1. The Reading is for the next class. Problems are for that day’s class. Problems for each week (MWF) are due the following Monday.

2. Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = or [C] c [D] d [A] a [B] b

3. Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = P A, P B, etc. not at equilibrium

4. Reaction quotient PCPDPCPD PAPBPAPB ab c d Q = Q will go to the value of K as the partial pressures go to equilibrium

5. Q vs K can predict where the reaction is in respect to equilibrium.

6. PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K aA + bB cC + dD

7. PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K aA + bB cC + dD Too much reactant, not enough product.

8. PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K Q > K aA + bB cC + dD

9. PCPDPCPD PAPBPAPB ab c d K = PCPDPCPD PAPBPAPB ab c d Q = Q < K Q > K aA + bB cC + dD Too much product, not enough reactant.

10. PCPDPCPD PAPBPAPB ab c d K = Q = Q > K aA + bB cC + dD Q < K Large vs c d PCPDPCPD c d PCPDPCPD ab PAPBPAPB ab PAPBPAPB

11. Q = c d PCPDPCPD ab PAPBPAPB

12. Exercise page 291 P 4 2 P 2 K = 1.39 @ 400 o C 1.40 mol P 4 1.25 mol P 2 Volume = 25.0 L

13. P 4 2 P 2 Q = (P P ) 2 2 P 4

14. P 4 2 P 2 Q = (P P ) 2 2 P 4 P = nRT V T = 673 K V = 25.0 L R = 0.0820578 L atm mol -1 K -1

15. P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = nRT V 4 = (1.4)(0.0820578)(673) 25.0 = 3.09 atm

16. P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = nRT V 2 = (1.25)(0.0820578)(673) 25.0 = 2.76 atm P P = nRT V 4 = (1.4)(0.0820578)(673) 25.0 = 3.09 atm

17. P 4 2 P 2 Q = (P P ) 2 2 P 4 P P = 2 2.76 atm P P = 4 3.09 atm = (2.76) 2 3.09 =

18. P 4 2 P 2 (P P ) 2 2 P 4 P P = 2 2.76 atm P P = 4 3.09 atm = (2.76) 2 3.09 = 2.46 Q = K = 1.39

19. P 4 2 P 2 (P P ) 2 2 P 4 P P = 2 2.76 atm P P = 4 3.09 atm = (2.76) 2 3.09 = 2.46 Q = K = 1.39 P 4 2 P 2

20. Converting between partial pressures and concentrations.

21. Converting between partial pressures and concentrations. Concentration = moles V

22. Converting between partial pressures and concentrations. Concentration = moles V PV = nRT

23. Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V PV = nRT

24. Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V = PAPA RT PV = nRT

25. Converting between partial pressures and concentrations. Concentration = moles V For gas ‘A’[A] = nAnA V = PAPA RT P A = RT[A] PV = nRT

26. P A = RT[A] 2 NO 2(g) N 2 O 4(g)

27. P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm

28. P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm P N O 24 /P ref (P NO 2 /P ref ) 2 = K

29. P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm 4 /P ref (P NO 2 /P ref ) 2 = K P N O 2 2 4 = RT[N 2 O 4 ]

30. P A = RT[A] 2 NO 2(g) N 2 O 4(g) P ref = 1 atm 4 /P ref (P NO 2 /P ref ) 2 = K P N O 2 2 4 = RT[N 2 O 4 ] P NO 2 = RT[NO 2 ]

31. 2 NO 2(g) N 2 O 4(g) 4 /P ref (P NO 2 /P ref ) 2 = K P N O 2 2 4 = RT[N 2 O 4 ] P NO 2 = RT[NO 2 ] [N 2 O 4 ](RT/P ref ) [NO 2 ] 2 (RT/P ref ) = [N 2 O 4 ] [NO 2 ] 2 x RT P ref ) ( K =

32. 2 NO 2(g) N 2 O 4(g) [N 2 O 4 ](RT/P ref ) [NO 2 ] 2 (RT/P ref ) = x K = [N 2 O 4 ] [NO 2 ] 2 [N 2 O 4 ] [NO 2 ] 2 = K RT P ref ) ( RT P ref ) (

33. [N 2 O 4 ] [NO 2 ] 2 = K RT P ref ) ( aA + bB cC + dD c = K [C] c [D] d PCPDPCPD PAPBPAPB ab d [A] a [B] b = K ( RT P ref ) a+b-c-d ?

34. [C] c [D] d [A] a [B] b = K ( RT P ref ) a+b-c-d Exercise page 297 CH 4 + H 2 O CO + 3 H 2 K = 0.172 [H 2 ]=[CO]=[H 2 O]= 0.00642 mol L -1

35. [C] c [D] d [A] a [B] b = K ( RT P ref ) a+b-c-d CH 4 + H 2 O CO + 3 H 2 K = 0.172 [H 2 ]=[CO]=[H 2 O]= 0.00642 mol L -1 (0.00642) 4 [CH 4 ](0.00642) = 0.172 ( RT P ref ) -2

36. [CH 4 ] = 0.172 ( RT P ref ) -2 CH 4 + H 2 O CO + 3 H 2 [CH 4 ] = K(RT) -2 = 3.153 x 10 -5 (0.00642) 3 3.153 x 10 -5 = 8.39 x 10 -2 mol L -1

37. Le Chatelier’s Principle

38. A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress.

39. Le Chatelier’s Principle Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation of the factor considered in the opposite direction.

40. Le Chatelier’s Principle Any system in chemical equilibrium, as a result of the variation in one of the factors determining the equilibrium, undergoes a change such that, if this change had occurred by itself, it would have introduced a variation of the factor considered in the opposite direction.

41. Le Chatelier’s Principle A system in equilibrium that is subjected to a stress reacts in a way that counteracts the stress. Stress = a factor affecting equilibrium

42. K = [C] c [D] d [A] a [B] b Anything causing a change in the concentration (or partial pressure) of a reactant or product is a stress.

43. Factors affecting equilibrium:

44. temperature

45. Factors affecting equilibrium: temperature pressure

46. Factors affecting equilibrium: temperature pressure volume

47. Factors affecting equilibrium: temperature pressure removal or addition of product volume

48. Factors affecting equilibrium: temperature pressure removal or addition of reactant volume removal or addition of product

49. Removal or addition of a reactant or product.

50. Removal or addition of a reactant or product. Increases or decreases concentration.

51. PAPA [A] PBPB [B] Removal or addition of a reactant or product. Increases or decreases concentration.

54. K = [B] [A]

55. I 2 + H 2 2 HI

56. P HI = 3.009 atm PIPI 2 = 0.4756 atm PHPH 2 = 0.2056 atm

57. I 2 + H 2 2 HI = 0.4756 atm = 0.2056 atm = 3.009 atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K

58. I 2 + H 2 2 HI = 0.4756 atm = 0.2056 atm = 3.009 atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K (3.009) 2 (0.4756)(0.2056) =

59. I 2 + H 2 2 HI = 0.4756 atm = 0.2056 atm = 3.009 atm P HI ()2)2 PIPI 2 PIPI 2 ()PHPH 2 PHPH 2 () = K (3.009) 2 (0.4756)(0.2056) = 92.60

60. I 2 + H 2 2 HI Increase P I 2 to 2.00 atm.

61. I 2 + H 2 2 HI K = 92.60 Start P 2.000 0.2056 3.009 PP 2 nd eq P Increase P I 2 to 2.000 atm

62. I 2 + H 2 2 HI K = 92.60 Start P 2.000 0.2056 3.009 PP 2 nd eq P Increase P I 2 to 2.000 atm -x -x +2x

63. I 2 + H 2 2 HI K = 92.60 Start P 2.000 0.2056 3.009 PP 2 nd eq P Increase P I 2 to 2.000 atm -x -x +2x 2.000-x 0.2056-x 3.009+2x

64. I 2 + H 2 2 HI K = 92.60 Start P 2.000 0.2056 3.009 PP 2 nd eq P Increase P I 2 to 2.000 atm -x -x +2x 2.000-x 0.2056-x 3.009+2x 92.6 = (3.009+2x) 2 (2.000-x)(0.2056-x) =

65. 92.6 = (3.009+2x) 2 (2.000-x)(0.2056-x) =

66. 92.6 = (3.009+2x) 2 (2.000-x)(0.2056-x) = (9.054 + 12.036x + 4x 2 ) (0.4112 - 2.2056x + x 2 ) = 92.6

67. 92.6 = (3.009+2x) 2 (2.000-x)(0.2056-x) = (9.054 + 12.036x + 4x 2 ) = 92.6 9.054 + 12.036x + 4x 2 = 92.6(0.4112 - 2.2056x + x 2 ) (0.4112 - 2.2056x + x 2 )

68. 92.6 = (3.009+2x) 2 (2.000-x)(0.2056-x) = (9.054 + 12.036x + 4x 2 ) = 92.6 (0.4112 - 2.2056x + x 2 ) (0.4112 - 2.2056x + x 2 ) 9.054 + 12.036x + 4x 2 = 38.08 - 204.24x + 92.6x 2 9.054 + 12.036x + 4x 2 = 92.6

69. 9.054 + 12.036x + 4x 2 = 38.08 - 204.24x + 92.6x 2 88.60x 2 - 216.276x + 29.023 = 0 x = -b  b 2 - 4 ac 2a abc =

70. 9.054 + 12.036x + 4x 2 = 38.08 - 204.24x + 92.6x 2 88.60x 2 - 216.276x + 29.023 = 0 x = -b  b 2 - 4 ac 2a abc = 0.142 or 2.30

71. Start P 2.000 0.2056 3.009 PP 2 nd eq P -x -x +2x 2.000-x 0.2056-x 3.009+2x 0.142 or 2.30x = I 2 + H 2 2 HI

72. Start P 2.000 0.2056 3.009 PP 2 nd eq P -x -x +2x 2.000-x 0.2056-x 3.009+2x 0.142 or 2.30x = I 2 + H 2 2 HI x = 2.30 would give a negative pressure for I 2

73. Start P 2.000 0.2056 3.009 PP 2 nd eq P -0.142 -0.142 0.284 1.858 0.0636 3.293 0.142 or 2.30x = I 2 + H 2 2 HI

74. Calculate K with new partial pressures.

75. Calculate K with new partial pressures. (3.293) 2 (1.858)(0.0636) =

76. Calculate K with new partial pressures. (3.293) 2 (1.858)(0.0636) = 91.77 K given = 92.60

77. Changing volume of system

78. V  1 P

79. Changing volume of system V  1 P If V is reduced, P increases. PV = nRT

80. Changing volume of system V  1 P If V is reduced, P increases. Le Chatlier’s principle requires that the equilibrium shift so that P decreases.

81. 2 NO 2(g) N 2 O 4(g)

82. 2 NO 2(g) N 2 O 4(g) An increase in pressure should favor an increase in the product, N 2 O 4.

83. 2 NO 2(g) N 2 O 4(g) An increase in pressure should favor an increase in the product, N 2 O 4. Each N 2 O 4 produced means a loss of two NO 2 molecules, a net loss of one molecule and a lower pressure.

84. 2 NO 2(g) N 2 O 4(g) A decrease in pressure favors a shift to NO 2.

85. Raising or lowering the temperature

86. Some reactions liberate heat to form products.

87. Raising or lowering the temperature Some reactions liberate heat to form products. This is an exothermic reaction.

88. Raising or lowering the temperature This is an endothermic reaction. Other reactions absorb heat to produce products.

89. Raising the temperature of an endothermic reaction will favor the formation of more product. R P

90. Raising the temperature of an exothermic reaction will favor the formation of more reactant. R P

91. NaOH (s) + H 2 O (l) Na + (aq) + OH - (aq) + H 2 O (l)

92. The solution of solid sodium hydroxide into water is exothermic.

93. K = [P] [R]

94. K = [P] [R] If a reaction is exothermic and the temperature is raised, K will decrease.

95. K = [P] [R] If a reaction is endothermic and the temperature is raised, K will increase.

96. Driving reactions to completion

97. Completion = 100% yield of product

98. Cl - (aq) + Ag + (aq) AgCl (s)

99. AgCl precipitates from the solution.

100. Cl - (aq) + Ag + (aq) AgCl (s) AgCl precipitates from the solution. As the AgCl precipitates, product is removed from solution.

102. Cl - (aq) + Ag + (aq) AgCl (s)

103. N 2 + 3 H 2 2 NH 3 All gases

104. N 2 + 3 H 2 2 NH 3 All gases exothermic

105. N 2 + 3 H 2 2 NH 3 All gases exothermic cool

106. N 2 + 3 H 2 2 NH 3 All gases exothermic cool

107. N 2 + 3 H 2 2 NH 3 Although a lower temperature favors more NH 3 formed, the lower temperature also leads to a very slow reaction.

108. N 2 + 3 H 2 2 NH 3 An increase in pressure should favor product.

109. N 2 + 3 H 2 2 NH 3 An increase in pressure should favor product.

110. N 2 + 3 H 2 2 NH 3 Ultimate solution: react at high Temperature to speed up reaction, cool until NH 3 becomes liquid. Remove from reaction vessel and repeat.

111. (time) CONCENTRATIONCONCENTRATION

112. Heterogeneous equilibrium

113. Involves at least two phases.

114. Heterogeneous equilibrium Involves at least two phases. What is the concentration of a pure liquid or a pure solid?

115. Concentrations are not a valid way to define a pure liquid or solid.

116. Concentrations are not a valid way to define a pure liquid or solid. Moles water Liters solvent =?

117. The concentration of a pure liquid or solid is defined as 1.

118. Law of Mass Action

119. 1. Gases enter equilibrium expressions as partial pressures in atmospheres.

120. Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L -1.

121. Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L -1. 3. Pure solids and liquids are represented by 1 at equilibrium, a dilute solvent is 1.

122. Law of Mass Action 1. Gases enter equilibrium expressions as partial pressures in atmospheres. 2. Dissolved species enter as concentrations in mol L -1. 3. Pure solids and liquids are represented by 1 at equilibrium, a dilute solvent is 1. 4. Partial pressures or concentrations of products appear in the numerator, reactants in the denominator. Each is raised to the power of its coefficient.

123. hemoglobin

124. Fe

125. heme Fe

126. Fe heme + O 2 Fe heme O 2

127. 4 heme Fe sites to bond O 2

128. Increase P O more Fe heme O 2 Fe heme + O 2 Fe heme O 2 2

129. 4 heme Fe sites to bond O 2 Fe heme O 2 Fe heme + O 2 Fe heme O 2 Increase P O by changing local atmosphere. 2 Increase P O more 2

130. 4 heme Fe sites to bond O 2 Decrease P O less Fe heme O 2 Fe heme + O 2 Fe heme O 2 2

131. 4 heme Fe sites to bond O 2 Decrease P O less Fe heme O 2 Fe heme + O 2 Fe heme O 2 2 Decrease P O 2 by going to higher altitude.

132. Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2

133. Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm Affinity for CO over O 2 factor of 200+.

134. Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm Affinity for CO over O 2 factor of 200+. CO 1 x 10 3 ppm < 1% P CO < 0.01 atm

135. Fe heme + CO Fe heme CO Fe heme + O 2 Fe heme O 2 POPO 2 = 0.20 atm CO 1 x 10 3 ppm < 1% P CO < 0.01 atm K heme-CO >> K heme-O 2