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PHASE SEPARATIONS
The evaporation of a liquid in a closed container
Particles will gain kinetic energy and break away from the surface of the liquid.
As the gaseous particles bounce around, some of them will hit the surface of the
liquid again, and be trapped there. There will rapidly be an equilibrium set up in
which the number of particles leaving the surface is exactly balanced by the number
rejoining it. In this equilibrium, there will be a fixed number of the gaseous
particles in the space above the liquid.
When these particles hit the walls of the container, they exert a pressure. This
pressure is called the saturated vapour pressure of the liquid.
Saturated vapour pressure and boiling point
A liquid boils when its saturated vapour pressure becomes equal to the external
pressure on the liquid.
If the liquid is in an open container and exposed to normal atmospheric pressure, the
liquid boils when its saturated vapour pressure becomes equal to 1 atmosphere (or
101325 Pa or 101.325 kPa or 760 mmHg). This happens with water when the
temperature reaches 100°C.
But at different pressures, water will boil at different temperatures. For example, at
the top of Mount Everest the pressure is so low that water will boil at about 70°C.
Depressions from the Atlantic can easily lower the atmospheric pressure so that
water will boil at 99°C - even lower with very deep depressions.
Raoult's Law
The vapour pressure of a solution with a non-volatile solute e.g. water with salt in it
is equal to the vapour pressure of the pure solvent at that temperature multiplied by
its mole fraction
In equation form, this reads: p = xsolv x P0solv
In this equation, Po is the vapour pressure of the pure solvent at a particular
temperature. xsolv is the mole fraction of the solvent.
You calculate this using: xsolv = moles of solvent / total number of moles
Suppose you had a solution containing 10 moles of water and 0.1 moles of sugar.
The total number of moles is therefore 10.1
 The mole fraction of the water is: xwater = 10 / 10.1 = 0.99
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Explanation of why Raoult's Law works
Remember that saturated vapour pressure is what you get when a liquid is in a
sealed container. An equilibrium is set up where the number of particles breaking
away from the surface is exactly the same as the number sticking on to the surface
again. Now suppose you added enough solute so that the solvent molecules only
occupied 50% of the surface of the solution.
If you reduce the number of solvent molecules on the surface, you are going to
reduce the number which can escape in any given time. The net effect of this is that
when equilibrium is established, there will be fewer solvent molecules in the vapour
phase – which means that the saturated vapour pressure would be lower.
Limitations of Raoult's Law
Raoult's Law only works for ideal solutions. An ideal solution is defined as one
which obeys Raoult's Law.
Examples of ideal mixtures
There is no such thing as an ideal mixture. However, some liquid mixtures get fairly
close to being ideal. These are mixtures of two very closely similar substances.
Common examples include:
 hexane and heptane
 benzene and methylbenzene
 propan-1-ol and propan-2-ol
Ideal mixtures and enthalpy change of mixing
If the temperature rises or falls when you mix the two liquids, then the mixture isn't
ideal.
Vapour pressure / composition diagrams
Suppose you have an ideal mixture of two liquids A and B. Each of A and B is
making its own contribution to the overall vapour pressure of the mixture. The
greater the mole fraction, the greater the vapour pressure exerted.
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Using a boiling point / composition diagram
B has the higher vapour pressure. That means that it will have the lower boiling
point. Remember this inverse relationship!!!!
If you boil a liquid mixture, you can find out the temperature it boils at, and the
composition of the vapour over the boiling liquid.
For example, in the next diagram, if you boil a liquid mixture C1, it will boil at a
temperature T1 and the vapour over the top of the boiling liquid will have the
composition C2.
All you have to do is to use the liquid composition curve to find the boiling point of
the liquid, and then look at what the vapour composition would be at that
temperature. Notice again that the vapour is much richer in the more volatile
component B than the original liquid mixture was.
The beginnings of fractional distillation
Suppose that you collected and condensed the vapour over the top of the boiling
liquid and reboiled it. You would now be boiling a new liquid which had a
composition C2. That would boil at a new temperature T2, and the vapour over the
top of it would have a composition C3. If you keep on doing this (condensing the
vapour, and then reboiling the liquid produced) you will eventually get pure B
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Please note
Raoult's Law only works for ideal mixtures. In these, the forces between the
particles in the mixture are exactly the same as those in the pure liquids. The
tendency for the particles to escape is the same in the mixture and in the pure
liquids. That is NOT TRUE in non-ideal mixtures.
Positive deviations from Raoult's Law
In mixtures showing a positive deviation from Raoult's Law, the vapour pressure of
the mixture is always higher than you would expect from an ideal mixture. The
deviation can be small or it may be large. Large deviations shown a very distorted
curve as shown below.
REMEMBER HIGHER VAPOUR PRESSURE MEANS LOWER BOILING
POINT!
Notice that mixtures over a range of compositions have higher vapour pressures
than either pure liquid. The maximum vapour pressure is no longer that of one of
the pure liquids.
Explanation
The fact that the vapour pressure is higher than ideal in these mixtures means that
molecules are breaking away more easily than they do in the pure liquids. That is
because the intermolecular forces between molecules of A and B are less than they
are in the pure liquids.
The classic example of a mixture of this kind is ethanol and water. This produces a
highly distorted curve with a maximum vapour pressure for a mixture containing
95.6% of ethanol by mass.
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Negative deviations from Raoult's Law
In exactly the same way, you can have mixtures with vapour pressures which are
less than would be expected by Raoult's Law. In some cases, the deviations are
small, but in others they are much greater giving a minimum value for vapour
pressure lower than that of either pure component.
Explanation
These are cases where the molecules break away from the mixture less easily than
they do from the pure liquids. New stronger forces must exist in the mixture than in
the original liquids. The example of a major negative deviation is a mixture of
nitric acid and water.
A large positive deviation from Raoult's Law: ethanol and water mixtures
If a mixture has a high vapour pressure it means that it will have a low boiling
point. The molecules are escaping easily and you won't have to heat the mixture
much to overcome the intermolecular attractions completely.
The implication of this is that the boiling point / composition curve will have a
minimum value lower than the boiling points of either A or B.
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Suppose you are going to distil a mixture of ethanol and water with composition C 1
as shown on the next diagram. It will boil at a temperature given by the liquid curve
and produce a vapour with composition C2. . If you reboil that, it will produce a new
vapour with composition C3.
You can see that if you carried on with this boiling-condensing-reboiling sequence,
you would eventually end up with a vapour with a composition of 95.6% ethanol. If
you condense that you obviously get a liquid with 95.6% ethanol.
This particular mixture of ethanol and water boils as if it were a pure liquid. It has a
constant boiling point, and the vapour composition is exactly the same as the liquid.
It is known as a constant boiling mixture or an azeotropic mixture or an
azeotrope.
A large negative deviation from Raoult's Law: nitric acid and water mixtures
Nitric acid and water form mixtures in which particles break away to form the
vapour with much more difficulty than in either of the pure liquids. In the case of
mixtures of nitric acid and water, there is a maximum boiling point of 120.5°C
when the mixture contains 68% by mass of nitric acid. That compares with the
boiling point of pure nitric acid at 86°C, and water at 100°C.
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Start with a dilute solution of nitric acid with a composition of C1 and trace through
what happens.
The vapour produced is richer in water than the original acid. If you condense the
vapour and reboil it, the new vapour is even richer in water. Fractional distillation of
dilute nitric acid will enable you to collect pure water from the top of the
fractionating column. As the acid loses water, it becomes more concentrated. Its
concentration gradually increases until it gets to 68% by mass of nitric acid. At that
point, the vapour produced has exactly the same concentration as the liquid, because
the two curves meet.
You produce a constant boiling mixture (or azeotropic mixture or azeotrope). If you
distil dilute nitric acid, that's what you will eventually be left with in the distillation
flask. You can't produce pure nitric acid from the dilute acid by distilling it.
Distilling nitric acid more concentrated than 68% by mass
This time you are starting with a concentration C2 to the right of the azeotropic
mixture.
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The vapour formed is richer in nitric acid. If you condense and reboil this, you will
get a still richer vapour. If you continue to do this all the way up the fractionating
column, you can get pure nitric acid out of the top.
As far as the liquid in the distillation flask is concerned, it is gradually losing nitric
acid. Its concentration drifts down towards the azeotropic composition. Once it
reaches that, there can't be any further change, because it then boils to give a vapour
with the same composition as the liquid. Distilling a nitric acid / water mixture
containing more than 68% by mass of nitric acid gives you pure nitric acid from the
top of the fractionating column and the azeotropic mixture left in the distillation
flask.
Immiscible liquids
If you have two immiscible liquids in a closed flask and keep everything still, the
vapour pressure you measure will simply be the vapour pressure of the one which is
floating on top. There is no way that the bottom liquid can turn to vapour. The top
one is sealing it in. For the purposes of the rest of this topic, we always assume that
the mixture is being stirred or agitated in some way so that the two liquids are
broken up into drops. At any one time there will be drops of both liquids on the
surface. That means that both of them contribute to the overall vapour pressure of
the mixture.
Total vapour pressure of the mixture
Assuming that the mixture is being agitated, then both of the liquids will be in
equilibrium with their vapours. The total vapour pressure is then simply the sum of
the individual vapour pressures:
. . . where po refers to the saturated vapour pressure of the pure liquid.
Notice that this is independent of the amount of each sort of liquid present. All you
need is enough of each so that both can exist in equilibrium with their vapour.
For example, phenylamine and water can be treated as if they were completely
immiscible.
At 98°C, the saturated vapour pressures of the two pure liquids are:
phenylamine
7.07 kPa
water
94.30 kPa
The total vapour pressure of an agitated mixture would just be the sum of these - in
other words, 101.37 kPa
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Boiling point of the mixture
Liquids boil when their vapour pressure becomes equal to the external
pressure. Normal atmospheric pressure is 101.325 kPa.
Compare that with the figure we have just got for the total vapour pressure of a
mixture of water and phenylamine at 98°C. Its total vapour pressure is fractionally
higher than the normal external pressure.
This means that such a mixture would boil at a temperature just a shade less than
98°C - in other words lower than the boiling point of pure water (100°C) and much
lower than the phenylamine (184°C).
Exactly the same sort of argument could be applied to any other mixture of
immiscible liquids.
Important conclusion
Agitated mixtures of immiscible liquids will boil at a temperature lower than the
boiling point of either of the pure liquids. Their combined vapour pressures are
bound to reach the external pressure before the vapour pressure of either of the
individual components get there.
Steam distillation
This exploits the concept shown above and allows for organic compounds which
would normally decompose at high temperatures, to be separated at lower
temperatures minimising decomposition.
Steam is blown through the mixture and the water and phenylamine turn to vapour.
This vapour can be condensed and collected.
The steam can be generated by heating water in another flask (or something
similar). As the hot steam passes through the mixture it condenses, releasing heat.
This will be enough to boil the mixture of water and phenylamine at 98°C provided
the volume of the mixture isn't too great. For large volumes, it is better to heat the
flask as well to avoid having to condense too much steam and increase the volume
of liquid in the flask too much.
Steam distillation allows separation of components which would decompose at high
temperatures usually organic compounds. Vacuum distillation operates via the use
of low pressures allowing components to be separated which would otherwise
decompose at high temperatures.
Some other applications of steam distillation
Steam distillation can be used to extract some natural products - for example, to
extract eucalyptus oil from eucalyptus, citrus oils from lemon or orange peel, and to
extract oils used in perfumes from various plant materials.
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Calculations involving steam distillation
The ratio of the moles of two liquids in the distillate is equal to the ratio of their
vapour pressures
e.g. phenylamine and water. NB A means phenylamine and W means water
nA
nW
=
If pA = 3.99 x 103 Pa
and pW = 9.70 x 104 Pa,
then ratio of the amounts =
4.11 x 10-2
pA
pW
If the amount of water produced is 1 g and the amount of phenylamine is 0.212 g,
and the molar mass of water is 18g, what is the molar mass of phenylamine?
Note nW =
mass of water (mW)
AND
molar mass of water (MW)
mA / (MA)
mW / MW
0.212 x
M
18
1
= 4.11 x 10-2

nA = mass of phenylamine (mA)
molar mass of phenylamine (M A)
0.212 / M
1/ 18
= 4.11 x 10-2
= 4.11 x 10-2
Therefore M = 0.212 x 18
4.11 x 10-2
= 92.8 g = 93 g
Solvent extraction
One solvent can be used to extract a solute from another solvent. This process is
called solvent extraction and this technique is very useful for extracting organic or
even inorganic solutes. For example, if iodine crystals are placed in water and then
a solvent like ether is added to the mixture and shaken in a separating funnel and
left for a few hours for equilibrium to be established, the majority of the iodine
crystals would be found in the non-aqueous layer. The ratio of the amounts of
solute found in each layer is a constant and this is called the partition coefficient.
partition coefficient (k) = CU CU = concentration of solute in upper layer
CL CL = concentration of solute in lower layer
Using the separating funnel, the lower layer is run off and collected and the solute
then separated from the solvent.
Sample calculation
The mass of iodine used in 0.9656 g and 50 cm3 each of water and ether was used in
the solvent extraction. After equilibrium was established, it was found that the
upper layer contained 0.0112 g and the lower layer contained 0.9544 g. What is the
partition coefficient for this system? CU / CL = 0.0112 / 0.9544 = 1.17 x 10-2
Examples of the application of distillation methods used in various industries
such as petroleum, rum and fragrance industries.
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Questions
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