The document discusses chemical equilibrium. It begins by defining chemical equilibrium as a state where the forward and reverse reaction rates are equal, but the reactions are still occurring dynamically. It also notes that at equilibrium, the concentrations or pressures of all species remain constant over time. The document then provides the definitions and expressions for equilibrium constants Kc and Kp, which relate the concentrations or pressures of reactants and products at equilibrium. It also discusses how equilibrium positions can be manipulated by changing conditions based on Le Chatelier's principle.

1. Chemical Equilibrium
When a system is at equilibrium, the forward and
reverse reaction are proceeding at the same rate
The concentrations of all species remains constant
over time, but both the forward and reverse
reaction never cease
That means this is a dynamic equilibrium, not
like a beach ball on a seal’s nose.
Equilibrium is signified with double arrows or
the equal sign
Symbol found in “Arrows” section of Insert
equations in Microsoft Office.

3. Dynamic Equilibrium

4. For given overall system composition
Always reach same equilibrium concentrations
Whether equilibrium is approached from
forward or reverse direction

5. Mass Action Expression

6. Mass Action Expression
Uses stoichiometric coefficients as exponent for
each reactant
For reaction: aA + bB
cC + dD
c
d
[C ] [D ]
Q=
b
a
[ A] [B ]
Reaction quotient
Numerical value of mass action expression
Equals “Q ” at any time, and
Equals “K ” only when reaction is known to be at
equilibrium
6

7. Calculation Results
Q = __[HI]2_ = same for all = K if all
[H2][I2] experiments at equilibrium
Expt
I
II
III
IV
[H2]
.0222
.0350
.0150
.0442
[I2]
.0222
.0450
.0135
.0442
[HI]
.156
.280
.100
.311
Kc
49.4
49.8
49.4
49.5

8. Equilibrium established

9. Equilibrium Law
At 440oC equilibrium law:
Kc = _[HI]2_ = 49.5
[H2][I2]
Equilibrium constant = Kc at given T
Use Kc as normally working with molarity
At equilibrium Q = Kc

10. Predicting Equilibrium Law

11. Predicting Equilibrium Law
Where only concentrations that satisfy this
equation are equilibrium concentrations
Numerator
Multiply [products] raised to their
stoichiometric coefficients
Denominator
Multiply [reactants] raised to their
stoichiometric coefficients
Kc = [products]f
[reactants]d

12. Check
3H2(g) + N2(g)
2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?

13. Check
3H2(g) + N2(g)
2NH3(g)
Kc = 4.26 × 108 at 25 °C
What is equilibrium law?
2
Kc =
[NH3 ]
3
[H2 ] [N2 ]
= 4.26 ´ 10
8

14. Example
In the reaction:
N2O4
Q = [NO2]2
[N2O4]
2 NO2

15. Check
Which of the following is the correct mass
action expression for the reaction:
Cu2+(aq) + 4NH3(aq)
[Cu(NH3)42+](aq)?
A.
B.
C.
Q =
Q=
[Cu(NH3 )2+ ]
4
[Cu2+ ][NH3 ]4
[Cu(NH3 )2+ ]
4
[Cu2+ ][NH3 ]
Q =
[Cu2+ ][NH3 ]4
[Cu(NH3 )2+ ]
4
D. none of these
15

16. Manipulating Equations

17. Reverse Equilibrium

18. Kc Related to Equations II

19. Kc Related to Equations III

20. Example

21. Adding Equations

22. Adding Equations

23. Adding Equations II

24. Adding Equations III
Adding equations gives the overall equation
and multiplying K values gives the overall
Kc:Kc =
[CO2] x [H2][O2]½ = 1.9 x 105
[CO][O2]½
[H2O]
Kc = (5.7 x 1045) x (3.3 x 10-41) = 1.9 x 105

25. Practice 1

26. Practice 2

27. Equilibrium Constant, Kc
Constant value equal to ratio of product
concentrations to reactant concentrations raised
to their respective exponents
Kc = [products]f
[reactants]d
Changes with temperature (van’t Hoff Equation)
Depends on solution concentrations
Assumes reactants and products are in solution

28. Equilibrium Constant, Kp
Based on reactions in which substances are
gaseous
Assumes gas quantities are expressed in
atmospheres in mass action expression
Use partial pressures for each gas in place of
concentrations
Ex. N2 (g) + 3 H2 (g)  2 NH3 (g)
Kp = _P2NH3
PN2.P3

29. How are Kp and Kc Related?
Start with ideal gas law
PV = nRT
Rearranging gives
æn ö
P = ç ÷ RT = MRT
èV ø
Substituting P/RT for molar concentration into Kc
results in pressure-based formula
∆n = moles of gas in product – moles of gas in reactant
Kp = Kc(RT )
Dn
Kp = Kc when Δn = 0
29

30. Check
Consider the reaction A(g) + 2B(g) 4C(g)
If the Kc for the reaction is 0.99 at 25ºC,
what would be the Kp?
A. 0.99
B. 2.0
C. 24.
D. 2400
E. None of these

31. Solution

32. Check 2

33. Solution 2

34. Practice 3

35. Heterogeneous Equilibrium
Homogeneous reaction/equilibrium
All reactants and products in same phase
Can mix freely
Heterogeneous reaction/equilibrium
Reactants and products in different phases
Can’t mix freely
Solutions are expressed in M
Gases are expressed in M
Governed by Kc

36. Heterogeneous Equilibrium II

37. Heterogeneous Equilibrium III
2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)
Kc = [Na2CO3(s)] [H2Og][CO2(g)] = [H2Og][CO2(g)]
[NaHCO3(s)]2
1 mol NaHCO3 , V = 38.9 mL
M = 1 mol/.0389 L = 25.7 M
2 mol NaHCO3 , V = 77.8 mL
M = 2 mol/.0778 L = 25.7 M

38. Check
Write equilibrium laws for the following:
Ag+(aq) + Cl– (aq)  AgCl(s)
Kc = ____1___
[Ag+][Cl-]
H3PO4(aq) + H2O(ℓ)  H3O+(aq) + H2PO4(aq)
Kc = [H3O+][H2PO4-]
[H3PO4]

39. Check
Given the reaction:
3Ca2+(aq) + 2PO43–(aq)
Ca3(PO4)2(s)
What is the mass action expression?
A.
Q =
B.
Q =
C.
Q =
D.
Q =
[Ca2+ ]3 [PO3- ]2
4
[Ca3 (PO 4 )2 ]
[Ca2+ ]3 [PO3- ]2
4
[1]
[Ca3 (PO 4 )2 ]
[Ca2+ ]3 [PO3- ]2
4
[1]
[Ca2+ ]3 [PO3- ]2
4
39

40. Large Kc
Kc >> 1
Rich in product
Goes toward complete
2SO2(g) + O2(g)  2SO3(g)
Kc = 7.0  1025 at 25 ° C

41. Small Kc
Kc << 1
Rich in reactant
Few products
H2(g) + Br2(g)  2HBr(g)
Kc = 1.4  10–21 at 25 °C

42. K near 1

43. Check
43

44. Equilibrium and Changes
Equilibrium positions
Combination of concentrations that allow Q = K
Infinite number of possible equilibrium
positions
Le Châtelier’s principle
System at equilibrium (Q = K) when upset by
disturbance (Q ≠ K) will shift to offset stress

45. Reaction Direction
Can us value of Q to predict reaction
direction
1. Qc > Kc reaction goes to left; reverse
2. Qc < Kc reaction goes to right; forward
3. Qc = Kc reaction at equilibrium.
4. If only reactants Q = 0; forward
5. If only reactants Q = ∞; reverse

46. Relationship Between Q and K
Q=K
Q<K
reaction at equilibrium
reactants go to products
Too many reactants
Must convert some reactant to product to move
reaction toward equilibrium: shift to right
Q>K
products go to reactants
Too many products
Must convert some product to reactant to move
reaction toward equilibrium: shift to left
46

47. Le Châtelier’s Principle
1.
2.
3.
4.
5.
Concentration
Pressure and volume
Temperature
Catalysts
Adding inert gas to system at constant
volume

48. 1. Effect of Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq)
CuCl42–(aq) + 4H2O
blue
yellow
Equilibrium mixture is blue-green
Kc 
[CuCl2  (aq )][H2 O]4
4
[Cu(H2 O)2  (aq )][Cl (aq )]4
4

Kc 
Kc
[H2 O]
4

[CuCl2  (aq )]
4
[Cu(H2 O)2  (aq )][Cl (aq )]4
4
Add excess Cl– (conc HCl)
Equilibrium shifts to products
Makes more yellow CuCl42–
Solution becomes green
48

49. 1. Effect of Change in Concentration
Cu(H2O)42+(aq) + 4Cl–(aq)
blue
CuCl42–(aq) + 4H2O
yellow
Kc 
2
[CuCl4 (aq )]
[Cu(H2 O)2  (aq )][Cl (aq )]4
4
Add Ag+
Removes Cl–:
Ag+(aq) + Cl–(aq)  AgCl(s)
Equilibrium shifts to reactants
Makes more blue Cu(H2O)42+
Solution becomes bluer
Add H2O?
49

50. Addition of Product
Add NO2(g) to N2O4(g) 2NO2(g)
System is disturbed so Q now > K; system
responds by converting more NO2(g) into
N2O4(g) until again Q = K but with larger
values for all concentrations.

51. Check
2SO2(g) + O2(g) → 2SO3(g)
Kc = 2.4 x 10-3 at 700 oC
Which direction will the reaction move if
0.125 moles of O2 is added to an
equilibrium mixture ?
A. Towards the products
B. Towards the reactants
C. No change will occur

52. Concentration Effect Summary
When changing concentrations of reactants
or products
Equilibrium shifts to remove reactants or
products that have been added
Equilibrium shifts to replace reactants or
products that have been removed

53. Effect of Change in P & V

54. Effect of Change in P & V
If Vexpect P
To reduce P, look at each side of reaction
Which has less moles of gas
Reactants = 3 + 1 = 4 moles gas
Products = 2 moles gas
Reaction favors products (shifts to right)

55. Effect of Change in P & V II
2NaHSO3(s)  NaSO3(s) + H2O(g) + SO2(g)
Kp = PH2O.PSO2
If you V of reaction,
Reactants: no moles gas = all solids
Products: 2 moles gas
V, causes P
Reaction shifts to left (reactants), as this
has fewer moles of gas

56. Effect of Change in Temperature
Cu(H2O)42+(aq) + 4Cl–(aq) + Heat 
blue
CuCl42–(aq) + 4H2O
yellow
Reaction endothermic
Adding heat shifts equilibrium toward products
Cooling shifts equilibrium toward reactants
56

57. Temperature Change
H2O(s)  H2O(ℓ)
H° =+6 kJ (at 0 °C)
Energy + H2O(s)  H2O(ℓ)
Energy is reactant
Add heat, shift reaction right
3H2(g) + N2(g)  2NH3(g) Hf°= –47.19 kJ
3 H2(g) + N2(g) 
2 NH3(g) + energy
Energy is product
Add heat, shift reaction left

58. Effect of T Change
 T shifts reaction in direction that produces
endothermic (heat absorbing) change
 T shifts reaction in direction that produces
exothermic (heat releasing) change
Changes in T change value of mass action
expression at equilibrium, so K changed

59. Effect of T Change II
K depends on T
T of exothermic reaction makes K smaller
More heat (product) forces equilibrium to
reactants
T of endothermic reaction makes K larger
More heat (reactant) forces equilibrium to
products

60. Catalysts
Catalysts lower Ea
Change in Ea has
equal effect on kf & kr
Result is no effect
on K

61. Addition of Inert Gas
Will not affect concentrations or partial
pressures of any components
Will not affect reaction

62. Learning Check:
Consider:
H3PO4(aq) + 3OH–(aq)
Q
3H2O(ℓ) + PO43–(aq)
3
[PO4 ]
 3
[OH ] [H3PO4 ]
What will happen if PO43– is removed?
 Q is proportional to [PO43– ]
  [PO43– ],  Q
 Q < K equilibrium shifts to right
62

63. Learning Check:
The reaction
H3PO4(aq) + 3OH–(aq)
3H2O(aq) + PO43–(aq)
is exothermic.
What will happen if system is cooled?
Q




heat
3
[PO4 ]
[OH ]3 [H3PO4 ]
Since reaction is exothermic, heat is product
Heat is directly proportional to Q
T,  Q
Q < K equilibrium shifts to right
63

64. Check
The equilibrium between aqueous cobalt ion and
the chlorine ion is shown:
[Co(H2O)6]2+(aq) + 4Cl–(aq) 
[Co(Cl)4]2–(aq) + 6H2O(ℓ)
blue
It is noted that heating a pink sample causes it to
turn violet. The reaction is:
A. endothermic B. exothermic
C. cannot tell from the given information
pink

65. Calculating Kc from [Equilibrium]
Most direct way.
Measure all reactant & product concentrations.
Kc always be the same as long as T constant
65
Tro: Chemistry: A Molecular Approach, 2/e

66. Equilibrium Calculations
For solution reactions, must use KC
For gaseous reactions, use either KP or KC
Calculate K from equilibrium [X] or PX
Find one or more equilibrium [X] or PX
from KP or KC

67. Kc From [X}s at Equilibrium
Use mass action expression to relate [X]
values
Can use:
Equilibrium [reactant]s & [product]s
From initial and one final [product] find
remaining {X}s then find Kc

68. Kc from Equilibrium [X]
1 N2O4(g)  2NO2(g)
If you place 0.0350 mol N2O4 in 1 L flask at
equilibrium, what is KC?
[N2O4]eq = 0.0292 M
[NO2]eq = 0.0116 M
Kc = _[NO2}2 = (0.0292)2 = 4.61 x 10-3
[N2O4]
(0.0292)

69. Practice 4
For the reaction: 2A(aq) + B(aq)  3C(aq)
the equilibrium concentrations are: A = 2.0 M, B
= 1.0 M and C = 3.0 M. What is the expected
value of Kc at this temperature?
A. 14
B. 0.15
C. 1.5
D. 6.75

70. From [X]i and [Y]f
Set up ICE table
Changes in same ratio as coefficients
[X]eq = [X]initial - [X]change
1. Assign x as coefficient, + for materials on side
it is going to or – for opposite
2. Solve for x, if 2nd order, take square root, use
quadratic equation or simplify if K very large
or very small.

71. Generic
Use concentration tables set up as follows:
A + B  2C
[A]
[B]
[C]
I
P
Q
R
C
-x
-x
+2x
E
P–x
Q–x
R + 2x
Kc = _ (R + 2x)2__
(P - x)(Q - x)

72. Example
2SO2(g) + O2(g)  2SO3(g)
1.000 mol SO2 and 1.000 mol O2 are placed in a
1.000 L flask at 1000 K. At equilibrium 0.925 mol
SO3 has formed. Calculate KC for this reaction.
[SO2]i = [O2]i = 1.00 mol = 1.00 M
1.00 L
[SO3]eq = 0.925 mol = 0.925 M
1.00 L

73. ICE Table
2SO2(g) + O2(g)  2SO3(g)
I 1.00
1.00
0.000
C -0.925
-0.462
+0.925
E 0.075
0.538
0.925
Kc = __[SO3]2__ = _(0.925)2__ 2.8 x 102
[SO2]2[O2] (.075)2(.538)

74. Practice 5
CH4(g) + H2O(g)  CO(g) + 3H2(g) Kc = 5.67
[CH4] = 0.400 M; [H2] = 0.800M; [CO] =0.300M
What is [H2O] at equilibrium?

75. Practice
H2(g) + I2(g)  2HI(g) at 425 °C
KC = 55.64
If one mole each of H2 and I2 are placed in a 0.500
L flask at 425 °C, what are the equilibrium
concentrations of H2, I2 and HI?
Kc = __[HI]2_= 55.64
[H2][I2]
[H2]i = [I2]i = 1.00 mol = 2.00 M
0.500 L

76. Solution
H2(g) +
I2(g) 
2HI(g)
I 2.00
2.00
0.00
C -x
-x
+2x
E 2.00 – x2.00 – x
2x
55.64 = ____(2x)2______ = ____(2x)2___
(2.00 – x)(2.00 – x) (2.00 – x)2
Take square root of both sides

77. Solution II

55.64 = 7.459 = __(2x)__
(2.00 – x)
Rearrange: (7.459)(2.00 – x) = 2x
14.918 – 7.459x = 2x
14.918 = 9.459x
x = 14.918 = 1.58
9.459
[H2]eq = [I2]eq = 2.00 – 1.58 = 0.42 M
[HI]eq = 2x = 2(1.58) = 3.16 M

78. Example 2
H2(g) + I2(g)  2HI(g) at 425 °C
KC = 55.64
If one mole each of H2, I2 and HI are placed in a
0.500 L flask at 425 °C, what are the equilibrium
concentrations of H2, I2 and HI?
Now have product as well as reactants initially
Kc = __[HI]2_= 55.64
[H2][I2]
H2]i = [I2]i = [HI]
1.00 mol = 2.00 M
0.500 L

79. Solution
H2(g) +
I2(g) 
2HI(g)
I 2.00
2.00
2.00
C -x
-x
+2x
E 2.00 – x2.00 – x
2.00 + 2x
55.64 = __(2.00 + 2x)2__ = (2.00 + 2x)2
(2.00 – x)(2.00 – x) (2.00 – x)2
Take square root of both sides

80. Solution

55.64 = 7.459 = (2.00 + 2x)_
(2.00 – x)
Rearrange: (7.459)(2.00 – x) = 2.00 + 2x
14.918 – 7.459x = 2.00 + 2x
12.918 = 9.459x
x = 12.918 = 1.37
9.459
[H2]eq = [I2]eq = 2.00 – 1.37 = 0.63 M
[HI]eq = 2.00 + 2(1.37) = 2.00 + 2.74 = 4.74 M

81. Practice 6
N2(g) + O2(g)
→ 2NO(g)
Kc = 0.0123 at 3900 oC
If 0.25 moles of N2 and O2 are placed in a
250 mL container, what are the equilibrium
concentrations of all species ?
A.
0.0526 M, 0.947 M, 0.105 M
B.
0.947 M, 0.947 M, 0.105 M
C.
0.947 M 0.105 M, 0.0526 M
D.
0.105 M, 0.105 M, 0.947 M

82. Practice
CH3CO2H(aq) + C2H5OH(aq)  CH3CO2C2H5(aq) +H2O(l)
acetic acid ethanol
ethyl acetate
KC = 0.11
An aqueous solution of ethanol and acetic acid,
each with initial concentration of 0.810 M, is
heated at 100 °C. What are the concentrations of
acetic acid, ethanol and ethyl acetate at
equilibrium?

83. Solution
Kc = __[CH3CO2C2H5]__
[CH3CO2H][C2H5OH]
CH3CO2H C2H5OH CH3CO2C2H5
I
0.810
0.810
0.000
C
-x
-x
+x
E
.810 – x
.810 – x
x
0.11 = ________x________
(.810 – x)(.810 – x)

84. Solution II
Rearranging:
0.11 x(0.6561 – 1.62x + x2) = x
Expand then write as quadratic equation
ax2 + bx + c = 0
0.07217 – 0.1782x + 0.11x2 - x = 0
0.11x2 – 1.1782x + 0.07217 = 0
x = -b  2 – 4ac)
(b
2a

85. Solution III
2
 (1.1782)  (1.1782)  4(0.11)(0.07217)
x 
2(0.11)
1.1782  (1.388)  (0.032) 1.1782  1.164
x 

0.22
0.22
This gives two roots: x = 10.6 and x = 0.064
Only x = 0.064 is possible
x = 10.6 is >> 0.810 initial concentrations
0.810 – 10.6 = negative concentration,
which is impossible
85

86. Solution IV
[CH3CO2C2H5]eq = x = 0.064 M
[CH3CO2H] = [C2H5OH]
= 0.810 – x = 0.810 – 0.064 = 0.746 M

87. Simplifications
If get trinomial use successive
approximations
If K is very small x is also small can drop x
in change part only
Valid if [X]I = 400x > K

88. More Problems
1. Finding Kc from change in reactant
concentrations.
PCl 3(g) + Cl2(g)  PCl5(g)
0.200 mol PCl3 and 0.100 mol Cl2 put in
1.00 L flask, at equilibrium found 0.120
mol PCl3. Means change in [PCl3] is 0.200 0.120 = 0.080 mol, change must be same for
other components. As volume is 1.0 L,
molarity has same value.

89. Example
Using Kc to calculate equilibrium
concentrations:
A flask contains a mixture of Br2 and
Cl2 ,both with an initial concentration of
0.0250 M. Find the equilibrium
concentration of all molecules from the
following equilibrium information.

90. Example Continued
[BrCl]
I
0
C
+2x
E
2x
Kc = [Br2][Cl2]
[BrCl]2
[Br2]
[Cl2]
.0250
.0250
-x
-x
.0250 – x .0250 - x
= (.0250 - x)2 = .145
(2x)2

91. Continued
Kc = [Br2][Cl2] = (.0250 - x)2 = .145
[BrCl]2
(2x)2
Take square root: (.0250 - x) = .381
2x
.0250 - x = .762 or .0250 = 1.762 x
x = .0250/1.762 = .0142 for Br2 & Cl2
BrCl = 2x = .0284 M

92. Practice
Find equilibrium concentrations with very
small Kc.
N2O + NO2  3NO Kc = 1.4 x 10-10
Start with .200 mol N2O & .400 mol NO2 in
4.00 L

93. Solution
[N2O] = .200 mol/4.00 L = 0.0500 M
[N2O] = .400 mol/4.00 L = 0.100 M
I
C
E
[N2O]
.0500
-x
.0500 – x
[NO2]
.100
-x
.100 – x
[NO]
0
+3x
3x

94. Problem 3 (II)
Kc =
[NO]3
=
[N2O][NO2]
(3x)3
=1.4 x10-10
(.0500 - x)(.100 - x)
Simplify reactant concentrations as x very
small:
27x3
= 1.4 x 10-10
(.0500)(.100)
x3 = 2.6 x 10-14 or x = 3.0 x 10 -5
[NO] = 3x = 9.0 x 10-5