CHAPTER 14: CHEMICAL
EQUILIBRIUM
Vanessa N. Prasad-Permaul
Valencia Community College
CHM 1046
1
Introduction
1. How far does a reaction proceed toward completion
before it reaches a state of chemical equilibrium?
2. Chemical equilibrium
a) The state reached when the concentrations of
reactants and products remain constant over time
b) A state in which the concentration of reactants
and products no longer change (net)
3. Equilibrium mixture
A mixture of reactants and products in the
equilibrium state
2
Introduction
4.
What are we interested in?
a) What is the relationship between the
concentration of reactants and products in an
equilibrium mixture?
b) How can we determine equilibrium
concentrations from initial concentrations?
c) What factors can be exploited to alter the
composition of an equilibrium mixture?
3
The Equilibrium State
 In previous chapters we have generally assumed
that chemical reactions result in complete
conversion of reactants to products
Many reactions do not go to completion!!
Example1:
4
The Equilibrium State
5
The Equilibrium State
The two experiments demonstrate that the
interconversion of N2O4 and NO2 is reversible and
that the same equilibrium state is reached starting
from either substance.
1. This is why we use a  instead of 
2. Since both NO2 and N2O5 are products and
reactants we will call the chemical on the left
reactants and on the right products.
3. All chemical reactions are reversible
6
The Equilibrium State
4.
We call a reaction irreversible when it proceed
nearly to completion
a. Equilibrium mixture contains almost all
products and almost no reactants
b. Reverse reaction is too slow to be detected
5.
In an equilibrium state the reaction does not stop
at particular concentrations of reactants and
products, the rates of the forward and reverse
reactions become equal.
Important: reaction does not stop
7
The Equilibrium State
6. Chemical equilibrium is a dynamic state in which
forward and reverse reactions continue at equal rates
so that there is no net conversion of reactants to
products
8
EXAMPLE 14.1:
CO(g) + 3H2(g)
CH4(g) + H2O(g)
When 1.000 mol CO & 3.000 mol H2 are placed in a 10.00 L vessel
@ 927oC and allowed to come to equilibrium; the mixture is found to
contain 0.387 mol H2O. What is the molar composition of the
equilibrium mixture? (How many moles of each substance are
present?)
CO(g) + 3H2(g)
Starting
Change
Equilibrium
1.000
-x
1.000 - x
3.000
-3x
3.000 - 3x
CH4(g) + H2O(g)
0
+x
x
0
+x
x = 0.387
CO = 1.000 – x = 1.000 – 0.387 = 0.613 mol
H2 = 3.000 – x = 3.000 – 3(0.387) = 1.839 mol
CH4 = x = 0.387 mol
EXERCISE 14.1:
CO(g) + H2O(g)
CO2(g) + H2(g)
Suppose there is a mixture containing 1.00 mol CO and 1.00 mol
H2O. When equilibrium is reached at 1000oC, the mixture contains
0.43 mol H2. What is the molar composition of the equilibrium
mixture?
The Equilibrium Constant, Kc
a A + bB  cC + dD
Kc = [C]c [D]d
[A]a [B]b
If we write the equation in the reverse direction
cC + dD  aA + bB
K’c = [A]a [B]b = 1
[C]c [D]d kc
11
The Equilibrium Constant, Kc
General equation:
aA + bB  cC + dD
Equilibrium equation: Kc = [C]c [D]d
[A]a [B]b
products
reactants
The substances in the equilibrium equation must
be gases or molecules and ions in solution:
NO SOLIDS! NO PURE LIQUIDS!
Kc units are omitted but you must say at what
temperature!
12
EXAMPLE 14.2:
Write the equilibrium-constant expression Kc for catalytic methylation
CO(g) + 3H2(g)
CH4(g) + H2O(g)
Kc = [CH4][H2O]
[CO][H2]3
Write the equilibrium-constant expression Kc for the reverse reaction
CH4(g) + H2O(g)
Kc =
CO(g) + 3H2(g)
[CO][H2]3
[CH4][H2O]
EXAMPLE 14.2:
Write the equilibrium-constant expression Kc the synthesis of NH3
N2(g) + 3H2(g)
Kc =
2NH3(g)
[NH3]2
[N2][H2]3
Write the equilibrium-constant expression Kc for the following rxn.
1/2N2(g) + 3/2H2(g)
Kc =
NH3(g)
[NH3]
[N2]1/2 [H2]3/2
EXERCISE 14.2:
Write the equilibrium-constant expression Kc for the following
reaction:
2NO2(g) + 7H2(g)
2NH3(g) + 4H2O(g)
Write the equilibrium-constant expression Kc for the following
reaction:
NO2(g) + 7/2H2(g)
NH3(g) + 2H2O(g)
The Equilibrium Constant Kp
Kp = equilibrium constant with respect to partial
pressures of reactants and products
a A + bB  cC + dD
Kp = (PC)c (PD)d
(PA)a (PB)b
Relationship between Kc and Kp
Kp = Kc(RT)n
16
Heterogeneous Equilibria
Introduction
1. So far we have been talking about
homogeneous equilibria, in which all reactants
and products are in a single phase (gas or
solution)
2. Heterogeneous equilibria are those in which
reactants and products are present in more
than one phase
17
Using the Equilibrium Constant
Introduction
Knowing the value of the equilibrium constant for a
chemical reaction lets us:
1. Judge the extent of the reaction
2. Predict the direction of the reaction
3. Calculate the equilibrium concentrations from
any initial concentrations
18
Using the Equilibrium Constant
The numerical value of the equilibrium constant for a reaction
indicates the extent to which reactants are converted to
products
1.
Large value for Kc > 103 reaction proceeds essentially to 100%
(mostly products)
2.
Small value for Kc < 10-3 reaction proceeds hardly at all before
equilibrium is reached (mostly reactants)
3.
If a reaction has an intermediate value of Kc = 103 to 10-3
a. Appreciable concentrations of both reactants and
products are present in the equilibrium mixture
19
EXAMPLE 14.3:
What is the value of Kc for the decomposition of HI at room temp.?
2HI(g)
Starting
Change
Equilibrium
0.800 M
-2x
4.00 mol – 2x
H2(g) + I2(g)
0
+x
x
Kc =
0
+x
x
CONC
4.00mol/5.00L = 0.800M
0.442mol/5.00L = 0.0884M
(0.800 – 2(0.0884))M = 0.623M
[H2] [I2]
[HI]2
= [0.0884M][0.0884M]
[0.623M]2
= 0.0201
EXERCISE 14.3:
Hydrogen sulfide, a colorless gas with a foul odor, dissociates on
heating:
2H2S(g)
2H2(g) + S2(g)
When 0.100 mol H2S was put into a 10.0L vessel and heated to
1132oC, the equilibrium mixture contained 0.0285 mol H2. What is the
value of Kc at this temperature?
EXAMPLE 14.4:
Quicklime (calcium oxide, CaO), is prepared by heating a source of
calcium carbonate CaCO3 such as limestone or seashells.
CaCO3(s)
CaO(s) + CO2(g)
Write the expression for Kc.
Kc = [CO2]
An equilibrium constant can also be written for a physical equilibrium.
H2O(l)
H2O(g)
Write the expression for Kc for the vaporization of water.
Kc = [H2O(g)]
EXERCISE 14.4:
The Mond process for purifying nickel involves the formation of
nickel tetracarbonyl Ni(CO)4, a volatile liquid, from nickel metal and
carbon monoxide. Carbon monoxide is passed over impure nickel to
form nickel carbonyl vapor, which, when heated, decomposes and
deposits pure nickel.
Ni(s) + 4CO(g)
Ni(CO)4(g)
Write the expression for Kc for this reaction
Predicting the direction of Reaction
Reaction Quotient = Qc
1. Not necessarily equilibrium concentrations, at some time,
t, snapshot of reaction
2. As time passes, Qc changes toward the value of Kc
3. When the equilibrium state is reached Qc = Kc
4. Qc allows us to predict the direction of reaction by
comparing the values of Kc and Qc
a) If Qc< Kc, net reaction goes from left to
right, (reactant to products)
b) If Qc > Kc, net reaction goes from right to
left, (products to reactants)
c) If Qc = Kc, no net reaction occurs
24
EXAMPLE 14.5:
A 50.0 L reaction vessel contains 1.00 mol N2, 3.00 mol H2, and 0.500
mol NH3. Will more ammonia be formed or will it dissociate when a
mixture goes to equilibrium @ 400oC? Kc is 0.500 @ 400oC.
N2(g) + 3H2(g)
Qc
=
=
2NH3(g)
[NH3]2
[N2][H2]3
[0.500mol/50.0L]2
[1.00mol/50.0L][3.00mol/50.0L]3
=
[0.0100]2
[0.0200][0.0600]3
Qc = 23.1 which is greater than Kc = 0.500
 the reaction will go to the left or ammonia will dissociate.
EXAMPLE 14.5:
A 10.0 L reaction vessel contains 0.0015 mol CO2 and 0.10 mol CO.
If a small amount of carbon is added to this vessel and the temperature
is raised to 1000oC, will more CO form?
CO2(g) + C(s)
The value of Kc is 1.17 at 1000oC
2CO(g)
EXAMPLE 14.6:
A gaseous mixture contains 0.30 mol CO, 0.10 mol H2 and 0.020 mol
H2O plus an unknown amount of CH4, in each liter. The mixture is at
equilibrium @ 1200K. What is the concentration of CH4 in this
mixture? Kc = 3.92
CO(g) + 3H2(g)
Kc
=
3.92 =
CH4(g) + H2O(g)
[CH4][H2O]
[CO][H2]3
[CH4][0.020mol/1.0L]
[0.30mol/1.0L][0.10mol/1.0L]3
= [CH4][0.020M]
[0.30M][0.10M]3
3.92[3.00 x 10-4M] = [CH4] = 0.059M
[0.020M]
EXAMPLE 14.6:
Phosphorus pentachloride gives an equilibrium mixture of PCl5, PCl3,
and Cl2 when heated.
PCl5(g)
PCl3(g) + Cl2(s)
A 1.00L vessel contains an unknown amount of PCl5 and 0.020 mol of
each PCl3 and Cl2 at equilibrium at 250oC.How many moles of PCl5
are in the vessel if Kc for this reaction is 0.0415 @ 250oC?
Altering an Equilibrium Mixture: Changes in Pressure and
Volume
In general Le Chatelier’s Principle predicts that:
1. An increase in pressure by reducing the
volume will bring about net reaction in
the direction that decreases the number
of moles of gas
2. A decrease in pressure by enlarging the
volume will bring about net reaction in
the direction that increases the number
of moles of gas.
29
Factors that Alter the Composition of an Equilibrium Mixture
Introduction
One of the principal goals of chemical synthesis is to maximize
the conversion of reactants to products while minimizing the
expenditure of energy.
1. Can be achieved if the reaction goes nearly to
completion at mild temperatures and pressures.
2. If the equilibrium mixture is high in reactants and poor
in products, the experimental conditions must be
changed.
3. Several factors can be exploited to alter the composition
of an equilibrium mixture.
A.
B.
C.
The concentration of reactants or products
The pressure and volume
The temperature
30
Le Chatelier’s Principle
Le Chatelier’s Principle
If a stress is applied to a reaction mixture at equilibrium, net
reaction occurs in the direction that relieves the stress
1. Stress means a change in the concentration, pressure,
volume, or temperature that disturbs the original
equilibrium
2. Reaction then occurs to change the composition of
the mixture until a new state of equilibrium is reached
3. The direction that the reaction takes (reactants to
products or products to reactants) is the one that
reduces the stress
31
Altering an Equilibrium Mixture:
Changes in Concentration
In general, when an equilibrium is disturbed by the
addition or removal of any reactant or product, Le
Chatelier’s principle predicts that:
1. The concentration stress of an added reactant or
product is relieved by net reaction in the direction
that consumes the added substance
2. The concentration stress of a removed reactant or
product is relieved by net reaction in the direction
that replenishes the removed substance
32
EXAMPLE 14.6:
Predict the direction of reaction when H2 is removed from a mixture
(lowering the concentration) in which the following equilibrium has
been established:
H2(g) + I2(g)
2HI(g)
When H2 is removed from the reaction mixture, HI will dissociate to
partially restore the H2 that was removed
 The reaction will go to the left
EXAMPLE 14.6:
Consider each of the following equilibria, which are disturbed as
indicated. Predict the direction of the reaction:
The following equilibria is disturbed by increasing pressure of carbon
dioxide:
CaCO3(s)
CaO(s) + CO2(g)
The following equilibrium is disturbed by increasing the concentration
of hydrogen:
2Fe(s) + 3H2O(g)
Fe2O3(s) + 3H2(g)
EXAMPLE 14.7:
Will the products increase, decrease or have no effect if the pressure is
increased:
CO(g) + Cl2(g)
COCl2(g)
Reaction decreases the number of molecules of gas  an increase in pressure
increases the amount of product and pushes the reaction to the right.
2H2S(g)
2H2(g) + S2(g)
Reaction increases the number of molecules of gas  an increase of pressure
decreases the amount of products and the reaction shifts to the left.
C(s)
+
S2(g)
CS2(g)
Reaction does not change the number of molecules. Only look at gas volumes when
decide the effect of pressure change on equilibrium composition. Pressure change
has no effect.
EXERCISE 14.7:
Can the amount of product be increased in each of the following
reactions by increasing the pressure? explain:
CO2(g) + H2(g)
4CuO(s)
2SO2(g)
CO(g) + H2O(g)
2Cu2O(s) + O2(g)
+
O2(g)
2SO3(g)
Altering the Equilibrium Mixture: Changes in Temperature
In general, the temperature dependence of the equilibrium
constant depends on the sign of H for the reaction
1.
The equilibrium constant for an exothermic reaction
(negative H) decreases as the temperature increases
2.
The equilibrium constant for an endothermic reaction
(positive H) increases as the temperature increases.
3.
H = standard enthalpy of reaction, enthalpy change
measured under standard conditions
4. Standard conditions = most stable form of a substance at 1
atm pressure and at a specified temperature, usually 25C;
1 M concentration for all substances
37
Altering the Equilibrium Mixture: Changes in Temperature
Le Chatelier’s Principle says that if heat is added to an equilibrium
mixture (increasing the temperature) net reaction occurs in the
direction that relieves the stress of the added heat.
1. For an endothermic reaction heat is absorbed by reaction
in the forward direction. The equilibrium shifts to the right at
the higher temperatures, Kc increases with increasing
temperature
2. For an exothermic heat is absorbed by net reaction in the
reverse direction, so Kc decreases with temperature, and
the reaction would flow to the left (reactants)
38
EXAMPLE 14.8:
Carbon monoxide is formed when carbon dioxide reacts with solid
carbon:
CO2(g) + C(s)
2CO(g) ; Ho = 172.5kJ
Is a high or low temperature more favorable to the formation of carbon
monoxide?
Increase products need to shift equilibrium to the right
Endothermic
The temperature needs to be raised
High temperature is more favorable to the formation of carbon
monoxide.
EXAMPLE 14.8:
Carbon monoxide is formed when carbon dioxide reacts with solid
carbon:
CO2(g) + H2(g)
CO(g) + H2O(g)
Is a high or a low temperature more favorable to the production of
carbon monoxide? Explain.
The Effect of a Catalyst on Equilibrium
A catalyst increases the rate of a chemical reaction by making
available a new, lower-energy pathway for conversion of
reactants to products.
1. Since the forward and reverse reaction pass through the
same
transition state, a catalyst lowers the activation
energy for both
2. The rates of the forward and reverse reactions increase by
the same factor
3. Catalyst accelerates the rate at which equilibrium is reached
4. Catalyst does not affect the composition of the equilibrium
mixture
41
The Effect of a Catalyst on Equilibrium
42
The Link Between Chemical Equilibrium and Chemical Kinetics
A+BC+D
Assuming that the forward and reverse reactions occur in a single
bimolecular step, elementary reactions, we can write the
following rate laws
Rate of forward reaction = kf [A] [B]
Rate of reverse reaction = kr [C] [D]
When t=0 [C] = [D] = 0
As A and B are converted to C and D the rate of the forward
reaction decreases and the rate of the reverse reaction is
increasing, until they are equal, chemical equilibrium
kf [A] [B] = kr [C] [D]
43
The Link Between Chemical Equilibrium and Chemical Kinetics
kf = [C] [D]
kr
[A] [B]
The right side of this equation is the equilibrium constant
expression for the forward reaction, which equals the
equilibrium constant Kc
Kc = [C] [D]
[A] [B]
Therefore the equilibrium constant is simply the ratio of
the rate constants for the forward and reverse reactions:
Kc = kf
kr
44
Example 1:
Which of the following is correct?
1.
Some reactions are truly not reversible
2.
All reactants go to all products in all reactions
3.
All reactions are reversible to some extent
4.
The rates of the forward and reverse reactions
will never be equal
45
Example 2:
Write the equilibrium equation for each of the
following reaction:
a)
N2(g) + 3 H2(g)  2 NH3(g)
b) 2 NH3(g)  N2(g) + 3 H2(g)
46
Example 3:
 The oxidation of sulfur dioxide to give sulfur trioxide is
an important step in the industrial process for
synthesis of sulfuric acid. Write the equilibrium
equation for each of the following reactions:
a)
b)
2 SO2(g) + O2(g)  2 SO3(g)
2 SO3(g)  2 SO2(g) + O2(g)
The following equilibrium concentrations were measured at 800
K:
[SO2] = 3.0 x 10-3 M
[O2] = 3.5 x 10-3 M
[SO3] = 5.0 x 10-2 M
Calculate the equilibrium constant at 800 K a and b
47
Example 4:
In the industrial synthesis of hydrogen, mixtures of
CO and H2O are enriched in H2 by allowing the
CO to react with steam. The chemical equation
for this so-called water-gas shift reaction is:
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 H2O, 6.12 atm of
CO2, and 20.3 atm H2?
48
Example 5:
When will kc = kp ?
1. 2 SO2(g) + O2(g)  2 SO3(g)
2.
CO(g) + H2O(g)  CO2(g) + H2(g)
3. N2(g) + 3 H2(g)  2 NH3(g)
49
Example 6:
Nitric oxide reacts with oxygen to give nitrogen
dioxide, an important reaction in the Ostwald
process for the industrial synthesis of nitric
acid:
2 NO(g) + O2(g)  2 NO2(g)
a)
If Kc = 6.9 x 105 @ 227C, what is the value of
Kp @ 227C?
b) If Kp = 1.3 x 10-2 @ 1000 K, what is the value of
Kc @ 1000 K?
50
Example 7:
For each of the following reactions, write the
equilibrium constant expression for
Kc
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)  Hg2Cl2(s)
51
Example 8:
Which of the following has a Heterogeneous
equilibria?
1. 2 SO2(g) + O2(g)  2 SO3(g)
2.
CO(g) + H2O(g)  CO2(g) + H2(g)
3. SiCl4(g) + 2 H2(g)  Si(s) + 4 HCl(g)
52
Example 9:
The equilibrium constant for the reaction
2 NO(g) + O2(g)  2 NO2(g)
is 6.9 x 105 @ 500 K. A 5.0 L reaction vessel at this
temperature was filled with 0.060 mol of NO, 1.0 mol
O2, and 0.80 mol NO2.
a) Is the reaction mixture at equilibrium? If not, in which
direction does the net reaction proceed?
b) What is the direction of the net 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
Example 10:
Consider the equilibrium for the water-gas shift reaction:
CO(g) + H2O(g)  CO2(g) + H2(g)
Use Le Chatelier’s principle to predict how the
concentration of H2 will change and what direction the
reaction will flow when the equilibrium is disturbed by:
1.
2.
3.
4.
Adding CO
Adding CO2
Removing H2O
Removing CO2
54
Example 11:
In the following reaction, if I take away CO, which
direction will the reaction proceed to equilibrium?
CO2(g) + H2(g) CO(g) + H2O(g)
Products 
2. Reactants 
1.
55
Example 12:
Which direction will the reaction flow if the
following equilibria is subjected to an increase in
pressure by decreasing the volume?
1.
CO(g) + H2O(g)  CO2(g) + H2(g)
2.
2 CO(g)  C(s) + CO2(g)
3.
N2O4(g)  2 NO2(g)
56
Example 13:
If I increase the pressure by decreasing the volume,
which direction will the reaction flow to reach
equilibrium?
C(s) + CO2(g)  2 CO(g)
Products 
2. Reactants 
1.
57
Example 14:
When air is heated at very high temperatures in an
automobile engine, the air pollutant nitric oxide
is produced by the reaction
N2(g) + O2(g)  2 NO(g)
H = 180.5 kJ
1. How does the equilibrium amount of NO
vary with an increase in temperature?
2. What direction is the net reaction flowing?
58
Example 15:
A platinum catalyst is used in automobile catalytic converters to
hasten the oxidation of carbon monoxide:
2 CO(g) + O2(g)  2 CO2(g)
H = -566 kJ
Suppose that you have a reaction vessel containing an
equilibrium mixture. Will the amount of CO increase,
decrease, or remain the same when:
1.
2.
3.
4.
5.
A platinum catalyst is added
The temperature is increased
The pressure is increased by decreasing the volume
The pressure is increased by adding argon gas
The pressure is increased by adding O2 gas
59
Example 16:
Nitric oxide emitted from the engines of supersonic transport planes can
contribute to the destruction of stratospheric ozone:
NO(g) + O3(g)  NO2(g) + O2(g)
This reaction is highly exothermic (E = -200 kJ), and its equilibrium
constant Kc is 3.4 x 1034 at 300 K
1.
Which rate constant is larger, kf or kr?
2.
The value of kf at 300 K is 8.5 x 106 M-1 s-1. What is the value of kr at the
same temperature?
3.
A typical temperature in the stratosphere is 230 K. Do the values of kf,
kr, and Kc increase or decrease when the temperature is lowered from
300 K to 230 K?
60
Example 17:
If I increase the temperature of reaction which
way will the reaction flow to equilibrium?
NO2(g) + O2(g)  NO(g) + O3(g)
H = 200 kJ
Products 
2. Reactants 
1.
61