1. Chapter 13 Chemical Equilibrium

2. Chapter 13 Table of Contents Copyright © Cengage Learning. All rights reserved 2 13.1The Equilibrium Condition 13.2The Equilibrium Constant 13.3Equilibrium Expressions Involving Pressures 13.4Heterogeneous Equilibria 13.5Applications of the Equilibrium Constant 13.6Solving Equilibrium Problems 13.7Le Châtelier’s Principle

3. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 3 Chemical Equilibrium The state where the concentrations of all reactants and products remain constant with time. On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation. The concept of chemical equilibrium is analogous to the flow of cars across a bridge connecting two cities.

4. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 4 Equilibrium Is: Macroscopically static Microscopically dynamic

5. Section 13.1 The Equilibrium Condition Return to TOC A Molecular Representation of 2NO 2 (g) → N 2 O 4 (g) Over Time in a Closed Vessel NO 2 N2O4N2O4

6. Section 13.1 The Equilibrium Condition Return to TOC Changes with Time for the Reaction H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) In a closed vessel at a high temperature

7. Section 13.1 The Equilibrium Condition Return to TOC Equilibrium of a Reaction of H 2 O and CO over Time H2OH2OCOH2H2 CO 2 H 2 O(g) + CO(g) H 2 (g) + CO 2 (g)

8. Section 13.1 The Equilibrium Condition Return to TOC Changes with Time in Reaction Rates of Forward and Reverse Reactions for H 2 O(g) + CO(g) H 2 (g) + CO 2 (g)

9. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 9 Changes in Concentration N 2 (g) + 3H 2 (g) 2NH 3 (g) Note that because of the reaction stoichiometry, H 2 disappears 3 times as fast as N 2 dose and NH 3 forms twice as fast as N 2 disappears.

10. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 10 Chemical Equilibrium Concentrations reach levels where the rate of the forward reaction equals the rate of the reverse reaction. The Equilibrium position-left, right, or somewhere is determined by many factors The initial concentrations The relative energies(reactants and products) The relative degree of organization (reactants and products) Try to achieve minimum energy and maximum disorder

11. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 11 Concept Check Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 O(g) to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. The concentrations of each product will increase, the concentration of CO will decrease, and the concentration of water will be higher than the original equilibrium concentration, but lower than the initial total amount.

12. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 12 Concept Check Consider an equilibrium mixture in a closed vessel reacting according to the equation: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) You add more H 2 to the flask. How does the concentration of each chemical compare to its original concentration after equilibrium is reestablished? Justify your answer. This is the opposite scenario of the previous slide

13. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 13 Consider the following reaction at equilibrium: jA + kB lC + mD A, B, C, and D = chemical species. Square brackets = concentrations of species at equilibrium. j, k, l, and m = coefficients in the balanced equation. K = equilibrium constant (given without units). j l k m [B][A] [D] [C] K =

14. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 14 Conclusions About the Equilibrium Expression Equilibrium expression for a reaction is the reciprocal of that for the reaction written in reverse. When balanced equation for a reaction is multiplied by a factor of n, the equilibrium expression for the new reaction is the original expression raised to the nth power; thus K new = (K original ) n. K values are usually written without units.

15. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Houghton Mifflin Company. All rights reserved. Chapter 13 | Slide 15 Table 13.1 Results of Three Experiments for the Reaction N 2 (g) + 3H 2 (g) -- 2NH 3 (g)

16. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 16 K always has the same value at a given temperature regardless of the amounts of reactants or products that are present initially. For a reaction, at a given temperature, there are many equilibrium positions but only one value for K.  Equilibrium position is a set of equilibrium concentrations.

17. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 17 K involves concentrations. K p involves pressures.

18. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 18 Example N 2 (g) + 3H 2 (g) 2NH 3 (g)

19. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 19 Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Equilibrium pressures at a certain temperature:

20. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 20 Example N 2 (g) + 3H 2 (g) 2NH 3 (g)

21. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 21 The Relationship Between K and K p K p = K(RT) Δn Δn = sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants. R = 0.08206 L·atm/mol·K T = temperature (in kelvin)

22. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 22 Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Using the value of K p (3.9 × 10 4 ) from the previous example, calculate the value of K at 35°C.

23. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 23 Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

24. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 24 Heterogeneous Equilibria Heterogeneous equilibria – involve more than one phase: 2KClO 3 (s) 2KCl(s) + 3O 2 (g) 2H 2 O(l) 2H 2 (g) + O 2 (g)

25. Section 13.4 Heterogeneous Equilibria Return to TOC Seven Sisters Chalk Cliffs in East Sussex, England The chalk is made up of compressed CaCO 3 skeletons of microscopic algae from the late Cretaceous Period

26. Section 13.4 Heterogeneous Equilibria Return to TOC Figure 13.6: (a) (b) Position of Equilibrium CaCO 3 (s) CaO(s) + CO 2 (g) K= 〔 CO 2 〕 The position of the equilibrium does not depend on the amounts of CaCO 3 (s) and CaO(s) present.

27. Section 13.4 Heterogeneous Equilibria Return to TOC (left) Hydrated Copper(II) Sulfate; (right) Water Applied to Anhydrous Copper(II) Sulfate

28. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 28 The position of a heterogeneous equilibrium does not depend on the amounts of pure solids or liquids present.  The concentrations of pure liquids and solids are constant. 2KClO 3 (s) 2KCl(s) + 3O 2 (g) PCl 5 (s) PCl 3 (l) + Cl 2 (g)

29. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 29 A value of K much larger than 1 means that at equilibrium the reaction system will consist of mostly products – the equilibrium lies to the right.  Reaction goes essentially to completion. The Extent of a Reaction

30. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 30 A very small value of K means that the system at equilibrium will consist of mostly reactants – the equilibrium position is far to the left.  Reaction does not occur to any significant extent. The Extent of a Reaction

31. Section 13.5 Applications of the Equilibrium Constant Return to TOC

32. Section 13.5 Applications of the Equilibrium Constant Return to TOC

33. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Houghton Mifflin Company. All rights reserved. Chapter 13 | Slide 33 (a).The boulder is thermodynamically more stable(lower potential energy) in position B than in position A but cannot get over hump H. (b).The reactant H 2,O 2 have a strong tendency to form H 2 O (lower potential energy). However, the large activation energy E a prevents the reaction at 25°C The magnitude of K for the reaction depends on ΔE, but the reaction rate depends on E a.

34. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 34 Concept Check If the equilibrium lies to the right, the value for K is __________. large (or >1) If the equilibrium lies to the left, the value for K is ___________. small (or <1)

35. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 35 Apply the law of mass action using initial concentrations instead of equilibrium concentrations. Reaction Quotient, Q

36. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 36 Q = K; The system is at equilibrium. No shift will occur. Q > K; The system shifts to the left.  Consuming products and forming reactants, until equilibrium is achieved. Q < K; The system shifts to the right.  Consuming reactants and forming products, to attain equilibrium. Reaction Quotient, Q

37. Section 13.5 Applications of the Equilibrium Constant Return to TOC Figure 13.8: Ratio at Equilibrium

38. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 38 Exercise Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #1: 6.00 M Fe 3+ (aq) and 10.0 M SCN - (aq) are mixed at a certain temperature and at equilibrium the concentration of FeSCN 2+ (aq) is 4.00 M. What is the value for the equilibrium constant for this reaction?

39. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 39 Set up ICE Table Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial6.00 10.00 0.00 Change – 4.00 – 4.00+4.00 Equilibrium 2.00 6.00 4.00 K = 0.333

40. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 40 Exercise Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #2: Initial: 10.0 M Fe 3+ (aq) and 8.00 M SCN − (aq) (same temperature as Trial #1) Equilibrium: ? M FeSCN 2+ (aq) 5.00 M FeSCN 2+

41. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 41 Exercise Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Trial #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq) 3.00 M FeSCN 2+

42. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 42 1)Write the balanced equation for the reaction. 2)Write the equilibrium expression using the law of mass action. 3)List the initial concentrations. 4)Calculate Q, and determine the direction of the shift to equilibrium. Solving Equilibrium Problems

43. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 43 5)Define the change needed to reach equilibrium, and define the equilibrium concentrations by applying the change to the initial concentrations. 6)Substitute the equilibrium concentrations into the equilibrium expression, and solve for the unknown. 7)Check your calculated equilibrium concentrations by making sure they give the correct value of K. Solving Equilibrium Problems

44. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 44 Exercise Consider the reaction represented by the equation: Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) Fe 3+ SCN - FeSCN 2+ Trial #19.00 M5.00 M1.00 M Trial #23.00 M2.00 M5.00 M Trial #32.00 M9.00 M6.00 M Find the equilibrium concentrations for all species.

45. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 45 Exercise (Answer) Trial #1: [Fe 3+ ] = 6.00 M; [SCN - ] = 2.00 M; [FeSCN 2+ ] = 4.00 M Trial #2: [Fe 3+ ] = 4.00 M; [SCN - ] = 3.00 M; [FeSCN 2+ ] = 4.00 M Trial #3: [Fe 3+ ] = 2.00 M; [SCN - ] = 9.00 M; [FeSCN 2+ ] = 6.00 M

46. Section 13.6 Solving Equilibrium Problems Return to TOC Example13.10 Copyright © Cengage Learning. All rights reserved 46 ＜ 〔 CO 2 〕 〔 H 2 〕 〔 CO 〕 〔 H 2 O 〕 Initial 1 1 1 1 Change -x -x +x +x equilibrium 1-x 1-x 1+x 1+x

47. Section 13.6 Solving Equilibrium Problems Return to TOC Example 13.11 Copyright © Cengage Learning. All rights reserved 47 ＜ 〔 H 2 〕 〔 F 2 〕 〔 HF 〕 Initial 2 2 2 Change -x -x +2x equilibrium 2-x 2-x 2+2x

48. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 48 Concept Check A 2.0 mol sample of ammonia is introduced into a 1.00 L container. At a certain temperature, the ammonia partially dissociates according to the equation: 2NH 3 (g) N 2 (g) + 3H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. K = 1.69

49. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 49 Concept Check A 1.00 mol sample of N 2 O 4 (g) is placed in a 10.0 L vessel and allowed to reach equilibrium according to the equation: N 2 O 4 (g) 2NO 2 (g) K = 4.00 x 10 -4 Calculate the equilibrium concentrations of: N 2 O 4 (g) and NO 2 (g). Concentration of N 2 O 4 = 0.097 M Concentration of NO 2 = 6.32 x 10 -3 M

50. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 50 If a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change.

51. Section 13.7 Le Châtelier’s Principle Return to TOC Figure 13.9: (a) (b) (c) Change in Equilibrium of a Mixture of N 2, H 2, and NH 3 N 2 increase(due to addition of N 2 ),less H 2, and more NH 3 than initial state.

52. Section 13.7 Le Châtelier’s Principle Return to TOC Volume of Container Reduced; Gaseous System Is Reduced colorless brown N 2 O 4 (g) 2NO 2 (g) b.V ↓ giving a greater concentration of N 2 O 4,NO 2 c.After few seconds,the coloris much lighter brown as the equilibrium shifts the brown NO 2 to colorless N 2 O 4,since the N 2 O 4 has the smaller number of molecules.

53. Section 13.7 Le Châtelier’s Principle Return to TOC Figure 13.10: (a) (b) (c) Change in Equilibrium of a Mixture of N 2, H 2, and NH 3 After a Decrease in Volume

54. Section 13.7 Le Châtelier’s Principle Return to TOC The Percent by Mass of NH 3 at Equilibrium in a Mixture of N 2, H 2 and NH 3 as a Function of Temperature and Total Pressure( +92kj ) V decrease → P increase →decrease the total number of gasous molecules in the system →NH 3 increase △ H is negative, T increase →shift to left →NH 3 decrease

55. Section 13.7 Le Châtelier’s Principle Return to TOC Observed Value of k for the Ammonia Synthesis Reaction as a Function of Temperature

56. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 56 Effects of Changes on the System 1.Concentration: The system will shift away from the added component. If a component is removed, the opposite effect occurs. 2.Temperature: K will change depending upon the temperature (endothermic – energy is a reactant; exothermic – energy is a product).

57. Section 13.7 Le Châtelier’s Principle Return to TOC Shifting the N 2 O 4 (g) → 2NO 2 (g) Equilibrium by Changing the Temperature N 2 O 4 (g) + 58kj 2NO 2 (g) 100 ℃ 0 ℃

58. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Houghton Mifflin Company. All rights reserved. Chapter 13 | Slide 58 Table 13.4 Shifts in the Equilibrium Position for the Reaction 58 kJ + N 2 O 4 (g) 2NO 2 (g)

59. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 59 Effects of Changes on the System 3.Pressure: a)The system will shift away from the added gaseous component. If a component is removed, the opposite effect occurs. b)Addition of inert gas does not affect the equilibrium position. c)Decreasing the volume shifts the equilibrium toward the side with fewer moles of gas.

60. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 60

61. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 61 Equilibrium Decomposition of N 2 O 4