L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.1
Two Component Systems
(A) Ideal Systems — System which obey Raoult’s Law
(1) Ideal solution
There are 3 categories of two—component liquid mixtures:
— Completely miscible liquids : e.g. water and ethanol
— partially miscible liquids : e.g. water and butanoic acid
— immiscible liquids : e.g. water + oil / benzene / organic solvent
The completely miscible liquids are homogeneous mixture or solutions. A model of ideal solution
can be developed and then the deviations of real and non-real solutions can also be shown.
A solution is said to be ideal if the intermolecular forces between its molecules were just the same
as those existing in the separate components of the solution, before these component are mixed
together into solution.
Note :
An ideal solution has the following properties:
(1) Intermolecular attractions between like molecules (A-A and B-B) and unlike molecules
(A—B) are equal .
There is thus little or no tendency for molecules to hinder or help the escape into the vapour
phase of molecules of the other component.
(2) There is no enthctlpy change on mixing of the two components.
(3) There is no volume change on mixing of the two components.
Ideal solutions obey Raoult’s Law.
Raoult’s Law 臘烏爾定律 states that the partial pressure PA of component A in a solution equals
the vapour pressure of pure component (PA0) multiplied by the mole fraction (XA) of the
component in the solution.
For an ideal solution consisting of two components A and B.
Raoult’s Law maybe expressed as
PA = XA PA0
and PB = XB PB0
The vapour of an ideal solution ma be considered to be an ideal gas, and Dalton’s law therefore
applies. Thus the total vapour pressure (PT) of the solution is given by
PT = PA + PB , hence
PT = XA PA0
+ XB PB0
Assumption of Raoult’s Law : the escaping tendency into vapour is proportional to the vapour
pressure.
Note :
The vapour above a mixture of liquids does not have the same composition as the liquid.
If XA(V) and XB(V) are the mole fractions of A and B in the vapour phase.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
X A(V )
X B (V )
Chpt.22: p.2
0
X A( l ) PA
P
 A 
PB X B ( l ) PB 0
If A is more volatile than B
PA0 > PB0
and
X A(V )
X B (V )

X A( l )
X B (l )
The vapour is richer than the liquid in A, the more volatile component.
Example :
At 80°C, the vapour pressure of benzene is 10.0x104 Nm-2 and that of methylbenzene is 4.0xl04 Nm-2 .
Estimate the mole fraction of methylbenzene in the vapour in equilibrium with a liquid mixture of
benzene and methylbenzene at 80 0C in which the mole fraction of methylbenzene is 0.60.
SOLUTION
<2> Graphical Representation of Raoult’s Law
The following figure illustrates graphically the vapour pressure of an ideal solution which obeys
Raoult’s law at all concentrations. -
(1) A plot of PA. and PB against XA and XB gives a straight line because the partial pressure of A
and B are ____________________________________________________________
(2) The total pressure PA+PB also gives a straight line when plotted against mole fraction of the
liquid.
(3) Few liquids show this kind of ideal behaviuor. However, the solution would exhibit ideal
behaviour nearly if the components are chemically alike.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.3
(4) Most mixtures (solution) contain one component which is more volatile (i.e. higher vapour
pressure) than the other.
e.g. X = a more volatile liquid, has higher escaping tendency
Y = a less volatile liquid, has lower escaping tendency
For the plot of vapour pressure- vapour composition, when a liquid of composition a is boiled, the
vapour in equilibrium with it. has a composition b - richer in the more volatile component X
For the plot of Boiling point-vapour composition, the vapour which is in equilibrium with the
liquid mixture at a particular temperature always has a higher proportion of the more volatile
component
Example
(i) A mixture of 70% Y is distilled. The temperature of the liquid rises until it boils at T1.
(ii) At T1 the boiling liquid has a composition of 70% Y whereas the vapour in equilibrium has a
composition of 46% Y.
At the same time, the more volatile component has 54% in vapour, which is richer than that in the
liquid (30%).
(iii) When the vapour cools down, it forms a distillate with the same composition as vapour (46%Y). It
this liquid is distilled again the new distillate boiling at T2 contains 20% of the less volatile
component Y only. The more volatile component becomes further enriched to 80%.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.4
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.5
(B) Non-ideal System — Two component systems
Raoult’s law is an idealization which best holds for very dilute solutions.
A high dilution reduces the differences in molecular structure of the components which give rise
to either attractive or repulsive interaction between them.
Non-ideal solutions are those solutions with strong deviation from the ideal solutions ( either
positive or negative) because of the strong interaction between the solute and solvent molecules.
(1) Positive deviation
The vapour pressure of a mixture is greater than that predicted for an ideal solution.
Reasons : (i) Attractions or bonding between molecules of component A and B (A—B) are weak.
(ii) Bonding of A—A and B—B types are broken on mixing the two components.
These bonds are often hydrogen. bonds.
As a result of these two factors, molecules of A and B have a greater tendency to escape from
the liquid phase into the vapour phase. The liquid mixture thus has a higher vapour
pressure.
Note : When the two components of such a system is mixed, a temperature fall is usually
observed.
(2) Negative deviation
The vapour pressure of a mixture is smaller than that predicted for an ideal solution.
Reasons :
Note :
A—B attractions or bonds are stronger than A—A and B—B attractions or bonds.
The molecules are consequently held back in the liquid phase and the vapour
pressure is lower.
When the two components of such a system is mixed, a temperature rise is usually
observed.
Table Liquid mixture which deviate from Raoult’s Law
Positive deviation
Negative deviation
benzene/ethanol
methanol/water
water/ethanol
nitric acid/water
ethyl ethanoate /ethanol
trichloromethane/propanone
ethanol/carbon disuiphide
When a mixture of two liquids shows very large positive deviation from Raoult’s law, the vapour
pressure-composition curve has minimum. A system with a very large negative deviation shows a
minimum.
The following figures shows the Liquid-vapour phase diagrams showing the positive and negative
deviation from Raoult’s law and Boiling temperature-composition curve for systems.
A maximum in the vapour pressure curves results in a minimum in the boiling temperature —
composition curve and vice versa.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.6
Example
At 50 0C, the vapour pressure of benzene (C6H6) is 273 torr and that of methylbenzene (C6H5CH3) is
95.0 torr. Assuming ideal behaviour, determine the total vapour pressure at 50°C of a mixture of 15.62g
of benzene and 73.70g of methylbenzene.
In an actual experiment the vapour pressure observed in the above system is 148 torr. Briefly explain
the origin of any difference.
SOLUTION
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.7
<2> Bonding Interpretation of Non-ideal behaviour in terms of Molecular Interactions
The vapour pressure of a liquid is a measure of the escaping tendency of the molecules within the
solution.
(a) Positive deviation
The vapour pressure of a mixture is greater- than that predicted for an ideal solution and the boiling
point is lower than expected.
Example : The system of ethanol and methylbenzene
In pure ethanol, the molecules are strongly polarized so that they are associated by hydrogen bonds.
When methylbenzene is introduced, the overall intermolecular attraction between the molecules are
weakened by breaking up many hydrogen bonds. Thus, the escaping tendency of the molecules is
enhanced and the vapour pressure rises and the boiling point falls.
Note : A fall of temperature is observed because of the
_____________________________________________________________________________
(b) Negative deviation
The vapour pressure of a mixture is smaller than that predicted for an ideal solution and the boiling
point is higher. than expected.
Example : The system of trichloromethane and ethyl ethanoate
In such system. new intermolecular attraction, the intermolecular hydrogen bonds is formed. This
attraction is stronger than the van der Waals.’ forces existing in molecules of either pure liquids.
Note : A rise of temperature is observed because of the
_____________________________________________________________________________
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.8
<3> Fractional Distillation
Fractional distillation has a number of important applications
(1) the manufacture of oxygen, nitrogen and the noble gases from liquid air.
(2) the refining of petroleum.
(3) the production of whisky and other alcoholic drinks.
The process can be carried out by using a fractionating column (in laboratory) or fractionating tower (in
industry).
Fractional distillation is equivualent 相等於 to a series of consecutive 連續 simple distillation, where the
condensed vapour from a previous distillation is used as the liquid for the next distillation.
At each point in a column, or at each plate in a tower, an equilibrium between liquid and vapour is set up.
The component with different boiling points in a mixture can be separated using fractional distillation.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.9
Fractional Distillation of an ideal solution
Example : mixture of benzene and methylbenzene (toluene)
(i)
A mixture with composition X boils at the temperature t1.
(ii) The vapour that comes out has a composition Y, and is condensed at some point during its assent in
the fractionating column.
(iii) By the heating effect of further hot vapour rising up. it boils again at t2 to form a vapour of
composition Z.
(iv) The liquid condensed from the vapour becomes increasingly richer in benzene (the more volatile
component)
(v) Pure benzene can be obtained at the top of the column if the column is sufficiently long and the
cooling surface area is sufficiently large. The distillate is collected at about 800C, the boiling point of
benzene.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.10
Fractional Distillation of solution showing a Negative deviation 負偏離 from the Raoult’s law
The following figure shows the boiling point against composition diagram for a solution showing
negative deviation from ideality.
(1) At the maximum point Z1 the liquid and the vapour have the same composition.
For a mixture having composition P is distilled, the vapour formed at t1 has exactly the same
composition as the liquid and no separation of the components is achieved.
(2) For mixture having composition a is heated, the vapour initially formed at t2 has a composition y
(richer in the more volatile compound A). If this vapour is then condensed and re-distilled, the
vapour formed will have composition x.
Carrying out this process in a fractionating column results finally in the separation of pure A in the
distillate, leaving the maximum, constant boiling mixture in the flask.
(3) For the solution has a composition to the right of Q, pure B would be obtained in the distillate,
leaving the maximum, constant boiling mixture “azeotrope 共沸混合物”.
(4) System showing positive deviation from ideality are separable into the minimum, constant boiling
mixture and the pure components.
L.S.T. Leung Chik Wai Memorial School
F.6 Chemistry
Chapter 22: Phase Equilibrium II
Chpt.22: p.11
Example
The diagram shows the boiling point composition variation of a two component at 1 atmospheric
pressure.
(a) Find from the graph the initial and final compositions of the vapour above a mixture M (mole
fraction XA = 0.4) when the mixture is distilled at 1 atmospheric pressure.
(b) Starting with this mixture M, determine the minimum number of evaporation-condensation cycle
(i.e. minimum number of distillation) required to produce a distillate in which XB is at least 0.9)
(c) Can this two component system be separated completely by fract1ional distillation? Explain the
answer briefly.