1. Chapter 13 Equilibrium

2. Unit Essential Question Z How do equilibrium reactions compare to other reactions?

3. Lesson Essential Question (sections 1-3) Z What is an equilibrium expression and how can an equilibrium expression be determined?

4. THE EQUILIBRIUM CONDITION Section 13.1

5. Reactions are Reversible Z Most reactions do NOT proceed to completion and then stop. Z Most reactions proceed in both the forward and reverse directions: A + B C + D ( forward) C + D A + B (reverse)

6. Reactions are Reversible Z A + B C + D ( forward) Z C + D A + B (reverse) Z Initially only A and B are present, so only the forward reaction is possible. Z As C and D are produced the reverse reaction speeds up, while the forward reaction slows down. Z Eventually the rates become equal.

7. Z Chemical equilibrium: concentrations of all species remain constant with time.

8. What is equal at Equilibrium? Z Rates are equal. Z Concentrations are not. Z Rates are determined by concentrations and activation energy. Z The concentrations do not change at equilibrium. Z Recall equilibrium is dynamic! a Reactions keep going at equilibrium! a A + B C + D

9. Section 1 Homework Z Pg. 614 # 9, 12

10. THE EQUILIBRIUM CONSTANT Section 13.2

11. Reaction Quotient, Q Z A ratio between the amount of reactants and products can be made at any time during the reaction. This is called the reaction quotient (Q). Z For aA + bB cC + dD the reaction quotient is: Q = [C] c [D] d [A] a [B] b Z Notice the effect of coefficients! Z Q may be written using concentrations (Q c ) or partial pressures (Q p ).

12. Equilibrium Constant, K Z The ratio between the amount of reactants and products at equilibrium is called the equilibrium constant (K). a K = [C] c [D] d [A] a [B] b a Again, may be K c or K p. a K has no units- it represents a ratio! Z Side note: P = CRT. Derived from ideal gas law since C = n/V (C = concentration). equilibrium expression

13. Equilibrium Constant, K Z Pure solids and liquids are NOT included in the equilibrium expression; their concentrations are assumed to be one. Why? a Concentrations of pure substances don’t change! Z Also, K is constant at any temperature. * If T changes then K will change!

14. Example Z Write the equilibrium expression (K p ) for the following reaction: SiH 4 (g) + 2Cl 2 (g) SiCl 4 (g) + 2H 2 (g) Remember to use partial pressures! K p = (P SiCl4 )(P H2 ) 2 (P SiH4 )(P Cl2 ) 2

15. Altering K Z If we write the reaction in reverse: cC + dD aA + bB Z Then the new equilibrium constant is: K ’ = [A] a [B] b = 1/K [C] c [D] d

16. Example Z At a given temperature, K = 1.3 x 10 -2 for the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) Z What is the value of K for the following: 2NH 3 (g) N 2 (g) + 3H 2 (g) K = 1/(1.3 x 10 -2 ) = 77

17. Altering K Z If we multiply the reaction by some number n: (aA + bB cC + dD)n naA + nbB ncC + ndD Z Then the new equilibrium constant is: K ’ = [C] nc [D] n d = ([C] c [D] d ) n = K n [A] na [B] nb ([A] a [B] b ) n

18. Example Z At a given temperature, K = 1.3 x 10 -2 for the following reaction: N 2 (g) + 3H 2 (g) 2NH 3 (g) Z What is the value of K for the following: 1/2N 2 (g) + 3/2H 2 (g) NH 3 (g) K = (1.3 x 10 -2 ) 1/2 = 0.11

19. Example- Calculating K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial Equilibrium Z [N 2 ] 0 =1.000 M [N 2 ] = 0.921M Z [H 2 ] 0 =1.000 M [H 2 ] = 0.763M Z [NH 3 ] 0 =0 M [NH 3 ] = 0.157M Z K = _ [NH 3 ] 2 _ = 0.0603 [N 2 ] [H 2 ] 3

20. Example- Calculating K Z N 2 (g) + 3H 2 (g) 2NH 3 (g) Initial Equilibrium Z [N 2 ] 0 = 0 M [N 2 ] = 0.399 M Z [H 2 ] 0 = 0 M [H 2 ] = 1.197 M Z [NH 3 ] 0 = 1.000 M [NH 3 ] = 0.203M Z K = 0.0602 Z K is the same no matter what the amount of starting materials.

21. What does K tell us? Z By looking at the size of K (and also Q), you can determine relative amounts of reactants and products present. Z Large values of K indicate greater amounts of products compared to reactants (forward reaction is favored/reaction lies to the right). Z Small values of K indicate more reactants are present (reverse reaction is favored/reaction lies to the left).

22. K c and K p  Values of K calculated from concentrations or partial pressures can be related: K p = K c (RT) Δn a R = ideal gas constant: 0.0821Latm/molK Make sure to choose the correct R value- look at units of P! Also T must be in K!  Δn = change in # moles of gas in the reaction: mol product – mol reactant

23. Example Z For the reaction C(s) + CO 2 (g) 2CO(g) K p =1.90 at 25°C. What is K c ? 1.90 = K c [(0.0821Latm/molK)(298K)] 1 K c = 0.0776

24. Sections 2&3 Homework Z Pg. 614 # 17(a&b), 18 (a&b),19(c&d), 21, 23, 25, 29

25. Lesson Essential Question (sections 5-7) Z How can equilibrium pressures and concentrations be calculated?

26. APPLICATIONS OF THE EQUILIBRIUM CONSTANT Section 13.5

27. Using Q Z Recall that it’s calculated the same way as the equilibrium constant, but for a system not at equilibrium (or if you’re not sure if a reaction is at equilibrium or not). Z By calculating Q you can compare the value to K to get an idea of where a reaction is in terms of reaching equilibrium.

28. What Q tells us: Z If Q<K a Not enough products (has not yet reached equilibrium). a Shift to the right to reach equilibrium. Z If Q>K a Too many products (has gone past equilibrium). a Shift to the left to reach equilibrium. Z If Q=K system is at equilibrium

29. Example Z For the reaction 2NO(g) N 2 (g) + O 2 (g) K = 2.4 x 10 3 at a certain temperature. If there are 0.024mol NO, 2.0mol N 2, and 2.6mol O 2 in a 1.0L flask, is the reaction at equilibrium? If not, in which direction will the reaction shift to reach equilibrium? Z Q = (2.0)(2.6) = 9.0 x 10 3 Q > K (0.024) 2 rxn. shifts left

30. Section 5 Homework Z Pg. 615 # 33(b&c), 34

31. SOLVING EQUILIBRIUM PROBLEMS Section 13.6

32. Procedure 1. Write balanced equation. 2. Write equilibrium expression. 3. Calculate Q, determine shift direction. 4. Start ICE table– determine I (initial concentrations). 5. Define change, C – add to I to find E (equilibrium concentrations). 6. Substitute E into expression and solve 7. Check values in expression, see if they equal K.

33. Example Z If K = 5.10 and there is 1.000mol of each substance present in a 1.000L flask, what are the equilibrium concentrations? CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) Z Initial concentrations: all 1.000M Z Determine Q: Q = (1.000M)(1.000M) = 1.000 Q<K, so (1.000M)(1.000M) rxn. shifts right

34. Example Cont. Z Now make your ICE table! Write ICE in a column below and to the left of the eqn. Z Use balanced equation coefficients and the direction the reaction shifts to find change. Z Add I and C to get E. Z ICE table helps you keep track of everything: CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) I 1.000 1.000 1.000 1.000 C -x -x +x +x E 1.000-x 1.000-x 1.000+x 1.000+x

35. Example Cont. Z Now plug the E values into the equilibrium expression and solve for x: 5.10 = [CO 2 ][H 2 ]_ = (1.000+x)(1.000+x) [CO] [H 2 O] (1.000-x)(1.000-x) 5.10 = (1.000 + x) 2 solve for x: x = 0.387 mol/L (1.000 – x) 2

36. Example Cont. Z [CO] & [H 2 O] = 1.000 – 0.387 = 0.613M Z [CO 2 ] & [H 2 ] = 1.000 + 0.387 = 1.387M Z Verify: K = (1.387)(1.387) = 5.12 (0.613)(0.613)

37. Practice Z K = 115 for a reaction where 3.000mol of all species were added to a 1.500L flask. What are the equilibrium concentrations? H 2 (g) + F 2 (g) 2HF(g) Q = 1.000, so rxn. will shift right. 115 = (2.000 + 2x) 2 => x = 1.528 (2.000 – x) 2 Z [H 2 ] = [F 2 ] = 0.472M ; [HF] = 5.056M

38. Using the Quadratic Formula Z You may have wondered when you would use this! Z Need it to solve some equilibrium problems (form: ax 2 + bx + c). Z Previous problems worked out nicely- just had to take the square root of both sides. Z Can use graphing calculator program (if you have it) to solve. Otherwise know it:

39. Example #1 – Quadratic Formula Z K = 115 for a reaction where 3.000mol of H 2 and 6.000mol F 2 were added to a 3.000L flask. What are the equilibrium concentrations? H 2 (g) + F 2 (g) 2HF(g) Z Q = _[HF] 2 _ [H 2 ][F 2 ] Z Reaction will shift right.

40. Example #1 Cont. H 2 (g) + F 2 (g) 2HF(g) I 1.000 2.000 0 C -x -x +2x E 1.000-x 2.000-x 2x K = 115 = _____(2x) 2 _____ (1.000-x)(2.000-x) (1.000-x)(2.000-x)(115) = 4x 2 (2.000 -3.000x+x 2 )(115) = 4x 2 230. – 345x + 115x 2 = 4x 2 111x 2 – 345x + 230. = 0 a = 111 b = -345 c = 230.

41. Example #1 Cont. Use quadratic formula: x = 345 ±√(-345) 2 – 4(111)(230.) 2(111) x = 2.14M and x = 0.968M Only one value of x is correct: [H 2 ] = 1.000M – x [H 2 ] = 1.000M – 2.14M = -1.14M [H 2 ] = 1.000M – 0.968M = 0.032M [F 2 ] = 2.000M – 0.968M = 1.032M [HF] = 2(0.968) = 1.936M a = 111 b = -345 c = 230. Can’t have a negative concentration!

42. What if K is very small? Z If K is < 10 -3 we can consider it to be negligible with respect to concentrations. a Can make calculations simpler. a Can ignore x only where it would make little change in the concentration. a MUST check to see if the calculated equilibrium concentration is less than 5% different from the initial – if it’s more than 5%, then you can’t disregard x.

43. Example #2- Small K Values Exactly 1.0mol NOC l is placed in a 2.0L flask. The value of K = 1.6x10 -5. What are the equilibrium concentrations? 2NOC l (g) 2NO(g) + C l 2 (g) *Rxn. will shift right b/c there are no products initially. *The equilibrium expression will be: K = [NO] 2 [C l ] = 1.6x10 -5 [NOC l ] 2

44. Example #2- Small K Values 2NOC l (g) 2NO(g) + C l 2 (g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *This would result in a complicated math equation to solve. *Simplification: K is small (<10 -3 ), so x is negligible only with respect to NOC l. -Small K means reaction barely proceeds right, so 0.50M NOCl won’t change noticeably.

45. Example #2- Small K Values 2NOC l (g) 2NO(g) + C l 2 (g) I 0.50 0 0 C -2x +2x +x E 0.50-2x 2x x *Simplification: 0.50 – 2x ≈ 0.50M 1.6x10 -5 = (2x) 2 (x) = _4x 3 _ (0.50) 2 (0.50) 2 x 3 = (1.6x10 -5 )(0.50) 2 => x = 0.010 4

46. Example #2- Small K Values Verify assumption that x is negligible with respect to NOC l : [NOC l ] = 0.50 – 2(0.010) = 0.48 *This must be less than 5% different from the initial concentration: 0.50-0.48 = 0.02 0.02/0.50 x 100 = 4% Assumption was ok!

47. Example #2- Small K Values Calculate the rest of the equilibrium concentrations: [NOC l ] = 0.50M (this was the simplification!) [NO] = 2x = 2(0.010) = 0.020M [C l ] = x = 0.010M

48. Section 6 Homework Z Pg. 616 # 45, 46, 48, 51

49. Lesson Essential Question Z How does Le Châtelier’s Principle effect the equilibrium of a reaction?

50. Le CHÂTELIER’S PRINCIPLE Section 13.7

51. Le Chatelier’s Principle Z If a change is applied to a system at equilibrium, the reactions will shift to reduce the change. Equilibrium is reestablished. Z Three types of change: a Concentration a Pressure a Temperature

52. Changes in Concentration Z Adding more of a substance will cause the reaction to shift in the direction that consumes it. Z Removing a substance will cause the reaction to shift in the direction that produces it. Z Example: N 2 O 4 (g) 2NO 2 (g) Z Adding NO 2 shifts the reaction to the left. Z How can you get the reaction to shift right?

53. Changes in Concentration Z What does this do to the amounts of reactants and products? Z The amounts of reactants and products will change in order to reestablish equilibrium. Z The ratio of reactants to products will be the same once equilibrium has been reestablished.

54. Changes in Pressure Z P can be changed by altering volume. a Note this is only applicable to gases! Z If V is decreased, P increases, and the reaction shifts in the direction that produces the least moles of gas. a Why? Z If V is increased, P decreases, and the reaction shifts in the direction that produces the most moles of gas. a Why? Z Equilibrium is reestablished in both cases. Fewer moles of gas = lower P. More moles of gas = higher P.

55. Example N 2 O 4 (g) 2NO 2 (g) Z What will happen if the volume of the container is decreased? Z If the volume of the container increases? Z Note: if there are equal # moles on both sides, changing P has no effect on equilibrium. Rxn. shifts left. Rxn. shifts right.

56. Change in Temperature Z Different from previous two changes! Z Doesn’t just change the equilibrium position, also changes the equilibrium constant K. Z The direction of the shift depends on whether it is exo- or endothermic. a It helps to think of E (heat) as a reactant or product.

57. Change in Temperature  Consider the same reaction previously examined: 58kJ + N 2 O 4 (g) 2NO 2 (g) Z Endothermic: think of heat as a reactant. Z Increasing T means increasing heat; reaction shifts right to consume additional heat. Z What happens if T decreases? a Reaction shifts left to generate more heat.

58. Practice N 2 (g) + 3H 2 (g) 2NH 3 (g) + heat Consider the above reaction. What happens if: Z Additional H 2 is added? Z T increases? Z V decreases? Z How could the reaction shift left other than increasing T? Rxn. shifts right Rxn. shifts left Rxn. shifts right

59. AP Practice Question At constant T, a change in V will NOT affect the moles of substances present in which of the following? a) H 2 (g) + I 2 (g) 2HI (g) b) CO(g) + Cl 2 (g) COCl 2 (g) c) PCl 5 (g) PCl 3 (g) + Cl 2 (g) d) N 2 (g) + 3H 2 (g) 2NH 3 (g)

60. AP Practice Question C(s) + H 2 O(g) CO(g) + H 2 O(g) endo. An equilibrium mixture of the reactants is placed in a sealed container at 150°C. The amount of the products may be increased by which of the following changes? a) Decreasing the V of the container. b) Raising T & increasing the V. c) Decreasing T of the container. d) Increasing V and adding 1mol of CO(g).

61. AP Practice Question 2CH 4 (g) + O 2 (g) 2CO(g) + 4H 2 (g) ΔH<0 In order to increase the value of the equilibrium constant, K, which of the following changes must occur? a) Increase T. b) Increase V. c) Decrease T. d) Add CO(g).

62. Section 7 Homework Z Pg. 617 #57, 59, 60