Chapter14

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Chapter 14
Chemical
Equilibrium

Describing Chemical Equilibrium
1. Chemical Equilibrium—A Dynamic Equilibrium
2. The Equilibrium Constant
3. Heterogeneous Equilibria; Solvents in
Homogeneous Equilibria
Using the Equilibrium Constant
4. Qualitatively Interpreting the Equilibrium
Constant
5. Predicting the Direction of Reaction
6. Calculating Equilibrium Concentrations
Contents and Concepts

Changing Reaction Conditions:
Le Châtelier’s Principle
7. Removing Products or Adding Reactants
8. Changing the Pressure and Temperature
9. Effect of a Catalyst

Learning Objectives
Describing Chemical Equilibrium
1. Chemical Equilibrium—A Dynamic
Equilibrium
a. Define dynamic equilibrium and chemical
equilibrium.
b. Apply stoichiometry to an equilibrium
mixture.

2. The Equilibrium Constant
a. Define equilibrium-constant expression
and equilibrium constant.
b. State the law of mass action.
c. Write equilibrium-constant expressions.
d. Describe the kinetics argument for the
approach to chemical equilibrium.
e. Obtain an equilibrium constant from
reaction composition.

2. The Equilibrium Constant (cont.)
f. Describe the equilibrium constant Kp;
indicate how Kp and Kc are related.State the
law of mass action.
g. Obtain Kc for a reaction that can be written
as a sum of other reactions of known Kc
values.

3. Heterogeneous Equilibria; Solvents in
Homogeneous Equilibria
a. Define homogeneous equilibrium and
heterogeneous equilibrium.
b. Write Kc for a reaction with pure solids or
liquids.
Using the Equilibrium Constant
4. Qualitatively Interpreting the Equilibrium
Constant
a. Give a qualitative interpretation of the
equilibrium constant based on its value.

5. Predicting the Direction of Reaction
a. Define reaction quotient, Q.
b. Describe the direction of reaction after
comparing Q with Kc.
c. Use the reaction quotient.

6. Calculating Equilibrium Concentrations
a. Obtain one equilibrium concentration given
the others.
b. Solve an equilibrium problem (involving a
linear equation in x).
c. Solve an equilibrium problem (involving a
quadratic equation in x).

Changing the Reaction Conditions;
7. Removing Products or Adding Reactants
a. State Le Châtelier’s principle.
b. State what happens to an equilibrium when
a reactant or product is added or removed.
c. Apply Le Châtelier’s principle when a
concentration is altered.

8. Changing the Pressure and Temperature
a. Describe the effect of a pressure change
on chemical equilibrium.
b. Apply Le Châtelier’s principle when the
pressure is altered.
c. Describe the effect of a temperature
change on chemical equilibrium.
d. Apply Le Châtelier’s principle when the
temperature is altered.
e. Describe how the optimum conditions for a
reaction are chosen.

9. Effect of a Catalyst
a. Define catalyst.
b. Compare the effect of a catalyst on rate of
reaction with its effect on equilibrium.
c. Describe how a catalyst can affect the
product formed.

Chemical reactions often seem to stop before they
are complete.
Actually, such reactions are reversible. That is,
the original reactants form products, but then the
products react with themselves to give back the
original reactants.
When these two reactions—forward and reverse—
occur at the same rate, a chemical equilibrium
exists.

The graph
shows how the
amounts of
reactants and
products
change as the
reaction
approaches
equilibrium.
CO(g) + 3H2(g)  CH4(g) + H2O(g)

This graph
shows how
the rates of
the forward
reaction and
the reverse
reaction
change as the
reaction
approaches
equilibrium.
CO(g) + 3H2(g)  CH4(g) + H2O(g)

Chemical equilibrium is the state reached by a
reaction mixture when the rates of the forward and
reverse reactions have become equal.

We can apply stoichiometry to compute the
content of the reaction mixture at equilibrium.

?
When heated PCl5, phosphorus
pentachloride, forms PCl3 and Cl2 as follows:
PCl5(g)  PCl3(g) + Cl2(g)
When 1.00 mol PCl5 in a 1.00-L container is
allowed to come to equilibrium at a given
temperature, the mixture is found to contain
0.135 mol PCl3. What is the molar
composition of the mixture?

We will organize this problem by using the
chemical reaction to set up a table of initial,
change, and equilibrium amounts.
Initially we had 1.00 mol PCl5 and no PCl3 or Cl2.
The change in each is stoichiometric:
If x moles of PCl5 react, then x moles of PCl3 and x
moles of Cl2 are produced.
For reactants, this amount is subtracted from the
original amount; for products, it is added to the
original amount.

PCl5(g)  PCl3(g) + Cl2(g)
Initial 1.00 mol 0 0
Change –x +x +x
Equilibrium 1.00 – x x x
We were told that the equilibrium amount of PCl3 is
0.135 mol. That means x = 0.135 mol.
We can now find the amounts of the other
substances.

Moles PCl5 = 1.00 – 0.135 = 0.87 mol
(2 decimal places)
Moles PCl3 = 0.135 mol
(given with 3 significant figures)
Moles Cl2 = 0.135 mol

The Equilibrium Constant, Kc
The equilibrium constant expression for a
reaction is obtained by multiplying the
concentrations of products, dividing by the
concentrations of reactants, and raising each
concentration term to a power equal to its
coefficient in the balanced chemical equation.
The equilibrium constant, Kc, is the value
obtained for the Kc expression when equilibrium
concentrations are substituted.

For the reaction
aA + bB  cC + dD
The equilibrium constant expression is
Kc =
[ ] [ ]
[ ] [ ]ba
dc
BA
DC

?
Methanol (also called wood alcohol) is made
commercially by hydrogenation of carbon
monoxide at elevated temperature and
pressure in the presence of a catalyst:
2H2(g) + CO(g)  CH3OH(g)
What is the Kc expression for this reaction?
[ ]
[ ] [ ]COH
OHCH
2
2
3
=cK

When we are given some information about
equilibrium amounts, we are able to calculate the
value of Kc.
We need to take care to remember that the Kc
expression uses molar concentrations.

?
Carbon dioxide decomposes at elevated
temperatures to carbon monoxide and
oxygen:
2CO2(g)  2CO(g) + O2(g)
At 3000 K, 2.00 mol CO2 is placed into a
1.00-L container and allowed to come to
equilibrium. At equilibrium, 0.90 mol CO2
remains.
What is the value of Kc at this temperature?

We can find the value of x.
2.00 – 2x = 0.90
1.10 = 2x
x = 0.55 mol
0.55 mol1.10 mol0.90 mol
x2x2.00 – 2xEquilibrium
+x+2x–2xChange
002.00 molInitial
O2(g)2CO(g) +2CO2(g) 
2x = 2(0.55) = 1.10 mol

2CO2(g)  2CO(g) + O2(g)
[ ] [ ]
[ ]2
2
2
2
CO
OCO
=cK
[ ] M1.10
L1.00
mol1.10
CO ==
[ ] M0.90
L1.00
mol0.90
CO2 ==
[ ] M0.55
L1.00
mol0.55
O2 ==
( ) ( )
( )2
2
0.90
0.551.10
=cK
0.82=cK

In a heterogeneous equilibrium, in the Kc
expression, the concentrations of solids and pure
liquids are constant (due to these substances’
constant density).
As a result, we incorporate those constants into the
value of Kc, thereby making a new constant, Kc′. In
other words, equilibrium is not affected by solids
and pure liquids as long as some of each is
present.
More simply, we write the Kc expression by
replacing the concentration of a solid or pure liquid
with 1.

?
Write the Kc expression for the following
reaction:
H2O(g) + C(s)  CO(g) + H2(g)
[ ] [ ]
[ ]OH
HCO
2
2
=cK

Given:
aA + bB  cC + dD; K1
When the reaction is reversed:
cC + dD  aA + bB; K2
The equilibrium constant expression is inverted:
K2 =
[ ] [ ]
[ ] [ ] 1
1
DC
BA
Kdc
ba
=

Given:
aA + bB  cC + dD; K1
When the reaction is doubled:
2aA + 2bB  2cC + 2dD; K2
The equilibrium constant expression, K2 , is the
square of the equilibrium constant expression, K1:
K2 =
[ ] [ ]
[ ] [ ] ba
dc
22
22
BA
DC [ ] [ ]
[ ] [ ]
2
BA
DC








= ba
dc
2
1K=

For the reaction
aA(g) + bB(g)  cC(g) + dD(g)
The equilibrium constant expressions are
Kc = and Kp = ba
dc
PP
PP
BA
DC[ ] [ ]
[ ] [ ]ba
dc
BA
DC

How are these related?
We know
From the ideal gas law, we know that
So,
A
A
[A]
V
n
=
RT
P
V
n A
A
A
=
RT
V
n
P
A
A
A =

When you express an equilibrium constant for a
gaseous reaction in terms of partial pressures, you
call it the equilibrium constant, Kp.
In general, the value of Kp is different from that of
Kc.
We will explore this relationship on the next slides.
Recall the ideal gas law and the relationship
between pressure and molarity of a gas:
RT
V
n
V
nRT
P 





==
nRTPV =

ba
dc
p
PP
PP
K
BA
DC
=
[ ] ( ) [ ] ( )
[ ] ( ) [ ] ( )bbaa
ddcc
RTRT
RTRT
BA
DC
=
[ ] [ ]
[ ] [ ]
( ) n
ba
dc
p RTK
∆
=
BA
DC
[ ] [ ]
[ ] [ ]
( )( )C D
A B
c d
c d a b
p a b
K RT
+ − −
=
Kp = Kc (RT)∆n

For catalytic methanation,
CO(g) + 3H2(g)  CH4(g) + H2O(g)
the equilibrium expression in terms of partial
pressures becomes
and
( )
( )2
2
RT
K
RTKK c
cp ==
−
3
HCO
OHCH
2
24
PP
PP
Kp =

?
The value of Kc at 227°C is 0.0952 for the
following reaction:
CH3OH(g)  CO(g) + 2H2(g)
What is Kp at this temperature?
Kp = 0.0952(RT)∆n
where
T = 227 + 273 = 500. K
R = 0.08206 L  atm/(mol 
K)
∆n = 2
Kp = 1.60 × 102

We can use the value of the equilibrium constant
in several ways.
First, we can qualitatively describe the content of
the reaction mixture by looking at the magnitude of
Kc.
Second, we can determine the direction in which a
reaction will proceed by comparing Kc to the value
of the reaction quotient, Q, which has the same
expression as Kc but uses nonequilibrium values.

Finally, we can determine equilibrium
concentrations given the initial concentrations and
the value of Kc.

When Kc is very large (>102
), the equilibrium
mixture is mostly products.
When Kc is very small (<10-2
), the equilibrium
mixture is mostly reactants.
When Kc approaches 1, the equilibrium mixture
contains appreciable amounts of both reactants
and products.

?
Kc = 0.82 for a reaction. Describe the
composition of the equilibrium mixture.
Because Kc < 100 and > 0.01, at equilibrium there
will be substantial amounts of both reactants and
products.

Reaction Quotient, Q
The reaction quotient has the same form as the
equilibrium constant, but uses initial concentrations
for its value.
When Kc > Q, the reaction proceeds to the right.
When Kc < Q, the reaction proceeds to the left.
When Kc = Q, the reaction is at equilibrium.

Qc must move toward Kc.
Here the
numerator must
increase; more
products must be
produced.
Here the
denominator must
increase; more
reactants must be
produced.

Calculating Equilibrium Concentrations
1. When all but one equilibrium concentration and
the value of Kc are known.
2. When the value of Kc and the initial
concentrations are known.
a. When the Kc expression is a perfect square:
solving a linear equation.
b. When the Kc expression is not a perfect
square: solving a quadratic equation.

?
Nickel(II) oxide can be reduced to the metal
by treatment with carbon monoxide.
CO(g) + NiO(s)  CO2(g) + Ni(s)
If the partial pressure of CO is 100. mmHg
and the total pressure of CO and CO2 does
not exceed 1.0 atm, will this reaction occur at
1500 K at equilibrium? (Kp = 700. at 1500 K.)

forward.proceedwillreactionThe
ppK Q>
2CO
CO
660.
6.60
100.
p
P
Q
P
= = =
mmHg100.CO =P
mmHg660.2CO =P

?
Nitrogen and oxygen form nitric oxide.
N2(g) + O2(g)  2NO(g)
If an equilibrium mixture at 25°C contains
0.040 M N2 and 0.010 M O2, what is the
concentration of NO in this mixture? The
equilibrium constant at 25°C is
1.0 × 10−30
.

[ ]
[ ] [ ]0.0100.040
NO
101.0
2
30
=× −
[ ] ( ) ( )
2 30
NO 1.0 10 0.040 0.010−
= ×
[ ] 342
104.0NO −
×=
[ ] M17
102.0NO −
×=
N2(g) + O2(g)  2NO(g)
[ ]
[ ] [ ]22
2
ON
NO
=cK

When the initial concentration and the value of Kc
are known, we return to the stoichiometric chart of
initial, change, and equilibrium (ICE) amounts or
concentrations to find the equilibrium
concentrations.

?
Hydrogen iodide decomposes to hydrogen
gas and iodine gas.
2HI(g)  H2(g) + I2(g)
At 800 K, the equilibrium constant, Kc, for this
reaction is 0.016.
If 0.50 mol HI is placed in a 5.0-L flask, what
will be the composition of the equilibrium
mixture in molarities?

2HI(g)  H2(g) + I2(g)
Initial 0.10 M 0 0
Change –2x +x +x
Equilibrium 0.10 – 2x x x
[ ] M0.10
L5.0
mol0.50
HI 0 ==

The Kc expression is, Kc =
Substituting:
Because the right side of the equation is a perfect
square, we can take the square root of both sides.
[ ][ ]
[ ] ( )2
2
2
20.1020.10
0.016
x
x
x
xx
−
=
−
=
[ ][ ]
[ ]2
22
HI
IH

Solving: 0.126(0.10 – 2x) = x
0.0126 – 0.252x = x
0.0126 = 1.252x
x = 0.010 M
Substituting:
[ ] [ ] Mx 0.010IH 22 ===
[ ] Mx 0.080.0200.1020.10HI =−=−=
( )x
x
20.10
0.126
−
=
( )2
2
20.10
0.016
x
x
−
=

When the Kc expression is not a perfect square,
the equation must be rearranged to fit the
quadratic format:
ax2
+ bx + c = 0
The solution is
a
acbb
x
2
42
−±−
=

?
N2O4 decomposes to NO2. The equilibrium
reaction in the gas phase is
N2O4(g)  2NO2(g)
At 100°C, Kc = 0.36.
If a 1.00-L flask initially contains 0.100 M
N2O4, what will be the equilibrium
concentration of NO2?

N2O4(g)  2NO2(g)
Initial 0.100 M 0
Change –x +2x
Equilibrium 0.100 – x 2x
Again, we begin with the table:

The Kc expression is, Kc =
Substitute:
Rearrange:
Substitute:
00.0360.364x2
=−+ x
[ ]
[ ]42
2
2
ON
NO
0.0360.364 −=== cba
[ ]
[ ] ( )x
x
x
x
−
=
−
=
0.100
4
0.100
2
0.36
22
( ) ( )( )
( )42
0.036440.360.36
2
−−±−
=x
02
=++ cbxax

Solve:
We eliminate the negative value because it is
impossible to have a negative concentration.
Substitute to find the equilibrium concentration of
NO2:
8
0.70560.36 ±
=
-
x
0.1050.045 ±−=x
0.150.06 −== xx
[ ] ( ) Mx 0.120.0622NO2 ===

?
Given:
H2(g) + F2(g)  2HF(g); Kc = 1.15 × 102
3.000 mol of each species is put in a
1.500-L vessel. What is the equilibrium
concentration of each species?
First calculate the initial concentrations:
[ ] [ ] [ ] M2.000
L1.500
mol3.000
HFFH 00202 ====

H2(g) + F2(g)  2HF(g)
Initial 2.000 M 2.000 M 2.000 M
Change –x –x +2x
Equilibrium 2.000 – x 2.000 – x 2.000 + 2x
[ ]
[ ][ ]22
2
FH
HF
=cK
[ ]
[ ]2
2
2
-2.000
22.000
101.15
x
x+
=×

[ ]
[ ]2
2
2
-2.000
22.000
101.15
x
x+
=×
[ ]
[ ]
2.000 2
10.72
2.000
x
x
+
=
−
( ) ( )10.72 2.000 2.000 2x x− = +
xx 22.00010.7221.44 +=−
19.44
1.528
12.72
x M= =
x12.7219.44 =

Now compute the equilibrium concentrations:
Double-check by substituting these equilibrium
concentrations into the Kc expression and solving.
The answer should be the value of Kc.
This is within round-off error.
[ ] [ ]2 2H F 2.000 2.000 1.528 0.47x M= = − = − =
[ ] Mx 5.063.062.00022.000HF =+=+=
[ ]
[ ] [ ]
2
2
2
22
2
101.16
(0.47)
(5.06)
FH
HF
×===cK

?
The value of Kc at 227°C is 0.0952 for the
following reaction:
CH3OH(g)  CO(g) + 2H2(g)
What is Kp at this temperature?
Kp = 0.0952(RT)∆n
T = 227 + 273 = 500. K
R = 0.08206 L  atm/(mol  K)
∆n = 2
Kp = 1.60 × 102

When a system in chemical equilibrium is
disturbed by a change in
• temperature,
• pressure, or
• concentration,
the system shifts in equilibrium composition in a
way that tends to counteract this change of
variable.

When a substance that is part of the equilibrium is
added to the mixture, the equilibrium shifts to use it
(in a direction that makes the substance a
reactant).
When a substance that is part of the equilibrium is
removed from the mixture, the equilibrium shifts to
produce it (in a direction that makes the substance
a product).

Changes in the partial pressure of substances that
are part of the equilibrium are handled in the same
way as adding or removing a substance.

?
The following reaction is at equilibrium:
COCl2(g)  CO(g) + Cl2(g)
a. Predict the direction of reaction when
chlorine gas is added to the reaction
mixture.
b. Predict the direction of reaction when
carbon monoxide gas is removed from
the mixture.

COCl2(g)  CO(g) + Cl2(g)
a. When we add Cl2, the reaction will shift in the
reverse direction to use it.
Note: reverse = left = .
b. When we remove CO, the reaction will shift in
the forward direction to produce it.
Note: forward = right = .

A change in the total pressure occurs because of a
change in the volume of the reaction container.
When the size of the container decreases, the
overall pressure increases. The reaction will shift
to reduce the pressure—that is, it will shift toward
the side of the reaction with fewer gas molecules.

When the size of the container increases, the
overall pressure decreases. The reaction will shift
to increase the pressure—that is, it will shift toward
the side with more gas molecules.
In the event that both sides of the equilibrium
reaction have the same number of moles of gas,
pressure has no effect on the equilibrium.

?
In which direction will each reaction shift
when the volume of the reaction container is
increased?
a. CO(g) + 2H2(g)  CH3OH(g)
b. 2SO2(g) + O2(g)  2SO3(g)
c. COCl2(g)  CO(g) + Cl2(g)

a. CO(g) + 2H2(g)  CH3OH(g)
This reaction shifts reverse = left = ←
b. 2SO2(g) + O2(g)  2SO3(g)
This reaction shifts reverse = left = ←
c. COCl2(g)  CO(g) + Cl2(g)
This reaction shifts forward = right = ←
When the container volume is increased, the total
pressure is decreased. Each system will shift to
produce more gas by shifting toward the side with
more moles of gas.

Changing the temperature changes the value of
the equilibrium constant.
Changing the temperature can also cause a shift in
the equilibrium.
The direction of each of these changes depends
on the sign of ∆Ho
.

For an endothermic reaction, ∆Ho
> 0 (positive), we
consider that heat is a reactant.
For an exothermic reaction, ∆Ho
< 0 (negative), we
consider that heat is a product.

For an endothermic reaction, increasing the
temperature increases the value of Kc.
For an exothermic reaction, increasing the
temperature decreases the value of Kc.
Decreasing the temperature has the opposite
effect.

In addition to the value of Kc, we can consider the
direction in which the equilibrium will shift.
When heat is added (temperature increased), the
reaction will shift to use heat.
When heat is removed (temperature decreased),
the reaction will shift to produce heat.

?
Given:
2H2O(g)  2H2(g) + O2(g); ∆H° = 484 kJ
Would you expect this reaction to be
favorable at high or low temperatures?
We rewrite the reaction to include heat:
Heat + 2H2O(g)  2H2(g) + O2(g)
When heat is added,
the reaction shifts forward = right = →.
The reaction is favorable at high temperatures.

?
A typical reaction that occurs in the process is
8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g)
Suppose the reaction is at equilibrium at 200°C,
then is suddenly cooled to condense the octane,
and then the remaining gases are reheated to
200°C. In which direction will the equilibrium shift?
The Fischer–Tropsch process for the
synthesis of gasoline consists of passing a
mixture of carbon monoxide and hydrogen
over an iron–cobalt catalyst.

This is essentially removing octane, a product.
This change causes the reaction to produce
octane by shifting forward = right = →.

?
A typical reaction that occurs in the Fischer–
Tropsch process is
8CO(g) + 17H2(g)  C8H18(g) + 8H2O(g)
In which direction will the equilibrium shift
when the pressure is increased?
When the overall pressure is increased, the
reaction will shift to reduce the pressure—that is,
the reaction shifts to fewer gas molecules. In this
case, the reaction will shift forward = right = →

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