1. The ICE table is as follows: H 2 I 2 HI 601

2. The ICE table is as follows: H 2 I 2 HI 602

3. The ICE table is as follows: H 2 I 2 HI 0.00623 – y 0.00414 – y 0.0224 + 2y 603

4. The ICE table is as follows: H 2 I 2 HI 0.00623 – y 0.00414 – y 0.0224 + 2y 604

5. Now think about any possible simplification. 605

6. Now think about any possible simplification. There is none! 606

7. Math Aside: The quadratic equation 607

8. Math Aside: The quadratic equation A quadratic equations looks like: 608

9. Math Aside: The quadratic equation A quadratic equations looks like: 609

10. Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form 610

11. Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form 611

12. Math Aside: The quadratic equation A quadratic equations looks like: The solutions (there are two of them) take the form Note that in an equilibrium problem, only one of the solutions will generate a physically acceptable solution to the problem. 612

13. We now need to solve 613

14. We now need to solve 614

15. We now need to solve 615

16. We now need to solve 616

17. We now need to solve Now solve the quadratic equation using with a = 50.3, b = -0.653, and c = 8.99x10 -4. 617

18. 618

19. y = 0.0114 M or y = 0.00157 M. 619

20. y = 0.0114 M or y = 0.00157 M. The first answer is physically impossible. 620

21. y = 0.0114 M or y = 0.00157 M. The first answer is physically impossible. Why? 621

22. y = 0.0114 M or y = 0.00157 M. The first answer is physically impossible. Why? The concentration of I 2 = 0.00414 – y = -0.00726 M which is physically impossible. We cannot have a negative concentration. 622

23. y = 0.0114 M or y = 0.00157 M. The first answer is physically impossible. Why? The concentration of I 2 = 0.00414 – y = -0.00726 M which is physically impossible. We cannot have a negative concentration. So the solution we seek is y = 0.00157 M. 623

24. At equilibrium: 624

25. At equilibrium: [H 2 ] = 0.00623 – 0.00157 = 0.00466 M 625

26. At equilibrium: [H 2 ] = 0.00623 – 0.00157 = 0.00466 M [I 2 ] = 0.00414 – 0.00157 = 0.00257 M 626

27. At equilibrium: [H 2 ] = 0.00623 – 0.00157 = 0.00466 M [I 2 ] = 0.00414 – 0.00157 = 0.00257 M [HI] = 0.0224 + 2x 0.00157 = 0.0255 M 627

28. At equilibrium: [H 2 ] = 0.00623 – 0.00157 = 0.00466 M [I 2 ] = 0.00414 – 0.00157 = 0.00257 M [HI] = 0.0224 + 2x 0.00157 = 0.0255 M Check: 628

29. At equilibrium: [H 2 ] = 0.00623 – 0.00157 = 0.00466 M [I 2 ] = 0.00414 – 0.00157 = 0.00257 M [HI] = 0.0224 + 2x 0.00157 = 0.0255 M Check: 629

30. There is a simple way to tell which way the reaction will proceed to equilibrium. 630

31. There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. 631

32. There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. 632

33. There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. If Q c > K c there must be too much HI in the mixture for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Q c becomes equal to K c. 633

34. There is a simple way to tell which way the reaction will proceed to equilibrium. Consider H 2(g) + I 2(g) 2 HI (g) and suppose all three gases are present at the start of the reaction. The reaction quotient is where the initial concentrations are employed. If Q c > K c there must be too much HI in the mixture for equilibrium to exist. Consequently, the reaction will proceed from right to left, until Q c becomes equal to K c. If Q c < K c, the reaction will proceed left to right. 634

35. In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved. 635

36. In many equilibrium problems, the equilibrium constant is very small. This allows a simplification of the mathematics, so that the general quadratic equation does not have to be solved. Example 3: 1.00 moles of N 2 is placed in a 1.00 liter vessel along with 1.10 moles of O 2 at 25 o C. Calculate the amount of NO formed at 25 o C. K c = 4.8 x 10 -31 at 25 o C for the reaction N 2(g) + O 2(g) 2 NO (g). 636

37. The ICE table is as follows: N 2 O 2 NO 637

38. The ICE table is as follows: N 2 O 2 NO 638

39. The ICE table is as follows: N 2 O 2 NO 639

40. The ICE table is as follows: N 2 O 2 NO 1.00 – y 1.10 – y 2y 640

41. The ICE table is as follows: N 2 O 2 NO 1.00 – y 1.10 – y 2y 641

42. Now 642

43. Now At this point, always stop and look for any possible simplifications. 643

44. Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. 644

45. Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – 645

46. Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it? 646

47. Now At this point, always stop and look for any possible simplifications. Notice that we do not have a perfect square on the right-hand side. In this case there is a mathematical simplification on the right-hand side of the equation – what is it? Notice that K c is very small, which means very little product will be formed. 647

48. Since y is very small, we try the approximations: 1.00 – y 1.00 1.10 – y 1.10 648

49. Since y is very small, we try the approximations: 1.00 – y 1.00 1.10 – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to 649

50. Since y is very small, we try the approximations: 1.00 – y 1.00 1.10 – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c 650

51. Since y is very small, we try the approximations: 1.00 – y 1.00 1.10 – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c 651

52. Since y is very small, we try the approximations: 1.00 – y 1.00 1.10 – y 1.10 We will check momentarily the quality of these approximations. The equilibrium constant expression simplifies to K c So y = 3.6 x 10 -16 M. 652

53. Since y = 3.6 x 10 -16 M the approximations 1.00 – y 1.00 1.10 – y 1.10 are justified. At equilibrium: [NO] = 2 y = 7.2 x 10 -16 M 653

54. When can approximations be made to simplify equilibrium constant expressions? 654

55. When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): 655

56. When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y 656

57. When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y and (B + y) B where B >> y 657

58. When can approximations be made to simplify equilibrium constant expressions? The only places where approximations can be made are (where A and B are concentrations): (A – y) A where A >> y and (B + y) B where B >> y Never in places like y (A – y) or 658

59. Le Châtelier’s Principle 659

60. Le Châtelier’s Principle We will examine the effect of the following on the equilibrium constant. 660