1. 1501

2. 1502

3. where Q is the reaction quotient. 1503

4. where Q is the reaction quotient. For a reaction system at equilibrium and Q = K c 1504

5. where Q is the reaction quotient. For a reaction system at equilibrium and Q = K c and hence (a very key result) 1505

6. For a reaction in which all the species are in the gas phase, 1506

7. It follows directly from the result 1507

8. It follows directly from the result for K c > 1 1508

9. It follows directly from the result for K c > 1 for K c < 1 1509

10. It follows directly from the result for K c > 1 for K c < 1 for K c = 1 In this case the position of the equilibrium does not favor either products or reactants forming. 1510

11. Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is -16.45 kJmol -1. Calculate the equilibrium constant for the reaction at 25.00 o C. In a certain experiment the initial concentrations are [H 2 ] = 0.00345 M, [N 2 ] = 0.00956 M, and [NH 3 ] = 0.00678 M. Predict the direction of the reaction. 1511

12. Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is -16.45 kJmol -1. Calculate the equilibrium constant for the reaction at 25.00 o C. In a certain experiment the initial concentrations are [H 2 ] = 0.00345 M, [N 2 ] = 0.00956 M, and [NH 3 ] = 0.00678 M. Predict the direction of the reaction. From we have 1512

13. Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is -16.45 kJmol -1. Calculate the equilibrium constant for the reaction at 25.00 o C. In a certain experiment the initial concentrations are [H 2 ] = 0.00345 M, [N 2 ] = 0.00956 M, and [NH 3 ] = 0.00678 M. Predict the direction of the reaction. From we have 1513

14. Example: The standard Gibbs energy for the reaction ½ N 2(g) + 3/2 H 2(g) NH 3(g) is -16.45 kJmol -1. Calculate the equilibrium constant for the reaction at 25.00 o C. In a certain experiment the initial concentrations are [H 2 ] = 0.00345 M, [N 2 ] = 0.00956 M, and [NH 3 ] = 0.00678 M. Predict the direction of the reaction. From we have 1514

15. Hence ln K c = 6.636 1515

16. Hence ln K c = 6.636 therefore K c = 762 1516

17. Hence ln K c = 6.636 therefore K c = 762 (Note almost one significant digit is lost in evaluating the exponential.) 1517

18. Hence ln K c = 6.636 therefore K c = 762 (Note almost one significant digit is lost in evaluating the exponential.) To predict the direction of the reaction, use 1518

19. Now 1519

20. Now 1520

21. Now = 342 1521

22. Now = 342 Since Q < K c the reaction will move in the direction to produce more NH 3. 1522

23. Employing 1523

24. Employing 1524

25. Employing = – 16.45 kJ mol -1 + 14.46 kJ mol -1 = – 1.99 kJ mol -1 1525

26. Employing = – 16.45 kJ mol -1 + 14.46 kJ mol -1 = – 1.99 kJ mol -1 Since < 0, the reaction takes place in the forward direction. 1526

27. Are diamonds forever? 1527

28. Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? 1528

29. Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? Tabulated data: S 0 (diamond) = 2.44 J K -1 mol -1 S 0 (graphite) = 5.69 J K -1 mol -1 1529

30. Are diamonds forever? In chemical jargon, is the transition C diamond C graphite spontaneous at room temperature? Tabulated data: S 0 (diamond) = 2.44 J K -1 mol -1 S 0 (graphite) = 5.69 J K -1 mol -1 for the reaction C graphite C diamond = 1.90 kJ 1530

31. For the reaction C diamond C graphite 1531

32. For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 1532

33. For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = - 1.90 x 10 3 J - 298 K (3.25 J K -1 ) 1533

34. For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = - 1.90 x 10 3 J - 298 K (3.25 J K -1 ) = - 1.90 x 10 3 J - 969 J = -2.87 kJ 1534

35. For the reaction C diamond C graphite = 1 mol x 5.69 J K -1 mol -1 – 1 mol x 2.44 J K -1 mol -1 = 3.25 J K -1 Using = - 1.90 x 10 3 J - 298 K (3.25 J K -1 ) = - 1.90 x 10 3 J - 969 J = -2.87 kJ Conclusion: The reaction C diamond C graphite is spontaneous! 1535

36. Electrochemistry 1536

37. Electrochemistry Two Key Ideas in this section: 1537

38. Electrochemistry Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate electricity. 1538

39. Electrochemistry Two Key Ideas in this section: (1) Use spontaneous redox chemistry to generate electricity. (2) Use electric current to drive non-spontaneous redox reactions. 1539

40. Oxidation-reduction reactions 1540

41. Oxidation-reduction reactions Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly. 1541

42. Oxidation-reduction reactions Oxidation number: A number equal to the charge an atom would have if its shared electrons were held completely by the atom that attracts them more strongly. Need to be able to distinguish reactions where there is a change in oxidation number (these are redox reactions) from reactions where there is no change in oxidation number. 1542

43. 1543

44. Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) 1544

45. Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) 1545 +2-2+ -+2-2+ -

46. Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s) 1546 +2-2+ -+2-2+ -

47. Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s) 1547 +2-2+ -+2-2+ - 00-+

48. Example: Na 2 SO 4(aq) + BaCl 2(aq) 2 NaCl (aq) + BaSO 4(s) There is no change in oxidation number, there is no electron transfer taking place in this reaction. Example: 2 Na (s) + Cl 2(g) 2 NaCl (s) In this example, there is a change in oxidation number, so electron transfer is taking place. 1548 +2-2+ -+2-2+ - 00-+

49. Half-Equations: 1549

50. Half-Equations: Na Na + + e - (1) 1550

51. Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) 1551

52. Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) 1552 - +0 0

53. Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) These are called the half-equations. 1553 - +0 0

54. Half-Equations: Na Na + + e - (1) 2e - + Cl 2 2 Cl - (2) These are called the half-equations. Multiply equation (1) by 2 and add equation (2) leads to the overall balanced equation. 2 Na + Cl 2 2 NaCl 1554 - +0 0

55. Oxidation: A process in which electrons are lost. Na Na + + e - 1555

56. Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl - 1556

57. Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl - Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl 2 in the above eq. 1557

58. Oxidation: A process in which electrons are lost. Na Na + + e - Reduction: A process in which electrons are gained. 2e - + Cl 2 2 Cl - Oxidizing Agent: A substance that gains electrons (in the reduction step). It undergoes a decrease in oxidation number. Example: Cl 2 in the above eq. Reducing Agent: A substance that donates electrons in a redox reaction (specifically in the oxidation step). It undergoes an increase in oxidation number. Example: Na in the above oxidation. 1558

59. Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step. 1559

60. Redox reaction: A reaction that can be split apart into an oxidation step and a reduction step. Example: 2 Na + Cl 2 2 NaCl 1560