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Ch. 18 Chemical Equilibrium
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Chapter 18
18.1 The Nature of Chemical Equilibrium
• List two everyday processes that can easily be
reversed and two that cannot.
• The freezing of water and the melting of ice
can be reversed
• The cooking of an egg or the lighting of a
match cannot be reversed.
• Reactions that “go to completion” refer to a
reaction in which all the reactants are converted
to products
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Chapter 18
Reversible Reactions
• But theoretically, every reaction can proceed in two
directions, forward and reverse.
• Essentially all chemical reactions are considered to
be reversible under suitable conditions.
• A chemical reaction in which the products can react
to re-form the reactants is called a reversible
reaction.

 2Hg(l ) + O2 (g )
2HgO(s ) 

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Chapter 18
• Chemical equilibrium is established when the rate
of its forward reaction equals the rate of its reverse
reaction
• When this happens the amounts of products
(concentrations) and reactants remain constant.
• Both reactions continue, but there is no net
change in their concentration
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Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
• products of the forward reaction favored, lies to the right
2SO2 (g ) + O2 (g ) 

 2SO3 (g )
• products of the reverse reaction favored, lies to the left

–


H2CO3 (aq ) + H2O(l) 
H
O
(
aq
)
+
HCO
 3
3 (aq )
• Neither reaction is favored

–


H2SO3 (aq ) + H2O(l ) 
H
O
(
aq
)
+
HSO
 3
3 ( aq )
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Chapter 18

 xC  yD
nA  mB 

• After equilibrium is reached, the individual concentrations
of A, B, C, and D undergo no further change if conditions
remain the same.
• i.e. : the ratio of their concentrations remains constant.
• The equilibrium constant is designated by the letter K.
•The expression
for K =
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Chapter 18
• The Equilibrium Constant Expression, K, is the ratio of
the product of the concentrations of substances formed at
equilibrium to the product of the concentrations of reacting
substances. Each concentration is raised to a power equal
to the coefficient of that substance in the chemical
equation.
• K is dependent on temperature and always determined
experimentally
• Only the concentrations of substances that can actually
change are included in K.
• So pure solids and liquids are omitted because their
concentrations cannot change.
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Chapter 18
What would be true if K = 1?
• Concentration of products = concentration of reactants
What would be true if K is greater than 1?
• Large K values indicate that the products are favored
(forward reaction)
What would be true if K is less than 1?
• Small K values indicate the reactants are favored (reverse
reaction)
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What does knowing the value of Keq help us predict?
• Whether the forward of the reverse reaction is favored
• Therefore, tell us if the reactants or products will be in
higher concentrations when equilibrium is established
*Does not tell us how quickly the reaction reaches the
equilibrium
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Examples of Equilibrium Constant Expressions
• Make sure you always balance an equation before
writing an equilibrium constant expression
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Examples of Equilibrium Constant Expressions
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Chapter 18
Sample Problem A
An equilibrium mixture of N2, O2 , and NO gases at
1500 K is determined to consist of 6.4  10–3 mol/L of
N2, 1.7  10–3 mol/L of O2, and 1.1  10–5 mol/L of NO.
What is the equilibrium constant for the system at
this temperature?
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Chapter 18
Section 1 The Nature of Chemical
Equilibrium
The Equilibrium Expression, continued
Sample Problem A Solution
Given: [N2] = 6.4  10–3 mol/L
[O2] = 1.7  10–3 mol/L
[NO] = 1.1  10–5 mol/L
Unknown: K
Solution:
The balanced chemical equation is

 2NO(g )
N2 (g ) + O2 (g ) 

The chemical equilibrium expression is
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Chapter 18
Section 18.2 Le Châtelier’s principle
• Imagine children playing on a seesaw.
• Five boys are sitting on one side and five girls on the
other, and the seesaw is just balanced.
• Then, one girl gets off, and the system is no longer at
equilibrium.
• One way to get the seesaw in balance again is for
one of the boys to move toward the girls’ side.
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Chapter 18
Section 2 Shifting Equilibrium
• Le Châtelier’s principle states that if a system at
equilibrium is subjected to a stress, the equilibrium is
shifted in the direction that tends to relieve the stress.
• This principle is true for all dynamic
equilibria, chemical as well as physical.
• Changes in pressure, concentration, and
temperature illustrate Le Châtelier’s
principle because they alter the equilibrium
position and thereby change the relative
amounts of reactants and products
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Chapter 18
Predicting the Direction of Shift: Pressure
• Change in pressure affects only equilibrium systems
in which gases are involved.
• An increase in pressure is an applied stress.
• The system can reduce the total pressure by
reducing the number of molecules.
• Therefore, increase in pressure favors the
reaction that produces fewer gas molecules
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Chapter 18
• Ex: Haber process for the synthesis of ammonia

 2NH3 (g )
N2 (g ) + 3H2 (g ) 

4 molecules of gas
2 molecules of gas
• When pressure is applied, the equilibrium will shift to
the right, and produce more NH3.
• A reduction in the total number of molecules
leads to a decrease in pressure.
Note: changes in pressure do not affect the value of the
equilibrium constant
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Chapter 18
Changes in Concentration
• An increase in the concentration of a reactant is a
stress on the equilibrium system.

C  D
A  B 

• To relieve the stress, some of the added A reacts with B
to form products C and D.
• Increase in concentration of one substance
shifts the equilibrium to the opposite side
(i.e. opposite side favored)
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Chapter 18
• Note: changes in concentration have no effect on
the value of the equilibrium constant.
• Such changes have an equal effect on the numerator and
the denominator of the chemical equilibrium expression.
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Chapter 18
Changes in Temperature
• Reversible reactions are exothermic in one direction
and endothermic in the other.
• The addition of energy in the form of heat shifts the
equilibrium so that energy is absorbed.
• An increase in temperature favors the
endothermic reaction
• A decreased in temperature (removal of energy)
favors the exothermic reaction
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Chapter 18
• A rise in temperature increases the rate of any reaction.
• In an equilibrium system, the rates of the opposing
reactions are raised unequally.
**The value of the equilibrium constant for a given
system is affected by the temperature.
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Chapter 18
Section 2 Shifting Equilibrium
• Haber Process: exothermic

 2NH3 (g ) + 92 kJ
N2 (g ) + 3H2 (g ) 

• A rise in temperature increases the rate of any reaction,
but the rates of the opposing reactions are raised
unequally.
• The value of the equilibrium constant for a given
system is affected by the temperature.
• A high temperature favors the decomposition of
ammonia, the endothermic reaction.
• At low temperatures, the forward reaction is too
slow to be commercially useful.
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Temperature Changes Affect an Equilibrium
System
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Chapter 18
Reactions That Go to Completion
• Some reactions involving compounds formed by the
chemical interaction of ions in solutions appear to go
to completion in the sense that the ions are almost
completely removed from solution.
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Chapter 18
3 Examples of Reactions that go to Completion
1. Formation of a Gas
H2CO3(aq)
H2O(l) + CO2(g)
• Reaction goes practically to completion because one
of the products, CO2, escapes as a gas if the
container is open to the air.
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Chapter 18
2. Formation of a Precipitate
Na (aq ) + Cl (aq ) + Ag (aq ) + NO3 (aq ) 
Na (aq ) + NO3 (aq ) + AgCl(s )
• If chemically equivalent amounts of the two solutes
are mixed, almost all of the Ag+ ions and Cl− ions
combine and separate from the solution as a
precipitate of AgCl.
• AgCl is only very sparingly soluble in water.
• The reaction thus effectively goes to completion because
an essentially insoluble product is formed.
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Chapter 18
3. Formation of a Slightly Ionized Product
• Neutralization reactions between H3O+ ions from
aqueous acids and OH− ions from aqueous bases
result in the formation of water molecules, which are
only slightly ionized.
H3O (aq ) + Na  (aq ) + Cl (aq ) + OH (aq ) 
Na (aq ) + Cl (aq ) + 2H2O(aq )
• Hydronium ions and hydroxide ions are almost entirely
removed from the solution.
• The reaction effectively runs to completion because the
product is only slightly ionized.
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Chapter 18
Common-Ion Effect
• The phenomenon in which the addition of an ion
common to two solutes brings about precipitation or
reduced ionization of a reactant
• Ex: hydrogen chloride gas is bubbled into a saturated
solution of sodium chloride.
HCl(g ) + H2O(l )  H3O (aq ) + Cl (aq )




NaCl(s ) 
Na
(
aq
)
+
Cl
(aq )

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Chapter 18
Common-Ion Effect, continued
• As the hydrogen chloride dissolves in sufficient
quantity, it increases the concentration of Cl− ions in
the solution, which is a stress on the equilibrium
system.
• The system can compensate by forming some solid
NaCl. The NaCl precipitates out, relieving the stress
of added chloride (reactions shifts to the left)




NaCl(s ) 
Na
(
aq
)
+
Cl
(aq )

• The new equilibrium has a greater concentration of
Cl− ions but a decreased concentration of Na+ ions.
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Chapter 18
Section 2 Shifting Equilibrium
Particle Model for the Common-Ion Effect
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Chapter 18
Solubility Product
• Soluble ionic compounds dissolve in water until they
are at equilibrium with their ions



AgCl(s ) 
Ag
(
aq)
+
Cl
(aq )

• The solubility product constant, Ksp, of a substance
is the product of the molar concentrations of its ions in
a saturated solution, each raised to the power that is
the coefficient of that ion in the balanced chemical
equation.
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
• For dissolution of the ionic compound AaBb (A cation, B
anion), the solubility-production expression is
AaBb (s) → aA + bB
Ksp = [A]a [B]b
• The equilibrium expression is written without including the
solid species (or water)
• The numerical value of Ksp helps us determine how
soluble a compound is in water
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
• Compounds with large Ksp values are more soluble
• Compounds with smaller Ksp values are less soluble
• The solubility of a solid is an equilibrium position that
represents the amount of the solid required to form a
saturated solution with a specific amount of solvent.
• It has an infinite number of possible values at a given
temperature and is dependent on other conditions,
such as the presence of a common ion.
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
• For a saturated solution of CaF2, the equilibrium
equation is
2
–


CaF2 (s ) 
Ca
(
aq
)

2F
(aq )

• The expression for the solubility product constant is
Ksp = [Ca] [F]2
• The solubility of CaF2 is 8.6  10−3 g/100 g of water at
25°C. Expressed in moles per liter this
concentration becomes 1.1  10−3 mol/L.
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Chapter 18
Section 3 Equilibria of Acids, Bases, and
Salts
Ionization Constant of Acids & Bases
• When we write the equiblium constant expression for
the ionization of an acid/base we call them
• Ka :acid ionization constant.
• Kb: base ionization constant.
• The acid ionization constant, Ka , is constant for a
specified temperature
• Weak acids, like CH3COOH, remainly largely
unionized as will have small Ka values.
• Strong acids, like HNO3, ionize completely and will
have larger Ka values.
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Chapter 18
Section 3 Equilibria of Acids,
Bases, and Salts
Ionization Constant of Water, Kw
• The self-ionization of water is an equilibrium reaction.




H2O(l ) + H2O(l ) 
H
O
(
aq
)
+
OH
(aq )
 3
• Equilibrium is established with a very low concentration
of H3O+ and OH− ions.
Kw=[H3O+][OH–] = 1.0  10-14
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Chapter 18
Section 4 Solubility Equilibrium
Determining Ksp for Reactions at Chemical
Equilibrium
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
• CaF2 dissociates to yield twice as many F− ions as Ca2+
ions.
[Ca2+] = 1.1  10−3 mol/L
[F− ] = 2.2  10−3 mol/L
Ksp  [Ca2 ][F ]2
Ksp = 5.3  10-9
• Calculations of Ksp ordinarily should be limited to
two significant figures.
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product Constants at 25°C
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
Sample Problem B
Calculate the solubility product constant, Ksp ,for
copper(I) chloride, CuCl, given that the solubility
of this compound at 25°C is 1.08  10–2 g/100. g
H2O.
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
Sample Problem B Solution
Given: solubility of CuCl = 1.08  10−2 g CuCl/100. g H2O
Unknown: Ksp
Solution:

 Cu (aq )  Cl– (aq )
CuCl(s ) 

Ksp=[Cu+][Cl–]
[Cu+] = [Cl–] = solubility in mol/L
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Chapter 18
Section 4 Solubility Equilibrium
Solubility Product, continued
Sample Problem B Solution, continued
1.09  10-3 mol/L CuCl
[Cu+] = [Cl–]=1.09  10-3 mol/L
Ksp = (1.09  10-3)(1.09  10-3) =
1.19  10-6
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Chapter 18
Section 4 Solubility Equilibrium
Calculating Solubilities
• The solubility product constant can be used to
determine the solubility of a sparingly soluble salt.
• How many moles of barium carbonate, BaCO3, can be
dissolved in 1 L of water at 25°C?
2
2–


BaCO3 (s ) 
Ba
(
aq
)

CO

3 (aq )
• The molar solubility of BaCO3 is 7.1  10−5 mol/L.
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Chapter 18
Section 4 Solubility Equilibrium
Calculating Solubilities, continued
Sample Problem C
Calculate the solubility of silver bromide, AgBr,
in mol/L, using the Ksp value for this compound.
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Chapter 18
Section 4 Solubility Equilibrium
Calculating Solubilities, continued
Sample Problem C Solution
Given: Ksp = 5.0 10−13
Unknown: solubility of AgBr
Solution:




AgBr(s ) 
Ag
(
aq)
+
Br
(aq )

[Ag+] = [Br−], so let [Ag+] = x and [Br−] = x
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Chapter 18
Section 4 Solubility Equilibrium
Limitations on the Use of Ksp
• The solubility product principle can be very useful
when applied to solutions of sparingly soluble
substances.
• It cannot be applied very successfully to solutions of
moderately soluble or very soluble substances.
• The positive and negative ions attract each other, and this
attraction becomes appreciable when the ions are close
together.
• Sometimes it is necessary to consider two equilibria
simultaneously.
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Chapter 18
Section 4 Solubility Equilibrium
Equilibrium
Calculations
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Chapter 18
Standardized Test Preparation
Multiple Choice

1. A chemical reaction is in equilibrium when

A.
forward and reverse reactions have ceased.

B.
the equilibrium constant equals 1.

C.
forward and reverse reaction rates are equal.

D.
No reactants remain.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
Standardized Test Preparation
Multiple Choice

1. A chemical reaction is in equilibrium when

A.
forward and reverse reactions have ceased.

B.
the equilibrium constant equals 1.

C.
forward and reverse reaction rates are equal.

D.
No reactants remain.
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Chapter 18
Standardized Test Preparation
Multiple Choice

2. Which change can cause the value of the equilibrium

constant to change?

A.
temperature

B.
concentration of a reactant

C.
concentration of a product

D.
None of the above
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
Standardized Test Preparation
Multiple Choice

2. Which change can cause the value of the equilibrium

constant to change?

A.
temperature

B.
concentration of a reactant

C.
concentration of a product

D.
None of the above
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Chapter 18
Standardized Test Preparation
Multiple Choice

3. Consider the following reaction:

 2CO(g )
2C(s ) + O2 (g ) 


The equilibrium constant expression for this reaction is

A.
C.
B.
D.


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Chapter 18
Standardized Test Preparation
Multiple Choice

3. Consider the following reaction:

 2CO(g )
2C(s ) + O2 (g ) 


The equilibrium constant expression for this reaction is

A.
C.
B.
D.


Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
Standardized Test Preparation
Multiple Choice

4. The solubility product of cadmium carbonate, CdCO3, is
1.0  10−12. In a saturated solution of this salt, the
concentration of Cd2+(aq) ions is

A.
5.0 . 10−13 mol/L.

B.
1.0 . 10−12 mol/L.

C.
1.0 . 10−6 mol/L.

D.
5.0 . 10−7 mol/L.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
Standardized Test Preparation
Multiple Choice

4. The solubility product of cadmium carbonate, CdCO3, is
1.0  10−12. In a saturated solution of this salt, the
concentration of Cd2+(aq) ions is

A.
5.0 . 10−13 mol/L.

B.
1.0 . 10−12 mol/L.

C.
1.0 . 10−6 mol/L.

D.
5.0 . 10−7 mol/L.
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Chapter 18
Standardized Test Preparation
Multiple Choice

5. Consider the following equation for an equilibrium
system:

 2Pb(s ) + CO2 (g ) + 2SO2 (g )
2PbS(
s ) + 3O2 (g ) + C(s ) 



Which concentration(s) would be included in the
denominator of the equilibrium constant expression?


A.
Pb(s), CO2(g), and SO2(g)

B.
PbS(s), O2(g), and C(s)

C.
O2(g), Pb(s), CO2(g), and SO2(g)

D.
O2(g)
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Chapter 18
Standardized Test Preparation
Multiple Choice

5. Consider the following equation for an equilibrium
system:

 2Pb(s ) + CO2 (g ) + 2SO2 (g )
2PbS(
s ) + 3O2 (g ) + C(s ) 



Which concentration(s) would be included in the
denominator of the equilibrium constant expression?


A.
Pb(s), CO2(g), and SO2(g)

B.
PbS(s), O2(g), and C(s)

C.
O2(g), Pb(s), CO2(g), and SO2(g)

D.
O2(g)
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Chapter 18
Standardized Test Preparation
Multiple Choice

6. If an exothermic reaction has reached equilibrium,

then increasing the temperature will

A.
favor the forward reaction.

B.
favor the reverse reaction.

C.
favor both the forward and reverse reactions.

D.
have no effect on the equilibrium.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Chapter 18
Standardized Test Preparation
Multiple Choice

6. If an exothermic reaction has reached equilibrium,

then increasing the temperature will

A.
favor the forward reaction.

B.
favor the reverse reaction.

C.
favor both the forward and reverse reactions.

D.
have no effect on the equilibrium.
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Chapter 18
Standardized Test Preparation
Multiple Choice



7. Le Châtelier’s principle states that
A.
at equilibrium, the forward and reverse reaction
rates are equal.

B.
stresses include changes in concentrations,
pressure, and temperature.

C.
to relieve stress, solids and solvents are omitted
from equilibrium constant expressions.

D.
chemical equilibria respond to reduce applied
stress.
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Chapter 18
Standardized Test Preparation
Multiple Choice



7. Le Châtelier’s principle states that
A.
at equilibrium, the forward and reverse reaction
rates are equal.

B.
stresses include changes in concentrations,
pressure, and temperature.

C.
to relieve stress, solids and solvents are omitted
from equilibrium constant expressions.

D.
chemical equilibria respond to reduce applied
stress.
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