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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
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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
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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!
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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.
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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
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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
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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=∆
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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
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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.
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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
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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.
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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
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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.
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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????????
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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
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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
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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
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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
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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
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1212
==
+−=−−−=
⎟⎠⎞
⎜⎝⎛ −−=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−
∆−=
−
−
pp
pxp
xp
TTRH
pp vap
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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
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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
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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
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3
2
4
2
2
1212
==
−⎟⎟⎠
⎞⎜⎜⎝
⎛=−
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−=⎟
⎠⎞
⎜⎝⎛
+⎟⎟⎠
⎞⎜⎜⎝
⎛−
∆−=
−−
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Vapour pressure
)log10ln(ln
10ln
10lnln
log
constant11lnln
'
'
1212
yxy
RH
B
RT
HkPaPA
TBAp
TTRH
pp
vap
vap
vap
=
∆=⇒
∆+=⇒
−=
+⎟⎟⎠
⎞⎜⎜⎝
⎛−
∆−=
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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.
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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
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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
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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)
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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
TTRH
dTTR
Hpp
dTRT
Hpd
dTRT
Hpd
vapT
T
vap
T
T
vapp
p
vap
+ constant
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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.
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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)
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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
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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
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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
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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