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Phase Equilibria (CH-203)

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 G m = G/n = H - TS Depends on the phase of the substance.

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Phase Equilibria (CH-203)

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  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 energy Gm = 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: DG = nGm(2) – nGm(1) = n{Gm(2) – Gm(1)} A spontaneous change occurs when DG < 0

  3. The Gibbs energy of transition from metallic white tin (a-Sn) to nometallic grey tin (b-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; 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 • Can usually ignore pressure dependence of G for condensed states • Can derive that, for a gas:

  8. 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

  9. 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.

  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 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 365.7 K and the normal boiling point is 388.4 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

  26. The vapour pressure of benzene in the range 042 oC can be expressed in the form log (p/kPa) = 7.0871 1785 K/ T. What is the enthalpy of vaporisation of liquid benzene?

  27. For benzene in the range 42100oC, B = 1687 K and A = 6.7795. Estimate the normal boiling point of benzene?

  28. 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 DtrsVdp = DtrsS dT T DtrsV dp = DtrsH dT dp/dT = DtrsH/(T DtrsV)

  29. 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

  30. 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.

  31. 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)

  32. 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

  33. 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

  34. 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

  35. 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

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