1. Chapter 13 Chemical Equilibrium: How can things that are moving seem to be standing still?

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 Objectives To define the Chemical Equilibrium To describe the equilibrium condition in terms of reactant and product concentration To describe the equilibrium condition in terms of forward and reverse reaction rates

4. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 4 A Gaseous Solute C = kP

5. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 5 Liquid/Vapor Equilibrium

6. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 6 Chemical Equilibrium So far, we have seen chemical equations written as: H 2 O(g) + CO(g)  H 2 (g) + CO 2 (g) Or stated as such that No Reaction occurs This may be an oversimplification, whereas… H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) Is a more thorough description of what is happening in virtually every chemical reaction. But, what do these symbols mean?

7. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 7 Chemical Equilibrium The double arrow sign,, represents an equilibrium condition. Equilibrium is: 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. Equilibrium is : Macroscopically static Microscopically dynamic

8. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 8 N 2 (g) + 3H 2 (g) 2NH 3 (g) Concentrations reach constant levels where the rate of the forward reaction equals the rate of the reverse reaction.

9. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 9 The Changes with Time in the Rates of Forward and Reverse Reactions Bees wake up and start leaving to collect honey…time passes. the # of bees leaving = the # of bees entering…equilibrium

10. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 10 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.

11. Section 13.1 The Equilibrium Condition Return to TOC Copyright © Cengage Learning. All rights reserved 11 Objectives Review To define the Chemical Equilibrium To describe the equilibrium condition in terms of reactant and product concentration To describe the equilibrium condition in terms of forward and reverse reaction rates

12. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 12 Objectives To define the equilibrium constant (K) To write the equilibrium constant expression for various chemical reactions

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 Write the equilibrium constant expression: H 2 O(g) + CO(g) H 2 (g) + CO 2 (g) 4NH 3 (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O(g)

15. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 15 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.

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.  K = [Products]  [Reactants] More Conclusions About the Equilibrium Expression

17. Section 13.2 Atomic MassesThe Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 17 Objectives Review To define the equilibrium constant (K) To write the equilibrium constant expression for various chemical reactions Work Session: Pg 613 # 1, 6, 12, 17& 18 a&b, 19

18. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 18 Objectives To write the Equilibrium Expression for gas reactions (Kp) To calculate K from Kp To see how the 5-Step Problem Solving Method can help you calculate without having to “KNOW” how to do a problem…

19. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 19 K involves concentrations. K p involves pressures.

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) Write the K and Kp expressions:

21. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 21 Example N 2 (g) + 3H 2 (g) 2NH 3 (g) Calculate Kp given the equilibrium pressures at a certain temperature:

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)

23. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 23 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) Using the value of K p (3.9 × 10 4 ) from the previous example, calculate the value of K at 35°C.

24. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 24 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.

25. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 25 The Power of the 5 Step! Take a look at the AP reference sheet… Even though you may only see an equation once, or perhaps never, and you may think that you forget or never learned how to use the equation….. The 5 Step is the “HOW” of working with equations… Being more familiar is more comfortable, however, take comfort in the Power of the 5 Step!

26. Section 13.3 The MoleEquilibrium Expressions Involving Pressures Return to TOC Copyright © Cengage Learning. All rights reserved 26 Objectives Review To write the Equilibrium Expression for gas reactions (Kp) To calculate K from Kp To see how the 5-Step Problem Solving Method can help you calculate without having to “KNOW” how to do a problem… Work Session pg 615 # 21, 26

27. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 27 Objectives 13.4 and 13.5 To define homogeneous and heterogeneous equilibria To understand why pure solids and liquids are left out of the equilibrium expression To evaluate the extent of a reaction based on the numerical value of K To use the Reaction Quotient (Q) to determine if a system is at equilibrium and what directional shift will occur to reach EQ

28. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 28 Homogeneous Equilibria Homogeneous equilibria – involve the same phase: N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq)

29. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 29 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)

30. Section 13.4 Heterogeneous Equilibria Return to TOC Copyright © Cengage Learning. All rights reserved 30 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. (Think of them as 1) 2KClO 3 (s) 2KCl(s) + 3O 2 (g)

31. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 31 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.  A + B  AB The Extent of a Reaction

32. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 32 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. A + B  No Rx The Extent of a Reaction

33. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 33 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)

34. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 34 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? Need [at Equilibrium] –Use ICE Table…

35. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 35 Set up ICE Table (Initial, Change, Equil’ Conditions) Fe 3+ (aq) + SCN – (aq) FeSCN 2+ (aq) Initial 6.00 10.00 0.00 Change- 4.00 – 4.00+4.00 Equilibrium 2.00 6.00 4.00 K = 0.333 (1:1 Stoich)

36. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 36 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) (Means you know K) Equilibrium: ? M FeSCN 2+ (aq) (ICE TABLE) (NEXT SLIDE)

37. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 37 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) (Means you know K)(ICE TABLE) I CECE 5.00 M FeSCN 2+

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 #3: Initial: 6.00 M Fe 3+ (aq) and 6.00 M SCN − (aq) Equilibrium: ? M FeSCN 2+ (aq) (ICE Set Up K) (Don’t solve) 3.00 M FeSCN 2+

39. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 39 Exercise (Start with the ICE, Keq, and Work Through the Algebra, no Quadratic needed) Consider the reaction represented by the equation (assume K = 0.333) : 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.

40. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 40 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

41. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 41 To determine if a system is at EQ, apply the law of mass action using initial (instantaneous) concentrations instead of equilibrium concentrations. In other words, calculate the Q expression and see if the conditions are the same as the K expression… Reaction Quotient, Q

42. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 42 Q = K; The system is at equilibrium. No shift will occur. Q > K;  Consuming products and forming reactants, until equilibrium is achieved.  The system shifts to the ……  ….left. Reaction Quotient, Q

43. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 43 Q < K;  Consuming reactants and forming products, to attain equilibrium.  The system shifts to the …………  ………right. Reaction Quotient, Q

44. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 44 For the synthesis of ammonia at 500 o C, Keq = 6.0 X 10 -2. Predict if the following system is in equilibrium. If not, indicate the direction in which the system will shift in order to reach eq’. a.[NH 3 ] = 1.0 X 10 -3 M; [N 2 ] = 1.0 X 10 -5 M; [H 2 ] = 2.0 X 10 -3 M a. 1.25 X 10 7 Left shift b.[NH 3 ] = 2.0 X 10 -4 M; [N 2 ] = 1.5 X 10 -5 M; [H 2 ] = 3.54 X 10 -1 M @ EQ’M c.[NH 3 ] = 1.0 X 10 -4 M; [N 2 ] = 5.0M; [H 2 ] = 1.0 X 10 -2 M 2.0 X 10 -3 Right Shift

45. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 45 http://www.usca.edu/chemistry/genchem/sigfig.htm

46. Section 13.5 Applications of the Equilibrium Constant Return to TOC Copyright © Cengage Learning. All rights reserved 46 Objectives Review 13.4 and 13.5 To define homogeneous and heterogeneous equilibria To understand why pure solids and liquids are left out of the equilibrium expression To evaluate the extent of a reaction based on the numerical value of K To use the Reaction Quotient (Q) to determine if a system is at equilibrium and what directional shift will occur to reach EQ Work Session: Pg 615 # 29, 31, 33, 37

47. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 47 Objectives To practice using the ICE method for solving Equilibrium Problems

48. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 48 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

49. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 49 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

50. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 50 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: NH 3 (g) N 2 (g) + H 2 (g) At equilibrium 1.00 mol of ammonia remains. Calculate the value for K. (Balance)(ICE) K = 1.69

51. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 51 Concept Check

52. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 52 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). (ICE, Concentration of N 2 O 4 = 0.097 M Concentration of NO 2 = 6.32 x 10 -3 M

53. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 53 Concept Check

54. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 54 Concept Check

55. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 55 Concept Check

56. Section 13.6 Solving Equilibrium Problems Return to TOC Copyright © Cengage Learning. All rights reserved 56 Objectives Review To practice using the ICE method for solving Equilibrium Problems Work Session: page 616 39, 41, 43*, 45 (setup only), 47, 51(setup only), 53 (setup only)

57. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 57 Objectives To understand Le Chatelier’s Principle To predict shifts in equilibrium HPhen (aq) H + (aq) + Phen - (aq) CLEAR PINK

58. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 58 Le Chatelier’s Principle 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.

59. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 59 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).

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

61. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 61

62. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 62 Equilibrium Decomposition of N 2 O 4

63. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 63 Predicting Shifts in Equilibria Which way will the equilibrium shift if: (don’t forget volume…) N 2 (g) + 3H 2 (g) 2NH 3 (g) HCN(aq) H + (aq) + CN - (aq) H = -12 kJ

64. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 64 Objectives To understand Le Chatelier’s Principle To predict shifts in equilibrium Work Session: pg 617 # 57, 59, 63

65. Section 13.7 Le Châtelier’s Principle Return to TOC Copyright © Cengage Learning. All rights reserved 65 Objectives