2. Chemical Equilibrium
The Equilibrium Condition
The Equilibrium Constant
Equilibrium Expressions Involving Pressures
Heterogeneous Equilibria
Applications of the Equilibrium Constant
Solving Equilibrium Problems
Le Chatelier’s Principle
3. Introduction
To this point, we have assumed that
reactions proceed to completion, that is,
until one of the reactants runs out.
Most reactions stop short of completion.
In fact, the system reaches chemical
equilibrium, the state where the
concentrations of all reactants and
products remain constant with time.
4. Introduction
Any chemical reactions carried out in a
closed vessel will reach equilibrium
Some reactions, the equilibrium position
favors the products so that the reaction
appears to go to completion. The
equilibrium position is said to lie “far to the
right”. This reflects the direction of the
products.
5. Introduction
Some reactions only occur to a slight
extent.
In this case, the equilibrium position is said
to lie “far to the left”. This reflects the
direction of the reactants.
7. The Equilibrium Condition
Equilibrium is not static but a highly dynamic
situation. On a molecular level many
molecules are moving back and forth
between reactants and products.
No net change in concentration of
reactants and products.
10. The Equilibrium Condition
H2O(g) + CO(g) ¬¾ H2(g) + CO2(g)
®
The equilibrium position
lies far to the right. This
reaction favors the
products. But the
reactants never reach a concentration of
zero.
11. The Equilibrium Condition
H2O(g) + CO(g)¬¾ H2(g) + CO2(g)
®
What would happen
if H2O(g) was added to
the system?
First, the forward reaction would increase,
then the reverse reaction would increase.
A new equilibrium would occur.
12. Characteristics of Chemical
Equilibrium
The equilibrium position is determined by
many factors:
initial concentrations.
relative energies of the reactants and
products.
relative degree of “organization” of the
reactants and products.
14. Law of Mass Action
The Law of Mass Action is a general description
of the equilibrium condition.
jA + kB ¬¾ lC + mD
®
l m
[C] [D]
K= j k
[A] [B]
The
square brackets indicate the
concentrations of the the reactants and
products at equilibrium. K is the equilibrium
constant.
15. Practice Problem #1
Write the equilibrium expression for the
following reaction:
4NH3(g) + 7O2(g) ¬¾ 4NO2(g) + 6H2O(g)
®
[NO2 ] [H 2O]
4 6
K=
[NH 3 ]4 [O2 ]7
16. The Equilibrium Constant
The value of the equilibrium constant at a
given temperature can be calculated if we
know the equilibrium concentrations of the
reaction components.
Equilibrium constants are typically given
without units.
17. Practice Problem #2
The following equilibrium concentrations
were observed for the Haber process for the
synthesis of ammonia at 127°C:
[NH3] = 3.1 x 10-2 mol/l
[N2] = 8.5 x 10-1 mol/l
[H2] = 3.1 x 10-3 mol/l
a) Calculate the value of K at 127°C for this
reaction.
3.8 x 104
18. Practice Problem #2
The following equilibrium concentrations
were observed for the Haber process for the
synthesis of ammonia at 127°C:
[NH3] = 3.1 x 10-2 mol/l
[N2] = 8.5 x 10-1 mol/l
[H2] = 3.1 x 10-3 mol/l
b) Calculate the value of the equilibrium
constant at 127°C for the reaction:
2NH3(g) ¬¾ N2(g) + 3H2(g)
®
2.6 x 10-5 (the reverse order reaction gives
the reciprocal of K)
19. Practice Problem #2
The following equilibrium concentrations
were observed for the Haber process for the
synthesis of ammonia at 127°C:
[NH3] = 3.1 x 10-2 mol/l
[N2] = 8.5 x 10-1 mol/l
[H2] = 3.1 x 10-3 mol/l
c) Calculate the value of the equilibrium
constant at 127°C for the reaction:
1
2 N2(g) + 2 H 2(g) « NH3(g)
3
1.9 x 102 (When the coefficients are ½ of
the balanced equation, new K = K1/2)
20. Equilibrium Expression
Summary
The equilibrium expression for a reaction is the
reciprocal of that for the reaction written in
reverse.
When the balanced equation for a reaction is
multiplied by a factor n, the equilibrium
expression for the new reaction is the original
expression raised to the nth power. Knew =Kon
K values are customarily written without units.
Law of mass action can describe reactions in
the solution and gas phase.
21. Equilibrium Expression
Summary
Theequilibrium expression and constant for a
reaction is the same at a given temperature,
regardless of the initial amounts of the
reaction components.
equilibrium concentrations will not always be
the same.
See Table 13.1 p600
22. Equilibrium Expression
A set of equilibrium concentrations is called
an equilibrium position.
There is only one equilibrium constant for
a particular system at a given
temperature, but there is an infinite
number of equilibrium positions.
23. Practice Problem #3
These results were Initial Equilibrium
collected for two [SO2] = 2.00 M 1.50 M
experiments involving
[O2] = 1.50 M 1.25 M
the reaction at 600°C
[SO3] = 3.00 M 3.50 M
between gaseous
sulfur dioxide and
oxygen to form Initial Equilibrium
gaseous sulfur [SO2] = 0.500 M 0.590 M
trioxide: [O2] = 0 M 0.0450 M
Show that the [SO3] = 0.350 M 0.260 M
equilibrium constant
is the same in both 4.36 and 4.32, within
experimental error.
experiments.
25. Pressure Equilibria
So far we have described equilibria involving
gases in terms of concentrations. Equilibria
involving gases also can be described with
pressure.
æ nö
PV = nRT, P = ç ÷ RT, P = CRT
èVø
C represents the molar concentration of the
gas.
jA + kB ¬¾ lC + mD
®
l m l m
CC PP
Kc = C D
j k
Kp = C D
j k
CC A B PP A B
26. Practice Problem #4
The reaction for the formation of nitrosyl chloride:
2NO(g) + Cl2(g) ¬¾ 2NOCl(g)
®
was studied at 25°C. The presurres at equilibrium
were found to be: NOCl =1.2 atm, NO = 5.0 x 10-2
atm, Cl2 = 3.0 x 10-1 atm. Calculate the value of
Kp for this reaction at 25°C.
1.9 x 103
27. Kc vs. Kp
Kp = Kc(RT)Δn
Δn is the sum of the coefficients of the
gaseous products minus the sum of the
coefficients of the gaseous reactants
jA + kB ¬¾ lC + mD
®
Δn=(l + m) – (j + k)
more moles of gas = more pressure
28. Practice Problem #5
Using the value of Kp obtained in Problem
#4, calculate the value of K at 25°C for the
reaction: 2NO(g) + Cl2(g) ¬¾ 2NOCl(g)
®
Kp =1.9 x 103
K = 4.6 x 104
30. Homogenous vs. Heterogeneous
Homogenous equlibria is where all the
reactants are in the same phase. Typically
gases
Heterogeneous equilibria involve more than
one phase.
The position of a heterogeneous equilibrium does
not depend on the amounts of pure solids or
liquids present.
Concentrations of pure solids and liquids cannot
change.
Concentrations of pure solids and liquids are not
included in the equilibrium expression for the
reaction
31. Example
CaCO3(s) ¬¾ CaO(s) + CO2(g)
®
[CO2 ]CCaO
K=
CCaCO3
K = [CO2 ]
This simplification only occurs with pure
solids or liquids and not solutions or gases.
32. Practice Problem #6
Write the expressions for K and Kp for the
following processes:
a) Solid phosphorus pentachloride decomposes
to liquid phosphorus trichloride and chloride gas.
b) Deep blue solid copper (II) sulfate
pentahydrate is heated to drive off water vapor
to form white solid copper (II) sulfate.
33. Practice Problem #6
a) Solid phosphorus pentachloride decomposes
to liquid phosphorus trichloride and chloride gas.
K = [Cl2] and Kp=PCl2
34. Practice Problem #6
b) Deep blue solid copper (II) sulfate
pentahydrate is heated to drive off water vapor
to form white solid copper (II) sulfate.
K = [H2O]5 Kp= (PH2O)5
36. Applications of the Equilibrium
Constant
Knowing the equilibrium constant for a
reaction allows us to predict several
important features of the reaction
The tendency for the reaction to occur
(but not the speed).
Whether or not a given set of
concentrations represents an equilibrium
condition.
37. Applications of the Equilibrium
Constant
If the reaction is not at equilibrium, we can
determine which way the reaction is moving by
taking the current law of mass action ratio and
comparing it to the equilibrium constant.
The ratio of non-equilibrium concentrations
gives us the reaction quotient, Q.
l m
[C] [D]
Q= j k
[A] [B]
38. Applications of the Equilibrium
Constant
To determine which direction a system will shift to
reach equilibrium, we compare the values of Q and
K.
Q=K. The system is at equlibrium; no shift will occur.
Q>K. The initial concentrations of product to initial
reactants is too large. To reach equilibrium, the
system must shift left, consuming products and
forming reactants.
Q<K. The ratio of initial concentrations of products
to initial concentrations of reactants is too small.
The system must shift to the right to form more
products.
40. Practice Problem #7
For the synthesis of ammonia at 500°C, the
equilibrium constant is 6.0 x 10-2. Predict the
direction in which the system will shift to reach
equilibrium in each of the following cases:
a) [NH3]=1.0x10-3M, [N2]=1.0x10-5M, [H2]=2.0x10-3M
b) [NH3]=2.00x10-4M, [N2]=1.50x10-5M, [H2]=3.54x10-1M
c) [NH3]=1.0x10-4M, [N2]=5.0M, [H2]=1.0x10-2M
a) Q>K, shift left. b)Q=K, no shift. c)Q<K, shift right.
41. Practice Problem #8
Dinitrogen tetroxide in its liquid state was used as
one of the fuels on the lunar lander for the NASA
Apollo missions. In the gas phase it decomposes to
gaseous nitrogen dioxide.
N2O4(g) ¬¾ 2NO2(g)
®
Consider an experiment in which gaseous N2O4
was placed in a flask and allowed to reach
equilibrium at a temperature where Kp=0.133. At
equilibrium, the pressure of N2O4 was found to be
2.71atm. Calculate the equilibrium pressure of
NO2(g).
.600 atm
42. The ICE Table
When initial concentrations and equilibrium
constants are known, but none of the equilibrium
positions are known it is helpful to write an ICE
table.
I:The Initial concentrations of products and
reactants
C: The Change in concentrations needed to
reach equilibrium is summarized in terms of
variables.
E: The Equilibrium values are summarized as a
combination of initial and change needed.
43. ICE Table Example:
Consider the reaction: N2(g) + 3H2(g) ¬¾ 2NH3(g)
®
K = 6.0 x 10-2 at 500°C.
The initial concentration of N2 is 3.0M and H2 is
2.0M. What are the equilibrium positions of this
reaction?
Initial (M) Change Equilibrium (M)
N2 3.0 -x 3.0 - x
H2 2.0 -3x 2.0 – 3x
NH3 0.0 +2x 2x
45. Practice Problem #9
At a certain temperature a 1.00 L flask initially
contained 0.298 mol PCl3(g) and 8.70x10-3 mol of
PCl5(g). After the system had reached equilibrium,
2.00 x 10-3 mol Cl2(g) was found in the flask.
Gaseous PCl5 decomposes according to the
reaction: PCl5(g) ¬¾ PCl3(g) + Cl2(g). Calculate
®
the equilibrium concentrations of all species and
the value of K.
>
46. Practice Problem #9
Initial: 0.298 mol PCl3(g). 8.70x10-3 mol of PCl5(g)in 1.00L
equil: 2.00 x 10-3 mol Cl2(g)
PCl5(g) ¬¾ PCl3(g) + Cl2(g)
®
Equilibrium expression: K =
[Cl2 ][PCl3 ]
ICE Table: [PCl5 ]
Initial (M) Change Equilibrium (M)
PCl5(g) 0.298 -2.00 x 10-3
PCl3(g) 8.70x10-3 +2.00 x 10-3
Cl2(g) 0.0 +2.00 x 10-3 2.00 x 10-3
k=8.96 x 10-2
47. Practice Problem #10
Carbon monoxide reacts with steam to
produce carbon dioxide and hydrogen. At
700 K the equilibrium constant is 5.10.
Calculate the equilibrium concentrations of
all species if 1.00 mol of each component is
mixed in a a 1.00L flask.
>
48. Practice Problem #10
Reaction:
CO(g) + H2O(g)¬¾ CO2(g) + H2(g), K= 5.10
®
Which way does the equilibrium need to go?
Q=
[CO2 ][H 2 ] Q=1.00 Q<K, shift right
[CO][H 2O]
ICE Table:
Initial (M) Change Equilibrium (M)
CO 1.0 -x 1.0 - x
H2O 1.0 -x 1.0 - x
CO2 1.0 +x 1.0 + x
H2 1.0 +x 1.0 + x
49. Practice Problem #10
Initial (M) Change Equilibrium (M)
CO 1.0 -x 1.0 - x
H2O 1.0 -x 1.0 - x
CO2 1.0 +x 1.0 + x
H2 1.0 +x 1.0 + x
(1.0 - x)(1.0 - x) (1.0 - x) 2
K = 5.10 = =
(1.0 + x)(1.0 + x) (1.0 + x)2
x = 0.387 mol/L
[CO] & [H2O] = .613M, [CO2] & [H2] = 1.387M
Double check K with expression.
50. Practice Problem #11
Assume that the reaction for the formation of
gaseous hydrogen fluoride from hydrogen and
fluorine has an equilibrium constant of 1.15 x 102 at
a certain temperature. In a particular experiment,
3.000 mol of each component was added to a
1.500 L flask. Calculate the equilibrium
concentration of all species.
>
51. Practice Problem #11
Reaction:
H2(g) + F2(g) ¬¾ 2HF(g), K=1.15 x 102
®
Which way does the equilibrium need to go?
Initial concentrations:
3.000mol/1.500L = 2.00 M
Q=
[HF]2 Q=1.00 Q<K, shift right
[H 2 ][F2 ]
ICE Table: Initial (M) Change Equilibrium (M)
H2 2.0 -x 2.0 - x
F2 2.0 -x 2.0 - x
HF 2.0 +2x 2.0 + 2x
52. Practice Problem #11
Initial (M) Change Equilibrium (M)
H2 2.0 -x 2.0 - x
F2 2.0 -x 2.0 - x
HF 2.0 +2x 2.0 + 2x
(2.000 + 2x)2
K = 1.15x10 2 =
(2.000 - x)2
x=1.528
[H2] & [F2]= 0.472 M
[HF] = 5.056 M
Check K with equilibrium values
54. Solving Equilibrium Problems
Strategy:
1. Write the balanced equation for the
reaction.
2. Write the equilibrium expression using
law of mass action.
3. List the initial concentrations.
4. Calculate Q, and determine the
direction of the shift needed for
equilibrium.
55. Solving Equilibrium Problems
Strategy continued:
5. Define the change needed to reach
equilibrium, and define the equilibrium
concentrations by applying the change
to the initial concentrations.
6. Substitute the equilibrium
concentrations into the equilibrium
expression, and solve for the unknown.
7. Check your calculated equilibrium
concentrations by making sure the give
the correct value of K.
56. Solving Equilibrium Problems
Typical systems do not produce an
expression that can be solved by taking the
square root of both sides.
To solve some expressions, we will use the
quadratic equation.
-b ± b 2 - 4ac
x=
2a
57. Equilibrium Example
Suppose for a synthesis of hydrogen fluoride from
hydrogen and fluorine, 3.000 mol H2 and 6.000 mol
F2 are mixed in a 3.000 L flask. Assume the
equilibrium constant for the synthesis reaction at
this temperature is 1.15x102. What are the
equilibrium concentrations of each component.
>
58. Equilibrium Example
1. Write the balanced equation for the reaction.
H2(g) + F2(g) ¬¾ 2HF(g)
®
2. What is the equilibrium expression?
[HF]2
K = 1.15x10 2 =
[H 2 ][F2 ]
3. What are the initial concentrations?
[H2] = 3.00mol/3.00 L = 1.000M
[F2] = 6.00mol/3.00 L = 2.000M
[HF]= 0
59. Equilibrium Example
4. What is Q?
Q does not need to be calculated in this
example. Since HF is not present initially, we
can assume that the reaction will shift to the
right to reach equilibrium.
5. What change is required to reach equilibrium?
Initial (M) Change Equilibrium (M)
H2 1.0 -x 1.0 - x
F2 2.0 -x 2.0 - x
HF 0.0 +2x 2x
60. Equilibrium Example
6. What is the value of K? (Use ICE in expression)
[HF]2 (2x)2
K= =
[H 2 ][F2 ] (1.000 - x)(2.000 - x)
collect terms and set = 0
(1.000 - x)(2.000 - x)(1.15x102 ) = 2x 2
ax2 + bx + c = 0
(1.11x102 )x 2 - (3.445x102 )x + (2.30x102 ) = 0
a=1.11x102, b=-3.45x102, c=2.30x102
61. Equilibrium Example
6. What is the value of K?
a=1.11x102, b=-3.45x102, c=2.30x102
-b ± b 2 - 4ac
x=
2a
Substituting these values give two answers for x:
x=2.14 mol/L and 0.968 mol/L
Both of these results are not valid; the changes in
concentration must be checked for validity
62. Equilibrium Example
6. What is the value of K?
x=2.14 mol/L or 0.968 mol/L
[H2] = 1.000 – x, [F2] = 2.000 – x, [HF] = 2x
[H2] = 3.2x10-2M, [F2] = 1.032M, [HF] = 1.936M
7. Check concentrations by substituting them into
the equilibrium expression.
63. Practice Problem #12
Assume that gaseous hydrogen iodide is
synthesized from hydrogen gas and iodine vapor
at a temperature where the equilibrium
constant is 1.00x102. Suppose HI at 5.000x10-1
atm, H2 at 1.000x10-2 atm, and I2 at 5.000x10-3
atm are mixed in a 5.000L flask. Calculate the
equilibrium pressures of all species.
64. Practice Problem #12
1. Write the balanced equation
2. What is the equilibrium expression?
3. What are the initial pressures?
4. What is the value of Q?
65. Practice Problem #12
5. What is the change required?
6. What is the value of Kp (& equilibrium pressures)?
7. Expression check.
PHI=4.29x10-1atm, PH2=4.55x10-2atm, PI2=4.05x10-2atm
66. Small Equilibrium Constants
Sometimes there are simplifications that can be
made to the math of some equilibrium
problems.
When reactions lie far to the left, the
equilibrium constants can be very small.
Changes in initial concentrations can be
negligible and partially disregarded.
67. Small K Example
Gaseous NOCl decomposes to form the gases
NO and Cl2. At 35°C the equilibrium constant is
1.6x10-5. In an experiment in which 1.0 mol NOCl
is placed in a 2.0 L flask, what are the
equilibrium concentrations?
1. What is the equation?
2NOCL(aq) ¬¾ 2NO(g) + Cl2(g)
®
2. What is the expression?
[NO]2 [Cl2 ]
K= 2
= 1.6x10 -5
[NOCl]
68. Small K Example
3. What are the initial concentrations?
[NOCl]=0.50M, [NO]=0, [Cl2] = 0
4. What is Q?
Direction must lie to the right for equilibrium.
5. What is the change required?
Initial (M) Change Equilibrium (M)
NOCl 0.50 -2x 0.50 - 2x
NO 0.0 +2x 2x
Cl2 0.0 +x x
69. Small K Example
6. What is the value of K (& concentrations)
Initial Change Equilibrium
(M) (M)
(2x)2 (x)
NOCl 0.50 -2x 0.50 - 2x K = 1.6x10 -5 =
(0.50 - 2x)2
NO 0.0 +2x 2x
Cl2 0.0 +x x
x must represent a relatively small number. In order
to simplify this expression, we can assume that:
0.50 – 2x ≅ .50
Therefore we can simplify the expression:
(2x)2 (x) 4x 3
1.6x10 -5 = 2
=
(0.50) (.50)2 x=1.0x10-2
x 3 = 1.0x10 -6
70. Small K Example
[NOCl] = .50 – 2x ≈ 0.50 M
[NO] = 2.0 x 10-2 M
[Cl2] = 1.0 x 10-2 M
7. Check the K expression.
72. Le Chatelier’s Principle
Several factors can control the position of a
chemical equilibrium.
Changes in Concentration
Temperature (removal or addition of energy)
Pressure
73. Le Chatelier’s Principle
Le Chatelier’s principle states that if a change is
imposed on a system at equilibrium, the position
of the equilibrium will shift in a direction that
tends to reduce that change.
It is important to realize that although changes
to the reaction may alter the equilibrium
positions, they do not alter the equilibrium
constant.
74. Le Chatelier’s Principle
Change in Concentration:
If a component (reactant or product) is added
to a reaction system at equilibrium (at
constant T and P or constant T and V), the
equilibrium position will shift in the direction
that lowers the concentration of that
component. If a component is removed, the
opposite effect occurs.
75. Practice Problem #13
Arsenic can be extracted from its ores by first
reacting the ore with oxygen (called roasting) to
form solid As4O6, which is then reduced using
carbon:
As4O6(s) + 6C(s) As4(g) + 6CO(g)
Predict the direction ¬¾ shift of the equilibrium
of ®
the
position in response to each of the following
changes in concentration.
a) Addition of carbon monoxide.
b) Addition or removal of carbon or tetrarsenic
hexoxide.
c) Removal of gaseous arsenic.
left shift, no effect, right shift
76. Le Chatelier’s Principle
Change in Pressure:
There are three ways to change the pressure
of a reaction system involving gaseous
components:
1. Add or remove a gaseous reactant or product.
2. Add an inert gas (one not involved in the
reaction).
3. Change the volume of the container
77. Le Chatelier’s Principle
Change in Pressure:
When an inert gas is added, there is no effect
on the equilibrium position. The addition of an
inert gas increases the total pressure but has
no effect on the concentrations of the
reactants or products.
The system will remains at the original equilibrium
position
78. Le Chatelier’s Principle
Change in Volume:
When the volume of the container holding a
gaseous system is reduced, the system
responds by reducing its own volume. This is
done by decreasing the total number of
gaseous molecules in the system.
A reaction will shift in order to reduce the
number of gas molecules.
79. Practice Problem #14
Predict the shift in equilibrium position that will
occur for each of the following processes when
the volume is reduced.
a. The preparation of liquid phosphorus
trichloride by the reaction.
P4(s) + 6Cl2(g) ¬¾ 4PCl3(l)
®
right shift
80. Practice Problem #14
Predict the shift in equilibrium position that will
occur for each of the following processes when
the volume is reduced.
b. The preparation of gaseous phosphorus
pentachloride according to the equation:
PCl3(g) + Cl2(g) ¬¾ PCl5(g)
®
right shift
81. Practice Problem #14
Predict the shift in equilibrium position that will
occur for each of the following processes when
the volume is reduced.
c. The reaction of phosphorus trichloride with
ammonia:
PCl3(g) + 3NH3(g) ¬¾ P(NH2)3(g) + 3HCl(g)
®
no effect
82. Le Chatelier’s Principle
Change in Temperature:
If energy is added or removed from a system in
equilibrium, the system will shift according to
the heat of the reaction.
Heat is a product in an exothermic reaction.
Heat is a reactant in an endothermic reaction.
The effect of temperature on equilibrium
changes the value of K because K changes with
temperature.
83. Practice Problem #15
For each of the following reactions, predict how
the value of K changes as the temperature is
increased.
a. N2(g) + O2(g) ¾ 2NO(g)
¬® ΔH = 181 kJ
b. 2SO2(g) + O2(g)¬¾ 2SO3(g) ΔH = -198 kJ
®
a) shift right b) shift left
