1. Calculating Equilibrium Composition  Example  Cl 2 (g) → 2Cl (g)

2. Calculating Equilibrium Composition  Example:  Cl 2 (g) → 2Cl (g)   Initially, n 0 moles of Cl 2 gas is placed in a closed reaction vessel.  The molecule partially dissociates into atoms.

3. Calculating Equilibrium Composition  Example  Cl 2 (g) → 2Cl (g)  Both are gasses, so use partial pressures rather than concentrations.  We know that at equilibrium we have a definite mixture (i.e., the composition of the mixture is not arbitrary), as the reactants and products are related.

4. Calculating Equilibrium Composition  What is the expression for the equilibrium quotient?

5. Calculating Equilibrium Composition  What is the expression for the equilibrium quotient?  We need expressions for the partial pressures at equilibrium.

6. Calculating Equilibrium Composition Initial No. of molesCl 2 = ?

7. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = ?

8. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0

9. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = ?

10. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 – 

11. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = ?

12. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2 

13. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2  Mole fractions at eq.Cl 2 = 

14. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2  Mole fractions at eq.Cl 2 = n 0 –  n 0 +  Cl = 2  /n 0 + 

15. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2  Mole fractions at eq.Cl 2 = n 0 –  n 0 +  Cl = 2  /n 0 +  Partial pressures at eq.Cl 2 = ?

16. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2  Mole fractions at eq.Cl 2 = n 0 –  n 0 +  Cl = 2  /n 0 +  Partial pressures at eq.Cl 2 = (n 0 –  n 0 +  )p 2Cl = (2  /n 0 +  )p

17. Calculating Equilibrium Composition Initial No. of molesCl 2 = n 0 2Cl = 0 No. of moles at eq.Cl 2 = n 0 –  Cl = 2  Mole fractions at eq.Cl 2 = n 0 –  n 0 +  Cl = 2  /n 0 +  Partial pressures at eq.Cl 2 = (n 0 –  n 0 +  )p 2Cl = (2  /n 0 +  )p Now express K p in terms of what we have.

18. Calculating Equilibrium Composition Partial pressures at eq.Cl 2 = (n 0 –  n 0 +  )p 2Cl = (2  /n 0 +  )p Now express K p in terms of what we have.  In terms of partial pressures. Which, in terms of  and n 0 is

19. Calculating Equilibrium Composition Partial pressures at eq.Cl 2 = (n 0 –  n 0 +  )p 2Cl = (2  /n 0 +  )p Now express K p in terms of what we have.  Which in terms of a and the total pressure is

20. Calculating Equilibrium Composition Partial pressures at eq.Cl 2 = (n 0 –  n 0 +  )p 2Cl = (2  /n 0 +  )p Now express K p in terms of what we have. This can now be expresses in terms of a and p only

21. Calculating Equilibrium Composition Thus, knowing K p and the total pressure, we can calculate the equilibrium composition of the mixture.

22. Relative stability of Gases, Liquids, and Solids

23.  Common experience:  Low T favours solids  High T favours gases  Similarly for high and low pressure.  Hear we study the conditions under which two (or even three) phases co-exist in equilibrium, at a given p and T.

24.  Phases: Solid, liquid, gases.  Gases exist in only one phase..

25.  Phases: Solid, liquid, gases.  Gases exist in only one phase..  Liquids primarily also exist in only one phase..

26.  Phases: Solid, liquid, gases.  Gases exist in only one phase..  Liquids primarily also exist in only one phase..  Exception: supercritical liquids

27.  Phases: Solid, liquid, gases.  Gases exist in only one phase..  Liquids primarily also exist in only one phase..  Exception: supercritical liquids.  Solids: can exist in several phases.  E.g., crystal structures..

28.  Water in a beaker, exists as a single phase.  Water and ice in a beaker = mixture of two distinct phases.

29. Conditions under which substances spontaneously form S, L, or G  Common experience  T reduces from 300 to 250 K.  Water (liquid) turns to ice (solid)  T increases from 300 to 400 K  Water turns to steam (gas)

30. Conditions under which substances spontaneously form S, L, or G  Solid CO 2 at room temperature.  Sublimes: Turns from solid to gas, with out going through a liquid phase.

31. Conditions under which substances spontaneously form S, L, or G  What determines which phase is favoured (most thermodynamically stable) at a given p and T?  What is the criterion for stability?

32. Conditions under which substances spontaneously form S, L, or G  The minimising of the Gibb’s energy. For a pure substance,  = chemical potential, n = mole fraction,

33. Conditions under which substances spontaneously form S, L, or G  The minimising of the Gibb’s energy. For a pure substance,

34. Conditions under which substances spontaneously form S, L, or G  As d  = dG m,

35. Conditions under which substances spontaneously form S, L, or G  Thus, the variation of  with p and T can be determined.

36. Conditions under which substances spontaneously form S, L, or G  S m and V m are always positive, thus;   decreases as T increases, and  Increases with increasing p.

37. Conditions under which substances spontaneously form S, L, or G  The entropy varies slowly with T ( as ln T),  Thus, over a limited T range,  a plot of  v. T at const. p is a straight line of negative slope.

38. Conditions under which substances spontaneously form S, L, or G  We know from experience that melting and boiling are endothermic.  Thus,  S =  H/T is positive for both of these constant T processes.  We also know that Gasses, liquids and solids all have positive heat capacities.

39. Conditions under which substances spontaneously form S, L, or G  Therefore,

40. Conditions under which substances spontaneously form S, L, or G  The entropy of a phase is the magnitude of the slope of  versus T.  Recall

41. Conditions under which substances spontaneously form S, L, or G Thus, the functional relationship between  and T for solids, liquids, and gasses (at a given p) can be expressed graphically. The stable state at any given T is the phase with the lowest .

42. Conditions under which substances spontaneously form S, L, or G Start in the solid phase and increase temperature. As T increases,  decreases with a certain slope.. Note the slopes for liquid and gas are greater. Therefore, they intersect. The points of intersection of the solid/liquid and the liquid/gas are the melting and boiling temperatures, respectively.

43. Conditions under which substances spontaneously form S, L, or G At the melting point (solid/liquid intersection) both phases exist in equilibrium. However, a further, but small increase in T results in complete melting. Why?

44. Conditions under which substances spontaneously form S, L, or G At the melting point (solid/liquid intersection) both phases exist in equilibrium. However, a further, but small increase in T results in complete melting. Why? The Liquid phase has a lower  at T m + dT than the solid phase.

45. Conditions under which substances spontaneously form S, L, or G At the melting point (solid/liquid intersection) both phases exist in equilibrium. However, a further, but small increase in T results in complete melting. Why? The Liquid phase has a lower  at T m + dT than the solid phase.

46. Conditions under which substances spontaneously form S, L, or G Similarly, at T b both liquid and gas coexist at eq. The system is a gas at T > T b.

47. Conditions under which substances spontaneously form S, L, or G Note: the progression from solid to liquid to gas cam be fully explained only by and and

48. Conditions under which substances spontaneously form S, L, or G What is we increase the temperature fast (too fast)?.

49. Conditions under which substances spontaneously form S, L, or G What is we increase the temperature fast (too fast)?. At a phase change, the system does not reach equilibrium, leading to super heating.

50. Conditions under which substances spontaneously form S, L, or G What is we increase the temperature fast (too fast)?. At a phase change, the system does not reach equilibrium, leading to super heating (bumping). Similarly, rapid cooling leads to supercooling (e.g., glass formation).

51. Conditions under which substances spontaneously form S, L, or G What happens as a function of p at constant T?

52. Conditions under which substances spontaneously form S, L, or G What happens as a function of p at constant T?

53. Conditions under which substances spontaneously form S, L, or G What happens as a function of p at constant T? Mostly, V m Solid < V m Liquid << V m Gas

54. Conditions under which substances spontaneously form S, L, or G What happens as a function of p at constant T? Mostly, V m Solid < V m Liquid << V m Gas Therefore,  versus T changes more rapidly. Therefore,  versus T changes more rapidly.

55. Conditions under which substances spontaneously form S, L, or G Note: V m gas >> V m liquid >> 0. Therefore, increasing p leads to an increase in the boiling point.

56. Conditions under which substances spontaneously form S, L, or G Note: if V m liquid > V m solid Increasing p leads to melting point elevation if V m liquid < V m solid Increasing p leads to melting point supression.