Download Topic guide 6.2: Phase equilibria

Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
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62
Phase equilibria
Many industrial processes involve several phases in equilibrium – gases, liquids,
solids and even different crystalline forms of the solid state. Predicting the
number of phases present under a given set of conditions is important in the
design of industrial processes and in predicting the behaviour of materials in
different sets of conditions. In this topic guide you will look in some detail at
processes such as distillation and crystallisation to illustrate how knowledge of
phase equilibria helps understand the limitations and design of these processes.
On successful completion of this topic you will:
•• understand the physical equilibria in one- and two-component systems (LO2).
To achieve a Pass in this unit you need to show that you can:
•• analyse pressure/temperature phase diagrams for simple, one-component
systems (2.1)
•• perform calculations using the Clapeyron and Clausius-Clapeyron
equations (2.2)
•• explain liquid-vapour equilibria and distillation in terms of vapour pressure/
composition and boiling point/composition plots for two-component
systems (2.3)
•• analyse phase diagrams for solid-liquid equilibria which form a simple
eutectic mixture (2.4).
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Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
1 Phase equilibria
Key term
Phase: A distinctive form of matter in
which physical properties are uniform
throughout. Phases can include the
familiar states of matter, such as solid,
liquid and gas, and also different
forms of matter in the same state, for
example, solutes in different solvents
and solids with different crystal
structures.
Carbon dioxide is a typical example of a substance that exists in three different
phases depending on the conditions of pressure and temperature to which it is
exposed.
Phase diagrams
Experimental data can be obtained for substances such as carbon dioxide in which
observations are made of the different phases that exist at various conditions of
pressure and temperature. From these observations and certain other key pieces
of data, for example, the triple point of the substance (the temperature and
pressure at which all three phases can exist), phase diagrams can be constructed.
The phase diagram for carbon dioxide is displayed in Figure 6.2.1:
•• it shows which of the three phases (solid, liquid or gas) is most
thermodynamically stable at different conditions of pressure and temperature
•• the solid lines in the diagram, where two phases meet, represent the
conditions of pressure and temperature at which these two phases are in
equilibrium
•• the triple point, where all three phases meet, is the point at which the three
phases are all in equilibrium.
Figure 6.2.1: The phase diagram
for this simple, one-component
system allows you to predict various
properties of carbon dioxide.
Pressure (bar)
Phase diagrams are often drawn with a logarithmic scale on the pressure axis (as in
Figure 6.2.1).
liquid
100
critical point
72.9
solid
10
5.185
1
triple point
vapour
0.1
–100 –78.4
–56.6
0
31.1 50
Temperature (°C)
Interpreting a phase diagram
Phase diagrams allow you to make deductions about the behaviour of the system.
•• The state of the substance at given conditions of temperature and pressure;
for example, the point with coordinates of 70 bar and 20 °C will fall within
the liquid phase. Notice that liquid carbon dioxide does not exist at any
temperature unless the pressure is greater than 5.185 bar.
•• The triple point, TP; in the case of carbon dioxide this occurs at 5.185 bar and
−56.6 °C.
•• The melting or boiling points at any given pressures; the ‘melting’ point at
1 bar is −78.4, but the phase transition that occurs is from solid directly to
vapour, a process called sublimation.
6.2: Phase equilibria
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Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
•• The changes that will happen when the system is heated. For example, when
carbon dioxide at 10 bar pressure is heated from −100 °C to 50 °C it begins as
a solid, melts to form a liquid and finally becomes a vapour. At 1 bar pressure,
the solid will sublime to form a vapour without passing through the liquid
phase.
The other significant point on the graph is the ‘critical point’ (CP) – the temperature
at the critical point (the critical temperature) is the highest temperature at which
the liquid can exist. Above this temperature, the substance exists as a single
uniform vapour phase. The pressure at the critical point is known as the critical
pressure, and if both the temperature and pressure of the substance exceed those
at the critical point, the substance is described as a supercritical fluid (see Case
study on page 4 for an application of this).
Other phase diagrams
1 Predict the most stable phase of
water at
(a) 0.1 bar and 200 K
(b) 100 bar and 400 K
(c) 0.001 bar and 600 K.
2 Deduce the conditions at which
the solid, liquid and vapour
phases will be in equilibrium.
3 What phase change will occur
when water at 0.001 bar is heated
from 200 K to 300 K?
4 What is the boiling point of water
at 10 bar?
critical point
100
liquid
10
1
0.1
0.01
0.001
100
n
tio
risa
o
p
va
fusion
Activity
1000
solid
vapour (gas)
subli
mat
ion
Figure 6.2.2: The phase
diagram for water.
Pressure (bar)
The phase diagram for water is shown in Figure 6.2.2.
200
triple point
300
400
500
600
700
Temperature (K)
Sulfur gives an example of a phase diagram where more than three phases are
included. This is because solid sulfur exists in two crystalline forms – rhombic and
monoclinic.
Figure 6.2.3: The phase diagram
for sulfur, showing the different solid
phases (rhombic and monoclinic)
that exist at different conditions.
Pressure (bar)
Four phases means that there will be three triple points, as shown in Figure 6.2.3.
3
1288
rhombic
liquid
1 Describe all the changes that
happen to a sample of rhombic
sulfur at 1 bar that is heated from
100 °C to 500 °C.
2 Which phases will be in
equilibrium at 151 °C and
1288 bar?
6.2: Phase equilibria
1bar
5×10–4
10–4
1
mo
no
clin
ic
Activity
2
gas
95.4
119.6
151
Temperature (°C)
3
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
The phase rule and degrees of freedom
You will notice in the phase diagram for sulfur (Figure 6.2.3) that, although
three different phases can exist in equilibrium at the triple points, there are no
conditions of pressure and temperature at which four different phases can be in
equilibrium.
This is predicted by the phase rule; this predicts the number of degrees of freedom
that can exist for a system at equilibrium.
The number of degrees of freedom is the number of physical variables that can
be varied without altering the number of phases at equilibrium – in the context
of simple phase diagrams, the physical variables can be taken as temperature and
pressure.
The phase rule states:
(1) F = C − P + 2
where F is the number of degrees of freedom
Activity
1 Choose a phase diagram from the
section above. Calculate a value
for the degrees of freedom at:
(a)a point on the phase
boundary, where two phases
are present
(b)the triple point, where three
phases are present.
Comment on the significance of
these answers.
2 Imagine a situation where four
phases are present. Calculate a
value for F and comment on what
this value might mean about
the feasibility of this situation
occurring.
C is the number of components in the system (i.e. the number of
different substances present)
P is the number of phases at equilibrium.
Consider sulfur at 200 °C and 1 atm:
•• this is a one-component system, so C = 1
•• according to the phase diagram, only liquid sulfur will be present under these
conditions, so P = 1
•• hence F = 1 − 1 + 2 = 2
•• this means that both temperature and pressure can be varied without altering
the number of phases.
If F = 1, it means that if, say, temperature is altered, then pressure must also be
altered to maintain the number of phases in equilibrium.
If F = 0, it means that no variable can be altered without altering the number of
phases in equilibrium.
Case study: Supercritical carbon dioxide?
When the temperature of a system exceeds the critical point, the liquid phase becomes a
supercritical fluid, in which it behaves like a gas but has the density of a liquid. Supercritical carbon
dioxide, which exists when the pressure is greater than 72.9 bar and the temperature is greater
than 31 °C, is now widely used as a solvent to replace more toxic and environmentally damaging
organic solvents in applications such as dry-cleaning and decaffeinating coffee beans.
•• Use the phase diagram in Figure 6.2.1 to predict whether carbon dioxide will exist as a
supercritical fluid under the following conditions: (a) 50 °C and 5 bar (b) 20 °C and 100 bar
(c) 50 °C and 100 bar.
6.2: Phase equilibria
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Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Portfolio activity (2.1)
Uranium hexafluoride (UF6) plays an important role in the enrichment of nuclear fuel. For the
enrichment process to work, UF6 must be in the gaseous state. Scientists will therefore make use of
phase diagrams to select conditions that will result in the gaseous phase being stable.
Use a suitable phase diagram to explain how they do this. In your answer:
•• find a pressure/temperature phase diagram for UF6
•• explain the features of this diagram, for example the meaning of the term triple point and the
degrees of freedom at the triple point
•• explain how you can use the phase diagram to predict the conditions needed for gaseous UF6.
Equations to describe phase boundaries
Some features of the phase diagrams shown above can be measured
experimentally; however, the phase boundary lines displayed in the diagrams are
calculated using theoretical ideas.
The equations that are used to construct the phase boundaries are the Clapeyron
and Clausius-Clapeyron equations. They have other valuable applications as well,
for example, in meteorology where the equation is applied to the processes that
cause the formation and dissipation of clouds.
Clapeyron equation
This is used in the cases of solid–solid or solid–liquid transitions, where the volume
change of the transition, ΔVtrans, remains relatively constant, regardless of changes
in pressure and temperature.
The Clapeyron equation describes how the slope of the phase boundary line
depends on temperature:
(2) Δp =
ΔHtrans
Δp ΔHtrans
=
. ΔT or
ΔT TΔVtrans
TΔVtrans
where Δp and ΔT are the changes in pressure and temperature
ΔHtrans is the enthalpy change of the transition (in joules)
ΔVtrans is the volume change of the transition (in m3: 1 m3 = 106 cm3).
6.2: Phase equilibria
5
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Pressure (bar)
Figure 6.2.4: The slope of the phase
ΔP
boundary curve, ΔT , is described
by the Clapeyron equation.
solid
P2
ΔP
P1
liquid
T1
ΔT
T2
Temperature (°C)
Slope of the phase boundaries
For a transition from solid to liquid, ΔH will be endothermic and will have a
positive value, and ΔV will usually be positive because volume usually increases
when a solid is melted.
Hence the slope of the transition line will be positive (see Figure 6.2.4) in the
phase diagrams for carbon dioxide (Figure 6.2.1) and sulfur (Figure 6.2.3).
Integrated form of the Clapeyron equation
By writing the Clapeyron equation in the form:
dp ΔH
=
dT TΔV
it follows that
Activity
When ice melts to form liquid water,
the volume decreases. The enthalpy
change for the melting of ice is
endothermic.
Use the Clapeyron equation to
comment on the slope of the
transition line between solid and
liquid on the phase diagram. Is this
consistent with what you observe
on the phase diagram shown in
Figure 6.2.2?
dp =
ΔH dT
ΔV T
And hence, integrating this yields the integrated form of the Clapeyron equation:
ΔH
(3) p2 – p1 =
(lnT1 – lnT2)
ΔV
Activity
Ice melts at 273.15 K at 1 bar pressure. Use the integrated form of the Clapeyron equation
(equation (3)) to predict the melting temperature of ice at 20 bar.
The molar volumes of ice and water at 273.15 K are 19.58 cm3 and 18.02 cm3, respectively, and the
enthalpy change of fusion (melting) is +6.03 kJ mol−1.
(Hint: Take 273.15 K as T1 and rearrange the equation to find ln T2 and hence T2. Be careful to
ensure that you convert the data provided into the appropriate units.)
Clausius-Clapeyron equation
For transitions such as solid–gas and liquid–gas, ΔV is no longer constant and the
Clapeyron equation is no longer valid.
However, by making several assumptions, a second equation can be derived, the
Clausius-Clapeyron equation.
6.2: Phase equilibria
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Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Take it further
Elements of Physical Chemistry (Atkins and de Paula, 2009) has a section explaining the derivation of
the Clausius-Clapeyron equation (pages 111–112).
The derivation of this equation assumes:
•• the molar volume of vapour is much greater than the molar volume of the solid or liquid so that
ΔV is approximately equal to the molar volume of the vapour
•• the vapour behaves as a perfect gas
•• the changes in pressure and temperature are small
•• ΔHvap remains constant over the range of pressure and temperature being studied.
Activity
Under standard conditions of
1 bar pressure, benzene boils at
353.25 K. At the boiling point of
any liquid, the vapour pressure
is equal to atmospheric pressure.
ΔHvap of benzene has a value of
+30.8 kJ mol−1.
Use the Clausius-Clapeyron equation
to calculate the vapour pressure of
benzene at 298.15 K.
The Clausius-Clapeyron equation is used to describe how the vapour pressure of a
liquid or gas changes when the temperature changes:
(4) lnp2 = lnp1 +
(5) ln
(
ΔHvap 1 1
–
T1 T2
R
(
p2 ΔHvap 1 1
=
–
T1 T2
p1
R
)
)
where R is the gas constant (R = 8.314 J K−1 mol−1)
T2 and T1 are temperatures in K.
If the vapour pressure, p1, at a temperature, T1, is given, then values of T1, T2 and p1
can be substituted into the equation to find p2, the vapour pressure of the liquid at
a new temperature, T2.
Portfolio activity (2.2 part 1)
Many metals have a volume increase of about 5% when they melt. Knowing the volume increase
on melting allows the Clapeyron equation to be used to predict how melting point will depend on
pressure.
Choose a metal for which data on molar volume, melting temperature and enthalpy changes of
fusion are readily available. Suitable examples could include copper, aluminium or zinc.
Explain how the Clapeyron equation can be used to calculate the pressure increase needed in
order to raise the melting temperature of your chosen metal by 20 K.
In your answer:
•• state the Clapeyron equation and explain the terms
•• find suitable data for your chosen metal
•• use the Clapeyron equation to predict the pressure needed to produce the required increase in
melting temperature.
Portfolio activity (2.2, part 2)
Choose a liquid important in industrial processes (for example, as a solvent or reagent). Suitable
examples could include ethanol or hexane. Find data relating to boiling point at 1 bar pressure and
enthalpy change of vapourisation.
Predict the vapour pressure of the liquid at a temperature of 20 °C below its boiling point.
In your answer:
•• state the Clausius-Clapeyron equation and explain the terms
•• find suitable data for your chosen liquid
•• use the Clausius-Clapeyron equation to predict the vapour pressure at the conditions stated.
6.2: Phase equilibria
7
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Case study
Hydrogen fuel cells are being developed as sources of power for cars and other vehicles. The fuel
cells generate electrical energy from the oxidation of hydrogen to water. Controlling the humidity
of the air stream used to remove the waste water from the cell is important to the efficient
operation of the cell as a high humidity is needed to prevent the membrane from drying out.
The Clausius-Clapeyron equation can be used to calculate the vapour pressure of water at the
operating conditions of the cell, which enables humidity at these conditions to be predicted.
Operating pressures of fuel cells vary from 1 to 6 bar, and operating temperatures are in the range
50 °C to 90 °C.
Calculate the vapour pressure of water in a fuel cell operating at 3 bar pressure and 65 °C (338 K).
The boiling point of water at 1 bar is 373 K. Take the ΔHvap of water as +40.5 kJ mol−1.
2 Binary mixtures
In the previous section you were concerned with one-component systems where
a single chemical substance was present, in a range of phases.
Key term
Binary mixture: A mixture
containing two different components,
for example hexane and heptane, or
water and ethanol.
In real industrial contexts such systems are rare. Many reactions involve solutions
of solutes in solvents or mixtures of volatile liquids. In this section you will look at
binary mixtures in which two components are present.
Phase diagrams of binary mixtures
Just as phase diagrams can be drawn to show the conditions of pressure and
temperature at which two phases are in equilibrium, temperature–composition
diagrams can be drawn to show the conditions at which the liquid and vapour
phases of the mixture will be at equilibrium.
The easiest way to do this is to simply measure the boiling point of different
mixtures of the liquids, with known proportions of each component.
The vapour formed from the boiling liquid can be collected and condensed. If this
liquid is then analysed, its composition can be determined, which in turn allows
the composition of the vapour to be deduced.
Figure 6.2.5: The boiling point
(Tb , lower curve) of a two-component
mixture varies with the composition
of the mixture. This graph also shows
the composition of the vapour in
equilibrium with the boiling liquid.
Temperature (°C)
The compositions of the liquid and the vapour formed by boiling the liquid are not
the same. You can see this in Figure 6.2.5.
100
composition of vapour
Tb (heptane) = 98 °C
90
80
composition of liquid
Tb (hexane) = 69 °C 70
60
50
40
6.2: Phase equilibria
0
0.25
0.5
0.75
1
Mole fraction of heptane
8
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Mole fraction
The composition of a mixture of two components, say, A and B, is usually defined
by the mole fraction of A, XA:
number of moles of A
(6) XA =
total number of moles of A + B
Activity
1 A mixture of hexane (Mr = 86.2) and heptane (Mr = 100.2) was made by mixing 14.2 g of hexane
with 8.0 g of heptane.
(a)Calculate the % by mass of heptane in the mixture.
(b)Calculate the mole fraction of heptane present in this mixture.
2 The boiling point of a mixture of hexane and heptane was measured as 75 °C. Use Figure 6.2.5
to estimate the mole fraction of heptane present.
3 A mixture of ethanol (Mr = 46) and water (Mr = 18) contains 95% ethanol by mass. Calculate the
mole fraction of ethanol. (Hint: Calculate the number of moles of ethanol and water in 100 g of
the mixture.)
In some situations, particularly when boiling point composition graphs are
used, you may find the composition represented by the percentage by mass of a
component, A:
mass of A
% by mass =
× 100
total mass of A and B
Composition of liquid and vapour phases
As mentioned above, Figure 6.2.5 makes it clear that the compositions of a liquid
and the vapour in equilibrium with it are not identical.
Activity
Use Figure 6.2.5 to estimate the composition of the vapour phase (in terms of the mole fraction of
heptane) in equilibrium with a mixture boiling at 80 °C. Comment on how the composition of this
phase compares with the composition of the liquid phase.
Compared to the composition of the liquid phase of a given boiling point, the
composition of the vapour phase in equilibrium with it will tend to contain a
greater mole fraction of the more volatile component.
Ideal mixtures
Key term
Partial vapour pressure: The partial
pressure due to the vapour of a single
component above a solid or liquid
mixture.
6.2: Phase equilibria
The mixture of hexane and heptane is a good example of a system that can be
regarded as an almost ideal mixture – one in which the attractive forces in the
mixture are identical to the attractive forces in the individual components.
Raoult’s law
An ideal mixture will obey Raoult’s law, which describes how the partial vapour
pressure of a component, A, depends on the composition of the mixture.
9
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
(7) PA = XA . PA°
where
PA is the partial vapour pressure of the component A
XA is the mole fraction of A
PA° is the partial vapour pressure of pure A.
It therefore follows that, because XB = 1 − XA, then
PB = (1 − XA) . PB°
The total vapour pressure of the system is therefore given by:
(8) P = PA + PB = PA . XA° + XB° . PB°
A vapour pressure–composition diagram for an ideal mixture will therefore be as
shown in Figure 6.2.6.
Vapour pressure
Figure 6.2.6: The partial vapour
pressures of the components of a
simple binary mixture are proportional
to the mole fraction of each of the
components in the mixture.
PB°
PA+ PB
PB
PA°
PA
0
0.25
0.5
0.75
1.0
Mole fraction of A
Activity
A mixture of methylbenzene and benzene behaves as an ideal mixture. Construct a graph to show
the variation in vapour pressure of the two components with composition at 20 °C.
The vapour pressure of pure benzene at 20 °C is 0.10 bar; the vapour pressure of pure
methylbenzene at 20 °C is 0.03 bar.
(Hint: use equation (7) to calculate vapour pressures of these liquids for various values of mole
fraction of benzene.)
Deviations from Raoult’s law
Non-ideal mixtures, that is, those that do not obey Raoult’s law, can show either a
positive or negative deviation from Raoult’s law (see Figure 6.2.7).
Positive deviation
Vapour pressure
Figure 6.2.7: A positive deviation
from Raoult’s law means that the
vapour pressure is greater than the law
predicts; a negative deviation means
that it is less than the law predicts.
P°B
Negative deviation
P°B
total vapour pressure
PA
P°A
P°A
PB
0
0.25
0.5
0.75
1.0
Mole fraction of A
0
0.25
0.5
0.75
1.0
Mole fraction of A
Positive deviations occur when the forces of attraction between molecules in the
mixture are weaker than the forces of attraction in the pure liquids.
6.2: Phase equilibria
10
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
The reverse is true for negative deviations – the forces of attraction in the mixture
are stronger than in the pure liquids.
If the deviations from ideal behaviour are significant, as in the graphs shown in
Figure 6.2.7, then the maximum vapour pressure will not occur for a pure sample
of the most volatile component but at a mixture with a particular composition.
These deviations from Raoult’s law have profound implications for distillation
processes; these processes are important in industries such as the oil industry.
Distillation
Distillation is used to separate liquid components of a system with different
boiling points.
Simple distillation
In simple distillation, a mixture of the two liquids is boiled and the vapour is then
condensed and collected. You will remember from the temperature composition
graphs at the start of this section that the composition of a boiling liquid and that
of the vapour in equilibrium with it are significantly different. This idea is crucial to
the understanding of distillation processes.
Figure 6.2.8: Distilling a pentaneoctane mixture produces a
distillate rich in pentane.
Temperature (°C)
For example, imagine a system containing pentane and octane. The temperature–
composition curve is shown in Figure 6.2.8 (notice that the composition is
expressed in terms of the percentage by mass of the octane component).
150
vapour
125
Tb (octane)
100
75
50
Tb (pentane)
25
Activity
Consider a pentane-octane mixture
that contains 80% by mass of
octane. Predict the percentage of
pentane in the distillate. Comment
on the effectiveness of this simple
distillation.
0
B
A
liquid
C
0
25
50
75
100
% by mass of octane
When a liquid containing 50% pentane by mass (point A) is boiled it will be in
equilibrium with a vapour (point B). On cooling this produces a liquid distillate
with the same composition as the vapour (point C). This liquid contains only
about 5% octane by mass and so is approximately 95% pentane. So this simple
distillation has been broadly successful.
Fractional distillation
If the difference in boiling points of two liquids in a mixture is less than about 60 °C,
or if the mixture being distilled is rich in the less volatile component, then simple
distillation will not be effective, as the distillate will only be slightly enriched in the
less volatile component. In these situations, fractional distillation is used.
In this process, a sequence of distillations occurs that result in a gradual
enrichment of the more volatile component in the distillate (see Figure 6.2.9).
6.2: Phase equilibria
11
Figure 6.2.9: In fractional distillation,
a series of distillations occur, gradually
producing a distillate that is richer and
richer in the more volatile component A.
Temperature (°C)
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
90
V2
80
50
V5
V1
90 bpB
L1
L2
70
60
vapour
V3
L3
V4
L5
0
100% A
0%
liquid
L4
80
20
60
40
40
60
20
80
0%
100% B
In Figure 6.2.9 a liquid with composition represented by point L1 is distilled,
resulting in a vapour with composition V1. This is condensed into a distillate L2,
which is redistilled to produce a vapour V2, and so on.
Fractional distillation in the laboratory and industry
Figure 6.2.10 shows the apparatus needed for fractional distillation in the
laboratory.
Figure 6.2.10: Fractional
distillation in the laboratory.
thermometer
water out
fractionating
column packed
with glass beads
(or similar)
water in
vent to fume
cupboard
mixture
product collects
heat
6.2: Phase equilibria
12
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Fractional distillation is achieved by passing the hot vapours through a vertical
column. As the vapours condense they drip back down the column to be
revapourised by more hot vapour rising up the column.
In the laboratory apparatus (shown in Figure 6.2.10), packing the column
with glass beads provides a large surface area on which condensation and
vapourisation can occur, increasing the number of distillation cycles that occur.
The vapour emerging from the top of the column will have undergone many
distillations and will be very rich in the most volatile components of the mixture.
In the case of industrial distillation of crude oil, distillates are tapped off at various
levels of the column. These distillates will have particular boiling point ranges and
will therefore be particularly rich in certain components of the crude oil mixture.
Take it further
Some useful detail about the design of these fractionating columns (in the laboratory and in
industry) can be found at http://www.chemguide.co.uk/physical/phaseeqia/idealfract.html.
Azeotropes
The explanations of the distillation processes described above are for mixtures
that display approximately ideal behaviour.
Activity
Explain why a large negative
deviation from Raoult’s law
can lead to a mixture having a
maximum boiling point at a certain
composition.
As noted earlier, some non-ideal mixtures show significant deviation from
Raoult’s Law. This can also have very significant implications for the boiling point–
composition curves:
•• A system with a large positive deviation from Raoult’s law forms mixtures with
anomalously high vapour pressure. The point on the graph with the highest
vapour pressure will have the lowest boiling point.
•• So, conversely, a system with a large negative deviation from Raoult’s law may
have a maximum boiling point at a certain composition.
Systems that display these maximum or minimum boiling points can form liquid
mixtures which, when distilled, form vapours with the same composition as the
original liquid mixtures. Mixtures with this property are described as azeotropes.
1. Negative deviation
2. Positive deviation
Temperature (°C)
Figure 6.2.11: Non-ideal mixtures can
form azeotropes: Graph 1 is a mixture
showing a negative deviation from
Raoult’s law that forms a low-boiling
azeotrope; Graph 2 is a mixture showing
a positive deviation from Raoult’s law
that forms a high-boiling azeotrope.
azeotropic mix
Tb (B)
Tb (B)
V1
vapour
V1
L1
V2
Tb (A)
V2
L1
liquid
L2
Tb (A)
L2
azeotropic mix L3
0% B
6.2: Phase equilibria
100% B
0% B
100% B
13
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Distillation of azeotropic mixtures
Look at Graph 1 in Figure 6.2.11, which shows a mixture which forms a low
boiling-point azeotrope.
Imagine starting with a mixture (L1) containing 80% B. Distillation produces a
vapour V1 that condenses to a liquid L2. If distillation continues, an azeotropic
mix will be obtained, at which the composition of the vapour Va and liquid La
will be identical.
Further distillation of this mixture will produce no further change and so it is
impossible to separate the components of this type of mixture by distillation.
Activity
Use Graph 2 of Figure 6.2.11 to describe the outcome of distillation for a system that forms a high
boiling-point azeotrope. (Hint: consider what happens when distillation is carried out on mixtures
with compositions L1, L2 and so on.)
Compare this outcome with that for a system that forms a low-boiling point azeotrope.
Distillation of binary mixtures – some examples
Activity: Methanol and water
Obtain some data for the boiling points of water-methanol mixtures.
Data is available from http://en.wikipedia.org/wiki/Methanol_(data_page), which is based on
the searchable database at http://www.cheric.org/research/kdb/hcvle/hcvle.php.
•• Select suitable sets of data for boiling point, mole fraction of liquid and mole fraction of vapour.
•• Plot these points on a temperature–composition graph. Draw suitable best fit curves.
•• Explain what would happen if a mixture of 80% methanol and 20% water was distilled.
Activity: Propanone
and water
Carry out a similar analysis on a
propanone-water mixture. Data
available at http://chestofbooks.
com/science/chemistry/
Distillation-Principles-AndProcesses/Chapter-XXI-Acetone.
html.
Portfolio activity (2.3)
1 Choose a system of two liquids that forms an ideal mixture. Examples could include
propan-1-ol mixed with either propan-2-ol or 2-methylpropan-2-ol. Explain how vapour
pressure–composition and boiling point–composition diagrams are helpful in understanding
the behaviour of this system and in explaining the process of distillation.
In your answer:
•• Find or produce a vapour pressure–composition diagram for your chosen system and
explain how this shows that the system behaves as an ideal mixture.
•• Find or plot a boiling point–composition diagram for your chosen system. Explain the
meaning of the two curves on the diagram.
•• Explain the process of distillation using this boiling point–composition diagram.
2 Choose a system of two liquids that form a mixture with significant deviation from ideal
behaviour. Examples could include methanol/chloroform or methanoic acid/water.
•• Find or produce a vapour pressure–composition diagram for your chosen system and
explain how this shows that the system deviates from ideal behaviour.
•• Find or plot a boiling point–composition diagram for your chosen system. Does the system
form an azeotropic mixture? If so, explain this term.
•• Explain any problems that might arise in the process of distillation using this boiling point–
composition diagram.
6.2: Phase equilibria
14
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
3 Solid-liquid equilibria
Just as a liquid mixture is in equilibrium with a vapour mixture at its boiling point,
so a mixture of two solids can be in equilibrium with a liquid mixture at its melting
point. Phase diagrams can be constructed and interpreted in a similar way to those
in the previous section.
This type of situation occurs when, for example, alloys are being formed, and so a
study of solid-liquid phase equilibria is important in metallurgy. Tiny changes in
the composition of an alloy can have a great effect on the key properties of the
alloy, such as strength or electrical conductivity.
Solid-liquid binary mixtures
Completely miscible binary mixtures
If two solids with very similar physical properties are being mixed (for example,
two metals being alloyed) then, not only will the liquids be miscible in all
proportions, but the solids may also be completely miscible in all proportions.
Case study
Molten aluminium-gallium alloys are being investigated as part of systems used to generate hydrogen
from water on a small scale. The two metals are miscible in all proportions even though the melting
temperatures of aluminium (660 °C) and gallium (30 °C) are very different. However, only alloys with
very low percentages of aluminium are molten at temperatures below 100 °C, which is a problem for
the hydrogen generation system as the aluminium is the active ingredient in these systems.
• Sketch out a graph similar to that in Figure 6.2.12 to show the phase diagram for this aluminiumgallium alloy. Consider an alloy which is in the liquid phase at 90 °C. Predict the maximum
percentage of aluminium present in such an alloy.
Figure 6.2.12: The phase diagram shows
the temperatures at which the solid and
liquid phases are formed when liquids
of different compositions are cooled.
Temperature
A phase diagram can be obtained by plotting cooling curves for mixtures with a
range of compositions, as shown in Figure 6.2.12.
Mixture containing 20% B + 80% A is cooled
Tm (A)
liquid
T1
T2
Tm (B)
mixture of liquid and solid
solid
0%
25%
50%
75%
100% B
Liquid mixtures of different compositions are cooled, and the temperatures at
which phase transitions occur are noted.
6.2: Phase equilibria
15
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
The phase transition from solid to liquid for a pure element occurs at a specific
temperature. However, the formation of a solid from this ideal liquid mixture
occurs over a range of temperatures, illustrated by the shaded sector in the graph
in Figure 6.2.12. At temperatures above the upper boundary of this sector, the
mixture exists as a liquid, below the lower boundary it is a solid, and within the
boundary it is a mixture of liquid and solid phases.
Eutectic systems
Figure 6.2.13: Phase diagram for
a system containing a mixture of
lead (Pb) and antimony (Sb). This
is called a eutectic system because
there is a specific composition of
the mixture (the eutectic mix) that
melts at a minimum temperature.
Temperature (°C)
A much more common situation with solid-liquid binary mixtures occurs when,
although the liquid forms are miscible in all proportions, the two solids are not.
In this case a very different temperature composition phase diagram is found, as
shown in Figure 6.2.13.
75% Sb
liquid
Tm (Pb) 328 °C
631 °C Tm (Sb)
liquid + crystals of Sb
liquid +
crystals of Pb
228 °C Tm (eutectic mixture)
eutectic mixture
crystals of Pb + Sb
0
25
50
75
100% Sb
In Figure 6.2.13 you can see that the mixture with a composition of about 12%
antimony has a melting/freezing point lower than any other composition. This
mixture is known as a eutectic mix.
A eutectic mix is also the composition at which melting occurs at a single fixed
temperature, the eutectic temperature. At all other compositions, melting (and
freezing) occurs over a range of temperatures.
In Figure 6.2.13, the section of the diagram to the left of the eutectic mix contains
both liquid and solid phases – but the solid is now pure lead. The section to the
right of the eutectic mix contains liquid and pure antimony.
Activity
Describe what happens when a liquid
with the composition 5% antimony
and 95% lead is cooled from 600 °C
to 228 °C.
Some interesting changes occur when liquid mixtures of various compositions
are cooled:
• A liquid with the eutectic composition (12% Sb) is cooled. It remains a liquid
until the eutectic melting temperature (about 228 °C) is reached. At this point
a solid forms with the eutectic composition (12% Sb, 88% Pb).
• A liquid containing 75% Sb is cooled. When it reaches about 550 °C, a solid
starts to form. This is pure antimony; the remaining liquid now has an
increased proportion of lead. As it cools further, more antimony crystallises
out and the proportion of lead increases, following the line that marks the
transition between the liquid and the liquid-crystal mixture. Eventually, this
liquid reaches the eutectic composition and so solid starts to form with the
eutectic composition.
So as these systems crystallise, the solid crystals that form will contain regions of
pure antimony or pure lead. These can often be observed as distinct microcrystals
in the structure.
6.2: Phase equilibria
16
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Practical work to obtain phase diagrams
Although some of the examples shown above relate to systems of metal mixtures
used in alloying, these can be difficult to study in the laboratory.
Take it further
Details of how to produce a simple temperature composition phase diagram from cooling curve
data is available at http://www.doitpoms.ac.uk/tlplib/phase-diagrams/cooling.php, provided
by the Cambridge University Department of Materials Science.
O
O
H3C
OH
N
H
Figure 6.2.14: Structures of acetanilide
(left) and benzoic acid (right).
Systems that can be more easily studied by the cooling curve method include
crystallisation of mixtures of acetanilide (melting point 114 °C) and benzoic
acid (melting point 122 °C). The structures of these two molecules are shown in
Figure 6.2.14.
Obtaining cooling curves can involve the use of quite large quantities of these
substances at temperatures that may produce significant amounts of vapour. It is
essential to know the hazards of any substance used in a cooling curve experiment
in solid, liquid and vapour form and to take appropriate precautions to minimise
the hazard.
Portfolio activity (2.4)
Obtain a phase diagram for a simple eutectic system by experimental investigation or research.
Suitable examples for which data is readily available include water/ethylene glycol or alloy
mixtures such as lead/tin or cadmium/bismuth.
Describe how the crystallisation of the system can be explained with reference to this diagram. In
your answer you should:
•• explain what is meant by a eutectic mixture
•• describe what happens to liquid mixtures of various compositions when they are cooled
•• describe the implications for the structure of the solid crystals formed by cooling.
Checklist
At the end of this topic guide you should be familiar with the following ideas:
 phase diagrams can be constructed for simple one-component systems showing the stable
phase at different conditions of temperature and pressure
 the variation of pressure with temperature for the phase boundaries in these phase diagrams
can be described by equations such as the Clapeyron and Clausius-Clapeyron equations;
these equations also involve quantities such as the volume change and enthalpy change of
the phase transition
 the vapour phase in equilibrium with a binary mixture of volatile liquids has a different
composition to that of the liquid phase; this can be shown using boiling point–composition
diagrams
 in ideal mixtures, the vapour pressure of each component is proportional to the mole fraction
of the component in the mixture; many pairs of liquids show deviation from this behaviour
 the process of distillation can be explained using boiling point–composition diagrams; the
diagrams also explain the difficulty of separating azeotropic mixtures using distillation
 similar melting point–composition diagrams can be constructed to show the behaviour
of solid/liquid equilibrium systems; the process of crystallisation and its limitations can be
explained using these diagrams.
6.2: Phase equilibria
17
Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and phase equilibria
Further reading
Elements of Physical Chemistry (Atkins and de Paula, 2009) has useful extra material on physical
equilibria in Chapter 5, including a mathematical derivation of the Clausius-Clapeyron equation
and its assumptions.
The Chemguide website (www.chemguide.co.uk) has an excellent series of pages on phase
equilibria. From the homepage, select the Physical Chemistry section and then use the Phase
Equilibria menu to select the relevant pages for this topic guide.
Acknowledgements
The publisher would like to thank the following for their kind permission to reproduce their
photographs:
Corbis: David Sutherland
All other images © Pearson Education
Every effort has been made to trace the copyright holders and we apologise in advance for any
unintentional omissions. We would be pleased to insert the appropriate acknowledgement in any
subsequent edition of this publication.
6.2: Phase equilibria
18