1. A more appropriate definition of K (for the same chemical reaction discussed previously) is With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless. 481

2. A more appropriate definition of K (for the same chemical reaction discussed previously) is With this definition, each concentration term is divided by the units of molarity, M, so K is then dimensionless. Because it is tedious to write the above equation, most people simply write to save writing time. 482

3. Relationship between Q and K 483 N 2 O 4(g) 2 NO 2(g)

4. 484

5. Different ways of expressing the equilibrium constant 485

6. Different ways of expressing the equilibrium constant Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase. 486

7. Different ways of expressing the equilibrium constant Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase. Example: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 487

8. Different ways of expressing the equilibrium constant Homogeneous equilibrium: A state of equilibrium between reactants and products in the same phase. Example: CH 3 CO 2 H (aq) CH 3 CO 2 - (aq) + H + (aq) 488

9. Example: For the gas phase reaction N 2 O 4(g) 2 NO 2(g) 489

10. Example: For the gas phase reaction N 2 O 4(g) 2 NO 2(g) When working with gaseous reactions, it is frequently more convenient to use different units. 490

11. For the preceding reaction 491

12. For the preceding reaction The subscript p on K indicates partial pressures are used in the mass action expression. 492

13. For the preceding reaction The subscript p on K indicates partial pressures are used in the mass action expression. is the equilibrium partial pressure of N 2 O 4. is the equilibrium partial pressure of NO 2. 493

14. For the preceding reaction The subscript p on K indicates partial pressures are used in the mass action expression. is the equilibrium partial pressure of N 2 O 4. is the equilibrium partial pressure of NO 2. In general, K p and K c (for the same reaction) are not equal. 494

15. Exercise: For the reaction a A (g) b B (g) find a connection between K p and K c. Assume the gases can be treated as ideal gases. 495

16. Exercise: For the reaction a A (g) b B (g) find a connection between K p and K c. Assume the gases can be treated as ideal gases. 496

17. Exercise: For the reaction a A (g) b B (g) find a connection between K p and K c. Assume the gases can be treated as ideal gases. Now from the ideal gas equation PV = n RT, 497

18. Exercise: For the reaction a A (g) b B (g) find a connection between K p and K c. Assume the gases can be treated as ideal gases. Now from the ideal gas equation PV = n RT, 498

19. Exercise: For the reaction a A (g) b B (g) find a connection between K p and K c. Assume the gases can be treated as ideal gases. Now from the ideal gas equation PV = n RT, Plug these two results into the expression for K p. 499

20. Hence: 500

21. Hence: 501

22. Hence: 502

23. Hence: 503

24. Hence: where 504

25. Hence: where and therefore 505

26. Sample Problems 506

27. Sample Problems Example: 2 NO (g) + O 2(g) 2 NO 2(g) At a temperature of 230 o C the concentrations of the various species are [NO] = 0.0542 M, [O 2 ] = 0.127 M, and [NO 2 ] = 15.5 M at equilibrium. Calculate the equilibrium constant K c at 230 o C. 507

28. Sample Problems Example: 2 NO (g) + O 2(g) 2 NO 2(g) At a temperature of 230 o C the concentrations of the various species are [NO] = 0.0542 M, [O 2 ] = 0.127 M, and [NO 2 ] = 15.5 M at equilibrium. Calculate the equilibrium constant K c at 230 o C. 508

29. Sample Problems Example: 2 NO (g) + O 2(g) 2 NO 2(g) At a temperature of 230 o C the concentrations of the various species are [NO] = 0.0542 M, [O 2 ] = 0.127 M, and [NO 2 ] = 15.5 M at equilibrium. Calculate the equilibrium constant K c at 230 o C. = 6.44 x 10 5 509

30. Example: The equilibrium constant K p for the reaction PCl 5 (g) PCl 3(g) + Cl 2(g) is 1.05 at 250 o C. If the equilibrium partial pressures of PCl 5 and PCl 3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl 2 ? 510

31. Example: The equilibrium constant K p for the reaction PCl 5 (g) PCl 3(g) + Cl 2(g) is 1.05 at 250 o C. If the equilibrium partial pressures of PCl 5 and PCl 3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl 2 ? 511

32. Example: The equilibrium constant K p for the reaction PCl 5 (g) PCl 3(g) + Cl 2(g) is 1.05 at 250 o C. If the equilibrium partial pressures of PCl 5 and PCl 3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl 2 ? This is shorthand for 512

33. Example: The equilibrium constant K p for the reaction PCl 5 (g) PCl 3(g) + Cl 2(g) is 1.05 at 250 o C. If the equilibrium partial pressures of PCl 5 and PCl 3 are 0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of Cl 2 ? This is shorthand for so that 513

34. and so 514

35. Heterogeneous Equilibria 515

36. Heterogeneous Equilibria A reaction often involves reactants that are not present in the same phase – this leads to a heterogeneous equilibrium. 516

37. Heterogeneous Equilibria A reaction often involves reactants that are not present in the same phase – this leads to a heterogeneous equilibrium. Example: Heating CaCO 3 in a closed vessel. 517

38. Heterogeneous Equilibria A reaction often involves reactants that are not present in the same phase – this leads to a heterogeneous equilibrium. Example: Heating CaCO 3 in a closed vessel. CaCO 3(s) CaO (s) + CO 2(g) 518

39. Heterogeneous Equilibria A reaction often involves reactants that are not present in the same phase – this leads to a heterogeneous equilibrium. Example: Heating CaCO 3 in a closed vessel. CaCO 3(s) CaO (s) + CO 2(g) 519

40. The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. 520

41. The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume. 521

42. The “concentration” of any pure solid is the ratio of the total number of moles present in the solid, divided by the volume of the solid. If part of the solid is removed, the number of moles of solid will decrease – but so will it volume. The ratio of moles to volume remains unchanged. 522

43. Consider the ratio: 523

44. Consider the ratio: 524

45. Consider the ratio: 525

46. Consider the ratio: If the temperature is held fixed, then [CaCO 3 ] is a constant. Similarly for [CaO], which is also a constant. 526

47. So we can rewrite 527

48. So we can rewrite in the form 528

49. So we can rewrite in the form Now since [CaCO 3 ] and [CaO] are both constant, the left-hand side of the preceding equation is constant. 529

50. So we can rewrite in the form Now since [CaCO 3 ] and [CaO] are both constant, the left-hand side of the preceding equation is constant. Now set 530

51. Hence K c = [CO 2 ] 531

52. Hence K c = [CO 2 ] Notice that terms involving pure solids do not appear in the final equilibrium constant expression. 532

53. Hence K c = [CO 2 ] Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions. 533

54. Hence K c = [CO 2 ] Notice that terms involving pure solids do not appear in the final equilibrium constant expression. This result generalizes to all chemical reactions. The corresponding expression for K p for the decomposition of CaCO 3 is: K p = 534

55. 535 The same concentration of CO 2 exists in both containers (provided the temperature is the same), even though the amounts of CaO and CaCO 3 are different.

56. Summary Comment ignored Concentration factors for pure solids and pure liquids are ignored in the expression for the equilibrium constant for a chemical reaction. 536

57. Multiple Equilibria 537

58. Multiple Equilibria Consider the two reactions: 538

59. Multiple Equilibria Consider the two reactions: A + B C + D C + D E + F 539

60. Multiple Equilibria Consider the two reactions: A + B C + D C + D E + F For the first reaction 540