Minimization of Gibbs Free Energy

1. Minimization of Gibbs Free Energy Chapter 4

2. What constitutes equilibrium? • Properties are constant with time • In Chapter 3 we determined that in order for a system to be at equilibrium • Temperature must be constant • Pressure must be constant

3. Temperature • Consider two phases • If the temperature is different, heat flows from hot to cold until equilibrium is reached. • When the temperature is the same, there is no heat transfer • Temperature is the driving force for heat transfer!

4. Pressure • Consider two gases, separated by a free moving piston • If the pressure in the gases is different, the piston will move until the pressure is the same – doing work • If the pressure is the same, the piston doesn’t move • Pressure is the driving force for boundary work!!

5. Air + water vapor Water + dissolved air Concentration • Consider two phases • If the composition of the phases is different, do they necessarily change until the concentrations are the same? • NO!!

6. Gibbs Free Energy • The two phases adjust until the Gibbs Free Energy of each phase is the same • Gibbs Free Energy is the driving factor in phase equilibria for a pure substance • It’s not quite that easy for multicomponent systems

7. What is the driving force????? • At equilibrium Gibbs Free Energy is a minimum – but that’s still not quite equivalent to T and P as driving forces • Chemical Potential • Driving Force in phase equilibria and chemical equilibria

8. Choose a system • In Chapter 3 we also determined that for a true equilibrium the system must be closed, and must have no moving parts (like a turbine) • Consider a piston cylinder device, at constant temperature and pressure for the following development

9. Steam Water Consider an internally reversible process – for example a piston cylinder device Quick Review

10. For a differential change in volume, the pressure is constant, and if the temperature is constant… rearranging Equation 4.4 – page 67

11. For any differential equilibrium change, chemical or physical or both, at constant T and P,

12. What does it mean? • At equilibrium, Gibbs Free Energy is a minimum The slope is = to zero

13. Let’s try an example • A large mass of steam and water are at equilibrium at 1 atm and 212 F • We allow one lbm of steam to condense, keeping the temperature and pressure constant Steam Q Water Find the change in Gibbs Free Energy

14. Dscondensation=-1.4446 Btu/lbm 0R Dhcondensation=-970.3 Btu/lbm Do the math… If you keep the temperature constant at 212 0F = 671.7 0R From the steam tables….

15. Does dG = 0 really mean that G is a minimum? • Not mathematically • If you consider irreversibilities however, you can show that dG >0 for any deviation from equilibrium • See page 68

16. We just saw in the example, that movement of a quantity of mass from one phase to another at equilibrium, does not result in a change in Gibbs Free Energy. • Our example was for a single component – but it applies to multicomponent systems too

17. The composition of the two phases is different at equilibrium Keep the temperature and pressure constant, and move the piston a differential amount m Let’s look at the air-water system again A differential quantity of mass moves into the gas phase Air + water vapor Water + dissolved air To cause this to happen, you’ll have to add some heat to the system

18. What is the corresponding Gibbs Free Energy change? Because we are at equilibrium, there will be no change in the Gibbs Free Energy of the system

19. G is a function of other properties • Using the state postulate, we can surmise that if we know two properties, we can find the value of g – the Gibbs Free energy per mole (or kg). • If we know how many moles there are of each component, we can find the total Gibbs Free Energy

20. Gibbs Free Energy We could have chosen any two state variables, but T and P are convenient

21. Take the derivative of G For Phase 1 But we know the change in T and P is 0, because we are at equilibrium!! For Phase 2

22. Now consider the value of dGsys For Phase 1 + = 0 The change in Gsys is zero too – we proved it just a few slides ago!! For Phase 2

23. All we have left are the derivatives with respect to n + = 0

24. + -dnb(1) -dna(1) = 0 We also know that whatever leaves phase one, goes into phase two So let’s simplify and

25. Combine terms This must be true for any number of species,(a,b,c, etc) and for any values of the molar deriviatives The only way that is possible, is if the values inside the brackets are 0

26. Partial molal Gibbs free energy The partial molal Gibbs Free Energy is the driving force we’ve been looking for. When it has the same value in each phase, we are at equilibrium The increase in Gibbs free energy per mole of i added to phase 1

27. Nomenclature Chemical Potential

28. At Equilibrium etc What if the chemical potentials aren’t equal? Then we aren’t at equilibrium, and matter will move from the high value to the low value

29. In our previous discussion we considered a non-reacting system, with multiple phases We can use the same approach for reacting systems The criterion of equilibrium applied to chemical reactions Single phase, in which a chemical reaction occurs Q Isothermal, constant pressure

30. Consider the special case of the reaction A ⇋B Pressure and temperature are constant, and the change in Gibbs Free energy is 0

31. From the stoichiometry A ⇋B We know that

32. Similarly, for A + B ⇋C Or the more general result for reacting systems

33. Some Consequences • Consider a system consisting of only one species, in two equilibrium phases – like the water system in the example

34. For a single pure species • And the Gibbs Free Energy per mole is the same in each phase

35. Some Other Relationships Remember that… So, taking the derivative gives But we know from the first Gibbs Relationship that… Now we can cancel out terms, leaving…

36. If you keep P constant… If you keep T constant… Which gives…

37. Derivatives on a per mole basis It’s obvious that the volume (v) is an absolute quantity, but so is the entropy (s). These expressions tell us that the change in Gibbs Free Energy with pressure at constant T is always positive … and that the change with temperature with constant P is always negative

39. Increasing Pressure Increasing Temperature

40. If you plot both the gas and liquid surface, they are at equilibrium where the surfaces intersect

41. The lower Gibbs value is the stable one. Therefore, at low temperatures and high pressures the liquid is stable

42. Enthalpy Gibbs Free Energy does not change during a phase change Entropy Gibbs Free Energy

43. Gibbs Free Energy Diagrams for Pressure Driven Changes Don’t worry – diamonds require a catalyst to change phase At room temperature, graphite is stable at low pressures, and diamond is stable at high pressures

44. Gibbs Free Energy Diagrams for Pressure Driven Changes The equilibrium point between diamond and graphite occurs at ~15 kbar, at 25 C

45. Diamond-graphite Equilibrium This is a plot of the equilibrium pressure, at different temperatures

46. Equilibrium occurs when the chemical potentials are equal Gibbs Free Energy Diagrams for Chemical Reactions n-butane⇋ isobutane

47. Equilibrium Constants • Law of mass action

48. For the equilibrium between normal and isobutane

49. Which matches our previous result Xiso= 0.82

50. LeChatelier’s Principle • Natural systems respond to changes in their external environment by internal adjustments to counteract those changes