This power point is based on the syllabus of TYBSc SEM V, unit 2, Paper 1, Mumbai university. This covers colligative properties, relative lowering of vapour pressure, elevation in boiling point, depressure in freezing point, Osmotic pressure, Chemical kinetics, various methods to find out these colligative properties, reverse osmosis, vant hoff factor, some numerical.

1. T. Y. B. Sc - SEM V
Unit II
(Chemical
Thermodynamics and
Chemical Kinetics)
Chemical Thermodynamics

2. Colligative properties
The properties of dilute solutions that depends
only on the number of solute molecules and
not on the type or chemical nature of the
solute.
Four colligative properties are –
1.Relative lowering of vapour pressure
2.Elevation in boiling point
3.Depression in Freezing point
4.Osmotic pressure

3. Relative lowering of vapour pressure
•When solute is dissolve in a solvent at a certain
temp. vapour pressure of a solvent lowers or
decreases
• Extent of decrease of vapour pressure depends on
the relative amount of the solute present in the
solution.
•Quantitative relationship between the lowering of
vapour pressure and the conc. of the solution can be
obtained by applying Raoult’s Law.
•Raoults Law
Partial vapour pressure of a constituent of a liquid
solution is equal to the product of its mole fraction X1
and its vapour pressure in the pure state

4. Relative lowering of vapour pressure
p = X1 P°
(1)
X1 = mole fraction of
solvent
P° = vapour pressure of
pure solvent
P = vapour pressure of
the solvent above a given
solution
Dissolution of a solute in a
solvent leads to a lowering
of the vapour pressure of
the-------

5. The magnitude of lowering of vapour pessure is given
as-
 ∆P = P° - P (2)
 From (1) and (2)
 ∆P = P° - P°X1
 = P° (1- X1 ) (X1 + X2 = 1)
 ∆P = P° X2 ( X2 = 1- X1 )
 X1 and X2 are mole fraction of solvent and solute
respectively

6. Thus the lowering of vapour pressure of the solvent depends
both on the vapour
pressure of the solvent and the mole fraction of solute in
solution. The relative
lowering of vapour pressure will be given as –
∆P = P° - P = P° X2 = X2 (3)
P° P° P°
Thus relative lowering of vapour pressure is equal to the
mole
fraction of the solute in solution
This is another statement of Raoult,s law.
From equation 3 we can conclude that the relative lowering of
vapour pressure
is totally independent of either nature of solvent or solute but

7. Experimental Determination of vapour
pressure
Dynamic method- (Gas saturation method)
•The apparatus consists of two sets of bulbs.
•The first set of three bulbs is filled with solution to half of their
capacity and second set of another three bulbs is filled with the
pure solvent.
•Each set is separately weighed accurately.
•Both sets are connected to each other and then with the
accurately weighed set of guard tubes filled with anhydrous
calcium chloride or some other dehydrating agents like P2O5,
conc. H2SO4 etc. The bulbs of solution and pure solvent are
kept in a thermostat maintained at a constant temperature.

8. Ostwald and Walker’s method

9. Experimental Determination of vapour
pressure
……Dynamic method
•In this method, a stream of fixed volume of dry air is
passed through
the pure solvent and finally an absorbent to absorb the
vapours of
the solvent. With aqueous solutions, anhydrous calcium
chloride is
used as the absorbent.
• As the air passes through the solution, it becomes
saturated upto
the vapour pressure of the solution.
• The actual loss of vapour is determined by weighing the
bulbs before
And after the air passed through them.

10. • Already saturated air with pressure of solvent upto pressure p
is passed through preweighted bulb containing the solvent
• VP of pure solvent p0 is greater then solution p,
•Air sample will take more vapours from pure solvent till it
become saturated
upto pressure p0
•Loss in mass of bulb of solution is W2, which is amount
required to saturate air from p to P0
∆W2 = p° - p
• Air is passed through a set of weighted u-tube, increase in
weight of U tube is
∆W1 + ∆W2
Vapour pressure required to saturate air from p to p°-
P° - P = ∆W2 = loss in mass of
solvent (bulb B)
P° ∆W1 + ∆W2 Gain in mass of U

11. Experimental Determination of
vapour pressure
2 Static Method- Diffrential method (manometric metho
Sugar solution
Water
Water molecules
Water molecules
Mercur
y
h = lowered vapour pressure

12. •In this method, the difference between the vapour pressures of
the
solvent and the solution is determined with the help of a
differential manometer.
• It consists of two bulbs which are connected to a manometer.
• One arm of the manometer is connected with the bulb, A,
containing the solvent and the
• other arm with the bulb, B, containing the solution.
The manometric liquid is an inert, non-volatile, low density
liquid such as P-bromonaphthalene, nbutylphthalate, etc.
• From the difference in the levels of the liquid in the two arms,
the difference in vapour pressure between the solvent and the
solution can be determined directly.

13. Static Method- Raoults Barometric
method
https://www.youtube.com/watch?v=XtXCLr
NuzY0

14. Elevation of boiling point
•Boiling point- The temp.
at which vapour pressure of
liquid is equal to the
atmospheric pressure.
• Vapour pressure of a
solution is lower than that of
the pure solvent, the
temperature
at which the solution
reaches atmospheric
pressure is higher i.e.
boiling point is raised.
• Difference between the
boiling point of solution and

15. •The elevation of boiling point can be easily interpreted from
the vapour temp. diagram
Elevation in boiling point-
∆Tb = T - T° (1)
[T° - b.p. of solvent, T = b.p. for solution]
•External pressure is P° (1 atm under normal condition)
At point E, solvent boils at T ° because at this temp. its
vapour pressure is P°, but solution attains this pressure P°
only at temp T (point F) hence boiling point of solution
corresponds to T which is higher than T° .
•By applying Claussius –Clapeyron equation and Raoults
law, it is possible to obtain a relation between elevation in
b.p. of the solution and concencentration.

16. Clausius Clapeyron equation - (relates variables in equilibrium
between phase of one component system )
ln P2 = ∆Hvap [ 1- 1 ]
P1 R T1 T2
Since point G(T ° , P) and F (T, P ° ) lie on the same curve CD,
applying C. C. equation to the solution –vapour equilibrium, the
expression becomes-
ln P° = ∆Hvap [ 1 - 1 ] (2)
P R T° T
(∆Hvap heat of vapourization per mole of solvent, P = Vapour Pressure
of solution at T° , P° = VapourPressure at T)
ln P° = ∆Hvap [T - T° ]
P R T° T
For dilute solutio, the difference between T and T° is small so T° ͟͠
T
ln P° = ∆Hvap [T - T° ]
P R T°
2

17. …..From equation
(1)
ln P° = ∆Hvap ∆Tb
(3)
P R T°
2
At temperature T°, the vapour pressure of the solution is P
(point G) whereas that of the pure solvent is P° (point E),
thus Raoults law could be applied
P = X1 P°
P = X1
P°
P = 1 - X2
P°
ln(P ) = ln (1 - X2)
P°

18. ln(P° ) = - ln (1 - X2) (4)
P
From (3) and (4)
ln(P° ) = - ln (1 - X2) = ∆Hvap ∆Tb
P R T°
2
ln (1 - X2) = - ∆Hvap ∆Tb (5)
R T°
2

19. For small value of X2
ln (1 - X2) ͟͠ - X2 (6)
from (5) and (6)
- X2 = - ∆Hvap ∆Tb
R T°
2
∆Tb = R T°
2 X2
(7)
∆Hvap
Application of equation 7 is to determine molecular weight of dissolved solute
X2 = n2 (8) ( n2 + n1 = n1 ,since for dil. Solution n1˃˃ n2 )
n2 + n1

20. ……From (7) and (8)
∆Tb = R T°
2 x n2
∆Hvap n1
= _R T°
2 x W2 / M2 M1 = M W of solvent, W1 =
Amount of solvent
∆Hvap W1 / M1 M2 = M W of solute, W2 = Amount
of solute
= R T°
2 x W2 / M2
∆Hvap W1
M1
= R T°
2 x ____ W2_____ (9) { lv = ∆Hv }
lv M2 x W1 M1
lv = Heat of vaporization per gram of the solvent, concentration of solution
is expressed in terms of its molality.

21. Equation 7 represents elevation in boiling point, since for a given solvent
∆Hvap ,
T° are constant the boiling point elevation is directly proportional to mole
fraction of solute and it is independent of nature of the solute, so
elevation in boiling point
is a colligative property.
W1 gram of solvent contains
W2 moles of solute in 1Kg of solvent will corresponds to
W2__ x 1000
M2
M2.W1
m = W2__ x 1000
M2.W1
W2__ = m_ (10)
M2.W1 1000
From (9) and (10)

22. ∆ ∆Tb = Kb
m Tb = Kb m
∆Tb = Kb m
(11)
Unit of Kb = K Kg mol-1
Boiling point elevation of any dil solution is directly
proportional to molality
Where Kb = __T°
2 R____
lv 1000
constant Kb is called as the molal boiling point elevation
constant or the molal elevation constant or
ebullioscopic constant.
Ebullioscopic constant- Elevation of boiling point
produced by dissolving one mole of solute in 1Kg of
solvent (i.e. 1 molal solution).

23. Molecular weight from boiling point
elevation
From equation (11)
∆Tb = Kb m
m = W2__ x 1000
M2.W1
∆Tb = Kb W2__ x 1000
M2.W1
M2 = Kb W2__ x 1000
∆Tb .W1

24. Freezing Point
Definition
Temperature at which the vapour pressure of solid is equal to the
vapour pressure of liquid.
Or
The temperature at which the liquid and its solid state coexist at
equilibrium, this condition is obtained when the vapour pressure
of solid and liquid are equal.
Depression of Freezing point
Solution has lower vapour pressure then pure solvent and hence
freezes at lower temp. than pure solvent. Thus there is depression
of freezing point of solvent when a non-volatile solute is dissolve
in it.
If T° is the freezing point of pure solvent and T is the freezing
point of solution then
∆Tf = T° - T (1) (T° > T)

26. From diagram at point B, the two forms (L,S) have same
vapour pressure, therefore T°, the temperature corresponds to
point B must be the freezing point of pure solvent. When a
solute is dissolve in a solvent, vapour pressure of solvent is
lowered. A new equilibrium is established at point E, where
vapour pressure of solvent of the solution and solid solvent
becomes identical.
The temp. T corresponds to the point E is the freezing point of
the solution.
Vapour pressure curve of the solution EF, always lies below the
vapour pressure curve of pure solvent, hence intersection of
vapour pressure curves of solution and solid solvent can occure
only at a point lower than T°. Therefore any solution must have
freezing point T, lower than that of the solvent T°.

27. Magnitude of ∆Tf depends both on the nature of the solvent and the
concentration of the solution.
Let
Ps = VP of solid and pure liquid solvent at T° (B)
P = VP of solid solvent and solution at temp. T (E)
P° = VP of pure supercooled liquid at temp. T (G)

33. Determination of Depression in freezing
point
1. Backmann Method

34. Construction
•Backman thermometer consisit of a thermometer bulb at the end
of a capillary tube which is connected to a mercury reservior
located at the top. The entire scale covers 6K.
By proper adjustment, initially mercury level should be on the
scale.
• Beckman apparatus consist of a freezing tube(A) with side arm
(C) through which a definite amount of solute can be added.
• A stopper with the Backman thermometer (B) and stirrer (D) is
fitted into the freezing tube.
•A guard tube (E) surrounds the tube to keep air space between A
andF. This helps to prevent rapid cooling of the contents of A.
•F is a wide vessel containing freezing mixture.
•The freezing mixture should be approximately 4-5 °C, below
freezing point
of pure solvent.

35. Working
•A known weight of pure solvent is placed in tube A.
•It is cooled with slow and continuous stirring
•This will lead to super cooling and the temp. of the solvent
will decrease by about 0.5 °C below its freezing point.
•Stirring is done vigorously when solid start separating and
the temp.
rises to the exact freezing point
• Once the temp. remain constant, it is noted as T°
•Same processor is repeated for solute.

36. Determination of Depression in freezing
point
•K. Rast (1922) used camphor and camphor derivatives as solvent for
cryoscopic work.
•This method can be used for determining the relative molecular masses of
those nonvolatile solutes which are soluble in molten camphor. Molal
depression constant of camphor is very high, i.e., 40.00 K kg mol-I.
• It means that when one mole of a solute is dissolved in one kilogram of
camphor, the depression in freezing point is 40°, which can be read using
ordinary thermometers.
•Small amount of camphor is thoroughly powdered and then introduced into
the capillary tube. Its melting point is then determined. (T°)
•A known mass of the solute is then mixed with 10 to 15 times its mass of
camphor and, the whole mixture is melted.
•After cooling, the solid mixture is thoroughly powdered and, its melting point
determined as described for camphor (T). The difference between the two
readings gives the depression of freezing point.
Tf = T° - T
2. Rast
Method

37. In this method we actually measure the depression in melting point of camphor.
However melting point of solid phase and freezing point of liquid phase of any subs

38. Numerical
1. A solution containing 0.514 g of a solute in 150g
of ethanol was found to boil at 351.34 K. If the
boiling point of ethanol is 351.3 K and and its molal
elevation constant is 1.19, calculate the molar
mass of the solute.
2. 0.498g of urea when dissolved in 25g of water
gave a boiling point elevation of 0.170C. If molar
mass of urea is 60, calculate the molal elevation
constant of water.

39. Osmosis and osmotic pressure
Semipermeable membrane – membrane that is permeable to
solvent molecules but not to the solute particles
Osmosis- flow of solvent from pure solvent to solution or from
lower to higher concentration.
Osmotic pressure (∏)- Pressure that must be applied to the
solution side to stop the inward flow of solvent.
Osmotic pressure is a colligative property and is related to the
activity of the solvent.
For dilute solution, the osmotic pressure is directly proportional
to the concentration of the solution.
Van’t Hoff equation
∏ = C R T

40. Derivation of Van’t Hoff equation
At equilibrium, the
chemical potential µ or
free energy per mole
of the solvent (A) will be
the same on both sides
of the membrane.
The solvent in pure
solvent compartment I is
at constant temperature
and is not subject to any
pressure change or
solute addition
Thus,
d (uA)pure solvent = 0

42. C is the concentration in terms of moles per liter of the solution or molarity of the solution.
This equation is valid only for dil. solution.

43. Measurement of Osmotic
Pressure

44. Measurement of Osmotic Pressure
Berkeley and Hartley's Method
Construction
•It consists of a porous tube A with a semipermeable membrane of copper
ferrocyanide deposited on its walls.
•The porous tube is fitted with a solvent reservoir on one side and a capillary
indicator (B) on the
other side.
•The porous tube containing the semipermeable membrane is filled with the
pure solvent and is surrounded by another tube (C) made of gun metal,
containing the solution whose osmotic pressure is to be measured.
• Due to osmosis, the solvent from the porous tube passes through the
semipermeable membrane into the solution.
•This movement of solvent particles is indicated by a fall in level in the
capillary indicator.

45. working
• The solvent is introduced in the inner tube through the
funnel, its level in the capillary act as an indicator.
• The funnel is disconnected from the inner tube by a stopcock
• The solution is taken in the jacket
• Due to osmosis, solvent from the cell passes into the
solution, and indicator level falls in the capillary
• Now applying pressure on the solvent in the outer tube, the
solvent is forced back into the cell
until the indicator shows the initial level.
• Applied pressure is measured by pressure gauge
•This is the osmotic pressure of the solution at the temperature
of experiment

46. Osmosis and reverse osmosis
Osmosis is the movement of
solvent molecules from the region
of pure solvent (area of low solute
concentration) towards the
solution (area of higher solute
concentration) through a semi-
permeable membrane.
Reverse osmosis (RO) If pressure
greater than osmotic pressure is
applied on the solution side, solvent
molecule will move in the reverse
direction, i.e. from solution to solvent
, this movement is known as
Reverse osmosis.

47. Application of reverse osmosis
•Potable water from brackish sources– Recent advances
have made the production of potable water from sea water
possible•
• Used in conjunction with ultrafiltration and ion exchange
to produce ultrapure water
• Extensively used to clean waste waters with small solutes
and high BOD– Starch recovery from potato processing
• Concentration of fruit and vegetable juices
– Superior flavour to those produced by heat concentration
– Lower energy requirements than evaporation processes
– Smaller plant cost and size
– Very high dissolved salt concentration
– High operating pressures

48. • To get ultrapure water for food processing and electronic indus
• To get pharmaceutical grade water
• For chemical, pulp and paper industry usable water
Desalination of sea water
Sea water under pressure is
introduced armed the hollow
fibres.
Fresh water is obtained from
inside the fibre. In actual set up
each unit contains more than
three million fibres together.Each
fibre is of about diameter of
human error.

49. Van’t Hoff factor
Colligative properties depends only upon the no of
particles of the solute but if the solute undergo
association or dissociation in solution, abnormal
molecular masses are obtained.
In order to account for these discrepancy Van’ t Hoff
introduced a factor “i” known as the van’t Hoff factor
i = Observed colligative property (actual)
Calculated colligative property(Expected)
For the Colligative properties discussed
i= (∆p/p°)obs = (∆Tb)obs = (∆Tf)obs = ∏obs
(∆p/p°)calc (∆Tb)calc (∆Tf)calc ∏cal
i=Actual no of particles present
No of particles expected to be present

50. i=1 – ideal behavior- urea in water
i<1 – solute particles associate – benzoic acid in benzene
i>1 – solute particles dissociate – sodium chloride in water

51. Numerical
1. The freezing point of cyclohexane is 279.5 K. A
solution of naphthalene (MM 128) is prepared in
cyclohexane by dissolving 0.65g of naphthalene
in 19.2g of cyclohexane. If cryoscopic constant
of cyclohexane is 20.1KKgmol-1. Calculate the
molecular weight of X.
2. A 4.0 % solution of a solute X in water has the
same boiling point as a 4.5 % solution of glucose
in water . If the molecular weight of glucose is
180, calculate the molecular weight of X.