1

Phase Equilibria (CH-203)

Phase transitions Change in phase without a change in chemical composition

Gibbs energy is at the centre of the discussion of transitions

Molar Gibbs energyGm = G/n = H - TS

Depends on the phase of the substance

2

A substance has a spontaneous tendency to change into a phase with the lowest

molar Gibbs energy

When an amount n of a substance changes from phase 1 (e.g. liquid) with molar Gibbs energy Gm(1) to phase 2 (e.g. vapour) with molar Gibbs energy Gm(2), the change in Gibbs energy is:

G = nGm(2) – nGm(1) = n{Gm(2) – Gm(1)}

A spontaneous change occurs when G < 0

3

The Gibbs energy of transition from metallic white tin (-Sn) to nometallic grey tin (-Sn) is +0.13 kJ mol-1 at 298 K. Which is the reference state of tin at this temperature?

White tin!

4

If at a certain temperature and pressure the solid phase of a substance has a lower molar Gibbs energy than its liquid phase, then the solid phase will be thermodynamically more stable and the liquid will (or at least have a tendency) to freeze.

If the opposite is true, the liquid phase is thermodynamically more stable and the solid will melt.

5

Proof-go back to fundamental definitions

G = H – TS; H = U + pV; U = q + wFor an infinitesimal change in G:

G + G = H + H – (T + T)(S + S) = H + H – TS – ST – TS – TSG = H – TS – STAlso can write: H = U + pV + Vp

U = TS – pV (dS = dqrev/T and dw = -pdV)

G = TS – pV + pV +Vp – TS – ST

G = Vp – ST Master Equations

6

Variation with pressure

Thus, Gibbs energy depends on:– Pressure– Temperature

We can derive (derivation 5.1 in textbook) that:

Gm = Vm p

=>Gm > 0 when p > 0

i.e. Molar Gibbs energy increases when pressure increases

7

Variation of G with pressure

Can usually ignore pressure dependence of G for condensed states

Can derive that, for a gas:

i

fm p

pRTG ln

8

Variation of G with temperature

Gm = –SmT

Gm= Gm(Tf) – Gm(Ti)

T = Tf – Ti

Can help us to understand why transitions occur

The transition temperature is the temperature when the molar Gibbs energy of the two phases are equal

The two phases are in EQUILIBIRIUM at this temperature

9

Variation of G with temperature

Gm = –SmT

Molar entropy is positive, thus an increase in T results in a decrease in Gm. Because:

Gm Sm

more spatial disorder in gas phase than in condensed phase, so molar entropy of gas phase is larger than for condensed phase.

10

Why do substances melt and vaporise?

At lower T solid has lowest Gm and thus most stable

As T increases, Gm of liquid phase falls below the solid phase and substance melts

At higher T, Gm of gas phase plunges below that of the liquid phase and the substance vaporises

11

Substances which sublime (CO2)

There is no temperature at which the liquid phase has a lower Gm than the solid phase.

Thus, as T increases the compound eventually sublimes into the gas phase.

12

Phase diagrams

Map showing conditions of T and p at which various phases are thermodynamically stable

At any point on the phase boundaries, the phases are in dynamic equilibrium

13

Phase boundaries

The pressure of the vapour in equilibrium with its condensed phase is called the vapour pressure of the substance.– Vapour pressure increases with temperature

because, as the temperature is raised, more molecules have sufficient energy to leave their neighbours in the liquid.

14

Vapour pressure of water versus T

15

Phase boundaries

Suppose liquid in a cylinder fitted with a piston.

Apply pressure > vapour pressure of liquid– vapour eliminated– piston rests on surface of liquid– system moves to liquid region of phase diagram

Reducing pressure????????

16

Question

What would be observed when a pressure of 7.0 kPa is applied to a sample of water in equilibrium with its vapour at 25oC, when its vapour pressure is 2.3 kPa?

Sample condenses entirely to liquid

17

Solid-Solid phase boundaries

Thermal analysis:– uses heat release

during transition

Sample allowed to cool and T monitored

On transition, energy is released as heat and cooling stops until transition is complete

18

Location of phase boundaries

Suppose two phases are in equilibrium at a given p and T.

If we change p, we must change T to a different value to ensure the two phases remain in equilibrium.

Thus, there must be a relationship between p that we exert and T we must make to ensure that the two phases remain in equilibrium

19

Location of phase boundaries Clapeyron equation (see derivation 5.4)

Clausius-Clapeyron equation (derivation 5.5)

TVT

Hp

trs

trs

constant11

lnln

ln

1212

2

2

TTR

Hpp

TRT

Hp

TRT

H

p

p

vap

vap

vap

Constant is

vapS/R

20

Example 1The vapour pressure of mercury is 160 mPa at 20°C. What is its vapour pressure at 50°C given that its enthalpy of vaporisation is 59.3 kJ mol-1?

Pa53.1

426096.0ln

258676.283258.1ln

)1016685.3(25.713283258.1ln

15.293

1

15.323

1

314.8

59300)10160ln(ln

constant11

lnln

2

2

2

42

32

1212

p

p

p

xp

xp

TTR

Hpp

vap

21

Example 2The vapour pressure of pyridine is 50.0 kPa at 365.7 K and the normal boiling point is 388.4 K. What is the enthalpy of vaporisation of pyridine?

1

5

4

1212

molkJ74.36

10922258.1

7063.0

)10598165.1(314.8

7063.0

7.365

1

4.388

1

314.850000

101325ln

constant11

lnln

H

Hx

xH

H

TTR

Hpp

vap

vap

vap

vap

vap

22

Example 3Estimate the boiling point of benzene given that its vapour pressure is 20.0 kPa at 35°C and 50.0 kPa at 58.8°C?

1

5

4

1212

molkJ74.32

1079854.2

91629.0

)10326709.2(314.8

91629.0

95.331

1

15.308

1

314.850000

20000ln

constant11

lnln

H

Hx

xH

H

TTR

Hpp

vap

vap

vap

vap

vap

23

Example 3 contd.

KT

Tx

xT

x

T

T

TTR

Hpp

vap

353

/110833193.2

1024517.31

1011977.4

15.308

1157.39386226.1

15.308

11

314.8

2.32745

20000

101325ln

constant11

lnln

2

23

3

2

4

2

2

1212

24

Vapour pressure

)log10ln(ln

10ln

10lnln

log

constant11

lnln

'

'

1212

yxy

R

HB

RT

H

kPa

PA

T

BAp

TTR

Hpp

vap

vap

vap

25

Vapour pressure

Substance A B/K Temperature range/°C

Benzene, C6H6(l) 7.0871 1785 0 to 42

6.7795 1687 42 to 100

Hexane, C6H14(l) 6.849 1655 10 to 90

Methanol, CH3OH(l) 7.927 2002 10 to 80

Methylbenzene, C6H5CH3(l) 7.455 2047 92 to 15

Phosphorus, P4(s, white) 8.776 3297 20 to 44

Sulfur trioxide, SO3(l) 9.147 2269 24 to 48

Tetrachloromethane, CCl4(l) 7.129 1771 19 to 20

* A and B are the constants in the expression log(p/kPa) A B/T.

26

27

The vapour pressure of benzene in the range 042 oC can be expressed in the form log (p/kPa) = 7.0871 1785 K/ T. What is the enthalpy of vaporisation of liquid benzene?

1molkJ17.34

10ln3145.8178510ln

)kPa/( log

H

HxxR

HB

T

BAp

vap

vap

vap

28

For benzene in the range 42100oC, B = 1687 K and A = 6.7795. Estimate the normal boiling point of benzene?

K38.353

16877796.60057.2

16877796.6)325.101log(

)kPa/( log

TT

T

T

BAp

29

Derivations

dGm = Vmdp – SmdTdGm(1) = dGm(2)

Vm(1)dp – Sm(1)dT = Vm(2)dp – Sm(2)dT

{Vm(2) – Vm(1)}dp = {Sm(2) – Sm(1)}dT

trsV dp = trsS dT

T trsV dp = trsH dT

dp/dT = trsH/(T trsV)

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Derivations: liquid-vapour transitions

dp/dT = vapH/(T vapV)

≈ vapH/{T Vm(g)} = vapH/{T (RT/p)}

(dp/p)/dT = vapH/(RT2)

d(ln p)/dT = vapH/(RT2)

122

1

2

2

ln

ln

2

111ln

ln

ln

2

1

2

1

2

1

TTR

HdT

TR

H

p

p

dTRT

Hpd

dTRT

Hpd

vapT

T

vap

T

T

vapp

p

vap

+ constant

31

Heat liquid in open vessel

As T is raised the vapour pressure increase.

At a certain T, the vapour pressure becomes

equal to the external pressure.

At this T, the vapour can drive back the

surrounding atmosphere, with no constraint on

expansion, bubbles form an boiling occurs.

32

Characteristic points

Remember: BP: temperature at which the vapour

pressure of the liquid is equal to the prevailing atmospheric pressure.

At 1 atm pressure: Normal Boiling Point (100°C for water)

At 1 bar pressure: Standard Boiling Point (99.6°C for water; 1 bar = 0.987atm, 1 atm = 1.01325 bar)

33

Heat liquid in closed vessel Vapour density increases until it equals that of

the liquid– surface between the two layers disappears– T is known as the critical temperature (TC)– vapour pressure at TC is critical pressure pC

– TC and pC together define the critical point If we exert pressure on a sample that is above

TC we produce a denser fluid No separation, single uniform phase of a

supercritical fluid occupies the container

34

Heat liquid in closed vessel

A liquid cannot be produced by the application of pressure to a substance if it is at or above its critical temperature

35

Triple point

There is a set of conditions under which three different phases coexist in equilibrium. The triple point. For water the triple point lies at 273.16 K and 611 Pa (0.006 atm).

The triple point marks the lowest T at which the liquid can exist

The critical point marks the highest T at which the liquid can exist

36

Summary• Thermodynamics tells which way a process will go

• Internal energy of an isolated system is constant (work

and heat). We looked at expansion work (reversible and

irreversible).

• Thermochemistry usually deals with heat at constant

pressure, which is the enthalpy.

• Spontaneous processes are accompanied by an increase

in the entropy (disorder?) of the universe

• Gibbs free energy decreases in a spontaneous process