18.3 Reversible Reactions and Equilibrium >
18.3 Reversible Reactions and Equilibrium >
Reaction Rates and Equilibrium
18.1 Rates of Reaction
18.2 The Progress of Chemical
Reactions
Fertilizers can increase
the amount of a crop per
unit of land. Most
fertilizers contain
ammonia or nitrogen
compounds made from
ammonia.
18.3 Reversible Reactions
and Equilibrium
18.4 Solubility Equilibrium
18.5 Free Energy and Entropy
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18.3 Reversible Reactions and Equilibrium > Reversible Reactions
2
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18.3 Reversible Reactions and Equilibrium > Reversible Reactions
•  Chemical reactions do not always progress in only
one direction. Some reactions are reversible.
Reversible Reactions
What happens at the molecular
level in a chemical system at
equilibrium?
3
& YOU
How did chemists help farmers
produce more food?
Chapter 18
1
CHEMISTRY
•  reversible reaction
Reactants ßà Products
4
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1
18.3 Reversible Reactions and Equilibrium > Reversible Reactions
18.3 Reversible Reactions and Equilibrium > Reversible Reactions
Reversible Reaction
2SO2(g) + O2(g) à 2SO3(g) (forward)
Establishing Equilibrium
2SO2(g) + O2(g) ß 2SO3(g) (reverse)
2SO2(g) + O2(g)
5
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18.3 Reversible Reactions and Equilibrium > Reversible Reactions
Dynamic Equilibrium
chemical equilibrium: when the rates of
the forward & reverse reactions are equal
•  No net change (does not mean NO
change)
•  Does NOT imply the same amount/
concentration of products and
reactants
2SO3(g)
6
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18.3 Reversible Reactions and Equilibrium > Reversible Reactions
What does dynamic mean?
Concentrations at Equilibrium
Although the rates of the forward and
reverse reactions are equal at equilibrium,
the concentrations of the components
usually are not.
7
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8
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2
18.3 Reversible Reactions and Equilibrium > Reversible Reactions
18.3 Reversible Reactions and Equilibrium > Reversible Reactions
Concentrations at Equilibrium
Concentrations at Equilibrium
The equilibrium position tells you whether the
forward or reverse reaction is more likely to
occur.
In principle, almost all reactions are reversible
to some extent under the right conditions.
•  In practice, however....
•  Suppose a single reactant, A, forms a single product,
B.
•  If no reactants can be detected, you can say that the
reaction has gone to completion, or is irreversible.
•  At equilibrium, the mixture contains 1% A and 99% B.
•  When no products can be detected, you can say that
no reaction has taken place.
•  The formation of B is said to be favored.
A
1%
9
B
99%
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Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
10
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Factors Affecting
18.3 Reversible Reactions and Equilibrium > Equilibrium:
Le Châtelier’s Principle
Le Châtelier’s Principle
The French chemist Henri Le Châtelier
(1850–1936) proposed what has come to
be called Le Châtelier’s principle:
Factors Affecting Equilibrium:
Le Châtelier’s Principle
If a stress is applied to a system in
dynamic equilibrium, the system changes
in a way that relieves the stress.
11
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3
Factors Affecting
18.3 Reversible Reactions and Equilibrium > Equilibrium:
Le Châtelier’s Principle
Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
Le Châtelier’s Principle
Concentration
Stresses that upset chemical equilibrium
include :
•  changes in the concentration of
reactants or products;
•  changes in temperature; and,
•  changes in pressure.
13
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Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
H2CO3(aq)
CO2(aq) + H2O(l)
< 1%
> 99%
•  The system has reached equilibrium.
•  The amount of carbonic acid is less than 1%.
14
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Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
Le Châtelier’s Principle
Le Châtelier’s Principle
Suppose carbon dioxide is added to the
system.
Suppose carbon dioxide is removed.
Add CO2
Direction of shift
H2CO3(aq)
CO2(aq) + H2O(l)
H2CO3(aq)
•  Increasing [CO2] causes the rate of the reverse
reaction to increase.
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Remove CO2
Direction of shift
CO2(aq) + H2O(l)
•  Decreasing [CO2] causes the rate of the reverse
reaction to decrease.
•  Adding a product to a reaction at equilibrium
favors the formation of reactants.
15
Add CO2
Direction of shift
•  Removing a product always favors the formation
of products.
16
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4
Factors Affecting
18.3 Reversible Reactions and Equilibrium >
H2CO3(aq)
18.3 Reversible Reactions and Equilibrium > Equilibrium:
Le Châtelier’s Principle
CO2(aq) + H2O(l)
Temperature
N2(g) + 3H2(g)
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Factors Affecting
18.3 Reversible Reactions and Equilibrium > Equilibrium:
Le Châtelier’s Principle
Pressure
Remove heat (cool)
Direction of shift
2NH3(g) + heat
In this exothermic reaction, heat is considered to be
a product.
•  During exercise, [CO2] in the blood increases. This shifts
the equilibrium toward reactants.
•  Increase of CO2 also triggers an increase in the rate of
breathing, removing more CO2 through the lungs.
•  Removing CO2 (a product) causes the equilibrium to shift
toward the products
•  this reduces the amount of H2CO3 because the
reactant is converted to products.
17
Add heat
Direction of shift
•  Increasing temperature = adding a product, shift
reaction to formation of reactants.
•  Cooling, or removing heat, shifts the equilibrium
position to the right, and the [product] increases.
18
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Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
Le Châtelier’s Principle
Pressure
When volume decreases, the pressure increases.
Equilibrium systems in which some
reactants and products are gases can
be affected by a change in pressure.
•  A shift will occur only if there are an
unequal number of moles of gas on each
side of the equation.
19
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Initial equilibrium
20
Equilibrium is
disturbed by an
increase in
pressure.
A new equilibrium
position is
established to favor
fewer molecules.
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5
Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
18.3 Reversible Reactions and Equilibrium >
Le Châtelier’s Principle
You can predict which way the equilibrium
position will shift by comparing the number of
molecules of reactants and products.
Add pressure
Direction of shift
Reduce pressure
Direction of shift
2NH3(g)
N2(g) + 3H2(g)
•  When 2 molecules of ammonia (product) form, 4 molecules
of reactants are used up.
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Affecting
18.3 Reversible Reactions and Equilibrium > Factors
Equilibrium:
2NH3(g) + heat
An increase in pressure and a decrease in
temperature would shift the equilibrium toward
the production of ammonia (product).
•  A shift toward ammonia (product) will reduce the number of
molecules.
21
& YOU
Fritz Haber and Karl Bosch figured out how
to increase the yield of ammonia when
nitrogen and hydrogen react. Their success
came from controlling the temperature and
pressure. In which direction did they adjust
each factor and why?
Pressure
N2(g) + 3H2(g)
CHEMISTRY
22
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.2
Le Châtelier’s Principle
Applying Le Châtelier’s Principle
Catalysts and Equilibrium
What effect will each of the following
changes have on the equilibrium position
for this reversible reaction?
Catalysts decrease the time it takes to
establish equilibrium.
PCl5(g) + heat
•  However, they do not affect the amounts of
reactants and products present at
equilibrium.
23
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a.
b.
c.
d.
24
PCl3(g) + Cl2(g)
Cl2 is added.
Pressure is increased.
Heat is removed.
PCl3 is removed as it forms.
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.2
18.3 Reversible Reactions and Equilibrium >
2 Solve Apply the concepts to this problem.
2 Solve Apply the concepts to this problem.
Analyze the effect of an increase in pressure.
Start with the addition of Cl2.
•  Reducing the number of
molecules that are gases
decreases the pressure.
•  Cl2 is a product.
•  Increasing the concentration of a product
shifts the equilibrium to the left.
PCl5(g) + heat
25
Add Cl2
Direction of shift
•  The equilibrium shifts to
the left.
PCl3(g) + Cl2(g)
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.2
PCl5(g) + heat
Increase pressure
Direction of shift
26
For a change in
pressure, compare the
number of molecules of
gas molecules on both
sides of the equation.
PCl3(g) + Cl2(g)
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18.3 Reversible Reactions and Equilibrium >
2 Solve Apply the concepts to this problem.
Sample Problem 18.2
2 Solve Apply the concepts to this problem.
Analyze the effect of removing heat.
Analyze the effect of removing PCl3.
•  The reverse reaction produces heat.
•  PCl3 is a product.
•  The removal of heat causes the
equilibrium to shift to the left.
•  Removal of a product as it forms causes
the equilibrium to shift to the right.
PCl5(g) + heat
27
Sample Problem 18.2
Remove heat
Direction of shift
PCl3(g) + Cl2(g)
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PCl5(g) + heat
28
Remove PCl3
Direction of shift
PCl3(g) + Cl2(g)
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7
18.3 Reversible Reactions and Equilibrium >
18.3 Reversible Reactions and Equilibrium >
In the following equilibrium reaction, in which
direction would the equilibrium position shift
with an increase in pressure?
In the following equilibrium reaction, in which
direction would the equilibrium position shift
with an increase in pressure?
4HCl(g) + O2(g)
4HCl(g) + O2(g)
2Cl2(g) +2H2O(g)
2Cl2(g) +2H2O(g)
Reducing the number of molecules that
are gases decreases the pressure. The
equilibrium will shift to the right.
29
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18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
30
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18.3 Reversible Reactions and Equilibrium >
Equilibrium Constants
What does the size of an equilibrium
constant indicate about a system at
equilibrium?
31
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32
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8
18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
Chemists express the equilibrium position as a
numerical value.
•  This value relates the amounts of reactants
to products at equilibrium.
aA + bB
The equilibrium constant (Keq) is the ratio of
product concentrations to reactant
concentrations at equilibrium.
aA + bB
cC + dD
•  From the general equation, each concentration
is raised to a power equal to the number of
moles of that substance in the balanced
chemical equation.
•  In this general reaction, the coefficients a,
b, c, and d represent the number of moles.
Keq =
33
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18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
cC + dD
34
[C]c x [D]d
[A]a x [B]b
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18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
The value of Keq depends on the temperature
of the reaction.
The size of the equilibrium constant
indicates whether reactants or products
are more common at equilibrium.
•  The flask on the left is in a dish of hot water.
•  The flask on the right is in ice.
Dinitrogen tetroxide is a
colorless gas. Nitrogen
dioxide is a brown gas.
N2O4(g)
35
2NO2(g)
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18.3 Reversible Reactions and Equilibrium > Equilibrium
Constants
18.3 Reversible Reactions and Equilibrium >
Expressing and Calculating Keq
The colorless gas dinitrogen tetroxide (N2O4)
and the brown gas nitrogen dioxide (NO2)
exist in equilibrium with each other.
The size of the equilibrium constant
indicates whether reactants or products
are more common at equilibrium.
•  When Keq has a large value, such as 3.1 x 1011, the
reaction mixture at equilibrium will consist mainly of
product.
N2O4(g)
•  When Keq has an intermediate value, such as 0.15 or
50, the mixture will have significant amounts of both
reactant and product.
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.3
38
UNKNOWN
Keq (algebraic expression) = ?
[NO2] = 0.030 mol/L
Keq (numerical value) = ?
Sample Problem 18.3
2 Calculate Solve for the unknowns.
Modify the general expression for the
equilibrium constant and substitute the known
concentrations to calculate Keq.
[N2O4] = 0.0045 mol/L
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18.3 Reversible Reactions and Equilibrium >
1 Analyze List the knowns and the unknowns.
KNOWNS
2NO2(g)
A liter of the gas mixture at equilibrium
contains 0.0045 mol of N2O4 and 0.030 mol
of NO2 at 10oC. Write the expression for the
equilibrium constant (Keq) and calculate the
value of the constant for the reaction.
•  When Keq has a small value, such as 3.1 x 10–11, the
mixture at equilibrium will consist mainly of reactant.
37
Sample Problem 18.3
•  Start with the general expression for
the equilibrium constant.
Place the concentration of
the product in the
[C]c x [D]d
numerator and the
Keq = [A]a x [B]b
concentration of the
•  Write the equilibrium constant
expression for this reaction.
reactant in the
denominator. Raise each
concentration to the power
equal to its coefficient in
the chemical equation.
[NO2]2
Keq = [N O ]
2 2
39
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40
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10
18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.3
18.3 Reversible Reactions and Equilibrium >
2 Calculate Solve for the unknowns.
3 Evaluate Does the result make sense?
Substitute the concentrations that are known
and calculate Keq.
•  Each concentration is raised to the correct
power.
(0.030 mol/L)2 (0.030 mol/L x 0.030 mol/L)
Keq = (0.0045 mol/L) =
(0.0045 mol/L)
•  The numerical value of the constant is
correctly expressed to two significant figures.
Keq = 0.20 mol/L = 0.20
•  The value for Keq is appropriate for an
equilibrium mixture that contains significant
amounts of both gases.
41
You can ignore the unit
mol/L; chemists report
equilibrium constants
without a stated unit.
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.4
42
H2(g) + I2(g)
Sample Problem 18.4
1 Analyze List the knowns and the unknown.
Find the concentrations of the reactants at
equilibrium. Then substitute the equilibrium
concentrations in the expression for the
equilibrium constant for this reaction.
KNOWNS
UNKNOWN
[H2] (initial) = 1.00 mol/L
Keq = ?
[I2] (initial) = 1.00 mol/L
[HI] (equilibrium) = 1.56 mol/L
2HI(g)
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18.3 Reversible Reactions and Equilibrium >
Finding the Equilibrium Constant
One mole of colorless hydrogen
gas and one mole of violet iodine
vapor are sealed in a 1-L flask and
allowed to react at 450oC. At
equilibrium, 1.56 mol of colorless
hydrogen iodide is present, together
with some of the reactant gases.
Calculate Keq for the reaction.
43
Sample Problem 18.3
44
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11
18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.4
18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.4
2 Calculate Solve for the unknown.
2 Calculate Solve for the unknown.
First find out how much H2 and I2 are consumed
in the reaction.
•  Calculate how much H2 and I2 remain in the
flask at equilibrium.
x + x = 1.56 mol
2x = 1.56 mol
x = 0.780 mol
mol H2 = mol I2 = (1.00 mol – 0.780 mol) = 0.22 mol
Let mol H2 used = mol I2 used = x.
The number of mol H2 and mol I2
used must equal the number of mol
HI formed (1.56 mol).
•  Write the expression
for Keq.
[HI]2
Keq = [H ] x [I ]
2
2
45
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.4
46
Keq =
1.56 mol/L x 1.56 mol/L
0.22 mol/L x 0.22 mol/L
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Sample Problem 18.4
3 Evaluate Does the result make sense?
•  Each concentration is raised to the correct
power.
Substitute the equilibrium concentrations of the
reactants and products into the equation and
solve for Keq.
Keq =
[C]c x [D]d
Keq = [A]a x [B]b
18.3 Reversible Reactions and Equilibrium >
2 Calculate Solve for the unknown.
(1.56 mol/L)2
0.22 mol/L x 0.22 mol/L
Use the general
expression for Keq as a
guide:
•  The value of the constant reflects the
presence of significant amounts of the
reactions and product in the equilibrium
mixture.
Keq = 5.0 x 101
47
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48
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12
18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.5
18.3 Reversible Reactions and Equilibrium >
Finding Concentrations at Equilibrium
Bromine chloride (BrCl) decomposes to form
bromine and chlorine.
2BrCl(g)
1 Analyze List the knowns and the unknowns.
Use the balanced equation, the equilibrium constant,
and the equilibrium constant expression to find the
unknown concentrations. According to the balanced
equation, when BrCl decomposes, equal numbers of
moles of Br2 and Cl2 are formed.
Br2(g) + Cl2(g)
At a certain temperature, the equilibrium
constant for the reaction is 11.1. A sample
of pure BrCl is placed in a 1-L container
and allowed to decompose. At equilibrium,
the reaction mixture contains 4.00 mol Cl2.
What are the equilibrium concentrations of
Br2 and BrCl?
49
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.5
UNKNOWN
[Br2] (equilibrium) = ? mol/L
Keq = 11.1
[BrCl] (equilibrium) = ? mol/L
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18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.5
2 Calculate Solve for the unknowns.
•  The volume of the container is 1 L, so
calculate [Br2] at equilibrium.
•  Rearrange the equation to solve for [BrCl]2.
[BrCl]2 =
4.00 mol
[Br2] =
= 4.00 mol/L
1L
[Br2] x [Cl2]
Keq
•  Substitute the known values for Keq, [Br2], and
[Cl2].
•  Write the equilibrium expression for the reaction.
[Br2] x [Cl2]
Keq =
[BrCl]2
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KNOWNS
[Cl2] (equilibrium) = 4.00 mol/L
50
2 Calculate Solve for the unknowns.
51
Sample Problem 18.5
[BrCl]2 =
4.00 mol/L x 4.00 mol/L
11.1
= 1.44 mol2/L2
52
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13
18.3 Reversible Reactions and Equilibrium >
Sample Problem 18.5
18.3 Reversible Reactions and Equilibrium >
2 Calculate Solve for the unknowns.
3 Evaluate Does the result make sense?
It makes sense that the equilibrium
concentration of the reactant and the
products are both present in significant
amounts because Keq has an intermediate
value.
•  Calculate the square root.
[BrCl] = 1.44 mol2/L2 = 1.20 mol/L
Use your
calculator to find
the square root.
53
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18.3 Reversible Reactions and Equilibrium >
54
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18.3 Reversible Reactions and Equilibrium >
HCl is formed when H2 and Cl2 react at
high temperatures.
H2(g) + Cl2(g)
Sample Problem 18.5
HCl is formed when H2 and Cl2 react at
high temperatures.
2HCl(g)
H2(g) + Cl2(g)
At equilibrium, [HCl] = 1.76 x 10–2 mol/L,
and [H2] = [Cl2] = 1.60 x 10–3 mol/L. What
is the value of the equilibrium constant?
2HCl(g)
At equilibrium, [HCl] = 1.76 x 10–2 mol/L,
and [H2] = [Cl2] = 1.60 x 10–3 mol/L. What
is the value of the equilibrium constant?
Keq =
[HCl]2
[H2] x [Cl2]
=
(1.60 x
(1.76 x 10–2 mol/L)2
mol/L) x (1.60 x 10–3 mol/L)
10–3
Keq = 121
55
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14