1

Phase Equilibria (CH-203)

Phase transitionsChange 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 (α-Sn) to nometallicgrey tin (β-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 – S∆T – T∆S – ∆T∆S

∆G = ∆H – T∆S – S∆TAlso can write: ∆H = ∆U + p∆V + V∆p

∆U = T∆S – p∆V (dS = dqrev/T and dw = -pdV)

∆G = T∆S – p∆V + p∆V +V∆p – T∆S – S∆T

∆G = V∆p – S∆TMaster 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 Gfor condensed states

Can derive that, for a gas:

i

fm p

pRTG ln=∆

8

Variation of G with temperature

∆Gm = –Sm∆T∆Gm= Gm(Tf) – Gm(Ti)

∆T = Tf – TiCan 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 = –Sm∆TMolar 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 stableAs T increases, Gm of liquid phase falls below the solid phase and substance meltsAt 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 stableAt 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 transitionSample allowed to cool and T monitoredOn transition, energy is released as heat and cooling stops until transition is complete

18

Location of phase boundariesSuppose two phases are in equilibrium at a given pand 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 boundariesClapeyron equation (see derivation 5.4)

Clausius-Clapeyron equation (derivation 5.5)

TVTHp

trs

trs ∆∆

∆=∆

( )

constant11lnln

ln

1212

2

2

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆−=

∆∆

=∆

∆∆

=∆

TTRH

pp

TRT

Hp

TRT

Hpp

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?

Pa53.1426096.0ln

258676.283258.1ln)1016685.3(25.713283258.1ln

15.2931

15.3231

314.859300)10160ln(ln

constant11lnln

2

2

2

42

32

1212

==

+−=−−−=

⎟⎠⎞

⎜⎝⎛ −−=

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆−=

−

−

pp

pxp

xp

TTRH

pp 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

molkJ74.3610922258.1

7063.0

)10598165.1(314.8

7063.0

7.3651

4.3881

314.850000101325ln

constant11lnln

−

−

−

=∆

∆−=−

−∆

−=

⎟⎠⎞

⎜⎝⎛ −

∆−=⎟

⎠⎞

⎜⎝⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆−=

H

Hx

xH

HTTR

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

molkJ74.321079854.2

91629.0

)10326709.2(314.8

91629.0

95.3311

15.3081

314.85000020000ln

constant11lnln

−

−

−

=∆

∆−=−

∆−=−

⎟⎠⎞

⎜⎝⎛ −

∆−=⎟

⎠⎞

⎜⎝⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆−=

H

Hx

xH

HTTR

Hpp

vap

vap

vap

vap

vap

23

Example 3 contd.

KTTx

xT

x

T

T

TTRH

pp vap

353/110833193.2

1024517.311011977.4

15.3081157.39386226.1

15.30811

314.82.32745

20000101325ln

constant11lnln

2

23

3

2

4

2

2

1212

==

−⎟⎟⎠

⎞⎜⎜⎝

⎛=−

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛−−=⎟

⎠⎞

⎜⎝⎛

+⎟⎟⎠

⎞⎜⎜⎝

⎛−

∆−=

−−

24

Vapour pressure

)log10ln(ln

10ln

10lnln

log

constant11lnln

'

'

1212

yxy

RH

B

RT

HkPaPA

TBAp

TTRH

pp

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 0−42 oCcan 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

TBAp

vap

vap

vap

28

For benzene in the range 42−100oC, 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

TBAp

29

DerivationsdGm = Vmdp – SmdT

dGm(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 dTT ∆trsV dp = ∆trsH dT

dp/dT = ∆trsH/(T ∆trsV)

30

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

TTRH

dTTR

Hpp

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 pointsRemember:

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 vesselVapour 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 fluidNo 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

l 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).

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

l 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