1. Chapter 14 Chemical Equilibrium Chemistry II

2. Speed of a chemical reaction is determined by kinetics How fast a reaction goes Extent of a chemical reaction is determined by thermodynamics How far a reaction goes

3. Hemoglobin protein (Hb) found in red blood cells that reacts with O 2 enhances the amount of O 2 that can be carried through the blood stream Hb + O 2 ⇌ HbO 2 the Hb represents the entire protein – it is not a chemical formula the ⇌ represents that the reaction is in dynamic equilibrium

4. Hemoglobin Equilibrium System Hb + O 2 ⇌ HbO 2 [Hb], [O 2 ], and [HbO 2 ] are all interdependent [Hb], [O 2 ], and [HbO 2 ] at equilibrium are controlled by the equilibrium constant, K eq the larger the value of K eq, the more product is found at equilibrium changing the concentration of any of these necessitates changing the others to reestablish equilibrium

5. O 2 Transport Hb + O2O2 ⇌ lungs, with high [O 2 ], the equilibrium shifts to combine Hb and O 2 to make more HbO 2 in the cells, with low [O 2 ] (muscles and organs use O 2 ), the equilibrium shifts to break down the HbO 2 and increase the [O 2 ] HbO 2 O 2 in lungs Hb HbO 2 O 2 in cells

6. Tro, Chemistry: A Molecular Approach6 HbF Hb Fetal Hemoglobin, HbF HbF + O 2 ⇌ HbFO 2 fetal hemoglobin’s K eq is larger than adult hemoglobin because HbF is more efficient at binding O 2, O 2 is transferred to the HbF from the mother’s hemoglobin in the placenta Hb + O2O2  HbO 2 O2O2 O2O2 Hb + O2O2  HbFO 2 O2O2

7. Tro, Chemistry: A Molecular Approach7 Oxygen Exchange between Mother and Fetus

8. Reaction Dynamics forward reaction: reactants  products therefore the [reactant] decreases and the [product] increases as [reactant] decreases, the forward reaction rate decreases reverse reaction: products  reactants assuming the products are not allowed to escape as [product] increases, the reverse reaction rate increases processes that proceed in both the forward and reverse direction are said to be reversible reactants ⇌ products

9. Hypothetical Reaction 2 Red ⇌ Blue Time[Red][Blue] 00.4000.000 100.2080.096 200.1900.105 300.1800.110 400.1740.113 500.1700.115 600.1680.116 700.1670.117 800.1660.117 900.1650.118 1000.1650.118 1100.1640.118 1200.1640.118 1300.1640.118 1400.1640.118 1500.1640.118 The reaction slows over time, But the Red molecules never run out! At some time between 100 and 110 sec, the concentrations of both the Red and the Blue molecules no longer change – equilibrium has been established. Notice that equilibrium does not mean that the concentrations are equal! Once equilibrium is established, the rate of Red molecules turning into Blue is the same as the rate of Blue molecules turning into Red

10. Hypothetical Reaction 2 Red ⇌ Blue

11. Reaction Dynamics Time Rate Rate Forward Rate Reverse Initially, only the forward reaction takes place. As the forward reaction proceeds it makes products and uses reactants. Because the reactant concentration decreases, the forward reaction slows. As the products accumulate, the reverse reaction speeds up. Eventually, the reaction proceeds in the reverse direction as fast as it proceeds in the forward direction. At this time equilibrium is established. Once equilibrium is established, the forward and reverse reactions proceed at the same rate, so the concentrations of all materials stay constant.

12. H 2 (g) + I 2 (g) ⇌ 2 HI(g) Concentration  Time  Equilibrium Established Since the [HI] at equilibrium is larger than the [H 2 ] or [I 2 ], we say the position of equilibrium favors products As the reaction proceeds, the [H 2 ] and [I 2 ] decrease and the [HI] increases Since the reactant concentrations are decreasing, the forward reaction rate slows down And since the product concentration is increasing, the reverse reaction rate speeds up Once equilibrium is established, the concentrations no longer change At equilibrium, the forward reaction rate is the same as the reverse reaction rate

13. Dynamic Equilibrium some reactions reach equilibrium only after almost all the reactant molecules are consumed – equilibrium favors the products other reactions reach equilibrium when only a small percentage of the reactant molecules are consumed – equilibrium favors the reactants

14. An Analogy: Pwad (left) ⇌ Pwad (right) Rules: 1. Each student wads up two paper wads. 2. You must start and stop as the timekeeper says. 3. Throw only one paper wad at a time. 4. If a paper wad lands next to you, you must throw it back.

15. Equal Number of Students on Each Side of the Classroom Time (s) LeftRight 0All0 5 10 15 20 Pwad (left) Pwad (right)

16. Most Students on the Left Side– 2 Students on the Right Side Time (s) LeftRight 00All 5 10 15 20 Pwad (left) ⇌ Pwad (right)

17. Most Students on the Left Side– 2 Students on the Right Side Time (s) LeftRight 0All0 5 10 15 20 Pwad (left) ⇌ Pwad (right)

18. Common Misconceptions Equilibrium means equal amounts of reactant and product. No - A reaction can be at equilibrium and have more paper wads on the product side of the room A reaction at equilibrium has stopped. No - The paper wads keep flying in both directions even after equilibrium is achieved Equilibrium can only be achieved by starting with reactants. No - Equilibrium can also be achieved when starting with all of the paper wads on the product side of the room.

20. The Equilibrium Constant [reactants] and [products] are not equal at equilibrium, there is a relationship between them Law of Mass Action: aA + bB ⇌ cC + dD, K eq (or K c ) is called the equilibrium constant Units vary from reaction to reaction, unitless in this case

21. The Equilibrium Constant so for the reaction: 2 N 2 O 5 ⇌ 4 NO 2 + O 2 Units: mol 3 /L 3

22. The Equilibrium Constant Significance of K eq K eq >> 1, when the reaction reaches equilibrium there will be more product than reactant molecules the position of equilibrium favors products K eq << 1, when the reaction reaches equilibrium there will be more reactant molecules than product molecules the position of equilibrium favors reactants

23. A Large Equilibrium Constant Remember: does not tell us about how fast, only how far

24. A Small Equilibrium Constant

25. Relationships between K and Chemical Equations reaction is written backwards, the equilibrium constant is inverted for the reaction aA + bB ⇌ cC + dD the equilibrium constant expression is: for the reaction cC + dD ⇌ aA + bB the equilibrium constant expression is:

26. Relationships between K and Chemical Equations coefficients of an equation are multiplied by a factor, K is raised to that factor for the reaction aA + bB ⇌ cC equilibrium constant expression is: for the reaction 2aA + 2bB ⇌ 2cC equilibrium constant expression is:

27. Relationships between K and Chemical Equations when you add equations to get a new equation, K eq for the new equation is the product of the equilibrium constants of the old equations for the reactions (1) aA ⇌ bB and (2) bB ⇌ cC the K expressions are: for the overall reaction aA ⇌ cC the K expression is:

28. K backward = 1/K forward, K new = K old n Ex 14.2 – Compute the equilibrium constant at 25°C for the reaction NH 3 (g) ⇌ 0.5 N 2 (g) + 1.5 H 2 (g) for N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g), K eq = 3.7 x 10 8 at 25°C K eq for NH 3 (g) ⇌ 0.5N 2 (g) + 1.5H 2 (g), at 25°C Solution: Concept Plan: Relationships: Given: Find: N 2 (g) + 3 H 2 (g)  2 NH 3 (g)K 1 = 3.7 x 10 8 K’K 2 NH 3 (g) ⇌ N 2 (g) + 3 H 2 (g) NH 3 (g) ⇌ 0.5 N 2 (g) + 1.5 H 2 (g)

29. Equilibrium Constant in Terms of Pressure Reactions Involving Gases the concentration of a gas in a mixture is proportional to its partial pressure aA(g) + bB(g) ⇌ cC(g) + dD(g) or

30. K c and K p when calculating K p, partial pressures are always in atm the values of K p and K c are not necessarily the same because of the difference in units the relationship between them is: Δn = c + d – (a + b) or no. moles of reactants minus no. Moles products When n = 0, K p = K c

31. Deriving the Relationship between K p and K c

32. Deriving the Relationship Between K p and K c for aA(g) + bB(g) ⇌ cC(g) + dD(g) substituting

33. Ex 14.3 – Find K c for the reaction 2 NO(g) + O 2 (g) ⇌ 2 NO 2 (g), given K p = 2.2 x 10 12 @ 25°C K has units L mol -1 since there are more moles of reactant than product, K c should be larger than K p, and it is K p = 2.2 x 10 12 atm -1 KcKc Check: Solution: Concept Plan: Relationships: Given: Find: KpKp KcKc 2 NO(g) + O 2 (g) ⇌ 2 NO 2 (g)  n = 2  3 = -1

34. Heterogeneous Equilibria: Reactions Involving Solids and Liquids aA(s) + bB(aq) ⇌ cC(l) + dD(aq) Concentrations of solids and liquids doesn’t change its amount can change, but the amount of it in solution doesn’t because it isn’t in solution solids and liquids are not included in the K eq expression

35. Heterogeneous Equilibria The amount of C is different, amounts of CO and CO 2 remain the same. C has no effect on the position of equilibrium.

36. Calculating Equilibrium Constants from Measured Equilibrium Concentrations To find K measure the amounts of reactants and products in a mixture at equilibrium may have different amounts of reactants and products, but the value of the equilibrium constant will always be the same as long as the temperature is kept constant

37. 37 Initial and Equilibrium Concentrations for H 2 (g) + I 2 (g) ⇌ 2HI(g) @ 445°C Initial Equilibrium Equilibrium Constant [H 2 ][I 2 ][HI][H 2 ][I 2 ][HI] 0.50 0.00.11 0.78 0.0 0.500.055 0.39 0.50 0.165 1.17 1.00.50.00.530.0330.934

38. Calculating Equilibrium Constants from Measured Equilibrium Concentrations Stoichiometry can be used to determine the equilibrium [reactants] and [products] if you know initial concentrations and one equilibrium concentration e.g.2A (aq) + B (aq) ⇌ 4C (aq) with initial concentrations [A] = 1.00 M, [B] = 1.00 M, and [C] = 0. You then measure the equilibrium concentration of C as [C] = 0.50 M. [A][B][C] initial molarity 1.00 0 change in concentration equilibrium molarity 0.50 +0.50-¼(0.50)-½(0.50) 0.75 0.88

39. Calculating Equilibrium Concentrations e.g.2A (aq) + B (aq) ⇌ 4C (aq) [A][B][C] initial molarity 1.00 0 change in concentration equilibrium molarity 0.50 +0.50-¼(0.50)-½(0.50) 0.75 0.88 Referred to as ICE table (initial, change, equilibrium) K eq = [C] 4 /[A] 2 [B] = 0.13

40. Ex 14.6  Find the value of K c for the reaction 2 CH 4 (g) ⇌ C 2 H 2 (g) + 3 H 2 (g) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M Construct an ICE table for the reaction for the substance whose equilibrium concentration is known, calculate the change in concentration +0.035 [CH 4 ][C 2 H 2 ][H 2 ] initial 0.1150.000 change equilibrium 0.035

41. Ex 14.6  Find the value of K c for the reaction 2 CH 4 (g) ⇌ C 2 H 2 (g) + 3 H 2 (g) at 1700°C if the initial [CH 4 ] = 0.115 M and the equilibrium [C 2 H 2 ] = 0.035 M use the known change to determine the change in the other materials add the change to the initial concentration to get the equilibrium concentration in each column use the equilibrium concentrations to calculate K c [CH 4 ][C 2 H 2 ][H 2 ] initial 0.1150.000 change equilibrium 0.035 +0.035 -2(0.035)+3(0.035) 0.045 0.105

42. The following data were collected for the reaction 2 NO 2 (g) ⇌ N 2 O 4 (g) at 100°C. Complete the table and determine values of K p and K c for each experiment.

45. The Reaction Quotient reaction mixture not at equilibrium; how can we determine which direction it will proceed? the answer is to compare the current concentration ratios to the equilibrium constant reaction quotient, Q for the gas phase reaction aA + bB ⇌ cC + dD the reaction quotient is:

46. The Reaction Quotient: Predicting the Direction of Change if Q > K, the reaction will proceed fastest in the reverse direction Q must decrease, the [products] will decrease and [reactants] will increase if Q < K, the reaction will proceed fastest in the forward direction Q must increase, the [products] will increase and [reactants] will decrease if Q = K, the reaction is at equilibrium the [products] and [reactants] will not change if a reaction mixture contains just reactants, Q = 0, and the reaction will proceed in the forward direction if a reaction mixture contains just products, Q = ∞, and the reaction will proceed in the reverse direction

47. 47 Q, K, and the Direction of Reaction

48. If Q = K, equilibrium; If Q K, reverse Ex 14.7 – For the reaction below, which direction will it proceed if P I2 = 0.114 atm, P Cl2 = 0.102 atm & P ICl = 0.355 atm for I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g), K p = 81.9 direction reaction will proceed Solution: Concept Plan: Relationships: Given: Find: I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g)K p = 81.9 QP I2, P Cl2, P ICl since Q (10.8) < K (81.9), the reaction will proceed to the right

49. Ex 14.8 If [COF 2 ] eq = 0.255 M and [CF 4 ] eq = 0.118 M, and K c = 2.00 @ 1000°C, find the [CO 2 ] eq for the reaction given. Units & Magnitude OK Check: Check: Round to 1 sig fig and substitute back in Solution: Solve: Solve the equilibrium constant expression for the unknown quantity by substituting in the given amounts Concept Plan: Relationships: Strategize: You can calculate the missing concentration by using the equilibrium constant expression 2 COF 2 ⇌ CO 2 + CF 4 [COF 2 ] eq = 0.255 M, [CF 4 ] eq = 0.118 M [CO 2 ] eq Given: Find: Sort: You’re given the reaction and K c. You’re also given the [X] eq of all but one of the chemicals K, [COF 2 ], [CF 4 ][CO 2 ]

50. A sample of PCl 5 (g) is placed in a 0.500 L container and heated to 160°C. The PCl 5 is decomposed into PCl 3 (g) and Cl 2 (g). At equilibrium, 0.203 moles of PCl 3 and Cl 2 are formed. Determine the equilibrium concentration of PCl 5 if K c = 0.0635

51. PCl 5 ⇌ PCl 3 + Cl 2 equilibrium concentration, M ?

52. Finding Equilibrium Concentrations When Given the Equilibrium Constant and Initial Concentrations or Pressures first decide which direction the reaction will proceed compare Q to K define the changes of all materials in terms of x use the coefficient from the chemical equation for the coefficient of x the x change is + for materials on the side the reaction is proceeding toward the x change is  for materials on the side the reaction is proceeding away from solve for x for 2 nd order equations, take square roots of both sides or use the quadratic formula may be able to simplify and approximate answer for very large or small equilibrium constants

53. [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations Construct an ICE table for the reaction determine the direction the reaction is proceeding since Q p (1) < K p (81.9), the reaction is proceeding forward

54. [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression and solve for x +2x xx xx 0.100  x 0.100+2x

55. [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute into the equilibrium constant expression and solve for x +2x xx xx 0.100  x 0.100+2x

56. [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium Ex 14.11  For the reaction I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations substitute x into the equilibrium concentration definition and solve +2x xx xx 0.100  x 0.100+2x 0.027 0.246 -0.0729 2(-0.0729)

57. [I 2 ][Cl 2 ][ICl] initial 0.100 change equilibrium -0.0729 Ex 14.11  For the reaction I 2 (g) + Cl 2 (g) ⇌ 2 ICl(g) @ 25°C, K p = 81.9. If the initial partial pressures are all 0.100 atm, find the equilibrium concentrations check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K 0.027 0.246 -0.0729 2(0.0729) K p (calculated) = K p (given) within significant figures

58. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (Hint: you will need to use the quadratic formula to solve for x)

59. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

60. 60 For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x

61. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.4980.00432 x = 0.00216 0.500  0.00216 = 0.498 [I 2 ] = 0.498 M 2(0.00216) = 0.00432 [I] = 0.00432 M  

62. Approximations to Simplify the Math when the equilibrium constant is very small, the position of equilibrium favors the reactants for relatively large initial [reactants], the [reactant] will not change significantly when it reaches equilibrium the [X] equilibrium = ([X] initial  ax)  [X] initial  we are approximating the equilibrium concentration of reactant to be the same as the initial concentration assuming the reaction is proceeding forward

63. Checking the Approximation and Refining as Necessary we can check our approximation afterwards by comparing the approximate value of x to the initial concentration if the approximate value of x is less than 5% of the initial concentration, the approximation is valid

64. Tro, Chemistry: A Molecular Approach64 [H 2 S][H 2 ][S 2 ] initial 2.50x10 -4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. Construct an ICE table for the reaction determine the direction the reaction is proceeding since no products initially, Q c = 0, and the reaction is proceeding forward

65. Tro, Chemistry: A Molecular Approach65 [H 2 S][H 2 ][S 2 ] initial 2.50x10 -4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. represent the change in the partial pressures in terms of x sum the columns to find the equilibrium concentrations in terms of x substitute into the equilibrium constant expression +x+x +2x 2x2x 2.50x10 -4  2x 2x2xx

66. 66 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx since K c is very small, approximate the [H 2 S] eq = [H 2 S] init and solve for x 2.50E-4

67. 67 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g)  2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2x2xx check if the approximation is valid by seeing if x < 5% of [H 2 S] init 2.50E-4 the approximation is not valid!!

68. 68 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx if approximation is invalid, substitute x current into K c where it is subtracted and re-solve for x new x current = 1.38 x 10 -5

69. [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx Substitute x current into K c where it is subtracted and re-solve for x new. If x new is the same number, you have arrived at the best approximation. x current = 1.27 x 10 -5 since x current = x new, approx. OK

70. Tro, Chemistry: A Molecular Approach70 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.50E-4  2x 2x2xx x current = 1.28 x 10 -5 substitute x current into the equilibrium concentration definitions and solve 2.24E-42.56E-51.28E-5

71. 71 [H 2 S][H 2 ][S 2 ] initial 2.50E-4 00 change equilibrium Ex 14.13  For the reaction 2 H 2 S(g) ⇌ 2 H 2 (g) + S 2 (g) @ 800°C, K c = 1.67 x 10 -7. If a 0.500 L flask initially containing 1.25 x 10 -4 mol H 2 S is heated to 800°C, find the equilibrium concentrations. +x+x +2x 2x2x 2.24E-4 2.56E-51.28E-5 check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated K to the given K K c (calculated) = K c (given) within significant figures

72. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? (use the simplifying assumption to solve for x)

73. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x since [I] initial = 0, Q = 0 and the reaction must proceed forward

74. Tro, Chemistry: A Molecular Approach74 For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x the approximation is valid!!

75. For the reaction I 2 (g) ⇌ 2 I(g) the value of K c = 3.76 x 10 -5 M/L at 1000 K. If 1.00 moles of I 2 is placed into a 2.00 L flask and heated, what will be the equilibrium concentrations of [I 2 ] and [I]? [I 2 ][I] initial0.5000 change-x-x+2x equilibrium0.500- x2x2x x = 0.00217 0.500  0.00217 = 0.498 [I 2 ] = 0.498 M 2(0.00217) = 0.00434 [I] = 0.00434 M 

76. Disturbing and Re-establishing Equilibrium once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same however if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established the new concentrations will be different, but the equilibrium constant will be the same unless you change the temperature

78. Le Ch âtelier’s Principle Le Châtelier's Principle: if a system at equilibrium is disturbed, the position of equilibrium will shift to minimize the disturbance disturbances all involve making the system open Concentration, temperature, volume, pressure

79. An Analogy: Population Changes When the populations of Country A and Country B are in equilibrium, the emigration rates between the two states are equal so the populations stay constant. When an influx of population enters Country B from somewhere outside Country A, it disturbs the equilibrium established between Country A and Country B. The result will be people moving from Country B into Country A faster than people moving from Country A into Country B. This will continue until a new equilibrium between the populations is established, however the new populations will have different numbers of people than the old ones.

80. The Effect of Concentration Changes on Equilibrium Adding reactant : aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance increases the amount of products until a new equilibrium is found that has the same K

81. The Effect of Concentration Changes on Equilibrium Removing product: aA + bB ⇌ cC + dD System shifts in a direction to minimize the disturbance will increase the amounts of products and decrease the amounts of reactants you can use this to drive a reaction to completion!

82. Disturbing Equilibrium: Adding or Removing Reactants after equilibrium is established, how will adding a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?

83. Disturbing Equilibrium: Adding Reactants adding a reactant initially increases the rate of forward reaction, but has no initial effect on the rate of reverse reaction. the reaction proceeds to the right until equilibrium is re-established. at the new equilibrium position, you will have more of the products than before, less of the non- added reactants than before, and less of the added reactant but not as little of the added reactant as you had before the addition at the new equilibrium position, [reactants] and [products] will be such that the value of K eq is the same

84. Disturbing Equilibrium: Adding or Removing Reactants after equilibrium is established, how will removing a reactant affect the rate of the forward reaction? How will it affect the rate of the reverse reaction? What will this cause? How will it affect the value of K?

85. Disturbing Equilibrium: Removing Reactants removing a reactant initially decreases the rate of the forward reaction, but has no initial effect on the rate of the reverse reaction. so the reaction is going faster in reverse the reaction proceeds to the left until equilibrium is re- established. at the new equilibrium position, you will have less of the products than before, more of the non-removed reactants than before, and more of the removed reactant but not as much of the removed reactant as you had before the removal at the new equilibrium position, the concentrations of reactants and products will be such that the value of the equilibrium constant is the same

86. Tro, Chemistry: A Molecular Approach86 The Effect of Adding a Gas to a Gas Phase Reaction at Equilibrium adding a gaseous reactant increases its partial pressure, causing the equilibrium to the right increasing its partial pressure increases its concentration does not increase the partial pressure of the other gases in the mixture adding an inert gas to the mixture has no effect on the position of equilibrium does not effect the partial pressures of the gases in the reaction

87. The Effect of Concentration Changes on Equilibrium N 2 O 4 (g) ⇌ NO 2 (g) Describe the effect of adding more NO 2 (g) to the following reaction:

88. The Effect of Concentration Changes on Equilibrium When NO 2 is added, some of it combines to make more N 2 O 4 Q = K Q > K Q = K

89. The Effect of Concentration Changes on Equilibrium N 2 O 4 (g) ⇌ NO 2 (g) Describe the effect of adding more N 2 O 4 (g) to the following reaction:

90. The Effect of Concentration Changes on Equilibrium When N 2 O 4 is added, reaction shifts to make more NO 2 N 2 O 4 (g) ⇌ NO 2 (g)

91. Summarizing If a chemical system is in equilibrium: Inc. [reactant] (Q < K) reaction shifts to right Inc. [product] (Q > K) reaction shifts to left Dec. [reactant] (Q > K) reaction shifts to left Dec. [product] (Q < K) reaction shifts to right

92. Effect of Volume Change on Equilibrium decreasing the size of the container increases the concentration of all the gases in the container increases their partial pressures if their partial pressures increase, then the total pressure in the container will increase according to Le Châtelier’s Principle, the equilibrium should shift to remove that pressure the way the system reduces the pressure is to reduce the number of gas molecules in the container when the volume decreases, the equilibrium shifts to the side with fewer gas molecules

93. 93 Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products. When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules – the reactant side. The Effect of Volume Changes on Equilibrium

94. Disturbing Equilibrium: Reducing the VolumeReducing the Volume for solids, liquids, or solutions, changing the size of the container has no effect on the concentration, therefore no effect on the position of equilibrium decreasing the container volume will increase the total pressure Boyle’s Law if the total pressure increases, the partial pressures of all the gases will increase  Dalton’s Law of Partial Pressures decreasing the container volume increases the concentration of all gases same number of moles, but different number of liters, resulting in a different molarity since the total pressure increases, the position of equilibrium will shift to decrease the pressure by removing gas molecules shift toward the side with fewer gas molecules at the new equilibrium position, the partial pressures of gaseous reactants and products will be such that the value of the equilibrium constant is the same

95. Summarizing If a chemical system is in equilibrium: Dec. volume causes reaction to shift in direction that has fewer moles of gas particles Inc. volume causes reaction to shift in direction that has the greater number of moles of gas particles If equal moles of gas on both sides, no effect Adding an inert gas has no effect

96. The Effect of Temperature Changes on Equilibrium Position exothermic reactions release energy and endothermic reactions absorb energy if we write Heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s Principle to predict the effect of temperature changes even though heat is not matter and not written in a proper equation

97. The Effect of Temperature Changes on Equilibrium for Exothermic Reactions for an exothermic reaction, heat is a product increasing the temperature is like adding heat according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat adding heat to an exothermic reaction will decrease the [products] and increase the [reactants] adding heat to an exothermic reaction will decrease the value of K how will decreasing the temperature affect the system? aA + bB ⇌ cC + dD + Heat

98. The Effect of Temperature Changes on Equilibrium for Endothermic Reactions for an endothermic reaction, heat is a reactant increasing the temperature is like adding heat according to Le Châtelier’s Principle, the equilibrium will shift away from the added heat adding heat to an endothermic reaction will decrease the [reactants] and increase the [products] adding heat to an endothermic reaction will increase the value of K how will decreasing the temperature affect the system? Heat + aA + bB ⇌ cC + dD

99. The Effect of Temperature Changes on Equilibrium

100. Not Changing the Position of Equilibrium - Catalysts catalysts provide an alternative, more efficient mechanism works for both forward and reverse reactions affects the rate of the forward and reverse reactions by the same factor therefore catalysts do not affect the position of equilibrium

101. Practice - Le Ch âtelier’s Principle 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g)  H° = -198 kJ How will each of the following changes affect the equilibrium concentrations of each gas once equilibrium is re-established? adding more O 2 to the container condensing and removing SO 3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding the inert gas helium to the container adding a catalyst to the mixture

102. Practice - Le Châtelier’s Principle 2 SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g) adding more O 2 to the container condensing and removing SO 3 compressing the gases cooling the container doubling the volume of the container warming the mixture adding helium to the container adding a catalyst to the mixture shift to SO 3 shift to SO 2 no effect