Mrs Khadijah Hanim Abdul Rahman
Week 14- 21&24 May 2012

Ability to evaluate problems concerning
chemical kinetics, phase diagrams and
electrochemistry
Definitions
The
Phase Rule
Two-component Systems:
a)
Vapour pressure Diagrams
b) Temperature-composition Diagrams
c)
Liquid-liquid Phase Diagrams
d) Liquid-solid Phase Diagrams
A phase of a substance is a form of matter that is uniform
throughout in chemical composition & physical state (solid,
liquid, gas).
 The number of phases in a system is denoted P.
 A solution of NaCl in water is a single phase (P=1).
 A slurry of ice & water is two-phase system (P=2).
 A calcium carbonate system undergoes thermal deposition –
two solid phase (CaCO3 & CaO) one gaseous phase (CO2)
so, Total: (P=3)
 An alloy of two metals is a two-phase system (P=2) if they are
immiscible but a single-phase system (P=1) if they are
miscible.


A phase transition is the spontaneous conversion of one phase into another
phase occurs at a characteristics temperature for a given pressure.

Constituent – a chemical species (an ion or a molecule) that is present.

A mixture of ethanol & water – Two constituents.

A solution of sodium chloride – Three constituents
(water, Na+ ions & Cl- ions).

Component – a chemically independent constituent of a system. Each phase
in the system may be considered to be composed of one or more
components. The number of components in the system must be the
minimum required to define all of the phases.
For example, in our system salt and water, we might have the components
Na, Cl, H, and O (four components), NaCl, H, and O (three components),
NaCl and HO (two components), or NaCl-H2O (one component). However,
the possible phases in the system can only consist of crystals of halite (NaCl),
H2O either liquid or vapor, and NaCl-H2O solution. Thus only two
components (NaCl and H2O) are required to define the system, because the
third phase (NaCl - H2O solution) can be obtained by mixing the other two
components


Phase Diagram – one of the most ways of presenting the physical changes of
state that a substance can undergo.

Phase rule – relation between the degree of freedom or
variance (F), the number of component (C) & the
number of phases at equilibrium (P) for a system of any
composition:
F CP2
 In
a system of solid sucrose in equilibrium with an
aqueous solution of sucrose:
 the system has two components (C) – water &
sucrose.
 the system has two phases (P) – solution & solid
sucrose
 the degree of freedom (F) = 2 – 2 + 2 = 2

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
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Variance (F) – the number of intensive variables (P, T &
mole fractions in each of the phases) that can be changed
independently without disturbing the number of phases in
equilibrium.
Phase rule is a general relation between the variance, F, the
no of components, C and no of phases at equilibrium, P for
a system of any composition:
F=C–P+2
In a single component, single-phase system (C=1, P=1), the
pressure and temperature may be changed independently
without changing the no of phases, so F=2.
If 2 phases are in equilibrium, (a liquid and its vapour for eg)
in a single component system (C=1, P=2), the temp (or
pressure) can be changed at will, but the change in temp (or
pressure) demands an accompanying change in pressure (or
temp) to preserve the no of phases in equilib. The variance
of the system has fallen to 1.
CaCO3 ( s)  CaO
2 (g)
( s)  CO


Phase 2
Phase1
Phase3
A phase is a homogenous portion of a system: There are
Three phases ( 2solid phase + 1 gaseous phase)
The system has 3 chemical species CaCO3, CaO and
CO2
The system has Three components (C=3).
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Vapour-pressure diagrams
The partial vapour pressures of the
components of an ideal solution of two
volatile liquids are related to the composition
of the liquid mixture by Roult’s law:
Where is the vapour pressure of pure A and
that of pure B. The total vapour pressure P
of the mixture is therefore
The composition of the liquid & vapour that are in
mutual equilibrium are not necessarily the same.
 The vapour should be richer in the more volatile
component.
 Dalton’s law: The mole fraction in the gas:
pA
pB
yA 
yB 
P
P
*
P

x
p
 Provided the mixture is ideal:
A
A
A
*
*
xA pA
P

x
p
B
B
B
y 

A


p  p  p xA
*
B
*
A
yB 1  y A
*
B
P  P  (P  P ) X A
*
B
*
A
*
B


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Figure shows the composition of the
vapour plotted against the
composition of liquid.
In all cases, yA>xA, that is the vapour
is richer than the liquid in the more
volatile component.
If B is non-volatile, so that at the
temperature of interest, then it
makes no contribution to the vapour
(yB=0).
p *A
1
p B*
x A p *A
yA  *
pB  p *A  pB* x A



Since we can relate the
composition of the liquid
to the composition of
vapour, we can also relate
the total vapor pressure to
the composition of the
vapour using this
equation:
*
A
*
B
*
B
p p
p *
p A  p  p*A y A


p *A
1
*
pB

In distillation, both vapour & the liquid compositions are of
equal interest – combine graph of pressure vs mole fraction of
A & graph total vapour pressure vs mole fraction of A the
graphs into one.
‘ a’ indicates the vapour
pressure of a mixture of
composition, xA
‘b’ indicates the
composition of the
vapour that is in
equilibrium with the
liquid at that pressure
Points that lie between the
two lines correspond to a
system in which there are
two phases present, one a
liquid and other a
vapour.
z A  xA
zA  yA

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

The changes to the system do not
affect the overall composition, so the
state of the system moves down the
vertical line (isopleth) that passes thru
a.
Until a1 is reached (when pressure has
been reduced to p1), the sample
consist of single liquid phase.
At a1, the liquid can exist in
equilibrium with its vapour. The
composition of the vapoour is given
by point a1’.
Line joining 2 points representing
phases in equilibrium, ‘tie-line’.
The composition of the liquid is the
same as initially (a1 lies on the
isopleth thru a), we can conclude that
at this P there is virtually no vapour
present, however the tiny amount of
vapour that is present has the
composition a1’.
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Now consider the effect of
lowering the pressure to P2:
taking the system to a
pressure and overall
composition represented by
point a2”.
This new P is below the
vapour pressure of original
liquid, so it vapourizes until
the vapour pressure of the
remaining liquid falls to P2.
The composition of such a
liquid must be a2.
The composition of the
vapour in equilibrium is given
by the point a2’ at the other
end of tie-line.
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Pressure is further reduced to P3,
The compositions of liquid and
vapour are represented by the
points a3 and a3’.
a3‘ corresponds to a system in
which the composition of the
vapour is the same as the overall
composition, we can conclude that
the amount of liquid present is
virtually 0, but the tiny amount of
liquid present has composition a3.
A further decrease in P, take the
system to point a4.
a4: only vapour present and its
composition is the same as the
initial overall composition of the
system.
Effect of lowering the pressure on a liquid
mixture of overall composition ‘a’
(a) A liquid in a container exist in
equilibrium with its vapour.
Liq phase
Vapour phase
(b) Drawing out piston – P change.
The composition of the phase
adjust as shown by the tie line in
phase diagram.(> vapour than liq)
Lowering P –
by drawing out a piston
(c) Piston is pulled so far out – all
liquid has vaporized and only
vapour is present. (P falls as the
piston is withdrawn and the point
on the phase diagram moves into
the one-phase region.



A point in the two-phase region of a
phase diagram indicates not only
qualitatively that both liquid & vapour are
present, but represents quantitatively the
relative amount of each.
to find the relative amounts of two
phases α & β that are in equilibrium –
measure the distances l α & lβ along the
horizontal line & then use lever rule.
Lever rule:
n l  n l
Amount of phase α
Amount of phase β

Temperature-composition diagram (to discuss
fractional distillation) :
 a phase diagram in which the boundaries show
the composition of the phases that are in
equilibrium at various temperatures (at a given P
[1atm]).
a) The distillation of mixtures:
 When liq composition a1 is heated, its boils
when reach T2 (so, Liq has comp a2 same as a1
& vapour has comp a’2) – vapour is reacher at
in the > volatile component A (the component
with lower boiling point)
 From the location of a2, we can state the vapour
comp at the boiling point and from the
location of tie line joining a2 to a’2 we can read
off the boiling temp, T2 of the original liq
mixture.

In simple distillation, the vapour is
withdrawn and condensed.

Technique used to separate a volatile
liquid from a non-volatile solute/solid,
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In fractional ditillation, the boiling and
condensation cycle is repeated
successively. Used to separate volatile
liquids.
We can follow the changes: by seeing
what happen when the 1st condensate of
composition a3 is reheated.
The mixture boils at T3 and yield vapour
composition a3’ which is even richer in
the more volatile component.
The vapour is withdrawn and the 1st drop
condenses to a liquid of composition a4.
The cycle is repeated until almost pure A
is obtained in the vapour and pure B
remains in the liquid.



The efficiency of a fractionating
column is expressed in terms of
the no. of theoretical plates: the
no of effective vaporization and
condensation steps that are
required to achieve a condensate
of given composition from a
given distillate.
To achieve the degree of
separation shown in (a) the
fractionating column must
correspond to 3 theoretical
plates.
To achieve the same separation
for the system shown in (b), in
which the components have
more similar partial pressures,
the fractionating column must be
designed to have 5 theoretical
plates.
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Although many liquids have tempcomposition phase diagrams resembling
the ideal version, there are deviations.
Deviations from ideality are not always so
strong as to lead to a max or min in the
phase diagram, but when they do there are
important consequences to ditillation.
Consider a liquid of composition a on the
right of the max in the figure.
The vapour (at a2’) of the boiling mixture
(at a2) is richer in A.
If that vapour is removed and the
remaining liquid will move to a
composition that is richer in B, such as
represented by a3, and the vapour that is
in equilibrium is a3’.
As that vapour is removed, the
composition of the boiling liquid shifts to
a4 and a4’.
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As evaporation proceeds, the
composition of remaining liquid
shifts towards B as A is drawn off.
The boiling point of the liquid
rises, and the vapour becomes
richer in B.
When so much of A has been
evaporated that the liquid has
reached composition b, the
vapour has the same composition
as the liquid.
Evaporation then occurs without
change of composition
The mixture is said to form
azeotrope.
When azeotrophic composition is
reached, distillation cannot
separate the 2 liquids because the
condensate has the same
composition
Example: HCl/water, which is
azeotrophic at 80% by mass of
water and boils unchanged at
108.6oC.
A high boiling azeotrope
(Maximum azeotrope) When
the liq of composition a is
distilled, the composition of
the remaining liq changes
towards b but not further
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Start with a mixture of composition
a1, and follow the changes in the
composition of the vapour that rises
through fractionating column.
The mixture boils at a2 to give
vapour composition of a2’.
This vapour condenses in the
column to a liquid of the same
composition, a3. liquid at
equilibrium with its vapour at a3’.
a3’ condenses higher up the tube to
give a liquid of the same
composition, a4.
The fractionation therefore shifts the
vapour towards the azeotrophic
composition at b.
Eg: ethanol/water which boils
unchanged when the water content
is 4% by mass and the temp is 78oC.
A low boiling azeotrope (Minimum
azeotrope). When the liq of
composition a is fractionally
distilled, vapour in eq in the
fractioning column moves towards
b and then remain unchange
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Phase separation of partially miscible liquids may
occur when temp is below the upper critical
solution temp OR above the lower critical
solution temp.
Partially miscible=liquids that do not mix in all
proportions at all temp.
Example: hexane and nitrobenzene.
The upper critical solution temp: is the highest
temp at which phase separation occurs.
The lower critical solution temp: temp below
which components not mix in all proportionswhich they form 2 phases.
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Small amount of a liq B is added to a sample of
another liq A at T’. Liq B dissolved completely
& binary system remains a single phase. As > B
is added, a stage comes at which no > dissolves.
(2 phases in eqm )The most abundant- consisting of A saturated
with B
The minor- trace of B saturated with A.
In the temp-composition diagram, the most
abundant is represented by the point a’ and the
minor is represented by point a”.
The relative abundances of the 2 phases are
given by Lever rule.
When more B is added, A dissolves in it slightly.
The compositions of the 2 phases in eqm
remain a’ and a”.
A stage is reached when so much of B is present
that it can dissolve all A and the system reverts to
a single phase.
The composition of the 2 phases at eqm varies
with temp. For hexane/nitrobenzene, raising the
temp increases their miscibility.
1 phase
2 phases

a)
b)
c)
A mixture of 50 g of hexane
(0.59 mol C6H14) & 50 g of
nitrobenzene (0.41 mol
C6H5NO2) was prepared at
290K.
What are the compositions of
the phases?
In what proportions /ratio do
they occur?
To what temperature must
the sample be heated in order
to obtain a single phase?
Lets denote hexane by H and
nitrobenzene by N
Composition of the phases
The compositions of phases in eqm are
given by the points where the tie line
representing the tem intersects the phase
boundary.
The point xN=0.41, T=290 K occurs in 2
phase region of the phase diagram
The horizontal line cuts the phase
boundary at xN=0.35 and xN=0.83 
composition of the 2 phases.
b) Proportions
Given by lever rule, the ratio of amounts
of each phase = ratio of distances lα and
lβ:
a)
That is, there are about 7 times more
hexane-rich phase than nitrobenzenerich phase.
c) Temp needed to obtain single phase
-
- The temp at which the components are
completely miscible is found by following
isopleth upwards and noting the temp at
which it enters the 1 phase region of the
phase diagram = 292K
Upper critical solution temp,
Tuc: the higest temp at which
phase seperation occurs.
 Above the upper critical
temp the 2 components are
fully miscible.
 Example of solid solution is
the palladium/hydrogen
system:
-1 phase: a solid solution of
hydrogen in palladium
-1 phase: palladium hydride
Up to 300oC but forms a
single phase at higher temp.
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Tlc: below which they mix in
all proportions and above
which they form 2 phases.
The lowest temp at which
phase separation occurs.
Example: water/triethylamine
At low temp the two
components are more
miscible because they form a
weak complex,
at higher temp the
complexes break up and the
two components are less
miscible.
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
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They occur because,
after the weak
complexes have been
disrupted, leading to
partial miscibility
The thermal motion at
higher temp
homogenizes the
mixture again, just in
the case of ordinary
partially miscible liquid.
Example: nicotine and
water, which are
partially miscible
between 61oC and
210oC.
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Knowledge of the temp-composition
diagrams for solid mixtures guides the
design of important industrial processes
such:
Manufacturing of liquid crystal display (LCD)
Semiconductors
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Eutectic composition:
Consider the 2 component liq (liq
solution of A & B)of composition a1.
(the changes that occur as the system is
cooled)
a1 to a2: system enters 2-phase region
(Liq + B). Pure solid B begins to come
out of solution and the remaining liq
becomes richer in A)
a2 to a3 : > of solid B forms and the
relative amounts of the solid & liq (in
eqm) are given by lever rule (equal
amount of each).the liq phase is richer in
A than before (b3) because some B has
been deposited.
a3 to a4 : less liq than at a3 (composition
e). This liq now freezes to give a 2-phase
system of pure solid B and pure solid A.
Eutetic composition –
the mixture with the
lowest melting point.

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
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The isopleth at e corresponds
to the eutectic composition,
the mixture with the lowest
melting point
A liquid with the eutectic
composition freezes at a single
temperature, without previously
depositing solid A or B.
A solid with the eutectic
composition melts, without change
of composition, at the lowest
temperature of any mixture.
E.g. Plumbum/ antimony,
tin/plumbum, Silicon/ aluminium,
benzene/napthalene,
chloroform/aniline
Eutetic composition –
the mixture with the
lowest melting point.

At 90oC, the vapour pressure of
methylbenzene is 53.3 kPa and that of 1,2dimethylbenzene is 20.0 kPa. What is the
composition of a liquid mixture that boils at
90oC when the pressure is 0.50 atm? What is
the composition of the vapor produced?

The vapour pressure of pure liquid A at 300 K
is 76.7 kPa and that of pure liquid B is 52.0
kPa. These two compounds form ideal liquid
and gaseous mixtures. Consider the
equilibrium composition of a mixture in
which the mole fraction of A in the vapour is
0.350. Calculate the total pressure of the
vapour and the composition of the liquid
mixture.
Benzene and toluene form nearly ideal solutions.
Consider an equimolar solution of benzene and
toluene are 9.9 kPa and 2.9 kPa, respectively. The
solution is boiled by reducing the external
pressure below the vapour pressure. Calculate
a)
The pressure when boiling begins
b) The composition of each component in the
vapour
c)
The vapour pressure when only a few drops of
liquid remains.
Assume that the rate of vaporization is low enough
for the temp to remain constant at 20oC.
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The following temperature/composition data
were obtained for a mixture of two liquids A and
B at 1.00 atm, where x is the mole fraction in the
liquid and y is the mole fraction in the vapour at
equilibrium.
θ/oC
125
130
135
140
145
150
xA
0.91
0.65
0.45
0.30
0.18
0.098
yA
0.99
0.91
0.77
0.61
0.45
0.25
The boiling points are 124oC for A and 155oC for
B. Plot the temperature/composition diagram for
the mixture.
what is the composition of the vapour in
equilibrium with the liquid composition (a)
xA=0.50 and (b) xB=0.33?

1-Butanol and chlorobenzene form a minimumboiling azeotropic system. The mole fraction of 1butanol in the liquid (x) and vapour (y) phases at 1.00
atm is given below for a variety of boiling
temperatures.
T/K
396.57 393.94 391.6
390.15 389.03 388.66 388.57
X
0.1065 0.1700 0.2646 0.3687 0.5017 0.6091 0.7171
y
0.2859 0.3691 0.4505 0.5138 0.5840 0.6409 0.7070
Pure chlorobenzene boils at 404.86 K
(a) Construct the chlorobenzene-rich portion of the
phase diagram from these data
(b) Estimate the temp at which a solution whose mole
fraction of 1-butanol is 0.3 begins to boil
(c) State the compositions and relative proportions of
the 2 phases present after a solution initially 0.300
1-butanol is heated to 393.94 K.
