Phase Equilibria and Colligative Properties.pdf

Fundamentals of Phase Equilibria
Phase Diagrams of One-Component Systems
Phase Diagrams of Two-Component Systems
Colligative Properties and Raoult’s Law
Osmosis, Osmotic Pressure, and Molecular Weight Determination

1. Phase Equilibria and
Colligative Properties:
Detailed Concepts,
Diagrams, and
Applications
UNDERSTANDING PHASE
RULES AND PROPERTIES
WITH PRACTICAL USES

2. Discussion Outline
•Fundamentals of Phase Equilibria
•Phase Diagrams of One-
Component Systems
•Phase Diagrams of Two-
Component Systems
•Colligative Properties and Raoult’s
Law
•Osmosis, Osmotic Pressure, and
Molecular Weight Determination

3. Fundamentals of
Phase Equilibria

4. Definition of Phases, Components, and Degrees of Freedom
Phases: Phases are distinct, homogeneous physical
parts of a system visible to the eyeor microscope.
Components: Components are chemically
independent substances that constitute the system.
Degrees of Freedom: Degrees of freedom are the
number of intensive variables that can change
independently without altering phase equilibrium.

5. Degrees of Freedom
DegreeofFreedomrepresentedbyFinthephaseRuleequation(F+P=C+2)
isdefinedasfollows:theleastnumberofvariablefactors(e.g.concentration,
pressureandtemperature)whichmustbespecifiedsothattheremainingvariables
arefixedautomaticallyandthesystemiscompletelydefined
AsystemwithF=0isknownasnonvariantorhavingnodegreeoffreedom.
AsystemwithF=1isknownasunivariantorhavingonedegreeoffreedom.
AsystemwithF=2isknownasbivariantorhavingtwodegreesoffreedom..

6. Gibbs Phase Rule and Its Significance
Gibbs Phase Rule Formula
The formula F = C -P + 2 defines the degrees of freedom in relation to
components and phases in a single-component system.
The formula F = C -P + 1 defines the degrees of freedom in relation to
components and phases in a two-component system.
Significance in Phase Coexistence
The rule helps predict how many variables can be controlled when multiple phases
coexist in equilibrium.

7. PhaseDiagramsofOne-ComponentSystems
SOLID
LIQUID
VAPOUR
TEMPERATURE (°C)
PRESSURE (ATM)
F = C - P + 2

8. Water: Phase Diagram
(1) The Curves OA, OB, OC
(2) The Triple Point O
(3) The Areas AOC, AOB,
BOC, and vapor phases
coexist in equilibrium.
A
O
B
C
Ice
Water
Water vapour
Temperature (°C)
Pressure (
atm
)
Triple point
Supercritical
water

9. Water: Phase Diagram
(1)The Curves OA, OB, OC
a) OA, the Vapour Pressure curve of Water.
Water-Water Vapour
b) OB, the Sublimation curve of Ice
Solid-Water Vapour
c) OC, the Fusion curve of Ice
Solid-Water
Along the cureve system has one degree of
freedom i.e., it is monovariant
A
O
B
C
Ice
Water
Water vapour
Temperature (°C)
Pressure (
atm
)
Triple point
Supercritical
water
F = C – P + 2
F = 1 – 2 + 2 = 1

10. Water: Phase Diagram
(2) The Triple Point O
The curves OA, OB and OC
meet at the triple point ‘O’
where all the three phases are
in equilibrium.

.
A
O
B
C
Ice
Water
Water vapour
Temperature (°C)
Pressure (
atm
)
Triple point
Supercritical
water
F = C– P + 2
F = 1 – 3 + 2 = 0
The system at the triple point is
nonvariant

11. Water: Phase Diagram
(3) The Areas AOC, AOB, BOC,
and vapor phases coexist in
equilibrium.
The areas or regions between the curves
show the conditions of temperature and
pressure under which a single phase–ice,
water or vapour is capable of stable
existence.
Thus each phase has 2 degrees of freedom
i.e., the system is bivariant.
A
O
B
C
Ice
Water
Water vapour
Temperature (°C)
Pressure (
atm
)
Triple point
Supercritical
water
F = C – P + 2
F = 1 – 1 + 2 = 2

12. Sulphur: Phase Diagram
(i) The six curves AB, BC, CD, BE, CE, EG
(ii) The three Triple points B, C, E
(iii) The four areas :
ABG marked ‘solid Rhombic’
BEC marked ‘solid Monoclinic’
GECD marked ‘liquid Sulphur’
ABCD marked ‘Sulphur vapour
(iv) Metastable (Point F) A metastable phase is a
state that is stable under certain conditions but has
a higher energy than a more stable, lower-energy
phase
A
E
B
C
Solid S
R
Temperature (°C)
Pressure (
atm
)
Liquid S
L
Vapour S
V
G
F

13. Sulphur: Allotropic Transformations and Phase Changes
Phase Diagram
The phase diagram of sulphur shows the temperature
and pressure conditions for different solid and liquid
phases.
Sulphur Allotropes
Sulphur exists in multiple allotropes that change
structure based on temperature variations.
Phase Transitions
Sulphur undergoes transitions between solid
allotropes, melting, and vaporization as temperature
increases.
A
E
B
C
Solid S
R
Temperature (°C)
Pressure (
atm
)
Vapour S
V
G
F

14. Carbon Dioxide: Phase Transitions Including the Triple Point
Triple Point (B)
Carbon dioxide’s triple point occurs at a specific temperature
and pressure where solid, liquid, and gas phases coexist in
equilibrium.
Sublimation Process (Curve AB)
At pressures below the triple point, carbon dioxide
sublimates directly from solid to gas without becoming
liquid.
Vapour Pressure curve (BC)
Liquid→solid under certain conditions.
Super critical fluid
A
E
B
C
Solid
Supercritical
fluid
Temperature (°C)
Pressure (
atm
)
Vapour
D
Liquid

15. Phase Diagrams of
Two-Component
Systems

16. Systems with Congruent
Melting Points (Ag-Pb)
Congruent Melting Concept
Congruent melting occurs when an intermediate
compound melts uniformly as a single phase without
decomposition.
Phase Diagram Characteristics
The phase diagram of Ag-Pb shows melting points
that correspond precisely to the stoichiometric
composition of the compound.
Composition
Temperature (
°
C)
100% Ag
%0 Pb
0% Ag
100% Pb
2.5 Ag
97.5 Pb
C
A
B
Solid Ag/
Solution (Pb/Ag)
Solution
(Pb/Ag)
Solid Pb/
Solution (Pb/Ag)
Solid Pb/Ag
303
961
327
Eutectic composition

17. Eutectic Systems: Concept
and Zn-Mg Example
Eutectic System Definition
Eutectic systems have a specific composition and
temperature with the lowest melting point in an alloy
system.
Zn-Mg System Example
The Zn-Mg alloy demonstrates eutectic behavior
with two solid phases coexisting at the eutectic
composition and temperature.
Composition
Temperature (
°
C)
0% Zn
100% Mg
100% Zn
0% Pb
C
A
B
Mg + liquid
Liquid
Zn + liquid
Mg + MgZn2
651
420
D
E
X
Mg + MgZn2
MgZn2 +
Liquid
575

18. Systems with Incongruent
Melting Points (NaCl-H2O)
Incongruent Melting Concept
Incongruent melting occurs when a compound
decomposes or changes phase instead of melting
directly.
NaCl-H2O System Behavior
The NaCl-H2O system exhibits hydrate formation
and dissociation during heating, illustrating
incongruent melting.
100% H
2O
0% NaCl
0% H
2O
100% NaCl
23.1
97.5 Pb
B
A
ICE +NaCl. 2H
2O
-21
0.0
0.15
ICE
+ Solution
NaCl. 2H
2O
+ Solution
NaCl
+ Solution
Temperature (
°
C)
Solution
C
D
E

19. Colligative
Properties and
Raoult’s Law
Lowering of vapour pressure, depression in freezing point, elevation in boiling point. Osmosis and osmotic
pressure. Experimental methods and determination of molecular weight.

20. Colligative Properties
A colligative property may be defined as one which
depends on the number of particles in solution and
not in any way on the size or chemical nature of the
particles.
Greek colligatus = Collected together

21. Colligative Properties
(1) Lowering of the Vapour Pressure
(2) Elevation of the Boiling Point
(3) Depression of the Freezing Point
(4) Osmotic Pressure
Vapour Pressure

22. Colligative Properties
(1) Lowering of the Vapour Pressure
(2) Elevation of the Boiling Point
(3) Depression of the Freezing Point
(4) Osmotic Pressure

23. Lowering of the Vapour Pressure Vapour Pressure
PsThe vapour pressure of pure solvent (P) is always greater than
the vapour pressure of the solution(Ps) when we add a non-
volatile solute to a pure solvent
??????� ∝
??????
�+ ??????
??????�=�
??????
�+ ??????
The vapour pressure of the solution is determined by the
number of molecules of the solvent present at any time on
the surface, which is proportional to the mole fraction (X).
Mole fraction of
pure solvent (P)
=
??????
??????+ ??????
=
??????
0+ ??????
=1
??????=�
??????
�+ ??????
??????ℎ�������,??????=�
??????�=??????
??????
�+ ??????
??????−??????�
??????
=
�
�+ �
Raoult’s law
P>Ps
n = number of moles of Solute, N = number of molecules of Solvent
?????? ∝??????
P

24. Detail Derivation
??????�=??????
??????
�+ ??????
1−
??????�
??????
=1−
??????
�+ ??????
??????−??????�
??????
=
�+ ??????−??????
�+ ??????
??????−??????�
??????
=
�+??????−??????
�+ ??????
∆??????
??????
=
�
�+ ??????
??????−??????�
??????
=
�
�+ ??????
∆?????? =?????? ×���� ��????????????�??????�� �� ������

25. Raoult’s Law and Ideal Solutions
??????−??????�
??????
=
�
�+ �
The relative lowering of the vapour
pressure of a dilute solution is equal to
the mole fraction of the solute present in
dilute solution.
?????? ∝??????
??????=??????
�×??????°
.
Ideal solutions perfectly follow Raoult’s law
with no deviations in vapor pressure
measurements.
Pressure
Mole fraction
�
�= 0 �
�=1
�
�= 1 �
�=0
??????°
�
??????°
�

26. Raoult’s Law and Ideal Solutions
Calculate the vapour pressure lowering caused by the addition of 100 g of sucrose
(mol mass = 342 g) to 1000 g of water, if the vapour pressure of pure water at 25°C
is 23.8 mm Hg.
??????−??????�
??????
=
∆??????
??????
=
�
�+ �
∆?????? = lowering of vapour pressure
p = vapour pressure of water = 23.8 mm Hg
n = moles of sucrose =
���
���
=0.292 moles
N = moles of water=
����
��
1= 55.5 moles
∆??????
23.8
=
0.292
0.292+55.5
Thus, the lowering of vapour
pressure is 0.125 mm Hg
∆?????? =0.125 ��

27. Determination of Molecular Mass from Vapour Pressure Lowering
The molecular mass of a nonvolatile solute can be determined by measuring the lowering of vapour
pressure (p – p
s) produced by dissolving a known weight of solute in a known weight of the solvent.
If w grams of solute is dissolved in W grams of the solvent, m and M are
molecular masses of the solute and solvent respectively, we have :
��.�� ����� �� ������=
�
�
=
�
�+ �
p – ps
�
=
�
�+ �
��.�� ����� �� �������=
�
�
=
�
�+ �
p – ps
�
=
�
�
�
�
+
�
�
p – ps
�
=
�
�
�
�
For a dilute solutions, the number of moles
(molecules) of solute (w/m), is very small
p – ps
�
=
��
��

28. Raoult’s Law and Ideal Solutions
The vapour pressure of ether (mol mass = 74) is 442 mm of Hg at 298 K. If 3g of a
compound A are dissolved in 50 g of ether at this temperature, the vapour pressure
falls to 426 mm of Hg. Calculate the molecular mass of A. Assume that the solution of
A in ether is very dilute.
??????−??????�
??????
=
��
��
?????? = 442 mm of Hg
??????
� = 426 mm of Hg
w = 3.0
W = 50 g
M = 74
�=
442 ×3×74
50×(442−426)
Molecular mass of A is 122.66
g
�=122.66 �

29. Elevation of the Boiling Point
Temperature (°C)
Vapour Pressure
Atmospheric pressure
T
ST
1T
2
When a pure solvent is heated, its vapour pressure increases and
when it equals the atmospheric pressure, it boils. The addition of a
non-volatile solute A lowers the vapour pressure and elevates the
boiling point, as the solution has to be heated to a higher
temperature to make its vapour pressure becomes equal to
atmospheric pressure.
∆??????=??????−??????�

30. Elevation of the Boiling Point
Temperature (°C)
Vapour Pressure
Atmospheric pressure
T
ST
1T
2
When a pure solvent is heated, its vapour pressure increases and
when it equals the atmospheric pressure, it boils. The addition of a
non-volatile solute A lowers the vapour pressure and elevates the
boiling point, as the solution has to be heated to a higher
temperature to make its vapour pressure becomes equal to
atmospheric pressure.
∆??????=??????−??????�

31. Elevation of the Boiling Point
Temperature (°C)
Vapour Pressure
Atmospheric pressure
T
ST
1T
2
When a pure solvent is heated, its vapour pressure increases
and when it equals the atmospheric pressure, it boils. The
addition of a non-volatile solute A lowers the vapour
pressure and elevates the boiling point, as the solution has
to be heated to a higher temperature to make its vapour
pressure becomes equal to atmospheric pressure.
∆??????=??????−??????�
P
S
P
1
P
2
AB C
E
D
��
��
=
��
��
??????�−??????�
??????�−??????�
=
??????−??????�
??????−??????�
∆??????∝
??????−??????�
??????
∆??????∝
��
��
∆??????=��
�
��
�
�

=Boiling point constant or Ebulioscopic constant of molal elevation constant;
�
� = mole fraction of solute
??????−??????�
??????
=
��
��
(??????������ ���)
∆??????=���
Since M (mol mass of solvent) is
constant

32. Depression of the Freezing Point
Temperature (°C)
Vapour Pressure
T
fT
1
T
2
The vapour pressure of a pure liquid changes with temperature
as shown by the curve ABC at B the freezing starts. Thus, point
B corresponds to the freezing point (T
f) of pure solvent.
The vapour pressure curves of a solution of a non-volatile solute
in the same solvent is similar to the vapour pressure curve of
the pure solvent and meets the freezing point curve at F and C,
indicating that T1 and T2 are freezing points of the solution,
respectively. The difference of the freezing point (∆??????) of the
pure solvent and the he solution is
∆??????=??????�−??????
A
B
C
F
P
s
P
1
P
2

33. Depression of the Freezing Point
Temperature (°C)
Vapour Pressure
T
fT
1
T
2
∆??????=??????
�−??????
A
B
C
F
P
s
P
1
P
2
E
D
��
��
=
��
��
??????
�−??????�
??????
�−??????�
=
??????−??????�
??????−??????�
∆??????∝
??????−??????�
??????
∆??????∝
��
��
∆??????=��
��
��
??????−??????�
??????
=
��
��
(??????������ ���)
∆??????=���
�
� = Freezing-point constant or Cryoscopic constant or Molal depression constant.
�
� = mole fraction of solute
Since M (mol mass of solvent) is
constant

34. Osmotic Pressure
(Greek Osmos = to push)
Semipermeable membrane
The flow of the solvent through a semipermeable
membrane from pure solvent to solution, or from a
dilute solution to a concentrated solution, is termed
Osmosis.

35. Osmotic Pressure
(Greek Osmos = to push)
Semipermeable membrane
The flow of the solvent through a semipermeable
membrane from pure solvent to solution, or from a
dilute solution to a concentrated solution, is termed
Osmosis.

36. Osmotic Pressure
Semipermeable membrane
Pressure
The hydrostatic pressure built up on the solution,
which stops the osmosis of pure solvent into the
solution through a semipermeable membrane, is called
Osmotic Pressure.

37. Osmotic Pressure
The osmotic pressure (π)
of a dilute solution is
inversely proportional to
the volume (V) containing
1 mole of the solute and
is directly proportional to
the absolute temperature
(T).
π∝
�
�
π∝??????
π V = R ?????? Where R is the Gas constant
If n moles of solute are dissolved in V litres of solution
π V = n R ??????
Van’t Hoff Equation for Solutions

38. Experimental methods and determination of molecular weight.
The boiling point of a solution containing 0.20 g of a substance A in 20.00 g of
ether is 0.17 K higher than that of pure ether. Calculate the molecular mass of X.
The Boiling point constant of ether per 1 Kg is 2.16 K.
∆??????=��
�
��
�.��=��
�.��
� ���
The boiling point constant of ether per 1 Kg is 2.16 K
The boiling point constant of ether per 1 g is 1000 X 2.16 K
�.��=
���� �.���.�
� ��
�.��=
���� �.���.�
� ��
�=
���� �.���.�
�.��× ��
The molecular mass of A =127.81 g

39. If the boiling point constant of ether per 1 Kg is 2.16 K, then the
boiling point constant of ether per 1 g is 1000 X 2.16 K. How?
�
�=
��
∆??????�
�
�=
�
���
�
°� ×�
�
�=
�
���
°�
�
�= g ×°�× mol
-1
�
�=
Kg
1000
°� mol
-1
����×��=�� °� mol
-1

40. Experimental methods and determination of molecular weight.
0.440 g of a substance A dissolved in 22.2 g of benzene lowered the
freezing point of benzene by 0.567 °C. Calculate the molecular mass of
the substance. (K
f = 5.12 °C mol
–1
)
∆??????=��
�
��
�.���=�
�
0.440 �
� 22.2 g
�=
���� ×5.12×0.440
�.���×22.2
The molecular mass of A =127.81 g
�.���=
����×5.12×0.440 �
� 22.2 g
�=178.9 g

41. Do Read
A. Bahl and G.D. Tuli, Essentials of Physical Chemistry by S. Chand Publication (2019, New
Delhi, 26thEdn.
Puri, Sharma and Pathania, Principles of Physical Chemistry. Vishal publishing house, (2018),
New Delhi 1stEdn.