1. CHEMICAL EQUILIBRIUM Chapter 16
ALL BOLD NUMBERED PROBLEMS

2. Properties of an Equilibrium
Equilibrium systems are
DYNAMIC (in constant motion) & REVERSIBLE

3. CoCl42- (aq) + 6 H2O (l)  Co(H2O)62+ (aq)
Cool
Heat
Blue to pink
CoCl42- (aq) H2O (l)  Co(H2O)62+ (aq)
Cool
Pink to blue
Co(H2O)62+ (aq)  CoCl42- (aq) H2O (l)
Heat

4. Chemical Equilibrium Fe3+ + SCN- FeSCN2+
Fe(H2O)63+
Fe(SCN)(H2O)52+
+ SCN-
+ H2O
colorless
red-orange

5. Chemical Equilibrium Fe3+ + SCN- FeSCN2+
After a period of time, the concentrations of reactants and products are constant.
The forward and reverse reactions continue after equilibrium is attained.

6. Examples of Chemical Equilibria
Phase changes such as H2O(s) H2O(liq)

7. Examples of Chemical Equilibria
Acid Base Equilibria
+Na2CO3
H3O+(aq) + 2CO32-(aq) OH-(aq) + 2HCO3-(aq)
+CO2
Color
Red
Purple
Violet
Blue
Blue-Green
Green pH 2 4 6 8 10 12

8. Examples of Chemical Equilibria
Formation of stalactites (ceiling) and stalagmites (floor)
CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)

9. Chemical Equilibria
CaCO3(s) + H2O(liq) + CO2(g) Ca2+(aq) + 2 HCO3-(aq)
At a given T and P of CO2, [Ca2+] and [HCO3-] can be found from the EQUILIBRIUM CONSTANT.

10. THE EQUILIBRIUM CONSTANT
For any type of chemical equilibrium of the type a A + b B c C + d D the following is a CONSTANT (at a given T).
If K is known, then we can predict concentrations of products or reactants.

11. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl in a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.
Solution
Set up a table of concentrations:
[NOCl] [NO] [Cl2]
Initial
Change
Equilibrium

12. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K.
Solution
Set of a table of concentrations:
[NOCl] [NO] [Cl2]
Initial
Change
Equilibrium
-0.66
+0.66
+0.33
1.34
0.33

13. Determining K K = [ NO ] Cl NOCl K = [ NO ] Cl NOCl (0.66) (0.33)
2 NOCl(g) NO(g) Cl2(g)
[NOCl] [NO] [Cl2]
Initial
Change
Equilibrium K = [ NO ] 2 Cl
NOCl K = [ NO ] 2 Cl
NOCl
(0.66)
(0.33)
(1.34)
0.080

14. Writing and Manipulating K Expressions
Solids and liquids NEVER appear in equilibrium expressions.
S(s) + O2(g) SO2(g) K =
[SO 2 ] [O

15. Writing and Manipulating K Expressions
Solids and liquids NEVER
appear in equilibrium
expressions.
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq) K =
[NH 4 +
][OH - ] 3

16. Writing and Manipulating K Expressions
Changing coefficients
S(s) /2 O2(g) SO3(g)
2 S(s) O2(g) SO3(g) K =
[SO 3 ] [O 2
3/2 K
new =
[SO 3 ] 2 [O K
new =
[SO 3 ] 2 [O (K
old )

17. Writing and Manipulating K Expressions
Changing direction
S(s) + O2(g) SO2(g)
SO2(g) S(s) + O2(g) K =
[SO 2 ] [O K
new = [O 2 ]
[SO K
new = [O 2 ]
[SO 1
old

18. Writing and Manipulating K Expressions
Adding equations for reactions
S(s) + O2(g) SO2(g) K1 = [SO2] / [O2]
SO2(g) + 1/2 O2(g) SO3(g)
K2 = [SO3] / [SO2][O2]1/2
NET EQUATION
S(s) /2 O2(g) SO3(g) K
net =
[SO 3 ] [O 2
3/2 1 •

19. Manipulating K Expressions
The equilibrium expression is tied to the equation from which it is written.
Changing the equation changes the expression and thus the numerical value of K.
Three general cases need to be considered.
1. If the equation is multiplied by a number, “a”, then the K is raised to the “a” power.
2. If the equation is reversed, then the new K is the reciprocal of the old K.
3. If two equations are added, the new K is the product of the two old K's.
Using these rules, new K's can be derived for the modified equations.
SAMPLE QUESTIONS

20. Writing and Manipulating K Expressions
Concentration Units
We have been writing K in terms of mol/L.
These are designated by Kc.
But with gases, P = (n/V)•RT = [ ] • RT
P is proportional to concentration, so we can write K in terms of P.
These are designated by Kp.
Kc and Kp may or may not be the same.
Kp = Kc(RT)Dn

21. Writing and Manipulating K Expressions
Practice Problems
Write the equation and the Kc expression for the formation of two moles of gaseous ammonia from the elements in the standard state.
Kc for this reaction at 25oC is 3.5x108.
Calculate Kp for this reaction.
Calculate Kc for the reaction forming the elements from one mole of ammonia gas.

22. The Meaning of K K = [NH ] [N ][H 3.5 x 10
We can tell if a reaction is product-favored or reactant-favored.
For N2(g) H2(g) NH3(g) K c =
[NH 3 ] 2 [N
][H
3.5 x 10 8
Concentration of products is much greater than that of reactants at equilibrium.
The reaction is strongly product-favored.

23. The Meaning of K For AgCl(s) Ag+(aq) + Cl-(aq)
Kc = [Ag+] [Cl-] = 1.8 x 10-5
Concentration of products is much less than that of reactants at equilibrium.
The reaction is strongly reactant-favored.
Ag+(aq) + Cl-(aq)
AgCl(s)
is product-favored.

24. Comparing Q and K
The relative magnitudes of Q and K tell us which direction the reaction will proceed to reach equilibrium.
If Q<K, not at equilibrium and Reactants > Products.
If Q=K, the system is at equilibrium.
If Q>K, not at equilibrium and Products > Reactants.

25. Using Q
Can tell if a reaction is at equilibrium. If not, which way it moves to approach equilibrium.

26. Using Q If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?H C
H—C—C—C—C—H
H—C—C—C—H
K =
n-butane
iso-butane
[iso]
[n]
= 2.5
If [iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium?
Which way does the reaction “shift” to approach equilibrium?

27. If Q = K, then system is at equilibrium.
Using Q
In general, all reacting chemical systems are characterized by their REACTION QUOTIENT, Q. Q =
product concentrations
reactant concentrations
If Q = K, then system is at equilibrium.
[iso] = 0.35 M and [n] = 0.15 M, are you at equilibrium? Q =
conc. of iso
conc. of n
0.35
0.15
2.3
Q (2.3) < K (2.5)
Reaction is NOT at equilibrium, so [Iso] must become ________ and [n] must ____________.
larger
decrease
SAMPLE QUESTIONS

28. Using K H2(g) + I2(g) 2 HI(g) K = [HI] [H ][I ] 55.3
PROBLEM: Place 1.00 mol each of H2 and I2 in a 1.00 L flask. Calculate the equilibrium concentrations.
H2(g) + I2(g) HI(g) K c =
[HI] 2 [H
][I ]
55.3

29. H2(g) + I2(g) HI(g), Kc = 55.3
Step 1. Set up table to define EQUILIBRIUM concentrations.
[H2] [I2] [HI]
Initial
Change
Equilib
where x is defined as amount of H2 and I2 consumed on approaching equilibrium.
-x -x +2x
1.00-x 1.00-x 2x

30. H2(g) + I2(g) 2 HI(g), Kc = 55.3 K = [2x] [1.00 - x][1.00 x] 55.3 7 .
Step 2. Put equilibrium concentrations into Kc expression. K c =
[2x] 2
[1.00 -
x][1.00 x]
55.3
Step 3. Solve Kc expression - take square root of both sides. 7 . 44 = 2x
1.00 - x

31. H2(g) + I2(g) 2 HI(g), Kc = 55.3 7 . 44 = 2x 1.00 - x
Step 3. Solve Kc expression - take square root of both sides. 7 . 44 = 2x
1.00 - x
x =
Therefore, at equilibrium
[H2] = [I2] = x = M
[HI] = 2x = M

32. Sample Problem
At a certain temperature 8.00 atm of H2S(g) comes to equilibrium with H2(g) and S2(g). The equilibrium pressure of the sulfur gas is 0.60 atm.
Write the equation for the decomposition of the hydrogen sulfide, calculate the equilibrium pressure for each gas, and calculate KP.
Solution

33. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)

34. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)K c =
[NO 2 ] [N O 4
at 298 K
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Step 1. Set up an equilibrium table
[N2O4] [NO2]
Initial
Change x
Equilib x 2x -x

35. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Step 2. Substitute into Kc expression and solve. K c =
0.0059
[NO 2 ] [N O 4
(2x)
(0.50 - x)
Rearrange: ( x) = 4x2
x = 4x2
4x x = 0
This is a QUADRATIC EQUATION
ax bx c = 0
a = 4 b = c =

36. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)
Solve the quadratic equation for x.
ax bx c = 0
a = 4 b = c = x = -b b 2 -
4ac 2a x = ( .
0059 ) 2 -
4(4)( )
2(4)
x = ± 1/8(0.046)1/2 = ± 0.027

37. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)= ( .
0059 ) 2 -
4(4)( )
2(4)
x = ± 1/8(0.046)1/2 = ± 0.027
x = or
But a negative value is not reasonable.
Conclusion
[N2O4] = x = M
[NO2] = 2x = M
More Sample Problems

38. EQUILIBRIUM AND EXTERNAL EFFECTS38
EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature, catalysts, and changes in concentration affect equilibria.
The outcome is governed by LE CHATELIER’S PRINCIPLE
“...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.”

39. EQUILIBRIUM AND EXTERNAL EFFECTS
Henri Le Chatelier
Studied mining engineering.
Interested in glass and ceramics.

40. Figure 16.6
50o C
0o C
NO2 / N2O4 is temperature dependent.

41. EQUILIBRIUM AND EXTERNAL EFFECTS
Temperature change ---> change in K
Consider the fizz in a soft drink CO2(g) + H2O(liq) CO2(aq) + heat
Decrease T. What happens to equilibrium position? To value of K?
K = [CO2] / P (CO2) K increases as T goes down because [CO2] increases and P(CO2) decreases.
Increase T. Now what?
Equilibrium shifts left and K decreases.

42. Temperature Effects on Equilibrium
N2O4 (colorless) + heat
2 NO2 (brown)
DHo = kJ K c =
[NO 2 ] [N O 4
Kc = at 273 K
Kc = at 298 K

43. EQUILIBRIUM AND EXTERNAL EFFECTS
Add catalyst ---> no change in K
A catalyst only affects the RATE of approach to equilibrium.
Catalytic exhaust system

44. NH3 Production Fritz Haber, 1909
N2(g) + 3 H2(g) NH3(g)
K = 3.5 x 108 at 298 K
K = at 723 K

45. EQUILIBRIUM AND EXTERNAL EFFECTS
Concentration changes
no change in K
only the position of equilibrium changes

46. Le Chatelier’s Principle
Adding a “reactant” to a chemical system.

47. Le Chatelier’s Principle
Removing a “reactant” from a chemical system.

48. Le Chatelier’s Principle
Adding a “product” to a chemical system.

49. Le Chatelier’s Principle
Removing a “product” from a chemical system.

50. Figure 16.7 (a) 7 n-butane molecules (b) 7 iso-butane are added
(c) What are the new equilibrium concentration?

51. Butane-Isobutane EquilibriumK =
[isobutane]
[butane]
2.5
isobutane

52. Butane Isobutane
Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5
butane
isobutane

53. Butane Isobutane Q = [isobutane] [butane] 1.25 0.50 + 1.50 0.63
Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5
Solution
Calculate Q immediately after adding more butane and compare with K. Q =
[isobutane]
[butane]
1.25
0.50 +
1.50
0.63
Q is LESS THAN K. Therefore, the reaction will shift to the ____________.
Right (products)

54. Butane Isobutane
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
Solution
Q is less than K, so equilibrium shifts right — away from butane and toward isobutane.
Set up concentration table
[butane] [isobutane]
Initial
Change
Equilibrium
- x + x
x x

55. Equilibrium has shifted toward isobutane.
Butane Isobutane
You are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane.
Solution K =
2.50
[isobutane]
[butane]
1.25 + x
2.00 -
x = 1.07 M
At the new equilibrium position,
[butane] = 0.93 M and [isobutane] = 2.32 M.
Equilibrium has shifted toward isobutane.

56. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)K c =
[NO 2 ] [N O 4
at 298 K
Increase P in the system by reducing the volume.

57. Nitrogen Dioxide Equilibrium N2O4(g) 2 NO2(g)K c =
[NO 2 ] [N O 4
at 298 K
Increase P in the system by reducing the volume.
In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P).
Therefore, reaction shifts LEFT and P of NO2 decreases and P of N2O4 increases.

58. Sample Problem 1. The temperature is increased.
Write the equation for the formation of two mole of ammonia gas from the elements in the standard state. DH for this reaction is kJ. Predict the change in K and the direction of shift of an equilibrium mixture when:
1. The temperature is increased.
2. The PT is increased by adding He.
3. The total volume is increased.
4.Hydrogen gas is removed.
5.Nitrogen gas is added.
6.The temperature is decreased.
7.The total volume is decreased.
3 H2 (g) + N2 (g) <--> 2 NH3 (g) kJ
smaller, left
none, right
none, left
larger, right

59. Practice Problems 1. 2 NO (g) + O2 (g) <--> 2 NO2 (g)
a) Write the Kp expression.
b) Write the Kc expression.
c) If Kc = 0.50 at 25oC, calculate Kp.
d) Determine Kc for the following reaction:
4 NO (g) O2 (g) <--> 4 NO2 (g)
e) If 2.5 moles of NO, O2, and NO2 are placed in a 5.0 L container, is the system at equilibrium? If not, will the reaction proceed to the left or to the right?
f) If 2.5 moles of NO and O2 are placed in a 5.0 L
container at 350oC, 1.1 moles of NO2 are present at equilibrium. Calculate Kc.

60. Practice Problems
moles of carbon dioxide is placed in a 0.50 L flask.
CO2 (g) <--> C (s) + O2 (g) Kc = 1.82 at 25oC
Calculate all equilibrium concentrations.
3. Calculate all equilibrium concentrations if the initial
concentration of Cl2 is .100.
Cl2 (g) <--> 2 Cl (g) Kc = 0.036
4. Calculate the equilibrium pressure of NO2 if the initial
pressure of N2O4 is 1.0 atm at 25o C.
N2O4 (g) <--> 2 NO2 (g) Kc = 0.11

61. Practice Problems
5. The initial concentrations of CO2 and H2 are M.
Calculate all equilibrium concentrations.
CO2 (g) + H2 (g) <--> CO (g) + H2O (g) Kc = 0.64
6. BaCO3(s) + 2H+(aq) <--> H2O(l) + CO2(g) +Ba2+(aq)
Explain the effects of the following on the given equation:
a) adding HCl b) adding BaCl2
c) adding H2O d) adding BaCO3
e) adding NaOH

62. Practice Problems Answers
1. a) b)
c) d) 0.25
e) left f) 1.6
2. 1.8, , 0.052
atm , 0.044, , 0.056
6. a) right b) left c) none d) none
e) right
P2 NO2
P2 NO P O2
Kc =
[NO2]2
[NO]2[O2]
Kp =
The End!!!

63. Sample Questions Write the expression.
H2 (g) O2 (g) <--> 2 H2O (g)
Kc =
[H2O]2
[H2]2[O2]

64. Sample Questions Write the expression.
H2 (g) O2 (g) <--> 4 H2O (g)
Kc =
[H2O]4
[H2]4[O2]2

65. Sample Questions Write the expression.
H2O (g) <---> 2 H2 (g) O2 (g)
Kc =
[H2] 2 [O2]
[H2O]2

66. Sample Questions 1. 2 H2 (g) + O2 (g) <--> 2 H2O (g) Kc = [H2O]2If
= 5.0
H2 (g) O2 (g) <--> 4 H2O (g)
Kc =
[H2O]4
[H2]4 [O2]2
= 25
H2O (g) <---> 2 H2 (g) O2 (g)
Kc =
[H2]2 [O2]
[H2O]2
= 0.20

67. Sample Questions
Write the equation and the Kc expression for the formation of two moles of gaseous ammonia from the elements in the standard state.
3 H2 (g) + N2 (g) <--> 2 NH3 (g)
Kc =
[NH3] 2
[H2] 3 [N2]

68. Sample Questions Kc for this reaction at 25oC is 3.5x108.
Calculate Kp for this reaction.
3 H2 (g) + N2 (g) <--> 2 NH3 (g)
Kp = Kc (RT)Dn
Kp = 3.5 x 108 (0.0821x 298)-2
Kp = x 105

69. Sample Questions
Calculate Kc for the reaction forming the elements from one mole of ammonia gas.
NH3 (g) <--> 3/2 H2 (g) + 1/2 N2 (g)
Kc =
[H2] 3/2 [N2]1/2
[NH3] = 1
(3.5 x 108)1/2
= 5.3 x 10-5

70. The Meaning of Q Q, Reaction Quotient
Same expression as the equilibrium expression.
Non-equilibrium concentration numbers may be used.
Used to determine if a system is at equilibrium and if not, which direction it will move to reach equilibrium.

71. Sample Questions
1. If [NO2] = and [N2O4] = 0.050, is the system at equilibrium? If not, in which direction does the reaction move to come to equilibrium?
N2O4 (g) <--> 2 NO2 (g) Kc = 0.36
[NO2] 2
[N2O4]
(0.050) 2
(0.050)
Q = = =
Q < Kc forward direction (to right, products)

72. Sample Questions 2. If [NO2] = 0.50 and [N2O4] = 0.40, is the
system at equilibrium? If not, in which
direction does the reaction move to come to equilibrium?
N2O4 (g) <--> 2 NO2 (g) Kc = 0.36
[NO2] 2
[N2O4]
(0.50) 2
(0.40)
Q = =
= 0.62
Q > Kc reverse direction (to left, reactants)

73. Sample Questions 3. If [NO2] = 0.50 and [N2O4] = 0.69, is the
system at equilibrium? If not, in which
direction does the reaction move to come to equilibrium?
N2O4 (g) <--> 2 NO2 (g) Kc = 0.36
[NO2] 2
[N2O4]
(0.50) 2
(0.69)
Q = =
= 0 .36
Q = Kc equilibrium

74. Determining K 2 NOCl(g) 2 NO(g) + Cl2(g)
Place 2.00 mol of NOCl is a 1.00 L flask. At equilibrium you find 0.66 mol/L of NO. Calculate K. (Review of slides 9-11.)
Solution
Set up a table of concentrations:
[NOCl] [NO] [Cl2]
Initial
Change
Equilibrium
- 0.66
+0.66
+ 0.33
1.34
0.33

75. Determining K K = [ NO ] Cl NOCl K = [ NO ] Cl NOCl (0.66) (0.33)
2 NOCl(g) NO(g) Cl2(g)
[NOCl] [NO] [Cl2]
Before
Change
Equilibrium K = [ NO ] 2 Cl
NOCl K = [ NO ] 2 Cl
NOCl
(0.66)
(0.33)
(1.34)
0.080

76. Sample Questions
1. PCl5 dissociates to produce PCl3 and Cl2. If 2.00 moles of PCl5 is placed in a 1.00 L flask it is found that 10.0% of the original PCl5 had dissociated, at equilibrium. Calculate Kc.
PCl <--> PCl Cl2
1.80
0.200
0.200
[PCl3][Cl2]
[PCl5]
(.200)(.200)
(1.80)
Kc = = =

77. Sample Questions The End!!!
2. If 1.50 moles of H2O are placed in a 3.00 L flask 1.20 moles of H2O remain at equilibrium. Calculate Kc.
2 H2O <--> H O2
Can we use only Moles?...
0.400
0.100
0.050
No….reason is all K’s are relative, that is since all reactions are done in different containers, moles and size of the container determine the K.
[H2]2 [O2]
[H2O]2
(.100)2(.050)
(0.400)2
Kc = = =
The End!!!

78. Sample Questions
3. When mole N2O and mole O2 are placed in a 1.00 L flask it is found that there is mole NO2 at equilibrium. Calculate Kc.
2 N2O O <--> NO2
0.010
0.041
0.020
[NO2]4
[N2O]2 [O2]3
(0.020)4
(.010)2(.041)3
Kc = =
= 23

79. Sample Questions
4. Some solid ammonium hydrogen sulfide is placed in a flask at 298 K. At equilibrium, the pressure was atm. Calculate Kp and Kc.
NH4HS <--> NH H2S some
- some
some - some
0.330
0.330
Kp = PNH3PH2S = (0.330)2 = 0.109
Kp = Kc(RT) = Kc (0.0821x298)2
Kc =

80. Sample Problem 2 H2S (g) <--> 2 H2 (g) + S2 (g) 8.00 0 0 - 1.20
+ 1.20
+ 0.60
6.80
1.20
0.60
PH22 PS2
PH2S2
(1.20)2 (0.60)
(6.80)2
Kp = =
= 0.019

81. Sample Problems
1. Calculate the equilibrium pressure of CO if 25 g COBr2 is placed in a 10.0 L flask at 298 K.
CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19
25 g mole
187.8 g
= 0.13 mole
P = nRT/V
P = (0.13 mol)( L atm/molK)(298K)/10.0 L)
P = 0.32 atm

82. Sample Problems
1. Calculate the equilibrium pressure of CO if 25 g COBr2 is placed in a 10.0 L flask at 298 K.
CO (g) + Br2 (g) <--> COBr2 (g) Kp = 0.19
+ x
+ x
- x x x x
PCOBr2
PCOPBr2
( x) x2
Kp = =
= 0.19
x = atm = PCO

83. Sample Problem
2. Calculate all equilibrium concentrations if moles of HI is placed in a 2.00 L flask.
2 HI <--> H I2 Kc = 0.018
- 2x
+ x
+ x x x x
[H2][I2]
[HI]2 x2
( x)2
Kc = =
= 0.018
x = M = [H2] = [I2]
[HI] = 0.20 M

84. Sample Problem
3. Calculate all equilibrium concentrations when the initial concentrations of SO2 and NO2 are
SO2 + NO2 <--> NO + SO Kc = 85.0
0.0500
- x
- x
+ x
+ x x x x x
[NO][SO3]
[SO2][NO2] x2
( x)2
Kc = =
= 85.0
x = M = [NO] = [SO3]
[SO2] = [NO2] = M