topic guide 6.2: phase equilibria · 2/6/2010 · 6.2 phase equilibria the phase rule and degrees...

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Unit 6: Physical chemistry of spectroscopy, surfaces and chemical and 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).

Phase equilibria6.2

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6.2: Phase equilibria

1 Phase equilibriaCarbon 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 diagramsExperimental 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.

Phase diagrams are often drawn with a logarithmic scale on the pressure axis (as in Figure 6.2.1).

Temperature (°C)

Pres

sure

(bar

)

5031.10–100 –78.4 –56.6

solid

vapour

liquid

0.1

15.185

10

72.9

100

triple point

critical point

Interpreting a phase diagramPhase 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.

Key termPhase: 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.

Figure 6.2.1: The phase diagram for this simple, one-component

system allows you to predict various properties of carbon dioxide.

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• 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

The phase diagram for water is shown in Figure 6.2.2.

vaporisation

subl

imat

ion

Temperature (K)

Pres

sure

(bar

)

100 200 300 400

solid

liquid

vapour (gas)

triple point

critical point

fusio

n

500 600 700

0.001

0.01

0.1

1

10

100

1000

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.

Four phases means that there will be three triple points, as shown in Figure 6.2.3.

Temperature (°C)

Pres

sure

(bar

)

12

3

95.4 119.6 151

rhombic

liquid

gas

mon

oclin

ic

10–4 5×10–4

1bar

1288

Activity1 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?

Figure 6.2.2: The phase diagram for water.

Figure 6.2.3: The phase diagram for sulfur, showing the different solid

phases (rhombic and monoclinic) that exist at different conditions.

Activity1 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?

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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

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.

Activity1 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.

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Portfolio activity (2.1)Uranium hexafluoride (UF

6) 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 UF

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• 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 boundariesSome 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

TΔVtrans

. ΔT or

ΔpΔT

= ΔH

trans

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).

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Temperature (°C)Pr

essu

re (b

ar)

T1

P2

P1

T2

solid

liquid

ΔP

ΔT

Slope of the phase boundariesFor 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 equationBy writing the Clapeyron equation in the form:

dpdT

= ΔH

TΔV it follows that

dp = ΔHΔV

dTT

And hence, integrating this yields the integrated form of the Clapeyron equation:

(3) p2 – p1 = ΔHΔV

(lnT1 – lnT2)

ActivityIce 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 T

2 and hence T

2. Be careful to

ensure that you convert the data provided into the appropriate units.)

Clausius-Clapeyron equationFor 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.

Figure 6.2.4: The slope of the phase boundary curve, ΔP

ΔT, is described

by the Clapeyron equation.

ActivityWhen 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?

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Take it furtherElements 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 • ΔH

vap remains constant over the range of pressure and temperature being studied.

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 + ΔHvap

R ( 1

T1

– 1T2)

(5) lnp2

p1

= ΔHvap

R ( 1

T1

– 1T2)

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.

ActivityUnder 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. ΔH

vap 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.

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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 ΔH

vap of water as +40.5 kJ mol−1.

2 Binary mixturesIn the previous section you were concerned with one-component systems where a single chemical substance was present, in a range of phases.

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 mixturesJust 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.

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.

Mole fraction of heptane0.25 0.5 0.75 10

composition of liquid

Tb (heptane) = 98 °C

Tb (hexane) = 69 °C

composition of vapour

40

50

60

70

80

90

100

Tem

pera

ture

(°C)

Key termBinary mixture: A mixture containing two different components, for example hexane and heptane, or water and ethanol.

Figure 6.2.5: The boiling point (T

b, 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.

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Mole fraction

The composition of a mixture of two components, say, A and B, is usually defined by the mole fraction of A, XA:

(6) XA = number of moles of A

total number of moles of A + B

Activity1 A mixture of hexane (M

r = 86.2) and heptane (M

r = 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 (M

r = 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:

% by mass = mass of A

total mass of A and B × 100

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.

ActivityUse 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 mixturesThe 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.

Key termPartial vapour pressure: The partial pressure due to the vapour of a single component above a solid or liquid mixture.

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(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.

Mole fraction of A

Vapo

ur p

ress

ure

0.25 0.5 0.75 1.00

PA°

PB°

PB

PA

PA+ PB

ActivityA 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 lawNon-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).

Mole fraction of A

Positive deviation Negative deviation

Vapo

ur p

ress

ure

0.25 0.5 0.75 1.00

P°A

P°B

PB

PA

Mole fraction of A0.25 0.5 0.75 1.00

P°A

P°B total vapour pressure

Positive deviations occur when the forces of attraction between molecules in the mixture are weaker than the forces of attraction in the pure liquids.

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.

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.

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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.

DistillationDistillation 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.

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).

% by mass of octane25 50 75 1000

liquid

Tb (octane)

Tb (pentane)

vapour

AB

C

0

25

50

75

100

125

150

Tem

pera

ture

(°C)

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).

Figure 6.2.8: Distilling a pentane-octane mixture produces a

distillate rich in pentane.

ActivityConsider 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.

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80 60 40 20 0%100%20 40 60 800% 100% B

liquid

V1

V2

V3

V4

V5 L5 L4

L3

L2

L1vapour

A0

90 90 bpB

80

70

60

50

Tem

pera

ture

(°C)

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.

fractionatingcolumn packedwith glass beads(or similar)

heat

water in

water out

thermometer

mixture

product collects

vent to fumecupboard

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.

Figure 6.2.10: Fractional distillation in the laboratory.

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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 furtherSome 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.

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.

2. Positive deviation1. Negative deviation

0% B 100% B

Tb (B)

Tb (A) liquid

vapour

azeotropic mix

azeotropic mix

L3

L2

L2

L1

L1

V1

V1

V2

V2

Tem

pera

ture

(°C)

0% B 100% B

Tb (B)

Tb (A)

ActivityExplain why a large negative deviation from Raoult’s law can lead to a mixture having a maximum boiling point at a certain composition.

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.

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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.

ActivityUse 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 L

1, L

2 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 waterObtain 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.

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.

Activity: Propanone and waterCarry out a similar analysis on a propanone-water mixture. Data available at http://chestofbooks.com/science/chemistry/Distillation-Principles-And-Processes/Chapter-XXI-Acetone.html.

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3 Solid-liquid equilibriaJust 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 eff ect on the key properties of the alloy, such as strength or electrical conductivity.

Solid-liquid binary mixturesCompletely 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 studyMolten 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 diff erent. 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 aluminium-gallium alloy. Consider an alloy which is in the liquid phase at 90 °C. Predict the maximum percentage of aluminium present in such an alloy.

A phase diagram can be obtained by plotting cooling curves for mixtures with a range of compositions, as shown in Figure 6.2.12.

0% 25% 50% 75% 100% B

liquid

mixture of liquid and solid

solid

Tm (B)

Tm (A) T1

T2

Mixture containing 20% B + 80% A is cooled

Tem

pera

ture

Liquid mixtures of diff erent compositions are cooled, and the temperatures at which phase transitions occur are noted.

Figure 6.2.12: The phase diagram shows the temperatures at which the solid and

liquid phases are formed when liquids of diff erent compositions are cooled.

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The phase transition from solid to liquid for a pure element occurs at a specifi c 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

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 diff erent temperature composition phase diagram is found, as shown in Figure 6.2.13.

25 50 75

75% Sb

100% Sb0

liquid

liquid + crystals of Sb

crystals of Pb + Sb

Tm (Pb) 328 °C

228 °C Tm (eutectic mixture)

631 °C Tm (Sb)Te

mpe

ratu

re (°

C)

eutectic mixtureliquid + crystals of Pb

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 fi xed 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.

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.

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 specifi c composition of

the mixture (the eutectic mix) that melts at a minimum temperature.

ActivityDescribe what happens when a liquid with the composition 5% antimony and 95% lead is cooled from 600 °C to 228 °C.

17



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 furtherDetails 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.

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.

ChecklistAt 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.

O

NH

H3C

O OH

Figure 6.2.14: Structures of acetanilide (left) and benzoic acid (right).

18



Further readingElements 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.

AcknowledgementsThe 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.

topic guide 6.2: phase equilibria · 2/6/2010 · 6.2 phase equilibria the phase rule and degrees...

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