1. Equilibrium UNIT 12

2. Overview  Concept of Equilibrium  Equilibrium constant  Equilibrium expression  Heterogeneous vs homogeneous equilibrium  Solving for equilibrium constant  Solving for K  Manipulating equilibrium constant  Using gases  Equilibrium Calculations  ICE method  ICE Stoichiometry  Reaction Quotient (Q)  Relationship K and Q  Le Chatlier’s Principle  Pressure  Volume  Concentration  Temperature  Catalysts

3. The Concept of Equilibrium  We’ve already used the phrase “equilibrium” when talking about reactions.  In principle, every chemical reaction is reversible... capable of moving in the forward or backward direction. 2 H 2 + O 2 2 H 2 O Some reactions are easily reversible... Some not so easy...

4. The Concept of Equilibrium  Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

5. The Concept of Equilibrium  As a system approaches equilibrium, both the forward and reverse reactions are occurring.  At equilibrium, the forward and reverse reactions are proceeding at the same rate.

6. A System at Equilibrium  Once equilibrium is achieved, the amount of each reactant and product remains constant.

7. A System at Equilibrium Rates become equalConcentrations become constant

8. Depicting Equilibrium  In a system at equilibrium, both the forward and reverse reactions are running simultaneously. We write the chemical equation with a double arrow:

9. Equilibrium Constant Forward reaction: Reverse reaction: Rate Law Rate law

10. Equilibrium Constant At equilibrium Rearranging gives: r

11. Equilibrium Constant  The ratio of the rate constants is a constant (as long as T is constant). The expression becomes

12. Equilibrium Expression (Law of Mass Action) To generalize, the reaction: Has the equilibrium expression: This expression is true even if you don’t know the elementary reaction mechanism.

13. Equilibrium Expression  [A], [B], [C], [D] = molar concentrations or partial pressures at equilibrium  Products = numerator and reactants = denominator  Coefficients in balanced equation = exponents

14. Equilibrium Expression  Equilibrium constant can have different subscripts  K c = concentration  Moles, liters, or molarity  K p = partial pressure  Atmospheres  K sp = solubility product  Reactant is a solid so only products are in expression

15. Equilibrium Expression  Does not depend on initial concentration of reactants and products  Solids and pure liquids are not included in equilibrium expression (only gases and solutions)  K eq does not have units (describes activity of reaction)  Depends only on the particular reaction and the temperature

16. Equilibrium Expression  The Concentrations of Solids and Liquids Are Essentially Constant  Therefore, the concentrations of solids and liquids do not appear in the equilibrium expression K c = [Pb 2+ ] [Cl − ] 2 PbCl 2 (s) Pb 2+ (aq) + 2 Cl − (aq)

17. Sample Exercise 1  Write the equilibrium expression for K c for the following reactions:

18. Sample Exercise 1 – Answers

19. Homogeneous vs Heterogeneous  Homogeneous equilibrium  Reactants and products are all in the same phase  Example: 2NOBr(g) ↔ 2NO(g) + Br 2 (g)  Heterogeneous equilibrium  Reactants and products are occur in different phases  Example: Pb Cl 2 (s) ↔ Pb 2+ (aq) + 2Cl - (aq)

20. Equilibrium - Summary  At equilibrium…  Concentrations of reactants and products no longer change with time  Neither reactants nor products can escape from the system  The equilibrium constant comes from a ratio of concentration

21. Solving for K eq K c, the final ratio of [NO 2 ] 2 to [N 2 O 4 ], reaches a constant no matter what the initial concentrations of NO 2 and N 2 O 4 are.

22. Solving for K eq = = 0.211 [0.0172] 2 [0.00140]

23. Solving for K eq This graph shows data from the last two trials from the table.

24. What does the value of K mean?  If K >> 1, the reaction is product-favored; contains mostly products at equilibrium If K << 1, the reaction is reactant-favored; contains mostly reactants at equilibrium.

25. Manipulating Equilibrium Constants  The equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

26. Manipulating Equilibrium Constants  The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

27. Manipulating Equilibrium Constants  If two reactions can be added together to create a third reaction, then the K eq for the two reactions can be multiplied together to get the K eq for the third reaction. If A + B ↔ CK eq = K 1 andC ↔ D + EK eq = K 2 thenA + B ↔ D + EK eq = K 1 K 2

28. Manipulating Equilibrium Constants (Summary)  The equilibrium constant in the reverse direction is the inverse of the equilibrium constant in the forward direction.  The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number.  The equilibrium constant for a net reaction made up of multiple steps is the product of the equilibrium constants for individual steps.

29. Equilibrium Constant and Gases  Because pressure is proportional to concentration for gases, the equilibrium expression can also be written in terms of partial pressures (instead of concentration): Mixed versions are also used sometimes:

30. Relationship between K c and K p From the ideal gas law we know that = Pressure in terms of concentration

31. Relationship between K c and K p  Substituting P=[A]RT into the expression for K p for each substance, the relationship between K c and K p becomes Where: K p = K c (RT)  n  n = (moles of gaseous product) – (moles of gaseous reactant) R = 0.0821 L∙atm/mol∙K GIVEN TO YOU ON AP CHEAT SHEET!!

32. Relationship between K c and K p  In the synthesis of ammonia from nitrogen and hydrogen, N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g). Given K c of 9.60 at 300°C, calculate K p.  n = (moles gaseous products – moles gaseous reactants) = 2-4 = -2 K p = K c (RT)  n K p = 9.60(0.0821×573K) -2 = 4.34×10 -3

34. ICE  At 1280 0 C the equilibrium constant (K c ) for the reaction is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Comparing initial concentrations to equilibrium concentrations

35. At 1280 0 C the equilibrium constant (K c ) for the reaction is 1.1 x 10 -3. If the initial concentrations are [Br 2 ] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium. Br 2 (g) 2Br (g) Let x be the change in concentration of Br 2 Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x [Br] 2 [Br 2 ] K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 Solve for x 14.4

36. K c = (0.012 + 2x) 2 0.063 - x = 1.1 x 10 -3 4x 2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x 4x 2 + 0.0491x + 0.0000747 = 0 ax 2 + bx + c =0 -b ± b 2 – 4ac  2a2a x = Br 2 (g) 2Br (g) Initial (M) Change (M) Equilibrium (M) 0.0630.012 -x-x+2x 0.063 - x0.012 + 2x x = -0.00178x = -0.0105 At equilibrium, [Br] = 0.012 + 2x = -0.009 Mor 0.00844 M At equilibrium, [Br 2 ] = 0.062 – x = 0.0648 M 14.4

37. 37 Example Problem: Calculate Concentration Note the moles into a 10.32 L vessel stuff... calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M 2 HI H 2 + I 2 Initial: Change: Equil: 0.242 M00 -2x+x+x 0.242-2xxx

38. 38 Example Problem: Calculate Concentration And yes, it’s a quadratic equation. Doing a bit of rearranging: x = 0.00802 or –0.00925 Since we are using this to model a real, physical system, we reject the negative Root. The [H 2 ] at equil. is 0.00802 M.

39. Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x

40. 40 Approximating If K eq is really small the reaction will not proceed to the right very far, meaning the equilibrium concentrations will be nearly the same as the initial concentrations of your reactants. 0.20 – x is just about 0.20 is x is really dinky. (Example: A million dollars minus a nickel is still a million dollars) If the difference between K eq and initial concentrations is around 3 orders of magnitude or more, go for it. Otherwise, you have to use the quadratic.

41. Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x More than 3 orders of mag. between these numbers. The simplification will work here.

42. Initial Concentration of I 2 : 0.50 mol/2.5L = 0.20 M I 2 2 I Initial change equil: 0.20 0 -x +2x 0.20-x 2x More than 3 orders of mag. between these numbers. The simplification will work here. x = 3.83 x 10 -6 M

43. Equilibrium Stoichiometry Given initial concentrations of reactants AND Equilibrium concentration of products (or vice versa)

44. Equilibrium Stoichiometry A closed system initially containing1.000 x 10 -3 M H 2 and 2.000 x 10 -3 M I 2 at 448  C is allowed to reach equilibrium. Analysis of the equilibrium mixture shows that the concentration of HI is 1.87 x 10 -3 M. Calculate K c at 448  C for the reaction: H 2 (g) + I 2 (g) 2 HI (g)

45. Equilibrium Stoichiometry [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change At Equilibrium 1.87 x 10 -3 ICE method: Write what we know.

46. Equilibrium Stoichiometry [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change+1.87 x 10 -3 At equilibrium 1.87 x 10 -3 Calculate the change in [HI]. [HI] Increases by 1.87 x 10 -3 M

47. Equilibrium Stoichiometry [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 1.87 x 10 -3 Use stoichiometry to solve for the change in concentration of [H 2 ] and [I 2 ] from the given change in [HI].

48. Equilibrium Stoichiometry [H 2 ], M[I 2 ], M[HI], M Initially1.000 x 10 -3 2.000 x 10 -3 0 Change-9.35 x 10 -4 +1.87 x 10 -3 At equilibrium 6.5 x 10 -5 1.065 x 10 -3 1.87 x 10 -3 Now calculate concentrations at equilibrium

49. …and, therefore, the equilibrium constant Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = 51 = (1.87 x 10 -3 ) 2 (6.5 x 10 -5 )(1.065 x 10 -3 )

50. The Reaction Quotient ( Q )  To calculate Q, one substitutes the initial concentrations of reactants and products into the equilibrium expression.  Q gives the same ratio the equilibrium expression gives, but for a system that is not at equilibrium. Q [A], [B], [C], [D] = initial molar concentrations or partial pressures

51. If Q = K, The system is at equilibrium.

52. If Q > K, There is too much product and the reaction proceeds backwards (equilibrium shifts to the left).

53. If Q < K, There is too much reactant, and the reaction proceeds forward (equilibrium shifts to the right).

54. Summary  Q uses same as equilibrium expression but with initial concentrations or pressures  Q = K, reaction is at equilibrium  Q > K, reaction goes backwards (shifts left)  Q < K, reaction goes forward (shifts right)

55. Le Châtelier’s Principle “If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”

56. Le Chatlier’s Principle 1. System starts at equilibrium. 2. A change/stress is then made to system at equilibrium.  Change in concentration  Change in volume  Change in pressure  Change in Temperature  Add Catalyst 3. System responds by shifting to reactant or product side to restore equilibrium.

57. Le Chatlier’s Principle – basic terms  When you take something away from a system at equilibrium, the system shifts in such a way as to replace some what you’ve taken away.  When you add something to a system at equilibrium, the system shifts in such a way as to use up some of what you’ve added.

58. Le Chatlier’s Principle - Concentration  A system shifts to offset the change in concentration of reactants or products

59. Le Chatlier’s Principle - Concentration  Adding reactants or products causes a shift in the other direction to offset the added reactants or products aA + bB cC + dD Add aA + bB cC + dD Add Shifts rightShifts left

60. Le Chatlier’s Principle - Concentration  Removing reactants or products causes a shift toward the removed reactants or products to offset the change aA + bB cC + dD Remove Shifts rightShifts left

61. Le Chatlier’s Principle – Volume & Pressure  Only matters when dealing with gases  Count the total moles of gas on each side of the equation to determine shift

62. Le Chatlier’s Principle – Volume & Pressure  The system shifts to remove gases and decrease pressure.  An increase in pressure favors the direction that has fewer moles of gas.  In a reaction with the same number of product and reactant moles of gas, pressure has no effect. 62

63. Le Chatlier’s Principle – Volume & Pressure CHANGE SHIFT Increase pressure Toward side with less moles of gas Decrease volume Toward side with less moles of gas Decrease pressure Toward side with more moles of gas Increase volume Toward side with more moles of gas A (g) + B (g) C (g)

64. Le Chatlier’s Principle – Temperature  Endothermic: heat can be considered as a reactant  A + heat  B + C  Exothermic: heat can be considered as a product  A + B  C + heat  Adding heat (i.e. heating the vessel) favors away from the increase  Removing heat (i.e. cooling the vessel), favors towards the decrease

65. Le Chatlier’s Principle – Temperature  Equilibrium constant, K, is temperature dependant  Temperature is only factor that can change the value of K ChangeExothermic Increase temperatureK decreases Decrease temperatureK increases Endothermic K increases K decreases

66. Le Chatlier’s Principle – Catalysts  Does not change K  Does not shift the position of an equilibrium system  System will reach equilibrium sooner

67. Catalysts increase the rate of both the forward and reverse reactions.

68. Equilibrium is achieved faster, but the equilibrium composition remains unaltered.

69. Le Chatelier Example #1 A closed container of ice and water is at equilibrium. Then, the temperature is raised. Ice + Energy  Water The system temporarily shifts to the _______ to restore equilibrium. right

70. Le Chatelier Example #2 A closed container of N 2 O 4 and NO 2 is at equilibrium. NO 2 is added to the container. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium. left

71. Le Chatelier Example #3 A closed container of water and its vapor is at equilibrium. Vapor is removed from the system. water + Energy  vapor The system temporarily shifts to the _______ to restore equilibrium. right

72. Le Chatelier Example #4 A closed container of N 2 O 4 and NO 2 is at equilibrium. The pressure is increased. N 2 O 4 (g) + Energy  2 NO 2 (g) The system temporarily shifts to the _______ to restore equilibrium, because there are fewer moles of gas on that side of the equation. left