Chapter 15
Principles of Chemical Equilibrium
Dr. Peter Warburton
peterw@mun.ca
http://www.chem.mun.ca/zcourses/1051.php
The equilibrium state
Chemical equilibrium is the state
reached when the concentrations of
the products and reactants remain
constant over time. The mixture of
reactants and products in the
equilibrium state is the equilibrium
mixture.
2
N2O4 (g)  2 NO2 (g)
We have used two directional arrows () to
show that this reaction does not go to
completion.
“The reaction occurs both
ways.”
3
Pitfall
The terms reactants and products are
arbitrary. We must always refer to a
balanced equation to be completely
understood.
N2O4 (g)  2 NO2 (g)
reactant
product
2 NO2 (g)  N2O4 (g)
reactant
product
4
Each reaction occurs at its own
rate as defined by its rate law.
One reaction will initially have a
faster rate than the other, and will
initially dominate the system. The
other reaction can be considered to
be dominated in the system
5
We saw in Kinetics that the rate of a
reaction decreases as time proceeds
because the concentrations of the
“reactants” decrease.
This is what we will see in the
dominant reaction.
6
What will happen to the dominated
reaction where the “reactant”
concentrations increase?
Its reaction rate will increase!
7
At some point in time the rate of
the forward reaction is THE SAME
as the rate of the reverse reaction.
This is a more correct means of
defining equilibrium.
8
Initially dominant reaction “slows down”
Initially dominated reaction “speeds up”
9
Equilibrium is a dynamic process
At
equilibrium
the rates of
the forward
and reverse
reactions
are the same,
BUT are not
equal to zero.
While no visible change is
occurring,
individual molecular events
are still occurring.
10
The equilibrium constant expression
The final concentrations will not always be the same
because we started with different numbers of N atoms
and O atoms in the last three experiments.
But ratio of [NO2]2 / [N2O4] is always the same!
11
Experiments 1 and 2
Start first experiment with no NO2 and 0.04 M N2O4
Start first experiment with 0.08 M NO2 and no N2O4
12
aA+bBcC+dD
The concentrations of all species in an
equilibrium mixture are related to each other
through the equilibrium constant equation in
terms of concentration.
This equilibrium equation applies only to this
specific balanced equation and the value of
the equilibrium constant, Kc must always be
stated at a specific temperature!
13
N2O4 (g)  2 NO2 (g)

NO2 
3
Kc 
 4.64 x 10 at 25C
N2O4 
2
If we change the temperature then the
equilibrium mixture will (most likely)
change. This means the equilibrium
constant will change. For this reaction
Kc = 1.53 at 127 C.
Notice there are no units for Kc!
14
Be careful!
Equilibrium constant equations (and therefore the
value of K) depend on the balanced equation
referenced.
A+BC+D
C+DA+B

C D
Kc 
A B
K
'
c

A B

C D
K’c DOES NOT EQUAL Kc, but rather
K’c = 1/Kc
15
Be careful!
Equilibrium constant equations (and
therefore the value of K) depend on the
balanced equation referenced.

C D
Kc 
A B
A+BC+D
2A+2B2C+2D

C D  CD 
Kc 
 

2
2
A B  AB 
2
'
2
2
K
2
c
16
Thermodynamic equilibrium constant Keq
The thermodynamic equilibrium
constant (Keq) equation takes the
same mathematical form as the
equilibrium constant equation in
terms of concentrations (Kc).
However, the composition of the
equilibrium mixture is expressed
in terms of activities.
17
Thermodynamic equilibrium constant Keq
Activities relate effective properties (like
concentration or pressure) of real
substances at given conditions in
comparison to the same substance acting
ideally at standard conditions.
Since activities make comparisons of the same
property type in ratio form, the property units
cancel out and all activities are unitless.
18
Thermodynamic equilibrium constant Keq
aA+bBcC+dD
a C  a D 
K eq 
a
b
a A  a B 
c
d
where ax = [X] / c0
(c0 is a standard concentration of 1 M)
or ax = Px / P0 (P0 is a standard pressure of 1 bar)
Note that ax = 1 for pure solids and liquids
19
Problem
For the reaction
CO (g) + 2 H2 (g)  CH3OH (g)
the equilibrium concentrations of CH3OH
and CO are found to be equal at 483 K. If
Kc = 14.5 at 483 K, what is the equilibrium
concentration of H2?
Answer: [H2] = 0.263 M
20
Problem
For the reaction
N2 (g) + 3 H2 (g)  2 NH3 (g)
Kc = 1.8 x 104 at a certain temperature.
What is the equilibrium concentration of H2 if
the equilibrium concentrations of N2 and NH3
are 0.015 M and 2.00 M respectively?
Answer: [H2] = 0.25 M
21
K for combined equilibria
If we can describe an overall equilibrium
reaction as the sum of two or more
other equilibrium processes, then the
equilibrium constant for the overall
reaction in the equilibrium constants of
the processes multiplied together.
Krxn = K1 x K2 x K3 …
22
Complex ions and solubility
AgCl(s) + 2 NH3(aq) → [Ag(NH3)2]+(aq) + Cl-(aq)
23
Complex ions and solubility
AgCl(s)  Ag+ (aq) + Cl- (aq)
Ag+ (aq) + NH3 (aq)  [Ag(NH3)]+ (aq)
[Ag(NH3)]+ (aq) + NH3 (aq)  [Ag(NH3)2]+ (aq)
K for the first reaction is 1.8 x 10-10
K for the second reaction is 2.0 x 103
K for the third reaction is 7.9 x 103
24
Complex ions and solubility
The sum of the three reactions is
AgCl(s) + 2 NH3(aq)  [Ag(NH3)2]+(aq) + Cl-(aq)
which will have an equilibrium constant that is
Krxn = K1 x K2 x K3
Krxn = 1.8 x 10-10 x 2.0 x 103 x 7.9 x 103
Krxn = 2.8 x 10-3
25
The equilibrium constant Kp
Gas phase equilibrium constants are
often expressed in terms of partial
pressures because they are
generally very easy to measure as a
function of the total pressure of the
system.
26
The equilibrium constant Kp
Recall, for an ideal gas A
PAV = nART
PA = (nART) / V
PA = (nA/V) RT
What is n / V? It is moles over volume,
which is concentration. So nA/V is [A] and
PA = [A] RT
27
N2O4 (g)  2 NO2 (g)

P 

P 
2
Kp
NO 2
N 2O 4
We can express an equilibrium constant
in terms of partial pressures because
they are related to concentrations!
Again, the equilibrium constant is unitless.
28
Kc and Kp are related
a A + b B  c C + d D (all are GASES!)


PC  PD 
C D
Kc 
or K p 
a
b
a
b
PA  PB 
A B
c
d
c
d


PC  PD 
[C]RT [D]RT
Kp 

a
b
a
b
PA  PB  [A]RT [B]RT
c
d
c
d
c
d


C D RT  RT 
RT  RT 

 Kc
a
b
a
b
a
b
RT  RT 
A B RT  RT 
c
d
c
d
29
Kc and Kp are related
Some of the RT terms in
c
d

RT  RT 
Kp  Kc
RT a RT b
will cancel each other out. In fact
RT  RT 
a
b
RT  RT 
c
d
(c  d) (a  b)
 (RT )
K p  K c (RT )(cd) (a  b)
Δn gas
 K c (RT )
where Δn gas  (c  d)  (a  b)
30
Kc and Kp are related
(c  d) (a  b)
K p  K c (RT )
Δn gas
 K c (RT )
where Δn gas  (c  d)  (a  b)
Use R = 0.08206 (Latm)(Kmol)-1
because it relates molarity to
pressure at a given
temperature.
31
N2O4 (g)  2 NO2 (g)
Dn = (2-1) = 1 so
(21)
Kp  Kc (RT)
 Kc (RT)
At 25 C Kc = 4.64 x 10-3
Kp = Kc(RT)
Kp = (4.64 x 10-3)(0.08206)(298)
note the lack of units and T is expressed in Kelvin!
Kp = 0.113 at 25 C
32
Problem
In the industrial synthesis of hydrogen often the
water-gas shift reaction is used
CO (g) + H2O (g)  CO2 (g) + H2 (g)
What is the value of Kp at 700 K if the partial
pressures in an equilibrium mixture at 700 K are
1.31 atm of CO
10.0 atm of water
6.12 atm of carbon dioxide, and
20.3 atm of hydrogen gas?
33
Problem answer
Kp = 9.48
34
Problem
In the industrial synthesis of nitric acid:
2 NO (g) + O2 (g)  2 NO2 (g)
If Kc = 6.9 x 105 at 227 C,
what is the value of Kp at this temperature?
If Kp = 1.3 x 10-2 at 1000 K,
what is the value of Kc at this temperature?
Answers: Kp at 227C = 1.7 x 104
and Kc at 1000 K = 1.1
35
Heterogeneous equilibria
Homogeneous equilibria occur in
systems where all compounds in the
equilibrium mixture are in the same
state.
Heterogeneous equilibria occur in
systems where some of the
chemicals of the equilibrium mixture
are in different states.
36
CaCO3 (s)  CaO (s) + CO2 (g)
Since one of the products is a gas, while the other
two compounds are solids, this is a
heterogeneous equilibrium.
Now, if we were to express the equilibrium
constant for this reaction, we would probably say

CaOCO 2 
"Kc"
CaCO3 
What is the concentration of a
solid, though?
37
What is the concentration of a solid
or liquid?
Concentration is moles per unit volume.
Also, density is mass divided by volume,
and molar mass is mass per number of
moles. So, for a pure substance
(mass / volume) / (mass / moles) = moles / volume
density / molar mass = (concentration)
38
What is the concentration of a solid
or liquid?
density / molar mass = (concentration)
Since both the density and molar mass of a pure
solid or liquid substance are constant, the
CONCENTRATION IS CONSTANT, and does not
change in a reaction as long as some of the
solid or liquid exists at all times.
This helps explain why the activities of
solids and liquids are equal to one!
39
CaCO3 (s)  CaO (s) + CO2 (g)
We choose not to include the
concentrations of solids and liquids in
the calculation of Kc!
K c  CO 2  " K c "
CaO
CaCO3 
The concentrations of the solids are “hidden”
inside the equilibrium constant.
If we look at the reaction in terms of pressure, then
Kp = (PCO2)
40
Thermodynamic equilibrium constant Keq
The activity of all pure solids and liquids
is one, and so solids and liquids have no
effect on the value of Keq
K eq 
a CaO a CO
a
 1a   

 a

1
2
CO 2
CO 2
CaCO3
41
CaCO3 (s)  CaO (s) + CO2 (g)
42
Problem
For each of the following reactions, write the
equilibrium constant expression for Kc. Where
appropriate, do the same for Kp and give the
relationship between Kc and Kp.
a) 2 Fe (s) + 3 H2O (g)  Fe2O3 (s) + 3 H2 (g)
b) 2 H2O (l)  2 H2 (g) + O2 (g)
c) SiCl4 (g) + 2 H2 (g)  Si (s) + 4 HCl (g)
d) Hg22+ (aq) + 2 Cl- (aq)  Hg2Cl (s)
43
Using the equilibrium constant
Judging the extent of a reaction: The
magnitude (size) of the constant K gives
an idea of the extent to which reactants
are converted to products.
We can make general statements about
the “completeness” of a given equilibrium
reaction based on the size of the value of
the equilibrium constant.
44
If the equilibrium constant is very large (>1000 for
instance), then the forward reaction is initially
very dominant and the reaction as written in the
balanced equation proceeds nearly to
completion before equilibrium is reached.
2 H2 (g) + O2 (g)  2 H2O (g)

H 2 O
47
Kc 
 2.4 x 10 at 500 K
2
H 2  O2 
2
45
If the equilibrium constant is very small (<10-3 for
instance), then the reverse reaction is initially
very dominant and the reaction as written in the
balanced equation barely proceeds at all before
equilibrium is reached.
2 H2O (g)  2 H2 (g) + O2 (g)

H 2  O 2 
 48
Kc ' 
 4.1 x 10 at 500 K
2
H 2O
2
46
If the equilibrium constant is between 10-3 and 103, this
means that the dominant reaction is not overpowering
the other reaction and we reach equilibrium somewhere
“in between” a state of “no reaction” and “completeness”.
Appreciable concentrations of all species are present in
the equilibrium mixture.
H2 (g) + I2 (g)  2 HI (g)

HI
Kc 
 57.0at 700K
H2 I2 
2
47
Predicting the direction of a reaction
If you put known concentrations of
products and reactants into the
equilibrium constant equation when you
know the system is NOT at equilibrium
you would get a value that does not
equal the equilibrium constant.
Can we use this value to tell which
reaction is dominant in this nonequilibrium system?
48
We define the reaction quotient Qc (or Qp
or Qeq) in exactly the same way we define
the equilibrium constant Kc (or Kp or Keq).

C D
Qc 
a
b
A B
c
d
When the system is not at equilibrium, then
Qc  K c
49
If Qc > Kc the reaction needs to create
more reactants (and use up products)
to get to equilibrium, so the reaction
will be going from right to left.
If Qc < Kc the reaction needs to create
more products (and use up reactants)
to get to equilibrium, so the reaction
will be going from left to right.
50
Figure
51
H2 (g) + I2 (g)  2 HI (g)

HI
Kc 
 57.0at 700K
H2 I2 
2
If [H2]t = 0.80 mol/L, [I2]t = 0.25 mol/L, and
[HI]t = 10.0 mol/L, then

HI
(10.0)2
Qc 

 500
H2 I2  (0.80)(0.25)
2
Qc  Kc, so the system is not at equilibrium.
Qc > Kc, the reaction will proceed from right to
left.
52
Problem
The equilibrium constant Kc for the reaction
2 NO (g) + O2 (g)  2 NO2 (g)
is 6.9 x 105 at 500 K. A 5.0 L reaction vessel at
this temperature was filled with 0.060 mol of NO,
1.0 mol of O2, and 0.80 mol of NO2.
a) Is the reaction mixture at equilibrium? If not,
which direction does the reaction proceed?
b) What is the direction of the reaction if the
initial amounts are 5.0 x 10-3 mol of NO, 0.20
mol of O2, and 4.0 mol of NO2?
53
Problem answer
a) Qc = 8.9 x 102. System is not at
equilibrium, and reaction will proceed right
since Qc < Kc.
b) Qc = 1.6 x 107. System is not at
equilibrium, and reaction will proceed left
since Qc > Kc.
54
Problem
In an earlier problem we saw the water-gas shift
reaction
CO (g) + H2O (g)  CO2 (g) + H2 (g)
where we calculated the value of Kp at 700 K to be
9.48. If we combine equal masses of all four
chemicals and let the system come to equilibrium,
which chemicals will have increased in quantity
and which will have decreased in quantity during
the reaction?
55
Problem answer
Qp  5.7 < Kp and reaction will proceed right
meaning we will increase CO2 (g) and H2 (g)
and decrease CO (g) and H2O (g).
56
Altering equilibrium conditions
We like to maximize
a product yield for a
reaction with a
minimum of energy
(and money) input.
If a reaction doesn’t
go to near
completion
we must adjust
experimental
conditions so the
reaction proceeds
as favourably as
possible!
57
Le Chatalier’s Principle
Three factors can be changed to affect
an equilibrium: the concentrations of the
chemicals involved, the pressure and/or
volume of the system, or the
temperature.
Le Chatalier’s Principle states that if a stress
is applied to a system at equilibrium, the
system will react in the direction that
minimizes the stress and brings the system
to a NEW equilibrium.
58
Changes in concentration
N2 (g) + 3 H2 (g)  2 NH3 (g)
Kc = 0.296 at 700 K.
59
The system re-established a NEW
equilibrium by reacting in such a way as to
decrease the stress to the system. Since
we have added a reactant (this is the
stress), the reaction should proceed
towards products to minimize the amount
of “extra” reactant added to the system.
60
In general, if we increase the
concentration of a reactant, the reaction
proceeds from reactants to products to
decrease the stress of added reactant to
our equilibrium system.
If we increase the concentration of a
product, the reaction proceeds from
products to reactants to decrease the
stress of added product to our
equilibrium system.
61
In the ammonia example of slide 59,
before we introduced more nitrogen (a
reactant) the reaction quotient was:

NH3 
(2.00)
Qc 

 0.296 K c
3
3
N 2 H 2  (0.50)(3.00)
2
2
The system was at equilibrium!
62
If we add 1.00 molL-1 nitrogen (a stress!)
to the original equilibrium system, the
reaction quotient will change and the
system will no longer be at equilibrium!

NH3 
(2.00)
Qc 

 0.0988 K c
3
3
N 2 H 2  (1.50)(3.00)
2
2
63
The reaction quotient is now less than the
equilibrium constant, meaning the reaction
must move from left to right to re-establish
equilibrium.
At the new equilibrium
[N2] = 1.31 molL-1
[H2] = 2.43 molL-1
[NH3] = 2.36 molL-1

NH3 
(2.36)2
Qc 

 0.296 K c
3
3
N 2 H 2  (1.31)(2.43)
2
64
Note that the [N2] in this new equilibrium
mixture is now lower than the 1.50 molL-1
we changed the concentration to after adding
N2 to the first equilibrium mixture.

NH3 
(2.36)
Qc 

 0.296 K c
3
3
N 2 H 2  (1.31)(2.43)
2
2
The system has reacted to minimize the
stress on the system by reducing the
amount of N2 to reach a new equilibrium!
65
Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
a) An equilibrium mixture for this reaction is
orange, which is the added colours of pale yellow
Fe3+ and the red FeNCS2+. SCN- is colorless.
66
Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
b) If we add FeCl3 to the solution, we
see the mixture gets more red,
meaning more FeNCS2+. Why?
67
Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
If we add KSCN to the solution, we
see the mixture gets more red,
meaning more FeNCS2+. Why?
68
Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
d) If we add H2C2O4 to the solution, we see the
mixture gets more yellow, meaning less FeNCS2+.
Why?
H2C2O4 (aq)  2 H+ (aq) + C2O42- (aq)
Fe3+(aq) + 3 C2O42- (aq)  [Fe(C2O4)3]3- (aq)
69
Fe3+ (aq) + SCN- (aq)  [FeNCS]2+ (aq)
e) If we add HgCl2 to the solution, we see the
mixture gets more yellow, meaning less FeNCS2+.
Why?
HgCl2 (s)  Hg2+ (aq) + 2 Cl- (aq)
Hg2+ (aq) + 4 SCN- (aq)  [Hg(SCN)4]2- (aq)
70
Problem
Consider the equilibrium for the water-gas shift
reaction:
CO (g) + H2O (g)  CO2 (g) + H2 (g)
Use Le Chatalier’s Principle to predict how the
concentration of H2 will change when the equilibrium
is disturbed by:
a) Adding CO
b) Adding CO2
c) Removing H2O
d) Removing CO2; also account for the change using
the reaction quotient Qc:
71
Problem
Calcination of limestone (decomposition of calcium
carbonate) occurs through the following reaction:
CaCO3 (s)  CO2 (g) + CaO (s)
After we establish this equilibrium system in a constant
volume container at a given temperature, what will be
the effect on equilibrium of
a)
b)
c)
d)
Doubling the amount of CO2
Doubling the amount of CaO
Removing half the CaCO3
Removing all the CaCO3
72
Effect of changes in pressure and volume
What happens when pressure is changed as
a result of a change in volume?
N2 (g) + 3 H2 (g)  2 NH3 (g) Kc = 0.296 at 700 K
Since PV = nRT then P = (nRT) / V
an increase in the volume decreases the
pressure of a system, or a decrease in
volume increases the pressure of the
system.
73
Pressure change due to volume change
Say we decrease the volume (and
increase the pressure) of our ammonia
formation equilibrium mixture. The stress
on the equilibrium is the increase in
pressure.
Le Chatalier’s Principle tells us the system
will respond by decreasing the pressure
of the system until a new equilibrium
mixture is achieved.
74
Pressure change due to volume change
Since the pressure is a direct result of the
number of moles of gas (more moles in a
given volume means more pressure), the
reaction will proceed in the direction
where the number of moles of gas is
decreased.
75
Pressure change through a change
in volume
In general, an increase in the pressure of
the system (caused by decreasing the
system volume!) causes the reaction to
shift to the side of the balanced equation
with less total moles of gas.
In general, a decrease in the pressure of
the system (caused by increasing the
system volume!) causes the reaction to
shift to the side of the balanced equation
with more total moles of gas.
76
Why is this the case?
Say we reduce the volume of our
ammonia equilibrium mixture by half.
We have doubled the
concentration of all gases!
Originally Qc

NH3 2
N 2 H 2 3
z2
 3  K c  0.29
xy
at 700K
2

NH3 
(2z)2
4z2
4 z2
1
new Qc 




(0.29) K c
3
3
3
3
N 2 H 2  (2x)(2y) (2x)(8y ) (2)(8)xy 4
Reaction will shift from left to right!
77
2 SO2 (g) + O2 (g)  2 SO3 (g)
78
Other things to note
If the total number of moles of gas on
the reactants side of a balanced equation
EQUALS
the total number of moles of gas on the
products side of a balanced equation,
then a
pressure change due to volume change
will not affect the equilibrium system.
79
Other things to note
We ALWAYS talked about pressure changes in
terms of volume change.
If we increase the pressure by adding inert
gas to the system, no equilibrium shift will be
seen because the partial pressures of the
gases in the equilibrium system have not
changed!
In other words, there is no actual stress being
placed on the equilibrium system
80
Problem
Does the number of moles of products
increase, decrease, or remain the same
when each of the following equilibria is
subject to an increase in pressure by
decreasing the volume?
a) CO (g) + H2O (g)  CO2 (g) + H2 (g)
b) 2 CO (g)  C (s) + O2 (g)
c) N2O4 (g)  2 NO2 (g)
81
Changes in temperature and equilibrium
Our reaction for the formation of ammonia
N2 (g) + 3 H2 (g)  2 NH3 (g) DH = -92.2 kJ.
We see as the
temperature
increases, the
value of Kc
decreases, so
the reaction
shifts towards
the reactants
with
increasing T
82
Is there some relationship between DH and Kc?
Yes!
N2 (g) + 3 H2 (g)  2 NH3 (g) + 92.2 kJ
We can think of heat as a “product” in the reaction.
As we increase the temperature of the system,
we increase the “concentration” of this
“product” and the reaction shifts from right to
left (towards the reactants).
The new equilibrium will have less products and
more reactants, giving a smaller value of Kc.
83
Is there some relationship between DH and Kc?
In an endothermic reaction, heat will be a
“reactant” so increasing the temperature
will shift the reaction from the left to the
right, increasing the value of Kc. Overall,
Exothermic reaction:
T  then KC 
Endothermic reaction:
T  then KC 
84
Problem
When air is heated at very high
temperatures in an engine, the air pollutant
nitric oxide is produced by the reaction
N2 (g) + O2 (g)  2 NO (g) DH = 180.5 kJ
How does the equilibrium amount of NO
vary with an increase in temperature?
85
Catalysis and equilibrium
Since both the forward and reverse
reactions pass through the same transition
state, a catalyst reduces the activation
energy for both the forward and reverse
reactions, by the same amount.
This increases the rates of both the
forward and reverse reactions by the same
amount!
86
87
Catalysis and equilibrium
Another way to think of it is that a
catalyst does not appear in the overall
balanced equation for a reaction
and therefore
it won’t appear in the equilibrium
constant equation
meaning
no change in the equilibrium constant
will be seen when you add a catalyst.
88
Problem
2 CO (g)  O 2 (g)
Pt



2 CO 2 (g) ΔH   566kJ
Suppose that you have a reaction vessel containing
an equilibrium mixture of all three species. Will the
amount of CO increase, decrease, or remain the
same when:
a) A platinum catalyst is added?
b) The temperature is increased?
c) The pressure is increased by decreasing the
volume?
d) The pressure is increased by adding argon gas?
e) The pressure is increased by adding O2 gas?
89
Equilibrium calculations
There are several different types of equilibriumbased calculations we can do:
T1) Find some equilibrium mixture data
from K, balanced equation and other
equilibrium mixture data (see slide 20)
T2) Find K from equilibrium mixture data
(see slide 33)
T3) Find K from some initial and equilibrium
mixture data and balanced equation
T4) Find equilibrium mixture data from initial
data, K, and the balanced equation
90
2 NO (g) + O2 (g)  2 NO2 (g)
Kc = 6.9 x 105 at 500 K (T1)
Say the system is at equilibrium and
[O2] = 1.0 mol/L and [NO2] = 0.80 mol/L
We can calculate [NO]!
NO2 
NO2 O2 
2
Kc

[NO] 
NO2 2 
K c O 2 
[NO] 
(0.64)
 4 mol


9.6
x
10
(6.9 x 105 )(1.0)
L
(0.80)2
(6.9 x 105 )(1.0)
91
2 NO (g) + O2 (g)  2 NO2 (g)
Kc = 6.9 x 105 at 500 K (T1)
Since concentrations are always positive,
we can throw out the negative answer.
The [NO] in the equilibrium mixture is
9.6 x 10-4 mol/L.
Let’s check this answer
Kc

NO2 2 
NO2 O 2 
(0.80)2
(9.6x104 ) 2 (1.0)
(0.64)
5


6.9
x
10
(9.3x107 )(1.0)
92
2 NO (g) + O2 (g)  2 NO2 (g)
Kc = 6.9 x 105 at 500 K (T1)
You might be asked to give your final
answer in moles (which means you
need to know the volume of your
container), or in grams (need to know
the container volume and the molar
mass)
93
Problem (T2)
Equilibrium is established at 1405 K for the
reaction
2 H2S (g)  2 H2 (g) + S2 (g)
in a 3.00 L reaction flask. If there are 0.11
mol S2, 0.22 mol H2, and 2.78 mol H2S in
the flask, what is Kc for the reaction at
1405 K?
Answer: Kc = 2.3 x 10-4
94
Problem (T3)
0.100 mol SO2 and 0.100 mol O2 are
introduced into an evacuated 1.52 L flask
at 900K. If the reaction is
2 SO3 (g)  2 SO2 (g) + O2 (g)
and 0.0916 mol of SO3 are found at
equilibrium then what is Kp for the reaction
at 900 K?
Answer: Kp = 2.2 x 10-2
95
Finding K
96
Problem (T4) #1
The H2/CO ratio in mixtures of carbon
monoxide and hydrogen (called synthesis
gas) is increased by the water-gas shift
reaction
CO (g) + H2O (g)  CO2 (g) + H2 (g)
which has an equilibrium constant Kc =
4.24 at 800 K. Calculate the equilibrium
concentrations of all species at 800 K if
only CO and H2O are present initially at
concentrations of 0.150 mol/L.
97
Problem (T4) #1 (ICE tables)
Using a balanced equation, we create a table of
Initial concentrations,
the Change in concentrations (based on
unknown quantities related by the
stoichiometry of the balanced equation),
and Equilibrium concentrations (sum of initial
concentration and change in concentration).
We can substitute our Equilibrium concentrations
into our equilibrium constant expression.
NOTE: If our system data are given as pressures,
we do exactly the same thing, but with pressures.
98
Problem (T4) #1
K c  4.24 
[CO2 ][H2 ]
(x)(x)
at 800 K so 4.24 
[CO][H2O]
(0.150 x)(0.150 x)
Taking the square root of both sides
4.24 
(x)(x)
x

(0.150 x)(0.150 x) 0.150 x
x
 2.059 
so  2.059 0.150 x   x
0.150 x
0.3089  2.059 x  x or - 0.3089  2.059 x  x
0.3089  3.059 x or - 0.3089  - 1.059 x
x  0.101mol L-1
or
x 0.292mol L-1
99
Problem (T4) #1
If we put both values of x back into all
our Equilibrium concentration
expressions, we’ll see one value of x
will give at least one negative
equilibrium concentration.
This isn’t physically possible!
Throw that value of x out and use
the other.
100
Problem (T4) #1
We can check our results by inserting these
equilibrium concentrations into the equilibrium
equation.
[CO2 ][H2 ] (0.101)(0.
101) 0.01020
Kc 


 4.25
[CO][H2O] (0.049)(0.
049) 0.002401
Our result is (within rounding error) the equilibrium
constant we were given, so our answers for the
equilibrium concentrations are correct.
101
Problem (T4) #2
The equilibrium constant Kp is 2.44 at 1000 K
for the reaction
C(s) + H2O (g)  CO (g) + H2 (g)
What are the equilibrium partial pressures
of H2O, CO, and H2 if the initial partial
pressures are PH2O = 1.20 atm, PCO = 1.00
atm, and PH2 = 1.40 atm?
102
Problem (T4) #2
Since this question starts with both reactants
and products in the initial mixture, it makes
sense to first check the reaction quotient to
see in which direction the reaction is going to
occur to reach equilibrium
Qp 
PCO PH 
P 
H 2O
2

1.001.40
1.20
 1.17
Since Q < K we expect to lose reactants and
gain products to get to equilibrium. This tells us
the signs of the changes that are occurring.
103
Problem (T4) #2
While this reaction quotient calculation step
isn’t absolutely necessary,
if we perform this and assign the correct
signs to our pressure changes, then we will
find that
any negative value of x we calculate
will not be physically possible.
104
Problem (T4) #2
(all in atm)
Initial press.
Press. change
Equil. press.
K p  2.44 
C(s)
N/A
N/A
N/A
PCO PH
P 
+
2
H2O (g)
1.20
-x
1.20 – x

CO (g)
1.00
+x
1.00 + x
+
H2 (g)
1.40
+x
1.40 + x
 at 1000K so 2.44 (1.00 x)(1.40 x)
(1.20 x)
H 2O
Rearranging, we get
2.44(1.20 x)  (1.00 x)(1.40 x)  0
[2.928 2.44 x] [1.40 2.40 x  x 2 ]  0
1.528 4.84 x  x  0
2
105
Problem (T4) #2
 (4.84) (4.84)2  4(1.00)(1.528)
 b  b 2  4ac
x
so x 
2a
2(1.00)
 (4.84) (4.84)2  4(1.00)(1.528)
x
2(1.00)
 (4.84) (4.84)2  4(1.00)(1.528)
or x 
2(1.00)
4.84 29.538
4.84 29.538
or x 
 2.00
 2.00
10.27
 .595
x
or x 
 2.00
 2.00
x  5.14atm or x  0.30atm
x
106
Problem (T4) #2
Our equilibrium partial pressures must all
be positive.
This only occurs for x = 0.30 atm
(the negative x value won’t work because
we have used the reaction quotient to help
set up our ICE table)
At equilibrium
PH2O = (1.20 atm – 0.30 atm) = 0.90 atm,
PCO = (1.00 atm + 0.30 atm) = 1.30 atm, and
PH2 = (1.40 atm + 0.30 atm) = 1.70 atm.
107
Problem (T4) #2
We should check our answer:
Kp

PCO PH
P 
H 2O
2


1.301.70
0.90
 2.46
which is the equilibrium constant we were
given, within rounding errors.
108
Problem (T4) #3
In a basic aqueous solution, chloromethane
undergoes a substitution reaction in which Cl- is
replaced by an OH-:
CH3Cl (aq) + OH- (aq)  CH3OH (aq) + Cl- (aq)
The equilibrium constant Kc is 1 x 1016.
Calculate the equilibrium concentrations of all
species in a solution prepared by mixing equal
volumes of 0.1 mol/L CH3Cl and 0.2 mol/L
NaOH.
109
Problem (T4) #3 answers
[CH3OH] = 0.05 mol/L,
[Cl-] = 0.05 mol/L
[OH-] = 0.05 mol/L,
[CH3Cl] = 5 x 10-18 mol/L.
110
111