tf
Tf
= 0oC - ∆Tf
= 0oC – 0.967oC
The freezing point of the solution (Tf )
=  0.976○C
∆Tb = boiling point of solution – boiling point
(100oC)
The boiling point of solution = 100.27○C
of solvent
What is the freezing point of a solution containing 478 g of
ethylene glycol (antifreeze) in 3202 g of water? The molar mass
of ethylene glycol is 62.01 g.
• DTf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg) =
478 g x
1 mol
62.01 g
= 2.41 m
3.202 kg solvent
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T f – Tf
Tf = T f – DTf = 0.00 0C – 4.48 0C = -4.48 0C
AZEOTROPIC MIXTURE
DEFINITION:
Mixture of two or more liquid substances in
such a ratio that the composition of the
mixture is not changed by simple
distillation.
• POSITIVE AZEOTROPES
-minimum boiling mixtures.
-Boiling point of an azeotrope is lower than
that of any of its constituents.
• NEGATIVE AZEOTROPES
- maximum boiling mixtures.
-Boiling point higher than that of any of its
constituents.
• HETEROAZEOTROPE.
Two liquids may be nearly immiscible but,
when their mixture is boiled, their vapor
consists of a fixed ratio of the two and coexists
with the two liquid phases.
• AZEOTROPIC
Combinations of solvents that do not form an
azeotrope when mixed in any proportion.
Phase diagram of a heteroazeotrope. Vertical axis is temperature,
horizontal axis is composition. The dotted vertical line indicates the
composition of the combined layers of the distillate whenever both
layers are present in the original mixture
NON-IDEAL SOLUTION
&
IDEAL SOLUTION
•Involves the intermolecular forces between molecules in solution are stronger
than those in pure liquid
•Therefore, vapour pressure of the solution is lower than vapour pressure of its
components or pure liquid.
•Example :
A
A
=
B
B
WEAKER THAN
A
B
•SO,the molecules in the solution have lower tendency to escape into vapour
phase.
•Therefore the process is EXOTHERMIC
Azeotrop
e
Nitric acid and water form mixtures in which
particles break away to form the vapour with
much more difficulty than in either of the
pure liquids.
That means that mixtures of nitric acid and water
can have boiling points higher than either of the
pure liquids because it needs extra heat to break
the stronger attractions in the mixture.
In the case of mixtures of nitric acid and
water, there is a maximum boiling point of
120.5°C when the mixture contains 68% by
mass of nitric acid. That compares with the
boiling point of pure nitric acid at 86°C, and
water at 100°C.
Notice the much bigger difference this time
due to the presence of the new ionic
interactions
USING THE DIAGRAM
Distilling dilute nitric acid
Start with a dilute solution of nitric acid
with a composition of C1and trace
through what happens.
As the acid loses water, it becomes more
concentrated. Its concentration gradually
increases until it gets to 68% by mass of nitric
acid. At that point, the vapour produced has
exactly the same concentration as the liquid,
because the two curves meet.
You produce a constant boiling mixture (or
azeotropic mixture or azeotrope). If you distil
dilute nitric acid, that's what you will eventually
be left with in the distillation flask. You can't
produce pure nitric acid from the dilute acid by
distilling it.
The vapour produced is richer in water than the
original acid. If you condense the vapour and
reboil it, the new vapour is even richer in water.
Fractional distillation of dilute nitric acid will
enable you to collect pure water from the top of
the fractionating column.
Distilling nitric acid more concentrated than 68% by mass
This time you are starting with a concentration C2 to the right of
the azeotropic mixture.
The vapour formed is richer in nitric
acid. If you condense and reboil this,
you will get a still richer vapour. If you
continue to do this all the way up the
fractionating column, you can get pure
nitric acid out of the top.
As far as the liquid in the distillation flask
is concerned, it is gradually losing nitric
acid. Its concentration drifts down
towards the azeotropic composition.
Once it reaches that, there can't be any
further change, because it then boils to
give a vapour with the same composition
as the liquid.
Distilling a nitric acid / water mixture
containing more than 68% by mass of
nitric acid gives you pure nitric acid from
the top of the fractionating column and
the azeotropic mixture left in the
Formed when the intermolecular forces between molecules in the mixture are
weaker than those in pure liquids.
A
A
=
B
B
STRONGER THAN
A
Vapour pressure of the solution is higher
than expected
B
The solution has a greater
tendency to evaporate or
escape into vapour
The process is endothermic
A large positive deviation from Raoult's
Law produces a vapour pressure curve
with a maximum value at some
composition other than pure A or B.
If a mixture has a high vapour pressure it
means that it will have a low boiling point
The molecules are escaping easily and you
won't have to heat the mixture much to
overcome the intermolecular attractions
completely.
The implication of this is that the boiling
point / composition curve will have a
minimum value lower than the boiling
points of either A or B.
AZEOTROPE
This particular mixture of ethanol and water
boils as if it were a pure liquid. It has a
constant boiling point, and the vapour
composition is exactly the same as the
liquid.
It is known as a constant boiling
mixture or an azeotropic mixture or
an azeotrope.
SUMMARY
Distilling a mixture of ethanol containing
less than 95.6% of ethanol by mass lets
you collect:
• A distillate containing 95.6% of ethanol in
the collecting flask (provided you are
careful with the temperature control, and
the fractionating column is long enough
• Pure water in the boiling flask.
Calculation to determine the solution is +ve or –ve deviation
If PT (actual) > P (calculated)
• -ve deviation
If PT (actual) = P (calculated)
• Ideal solution
If PT (actual) < P (calculated)
• +ve deviation
EXAMPLE 1:
Given that P⁰A is 50 kPa and P⁰B is 25 kPa.XA is 0.3
A solution of A and B has a vapour pressure of 60 kPa. Determine whether the
solution is ideal or not?
SOLUTION:
PA=XA (P⁰A)
PB =XB(P⁰B)
PT (calculated)=PA+PB
=(0.3)(50)
=(1-0.3)(25)
=15kPa+17.5kPa
=15 kPa
=(0.7)(25)
=32.5 kPa
=17.5 kPa
PT(actual) > PT(calculated)
Therefore, it is a non-ideal solution and has a +ve deviation (:
EXAMPLE 2:
Given that P⁰A is 32 kPa and P⁰B is 60 kPa
A solution has a vapour pressure 50 kPa contains 12 mol of A and 15 mol of B.
Does this solution behave ideally?
SOLUTION:
PA=XA(P⁰A)
=[12÷(12+15)] [32]
=14.22 kPa
PB=XB(P⁰B)
=[15(12+15)] [60]
=33.60 kPa
As PT(actual) > PT(calculated)
Therefore, the solution is non-ideal solution.
The answer is, NO,the solution doesn’t behave ideally.
PT(calculated)=14.22+33.60
=47.8 kPa
Principle in simple distillation & fractional
distillation of a Binary Liquid Mixture
Prepared By :
• Noor Yussuf b. Othman
• Nur Alya bt. Razak
• What is Simple Distillation??
> The process of heating and condensing
to purify substances
SIMPLE DISTILLATION
• Is used to separate a mixture of liquid with
difference in boiling point
• It is less effective than fractional distillation as the
distillate is still a mixture of both components
• The principle is to conduct multiple, little
distillations . Each vaporisation or condensation
cycle will result in the new vapour phase being
enriched in the lower boiling point (higher vapour
pressure component)
• By using more of these cycles, two components
with different boiling points can be separated.
Simple distillation from pure
water from salt water.
1. The impure liquid is heated in distillation flask. Since
water vaporise at much lower temperature than salt,
pure water vapour is formed.
2. Water vapour is cooled in condenser and condenses
back into liquid water which is collected in another
flask. The collected water is pure water and does not
contain any salt.
3. The liquid obtained through condensation of vapour in
the distillation process is known as distillate.
4. The salt behind in distillation flask is known as residue.
• Fractional distillation
• Definition : separation of a mixture into its
component parts or fractions
Fractional distillation
• Separates chemical compounds by their
boiling points by heating them to a
temperature at which several fractions of the
compound will evaporate
• Generally the components parts boil at less
than 25’C from each other under one
atmospheric pressure
• A fractionating column is inserted between
the distillation flask and the distillation head
• It provides a large surface area in which
mixture can be continuously vaporized and
condensed
Principle of fractional distillation
• As the vapours ascend the column from the
boiling mixture below, the high boiling
components are condensed and returned to the
flask, the ascending column of vapour thus being
steadily ‘scrubbed’ by the descending column of
liquid condensate
• The ascending column of the vapour becomes
therefore steadily richer in the lowest boiling
component, and the descending column of
condensate steadily richer in the highest boiling
component
Boiling point – composition diagram
T
B
T
w
T
x
TA
Y
X
W
Boiling point-composition diagram
• A has a lower boiling point so A is more
volatile than B.
• Assume we have a liquid mixture with
composition W. This mixture will boil at Tw and
give a vapour of composition X, which is richer
in A.
• If this vapour of composition of X is
condensed, it will form a liquid which also has
composition X.
Boiling point-composition diagram
• This new liquid will boil at temperature Tx (which
is lower than Tw ) to give a vapour of composition
Y, which condenses to give a liquid of
composition Y. This liquid is even richer in A
• The coiling and condensation cycle can be
repeated until a vapour consisting of pure A is
abtained which can be condensed into a liquid of
pure A.
• Pure A is the distillate. The less volatile liquid
remaining in the flask is pure B and is known as
the residue.
Vapour pressure-composition diagram
Vapour pressure-composition diagram
• Assume you have an ideal mixture of two
liquids, A and B.
• Each of A and B is making its own contribution
to the overall vapour pressure of the mixture.
• Suppose you double the mole fraction of A in
the mixture. According to Raoult’s Law, you
will double its partial vapour pressure. If you
triple the mole fraction, its partial vapour
pressure will triple, and so on.
Vapour pressure-composition diagram
• In other words, the partial vapour pressure of A
at a particular temperature is proportional to its
mole fraction.
• If you plot a graph of the partial vapour pressure
of A against its mole fraction, you will get a
straight line sloping up.
• Now we will plot the line for B. The mole fraction
of B falls as A increases so the line will slope
down.
• As the mole fraction of B falls, its vapour pressure
will fall at the same rate.
Vapour pressure-composition diagram
• Notice that the vapour pressure of pure B is
way higher than A. That means that molecules
must break away more easily form the
sureface of B than A. B is more volatile than A.
• To get the total vapour pressure, you need to
add the values of A and B together at each
composition. The net effect of that is to give
you a straight line as shown in the diagram.