1. Phase Equilibria (CH-203)
Spectroscopy handout
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 energy
Gm = G/n = H - TS
Depends on the phase of the substance
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2. DG = nGm(2) – nGm(1) = n{Gm(2) – Gm(1)}
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:
DG = nGm(2) – nGm(1) = n{Gm(2) – Gm(1)}
A spontaneous change occurs when DG < 0
A substance has a spontaneous tendency to change into a phase with the lowest molar Gibbs energy

3. The Gibbs energy of transition from metallic white tin (a-Sn) to nometallic grey tin (b-Sn) is 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; DU = Dq + Dw
For an infinitesimal change in G:
G + DG = H + DH – (T + DT)(S + DS)
= H + DH – TS – SDT – TDS – DTDS
DG = DH – TDS – SDT
Also can write: DH = DU + pDV + VDp
DU = TDS – pDV (dS = dqrev/T and dw = -pdV)
DG = TDS – pDV + pDV +VDp – TDS – SDT
DG = VDp – SDT
Master Equations

6. Variation with pressure
Thus, Gibbs energy depends on:
Pressure
Temperature
We can derive (derivation 5.1 in textbook) that:
DGm = Vm Dp
=>DGm > 0 when Dp > 0
i.e. Molar Gibbs energy increases when pressure increases

7. Variation of G with pressure
Spectroscopy handout
Variation of G with pressure
Can usually ignore pressure dependence of G for condensed states
Can derive that, for a gas:
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8. Variation of G with temperature
Spectroscopy handout
Variation of G with temperature
DGm = –SmDT
DGm= Gm(Tf) – Gm(Ti)
DT = 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
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9. Variation of G with temperature
Spectroscopy handout
Variation of G with temperature
DGm = –SmDT
Molar entropy is positive, thus an increase in T results in a decrease in Gm.
Because:
DGm  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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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)
Spectroscopy handout
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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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. Sample condenses entirely to liquid
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 Dp that we exert and
DT 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)
Constant is
DvapS/R

20. Example 1
The 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?

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

22. Example 3
Estimate 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?

23. Example 3 contd.

24. Vapour pressure

25. Vapour pressure

27. The vapour pressure of benzene in the range 042 oC can be expressed in the form log (p/kPa) = 1785 K/ T. What is the enthalpy of vaporisation of liquid benzene?

28. For benzene in the range 42100oC, B = 1687 K and A = 6. 7795
For benzene in the range 42100oC, B = 1687 K and A = Estimate the normal boiling point of benzene?

29. dp/dT = DtrsH/(T DtrsV)
Derivations
dGm = 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
DtrsV dp = DtrsS dT
T DtrsV dp = DtrsH dT
dp/dT = DtrsH/(T DtrsV)

30. Derivations: liquid-vapour transitions
dp/dT = DvapH/(T DvapV)
≈ DvapH/{T Vm(g)} = DvapH/{T (RT/p)}
(dp/p)/dT = DvapH/(RT2)
d(ln p)/dT = DvapH/(RT2)
+ 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 = 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 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