AN INTRODUCTION TO SATURATED VAPOUR

PRESSURE

Evaporation of the Liquid in an Open Container• The average energy of the

particles in a liquid is governed by the temperature

• The higher the temperature, the higher the average energy. But within that average, some particles have energies higher than the average, and others have energies lower than the average.

• Water molecules are simply breaking away from the surface layer.

• The energy which is lost as the particles evaporate is replaced from the surroundings.

• As the molecules in the water jostle with each other, new molecules will gain enough energy to escape from the surface.

Evaporation of the Liquid in an Closed Container

• first the molecules start to evaporate and hit the side of the container and cool down.

• After some time, the rate of evaporation and condensation becomes equal and the closed system reaches equilibrium.

• When these particles hit the walls of the container, they exert a pressure. This pressure is called the saturated vapour pressure (also known as saturation vapour pressure) of the liquid.

The variation of saturated vapour pressure with temperature

The effect of temperature on the equilibrium between liquid and vapour. This can be seen in two ways:

• If you increase the temperature, you are increasing

the average energy of the particles present.

• That means that more of them are likely to have enough

energy to escape from the surface of the liquid. That will

tend to increase the saturated vapour pressure.

• OR:When the space above the liquid is saturated with vapour particles, you have this equilibrium occurring on the surface of the liquid:

The forward change (liquid to vapour) is endothermic. It needs heat to convert the liquid into the vapour.

According to Le Chatelier, increasing the temperature of a system in a dynamic equilibrium favours the endothermic change. That means that increasing the temperature increases the amount of vapour present, and so increases the saturated vapour pressure.

The effect of temperature on the saturated vapour pressure of water

Saturated vapour pressure and boiling

point

•A liquid boils when its saturated vapour

pressure becomes equal to the external

pressure on the liquid. When that

happens, it enables bubbles of vapour to

form throughout the liquid - those are the

bubbles you see when a liquid boils.

Sublimation

Solids can also lose particles from their surface to form a vapour, except that in this

case we call the effect sublimation rather than evaporation. Sublimation is the direct

change from solid to vapour (or vice versa) without going through the liquid stage.

In most cases, at ordinary temperatures, the saturated vapour pressures of solids

range from low to very, very, very low. The forces of attraction in many solids are too

high to allow much loss of particles from the surface.

However, there are some which do easily form vapours. For example, naphthalene

(used in old-fashioned "moth balls" to deter clothes moths) has quite a strong smell.

Molecules must be breaking away from the surface as a vapour, because otherwise

you wouldn't be able to smell it.

Another fairly common example is solid carbon dioxide - "dry ice". This never forms a

liquid at atmospheric pressure and always converts directly from solid to vapour.

That's why it is known as dry ice.

Saturated vapour pressure and solids

PHASE DIAGRAMS OF PURE SUBSTANCESPhase diagrams

A phase diagram lets you work out exactly what phases are present at any given temperature and pressure. In the cases we'll be looking at on this page, the phases will simply be the solid, liquid or vapour (gas) states of a pure substance.

This is the phase diagram for a typical pure substance.

These diagrams (including this one) are nearly always drawn highly distorted in order to see what is going on more easily. There are usually two major distortions. We'll discuss these when they become relevant.

If you look at the diagram, you will see that there are three lines, three areas marked "solid", "liquid" and "vapour", and two special points marked "C" and "T".

The three areas

These are easy! Suppose you have a pure substance at three different sets of conditions of temperature and pressure corresponding to 1, 2 and 3 in the next diagram.

Under the set of conditions at 1 in the diagram, the substance would be a solid because it falls into that area of the phase diagram. At 2, it would be a liquid; and at 3, it would be a vapour (a gas).

Normal melting and boiling points

The normal melting and boiling points are those when the pressure is 1 atmosphere. These can be found from the phase diagram by drawing a line across at 1 atmosphere pressure.

The phase diagram for water

The phase diagram for carbon dioxide

RAOULT'S LAW AND NON-VOLATILE SOLUTESRaoult's Law

There are several ways of stating Raoult's Law, and you tend to use slightly different

versions depending on the situation you are talking about. You can use the simplified

definition in the box below in the case of a single volatile liquid (the solvent) and a

non-volatile solute.

The vapour pressure of a solution of a non-volatile solute is equal to the vapour pressure of the pure solvent at that temperature multiplied by its mole fraction. In equation form, this reads:

In this equation, Po is the vapour pressure of the pure solvent at a particular

temperature.

xsolv is the mole fraction of the solvent. That is exactly what it says it is - the fraction

of the total number of moles present which is solvent.

A simple explanation of why Raoult's Law works

Limitation of Raoults law

• Raoult's Law only works for ideal solutions. An ideal solution is defined as one which obeys Raoult's Law.

• Only very dilute solution obeys this laws.• The forces of attraction between solvent

and solute are exactly the same as between the original solvent molecules

Raoult's Law and melting and boiling points

Phase diagram water and a Non volatile substance

Because of the changes to the phase diagram, you can see that:

the boiling point of the solvent in a solution is higher than that of the pure solvent; the freezing point (melting point) of the solvent in a solution is lower than that of the pure

solvent.

We have looked at this with water as the solvent, but using a different solvent would make no difference to the argument or the conclusions.

The only difference is in the slope of the solid-liquid equilibrium lines. For most solvents, these slope forwards whereas the water line slopes backwards.

RAOULT'S LAW AND IDEAL MIXTURES OF LIQUIDS

• Raoult's Law and how it applies to mixtures of two volatile liquids

• Examples of ideal mixtures

• There is actually 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.

• Commonly quoted examples include:

• hexane and heptane

• benzene and methylbenzene

• propan-1-ol and propan-2-ol

Ideal mixtures and intermolecular forces

In a pure liquid, some of the more energetic molecules have enough energy to overcome the intermolecular attractions and escape from the surface to form a vapour.

The smaller the intermolecular forces, the more molecules will be able to escape at any particular temperature.

If you have a second liquid, the same thing is true. At any particular temperature a

certain proportion of the molecules will have enough energy to leave the surface.

In an ideal mixture of these two liquids, the tendency of the two different sorts of

molecules to escape is unchanged.

You might think that the diagram shows only half as many of each molecule

escaping - but the proportion of each escaping is still the same. The diagram is for a

50/50 mixture of the two liquids. That means that there are only half as many of each

sort of molecule on the surface as in the pure liquids. If the proportion of each

escaping stays the same, obviously only half as many will escape in any given time.

If the red molecules still have the same tendency to escape as before, that must mean that the intermolecular forces between two red molecules must be exactly the same as the intermolecular forces between a red and a blue molecule.

If the forces were any different, the tendency to escape would change.

Exactly the same thing is true of the forces between two blue molecules and the forces between a blue and a red. They must also be the same otherwise the blue ones would have a different tendency to escape than before.

If you follow the logic of this through, the intermolecular attractions between two red molecules, two blue molecules or a red and a blue molecule must all be exactly the same if the mixture is to be 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 -

as we've seen above.

• Let's focus on one of these liquids - A, for example.

• Suppose you double the mole fraction of A in the mixture (keeping the

temperature constant). According to Raoult's Law, you will double its partial

vapour pressure. If you triple the mole fraction, its partial vapour pressure

will triple - and so on.

• In other words, the partial vapour pressure of A at a particular temperature

is proportional to its mole fraction. If you plot a graph of the partial vapour

pressure of A against its mole fraction, you will get a straight line.

Now we'll do the same thing for B - except that we will plot it on the same set of axes. The mole fraction of B falls as A increases so the line will slope down

rather than up. As the mole fraction of B falls, its vapour pressure will fall at the same rate.

Notice that the vapour pressure of pure B is higher than that of pure A. That means that molecules must break away more easily from the surface of B than of A. B is the more volatile liquid.To get the total vapour pressure of the mixture, you need to add the values for A and B together at each composition. The net effect of that is to give you a straight line as shown in the next diagram.

Boliling point / composition diagrams

• If a liquid has a high vapour pressure at a particular temperature, it

means that its molecules are escaping easily from the surface.

• If, at the same temperature, a second liquid has a low vapour

pressure, it means that its molecules aren't escaping so easily.

• What does that imply about the boiling points of the two liquids?

• There are two ways of looking at this.

• Either:

• If the molecules are escaping easily from the surface, it must mean that the

intermolecular forces are relatively weak. That means that you won't have to supply

so much heat to break them completely and boil the liquid.

• The liquid with the higher vapour pressure at a particular temperature is the one with

the lower boiling point.

• Or:

• Liquids boil when their vapour pressure becomes equal to the external pressure. If a

liquid has a high vapour pressure at some temperature, you won't have to increase

the temperature very much until the vapour pressure reaches the external pressure.

On the other hand if the vapour pressure is low, you will have to heat it up a lot more

to reach the external pressure.

• The liquid with the higher vapour pressure at a particular temperature is the one with

the lower boiling point.

Constructing a boiling point / composition diagram

• To make this diagram really useful (and finally get to the phase

diagram we've been heading towards), we are going to add another

line. This second line will show the composition of the vapour over

the top of any particular boiling liquid.

• If you boil a liquid mixture, you would expect to find that the more

volatile substance escapes to form a vapour more easily than the

less volatile one.

• That means that in the case we've been talking about, you would

expect to find a higher proportion of B (the more volatile component)

in the vapour than in the liquid. You can discover this composition by

condensing the vapour and analysing it. That would give you a point

on the diagram.

Using the phase 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.

• 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. You can see that

we now have a vapour which is getting

quite close to being pure B. If you keep on

doing this (condensing the vapour, and

then reboiling the liquid produced) you will

eventually get pure B.

FRACTIONAL DISTILLATION OF IDEAL MIXTURES OF LIQUIDS

Fractional distillation in the lab-The apparatus

Some notes on the apparatus

• The fractionating column is packed with glass beads (or something similar) to

give the maximum possible surface area for vapour to condense on. You will

see why this is important in a minute. Some fractionating columns have spikes

of glass sticking out from the sides which serve the same purpose.

• If you sketch this, make sure that you don't completely seal the apparatus.

There has to be a vent in the system otherwise the pressure build-up when

you heat it will blow the apparatus apart.

• In some cases, where you are collecting a liquid with a very low boiling point,

you may need to surround the collecting flask with a beaker of cold water or

ice.

• The mixture is heated at such a rate that the thermometer is at the

temperature of the boiling point of the more volatile component. Notice that the

thermometer bulb is placed exactly at the outlet from the fractionating column.

Relating what happens in the fractionating column to the phase diagram

Fractional distillation industrially

• The column contains a number of trays that the liquid collects on as the vapour condenses. The up-coming hot vapour is forced through the liquid on the trays by passing through a number of bubble caps. This produces the maximum possible contact between the vapour and liquid. This all makes the boiling-condensing-reboiling process as efficient as possible.

• The overflow pipes are simply a controlled way of letting liquid trickle down the column.

• If you have a mixture of lots of liquids to separate (such as in petroleum fractionation), it is possible to tap off the liquids from some of the trays rather than just collecting what comes out of the top of the column. That leads to simpler mixtures such as gasoline, kerosene and so on.

NON-IDEAL MIXTURES OF LIQUIDS

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• behave just like ideal mixtures

as far as distillation is concerned

• But some liquid mixtures have very large positive deviations from Raoult's Law, and in these cases, the curve becomes very distorted.

• 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.

Explaining the deviations• 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.

• You can see this when you mix the liquids. Less heat is evolved when the new attractions are set up than was absorbed to break the original ones.

• Heat will therefore be absorbed when the liquids mix.

• The enthalpy change of mixing is endothermic.

• 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.

Negative deviations from Raoult's Law

• the deviation are much

greater giving a minimum

value for vapour pressure

lower than that of either

pure component

Explaining the deviations

• 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.

• You can recognise this happening because heat is evolved when you mix

the liquids - more heat is given out when the new stronger bonds are made

than was used in breaking the original weaker ones.

• Many (although not all) examples of this involve actual reaction between the

two liquids.

• The example of a major negative deviation that we are going to look at is a

mixture of nitric acid and water. These two covalent molecules react to give

hydroxonium ions and nitrate ions.

• You now have strong ionic attractions involved.

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.

• In the case of mixtures of ethanol and water, this minimum occurs with 95.6% by mass of ethanol in the mixture. The boiling point of this mixture is 78.2°C, compared with the boiling point of pure ethanol at 78.5°C, and water at 100°C.

• You might think that this 0.3°C doesn't matter much, but it has huge implications for the separation of ethanol / water mixtures.

the boiling point / composition curve for ethanol / water mixtures

Distillation of an ethanol and water Mixture

Notes

• What happens if you reboil that liquid?

• The liquid curve and the vapour curve meet at that point. The vapour produced will have that same composition of 95.6% ethanol. If you condense it again, it will still have that same composition.

• You have hit a barrier. It is impossible to get pure ethanol by distiling any mixture of ethanol and water containing less than 95.6% of ethanoll.

• 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.

To summarise• Distilling a mixture of ethanol containing less than 95.6% of ethanol by mass

lets you collect:

• a distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control, and the fractionating column is long enough);

• pure water in the boiling flask.

What if you distil a mixture containing more than 95.6% ethanol?• Work it out for yourself using the phase diagram, and starting with a

composition to the right of the azeotropic mixture. You should find that you get:

• a distillate containing 95.6% of ethanol in the collecting flask (provided you are careful with the temperature control, and the fractionating column is long enough);

• pure ethanol in the boiling flask.

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. You can see this from the vapour pressure / composition curve.

• Mixtures of nitric acid and water can have boiling points higher than either of the pure liquids because it needs extra heat to break the stronger attractions in the mixture.

• 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.

• Notice the much bigger difference this time due to the presence of the new ionic interactions.

vapour pressure / composition curve of Nitric Acid and water

Distilling dilute nitric acid

• 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

• 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.