1. The study of reactions that occur in both directions.
Chemical Equilibrium
The study of reactions that occur in both directions.

2. So far with reactions…. Looked at reactions that go to completion
Used stoichiometry for calculation of many quantities
Looked at spontaneity and rates of reactions

3. Now…
Reactions can be reversible
They reach a state of equilibrium

4. Examples Vapor pressure Evaporation begins
Over time, the system undergoes evaporation and condensation at the same rate

5. Examples Dissolving and Crystallization
A system could have an equal amount of precipitate (changes states at an equal rate)
System of iron (II) dichromate
at equilibrium

6. Examples NO2 (g) + NO2 (g) N2O4 (g) NO2 Equilibrium!! N2O4
NO2 = dark brown, N2O4 = colorless
Ultimately ends up somewhere in between
Equlibrium Change in Action

7. Equilibrium Defined
Concentration of products and reactants remain constant over time
The reaction is reversible (can go both directions)
The rate of forward reaction equals the rate of the reverse reaction
Dynamic!! (looks the same when taking a snapshot, but constantly moving back and forth)

8. Demo, then Graph of Equilibrium
Time

9. Explaining the Graph What’s happening to the products?
What’s happening to the reactants?
Which one is favored?
Different reactions have different equilibria…
[P]
Time

10. On a molecular level…
Where does the following reach equilibrium?

11. Equilibrium Expressions
Idea by Guldberg and Waage (1864)
Called the Law of Mass Action
Given a reaction:
aA + bB  cC + dD

12. Equilibrium Expressions
aA + bB  cC + dD
MUST use
concentrations of
products over reactants K =
[C]c [D]d
[A]a [B]b
coefficients of balanced
equation become
exponents
Equilibrium
constant

13. Equilibrium ExpressionsK =
[C]c [D]d
[A]a [B]b
Works only for GASES and IONS
No pure solids or liquids included
Example
C3H8 (g) + O2 (g)  CO2 (g) + H2O (g) K =
[CO2]3 [H2O]4
[C3H8] [O2]5

14. Equilibrium Expressions
What if it has solids or liquids?
Called heterogeneous equilibrium
The concentration of solids and liquids is assumed to always remain constant, so they are not included…
Example: Ca (s) + O2 (g)  CaO (s)

15. Equilibrium Expressions
More examples
H2 (g) + I2 (s)  2 HI (g)
CuSO45 H2O (s)  CuSO4 (s) + 5 H2O (g)
N2 (g) + 3 H2 (g)  2 NH3 (g)

16. Values of K Equilibrium constant, K, is found by: [products]
[reactants]
If K = 1…..
equal ratio of products and reactants
If K > 1 ….
reaction favors products
If K < 1 ….
reaction favors reactants

17. What changes K? Change the temperature.
Equilibrium is temperature dependent.

18. What changes K? Change the reaction
Look at the original reaction and see the change that was made to it.
Whatever you do to the reaction, you do to the power of K
Example
2 NO(g) + O2 (g)  2 NO2 (g) K= 4.67 x 1013
Change the coefficients
NO(g) + 1/2 O2 (g)  NO2 (g)
What is the change?
K’ = K1/2 = (4.67 x 1013)1/2 = 6.83 x 106

19. What changes K? Change the reaction
Look at the original reaction and see the change that was made to it.
Whatever you do to the reaction, you do to the power of K
Example
2 NO(g) + O2 (g)  2 NO2 (g) K= 4.67 x 1013
Reverse the reaction
2 NO2 (g)  2 NO (g) + O2 (g)
What is the change?
K’ = 1/K = 1/(4.67 x 1013) = 2.14 x 10-14

20. Equilibrium Constant Can also find from reaction mechanism OVERALL:
N2O(g) + 3/2 O2 (g)  2 NO2 (g) Kc = ??
Steps:
N2O(g) + 1/2 O2 (g)  2 NO (g) Kc1 = 1.7 x 10-13
2 NO(g) + O2 (g)  2 NO2 (g) Kc2 = 4.67 x 1013
To find Kc?
Kc = Kc1 x Kc2

21. Equilibrium including Gases
Look at partial pressures instead of molarities
aA (g) + bB (g)  cC (g) + dD (g) Kp =
(PC)c (PD)d
(PA)a (PB)b

22. Equilibrium including Gases
If it says Kp, you must use pressures
Given moles and L, must use temperature and PV = nRT to get pressures of each species
K or Kc is still molarity

23. Practice Problems Write equilibrium expressions for:
a) 2 O3(g)  3 O2(g)
b) 2 NO(g) + Cl2(g)  2 NOCl(g)
c) BaSO4(s)  Ba+2(aq) + SO4-2(aq)

24. Practice Problems Changing K by changing the equation
Original equation:
2 NO (g) + O2 (g)  2 NO2 (g) Keq =
NO (g) + ½ O2 (g)  NO2
2 NO2 (g)  2 NO (g) + O2 (g)

25. Equilibrium Calculations

26. Value of the Equilibrium Constant
K tells where the equilibrium lies
How likely (to what extent) the reaction is to occur
Does not give rate (that’s KINETICS)
Three possibilities
K = 1
Equal ratio of products to reactants
Equilibrium lies in the “middle”

27. Value of the Equilibrium Constant
K tells where the equilibrium lies
How likely (to what extent) the reaction is to occur
Does not give rate (that’s KINETICS)
Three possibilities
K >1
Bigger number on top, so…
More products
Equilibrium favors the PRODUCTS (to the right)

28. Value of the Equilibrium Constant
K tells where the equilibrium lies
How likely (to what extent) the reaction is to occur
Does not give rate (that’s KINETICS)
Three possibilities
K <1
Bigger number on bottom, so…
More reactants
Equilibrium favors the REACTANTS (to the left)

29. Reaction Quotient (Q)
Where you are at
RIGHT NOW!!!

30. Reaction Quotient (Q) Q Snapshot of reaction at some point
Same calculation as Kc
They will give you the K of the reaction
Compare Q and K Q =
[C]c [D]d
[A]a [B]b

31. Reaction Quotient (Q) Q = 1.5 atm Example: [N2O4] = [NO2]2
2 NO2 (g)  N2O4 (g) Kp = 100°C
0.200 moles of NO2 are mixed with moles of N2O4 in a 4.00 L flask at 100°C. Q =
[N2O4]
[NO2]2
= 1.5 atm

32. Reaction Quotient (Q) Where am I?? Compare Q to K to see where you are
Kp = 11, Q = 1.5
Compare Q to K to see where you are
3 situations
Q = K
At EQUILIBRIUM!!

33. Reaction Quotient (Q) Where am I?? Compare Q to K to see where you are
Kp = 11, Q = 1.5
Compare Q to K to see where you are
3 situations
Q > K
Bigger number, so…
Favors products
Must shift left (towards reactants to get to equilibrium)

34. Reaction Quotient (Q) Where am I?? Compare Q to K to see where you are
Kp = 11, Q = 1.5
Compare Q to K to see where you are
3 situations
Q < K
Smaller number, so…
Favors reactants
Must shift right (towards products to get to equilibrium)

35. IMPORTANT PROBLEM Calculate concentrations at equilibrium
Use an ICE table

36. ICE Problems Steps for doing an ICE problem Write a balanced equation
Write your equilibrium expression
Calculate Q and see which way it will shift/change to reach equilibrium (usually don’t have to do this if just given reactants initially)

37. ICE Problems I = initial concentration Set up ICE
+ C = change in concentration
E = equilibrium concentration

38. ICE Problems Solve for x Find equilibrium concentrations
Check by plugging in equilibrium concentrations (it should equal K – or really close)

39. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
Carbon monoxide reacts with steam to produce carbon dioxide and hydrogen. At 700 K the equilibrium constant is Calculate the equilibrium concentrations of all species if mol of each component is mixed in a L flask.

40. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10, [] of all is M (NOT at equilibrium)
Write a balanced equation.
CHECK!

41. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10, [] of all is M (NOT at equilibrium)
Write the equilibrium expression.
Also expression for Q…..
K =
[CO2] [H2]
[CO] [H2O]

42. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10, [] of all is M (NOT at equilibrium)
Solve for Q and see where you are.
Q =
[1.000] [1.000]
= 1.000
Q is LESS than K, so we are on the reactants side,
and we need to shift to products.

43. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
Set up ICE!!!!! I +C E
-x x x x
x x x x

44. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10
Plug into expression, solve for x
5.10 =
( x)( x)
(1.000 – x)(1.000 – x)
( x)2
(1.000 – x)2 =
Nice!! Makes this
much less ugly

45. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10
Plug into expression, solve for x
5.10 =
( x)
x = 0.387
(1.000 – x)
Just solved for the CHANGE

46. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
x = 0.387
Find equilibrium concentrations.
[CO] = x = M
[H2O] = – x = M
[CO2] = x = M
[H2] = x = M

47. ICE Example – fairly simple
CO (g) + H2O (g)  CO2 (g) + H2 (g)
K = 5.10
Check your equilibrium constant.
K =
[1.387] [1.387]
[0.613] [0.613]
= 5.10
WOO HOO!!

48. Given Equilibrium Condition
Sulfur trioxide decomposes in a sealed container: 2 SO3 (g)  2 SO2 (g) + O2 (g). Initially, the vessel is charged at 1000 K with SO3 (g) at a partial pressure of atm. At equilibrium, the partial pressure of SO3 (g) is atm. Calculate the Kp at 1000 K.

49. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
SO3 is initially atm, and atm at equilibrium
Write a balanced equation.
CHECK!

50. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
SO3 is initially atm, and atm at equilibrium
Write the equilibrium expression.
Don’t need to do Q since starting with just reactants
K =
[SO2]2 [O2]
[SO3]2

51. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
Set up ICE!!!!! I +C E
-2x x x
x x
Stoichiometric
relationship!!

52. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
Solve for x
0.500 – 2x = 0.200
2x = 0.300
x = 0.150

53. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
Find equilibrium pressures.
[SO3] = atm (given)
[SO2] = 2x = atm
[O2] = x = atm

54. Given Equilibrium Condition
2 SO3 (g)  2 SO2 (g) + O2 (g)
Solve for your equilibrium constant.
K =
[0.300]2 [0.150]
[0.200]2
= 0.338

55. ICE Example – Uglier…
Suppose mol of H2 and mol of F2 are mixed in a L flask. Assume Kc = 1.15 x 102. Find the equilibrium concentration of each species.

56. K = ICE Example – Uglier… Write a balanced equation.
H2 (g) + F2 (g)  2 HF (g)
Write the equilibrium expression.
K =
[HF]2
[H2] [F2]

57. ICE Example – Uglier… Solve for Q and see where you are.
Don’t need to!! The reaction only has H2 and F2, so it hasn’t even started yet. It HAS to start making products.

58. ICE Example – Uglier… I +C E Set up ICE!!!!! (need molarities….)
H2 (g) + F2 (g)  2 HF I +C E
Stoichiometric
relationship!!
-x x
+2x
x x x

59. ICE Example – Uglier… H2 + F2  2 HF 1.15 x 102 = K = 1.15 x 102
Plug into expression, solve for x
Rearrange and solve with the
QUADRATIC EQUATION
1.15 x 102 =
(2x)2
(1.000 – x)(2.000 – x)

60. ICE Example – Uglier… H2 + F2  2 HF K = 1.15 x 102
Plug into expression, solve for x
111 x2 – 345 x = 0
x = OR x = 0.968
Doesn’t make sense,
H2 was only M to begin with…
This one looks good!!!
We’ll use it!!!!

61. ICE Example – Uglier… H2 + F2  2 HF x = 0.968
Find equilibrium concentrations.
[H2] = – = M
[F2] = – = M
[HF] = 0 + 2(0.968) = 1.94 M

62. K = ICE Example – Uglier… = 115 H2 + F2  2 HF YES!!!! K = 1.15 x 102
Check your equilibrium constant.
K =
[1.94]2
[0.032] [1.032]
= 115
YES!!!!

63. ICE Example – A nicer way …
Gaseous NOCl decomposes to form the gases NO and Cl2. At 35°C, the equilibrium constant is 1.6 x In an experiment where 1.0 mol NOCl is placed in a 2.0 L flask, what the equilibrium concentrations?

64. ICE Example – A nicer way…
Write a balanced equation.
2 NOCl (g)  2 NO (g) + Cl2 (g)
Write the equilibrium expression.
K =
[NO]2 [Cl2]
[NOCl]2

65. ICE Example – A nicer way…
Solve for Q and see where you are.
Don’t need to!! The reaction only has H2 and F2, so it hasn’t even started yet. It HAS to start making products.

66. ICE Example – A nicer way…
Set up ICE!!!!! (need molarities….)
2 NOCl (g)  2 NO (g) + Cl2 (g) I +C E
- 2x x x
0.50 – 2x x x

67. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
K = 1.6 x 10-5
Plug into expression, solve for x
1.6 x 10-5 =
(2x)2 (x)
(0.50 – 2x)2
Ewwww……

68. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
K = 1.6 x 10-5
Plug into expression, solve for x
1.6 x 10-5 =
(2x)2 (x)
(0.50 – 2x)2
Assume that since K is small, the change
in the reactants will be small enough to
assume it is ZERO

69. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
K = 1.6 x 10-5
Plug into expression, solve for x
1.6 x 10-5 =
4x3
(0.50)2
x = 1.0 x 10-2
MUCH nicer…..

70. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
K = 1.6 x 10-5
Plug into expression, solve for x
x = 1.0 x 10-2
We should check the approximation using the 5% rule
x should be less than 5% of the initial concentration

71. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
x = 1.0 x 10-2 = 0.010
Find equilibrium concentrations.
[NOCl] = 0.50 M
[NO] = 2x = 2(0.010) = M
[Cl2] = x = M

72. ICE Example – A nicer way…
2 NOCl (g)  2 NO (g) + Cl2 (g)
K = 1.6 x 10-5
Check your equilibrium constant.
K =
[0.020]2 [0.010]
[0.50]2
= 1.6 x 10-5
SUPER!!!

73. Solubility & Equilibrium
Ksp

74. Solubility & Equilibrium
The beginning
XY(s) ==> X+ + Y-

75. Solubility & Equilibrium
The beginning
XY(s) ==> X+ + Y-
As time continues the [ions] begins to increase.

76. Solubility & Equilibrium
Greater the chance that they will collide and reform
XY(s) <== X+ + Y-
When it becomes saturated it reaches equilibrium.
XY(s) <==> X+ + Y-

77. Solubility & Equilibrium
Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
 Solubility = molar solubility = mol/L
Be careful ~ sometimes given g/L or have to solve
for g/L

78. Solubility & Equilibrium
Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
1) Just ions, no solids

79. Solubility & Equilibrium
Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
2) It doesn’t matter if you have stuff on the bottom.

80. Solubility & Equilibrium
Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
3) Not solubility - that’s an equilibrium position(Q) (will more dissolve or not?)

81. Solubility & Equilibrium
Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
4) Equilibrium so remember ICE

82. Ksp Calculations
The solubility of copper (I) bromide is 2.0 x 10-4 mol/L at 25 C. Calculate Ksp.

83. Ksp Calculations
The solubility of copper(I) bromide is 2.0 x 10-4 mol/L at 25°C. Calculate Ksp.
CuBr(s) <==> Cu Br-1 I
C x x
E x x 10-4

84. Ksp Calculations
The solubility of copper(I) bromide is 2.0 x 10-4 mol/L at 25°C. Calculate Ksp.
CuBr(s) <==> Cu Br-1
Ksp = [2.0 x10-4][2.0 x 10-4]
= 4.0 x (no units for K)

85. Ksp Calculations – Your Turn
The solubility of silver phosphate is x 10-5 M. Calculate its Ksp.

86. Ksp Calculations Cu(IO3)2(s) <==> Cu+2 + 2 IO3- I
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3- I

87. Ksp Calculations Cu(IO3)2(s) <==> Cu+2 + 2 IO3- I 0 0 C
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3- I C

88. Ksp Calculations Cu(IO3)2(s) <==> Cu+2 + 2 IO3- I 0 0 C +x +2x E
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3- I
C x x E

89. Ksp Calculations Cu(IO3)2(s) <==> Cu+2 + 2 IO3- I 0 0 C +x +2x
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3- I
C x x
E x x

90. Ksp Calculations Cu(IO3)2(s) <==> Cu+2 + 2 IO3-
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
Ksp = [Cu+2][IO3-]2

91. Ksp Calculations
The Ksp value of copper(II) iodate is x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
Ksp = [Cu+2][IO3-]2
Ksp = [x][2x]2

92. Ksp Calculations
The Ksp value of copper(II) iodate is 1.4 x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
Ksp = [Cu+2][IO3-]2
Ksp = [x][2x]2
1.4 x 10-7 = [x][2x]2

93. Ksp Calculations
The Ksp value of copper(II) iodate is 1.4 x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
Ksp = [Cu+2][IO3-]2
Ksp = [x][2x]2
1.4 x 10-7 = [x][2x]2
1.4 x 10-7 = 4x3

94. Ksp Calculations x = 3.3 x 10-3 mol/L
The Ksp value of copper(II) iodate is 1.4 x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
Ksp = [Cu+2][IO3-]2
Ksp = [x][2x]2
1.4 x 10-7 = [x][2x]2
1.4 x 10-7 = 4x3
x = 3.3 x 10-3 mol/L

95. Ksp Calculations
The Ksp value of copper(II) iodate is 1.4 x 10-7 at 25°C. Calculate its solubility.
Cu(IO3)2(s) <==> Cu IO3-
[Cu+2] = 3.3 x 10-3 mol/L
[IO3-] =2(3.3 x 10-3 mol/L) = 6.6 x 10-3 mol/L

96. Ksp Calculations – Your Turn
The Ksp of mercury (II) sulfide is x Find its molar solubility.

97. Ksp Calculations What does it all mean?
According to Ksp, EVERYTHING is at least very, very, very, very slightly soluble.
Looking at the Ksp can help you figure out the solubility of compounds compared to each other – with a couple key rules about doing it!

98. Ksp Calculations What does it all mean?
Salts that produce the same # of ions can be compared to see which one is the most/least soluble.
AgBr vs AgI vs AgCl
Ksp values 5.0x x x10-10
Appendix A25

99. Ksp Calculations What does it all mean?
If the salts break into different # of ions, you can’t just look. Must calculate.

100. Ksp Calculations 2 example problems: Which is more soluble:
copper (II) carbonate or cadmium carbonate?
Ksp = 2.5 x Ksp = 5.2 x 10-12

101. Ksp Calculations 2 example problems: Which is more soluble:
tin (II) hydroxide or strontium phosphate?
Ksp = 3.0 x Ksp = 1.0 x 10-31