Petrucci • Harwood • Herring • Madura
GENERAL
CHEMISTRY
Ninth
Edition
Principles and Modern Applications
Chapter 15: Principles of Chemical Equilibrium
Philip Dutton
University of Windsor, Canada
Prentice-Hall © 2007
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General Chemistry: Chapter 15
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Contents
15-1
15-2
15-3
15-4
15-5
15-6
15-7
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Dynamic Equilibrium
The Equilibrium Constant Expression
Relationships Involving Equilibrium Constants
The Magnitude of an Equilibrium Constant
The Reaction Quotient, Q: Predicting the Direction of a
Net Change
Altering Equilibrium Conditions:
Le Châtelier’s Principle
Equilibrium Calculations: Some Illustrative Examples
General Chemistry: Chapter 15
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15-1 Dynamic Equilibrium
 Equilibrium – two opposing
processes taking place at
equal rates.
H2O(l)
NaCl(s)
H2O(g)
H2O
CO(g) + 2 H2(g)
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NaCl(aq)
CH3OH(g)
General Chemistry: Chapter 15
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Dynamic Equilibrium
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General Chemistry: Chapter 15
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15-2 The Equilibrium Constant Expression
 Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g)
CH3OH(g)
k-1
k1
CH3OH(g)
CO(g) + 2 H2(g)
CO(g) + 2 H2(g)
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k1
k-1
CH3OH(g)
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Three Approaches to the Equilibrium
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General Chemistry: Chapter 15
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Three Approaches to Equilibrium
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General Chemistry: Chapter 15
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Three Approaches to Equilibrium
CO(g) + 2 H2(g)
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k1
k-1
CH3OH(g)
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The Equilibrium Constant Expression
k1
Forward: CO(g) + 2 H2(g) → CH3OH(g)
k-1
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = Rrvrs
CO(g) + 2 H2(g)
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
k1
k-1
CH3OH(g)
k1[CO][H2]2 = k-1[CH3OH]
k1
k-1
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=
[CH3OH]
[CO][H2
]2
= Kc
General Chemistry: Chapter 15
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General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc=
Thermodynamic
Equilibrium constant = Keq=
=
[G]g[H]h ….
[A]m[B]n ….
(aG)g(aH)h ….
(aA)a(aB)b ….
(G)g(H)h ….
[G]g[H]h ….

m
n
….
(A)a(B)b ….
[A] [B]
1
under ideal conditions
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General Chemistry: Chapter 15
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15-3 Relationships Involving the
Equilibrium Constant
 Reversing an equation causes inversion of K.
 Multiplying by coefficients by a common factor
raises the equilibrium constant to the
corresponding power.
 Dividing the coefficients by a common factor
causes the equilibrium constant to be taken to
that root.
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General Chemistry: Chapter 15
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Combining Equilibrium Constant
Expressions
N2O(g) + ½O2
N2(g) + ½O2
N2(g) + O2
2 NO(g) Kc= ?
N2O(g)
2 NO(g)
Kc(2)=
2.710+18
Kc(3)= 4.710-31
[N2O]
=
[N2][O2]½
[NO]2
=
[N2][O2]
[NO]2
[NO]2 [N2][O2]½
1
-13
Kc=
=
K
=
=
1.710
c(3)
[N2O][O2]½ [N2][O2] [N2O]
Kc(2)
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General Chemistry: Chapter 15
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Gases: The Equilibrium Constant, KP
 Mixtures of gases are solutions just as liquids are.
 Use KP, based upon activities of gases.
2 SO2(g) + O2(g)
aSO3 =
PSO3
P
2 SO3(g)
aSO2 =
KP =
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KP =
PSO2
(aSO2)2(aO2)
aSO3 =
P
(PSO3 )2
(aSO3)2
PO2
P
P
(PSO2)2(PO2)
General Chemistry: Chapter 15
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Gases: The Equilibrium Constant, KC
 In concentration we can do another substitution
2 SO2(g) + O2(g)
[SO3]=
nSO3
V
=
PSO3
[SO2]=
RT
2 SO3(g)
PSO2
RT
[O2] =
PO2
RT
PX
(aX ) =
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[X]
=
c◦
RT
c◦
General Chemistry: Chapter 15
PX = [X] RT
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Gases: The Equilibrium Constant, KC
2 SO2(g) + O2(g)
KP =
(aSO3)2
(aSO2)2(aO2)
= P
=
= P
RT
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2 SO3(g)
(PSO3 )2
PX = [X] RT
(PSO2)2(PO2 )
([SO3] RT)2
([SO2] RT)2([O2] RT)
([SO3])2
([SO2])2([O2])
=
General Chemistry: Chapter 15
KC
P
Where P = 1 bar
RT
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An Alternative Derivation
2 SO2(g) + O2(g)
PSO3
Kc =
Where P = 1 bar
2
2
PSO3
[SO3]
RT
=
=
RT
2
2
2
[SO2] [O2]
PSO2 PO2
PSO2 PO2
RT
Kc = KP(RT)
In general terms:
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2 SO3(g)
RT
KP = Kc(RT)-1
KP = Kc(RT)Δn
General Chemistry: Chapter 15
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Pure Liquids and Solids
 Equilibrium constant expressions do not contain
concentration terms for solid or liquid phases of a
single component (that is, pure solids or liquids).
C(s) + H2O(g)
CO(g) + H2(g)
PCOPH2
[CO][H2]
(RT)1
Kc =
=
[H2O]2
PH2O2
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General Chemistry: Chapter 15
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Burnt Lime
CaCO3(s)
Kc = [CO2]
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CaO(s) + CO2(g)
KP = PCO2(RT)
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15-4 The Significance of the Magnitude of
the Equilibrium Constant.
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General Chemistry: Chapter 15
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15-5 The Reaction Quotient, Q: Predicting
the Direction of Net Change.
CO(g) + 2 H2(g)
k1
k-1
CH3OH(g)
 Equilibrium can be approached various ways.
 Qualitative determination of change of initial
conditions as equilibrium is approached is needed.
Qc =
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[G]tg[H]th
[A]tm[B]tn
At equilibrium Qc = Kc
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Reaction Quotient
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General Chemistry: Chapter 15
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15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
 When an equilibrium system is subjected to a
change in temperature, pressure, or concentration
of a reacting species, the system responds by
attaining a new equilibrium that partially offsets
the impact of the change.
What happens if we add SO3 to this equilibrium?
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General Chemistry: Chapter 15
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Le Châtelier’s Principle
2 SO2(g) + O2(g)
k1
k-1
2 SO3(g)
[SO3]2
Q=
= Kc
2
[SO2] [O2]
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General Chemistry: Chapter 15
Kc = 2.8102 at 1000K
Q > Kc
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Effect of Condition Changes
 Adding a gaseous reactant or product changes Pgas.
 Adding an inert gas changes the total pressure.
 Relative partial pressures are unchanged.
 Changing the volume of the system causes a change
in the equilibrium position.
nSO3
2
2
[SO3]
nSO3
V
Kc =
=
=
V
2
2
2
[SO2] [O2]
nSO2 nO2
nSO2 nO2
V
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V
General Chemistry: Chapter 15
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Effect of Change in Volume
[G]g[H]h
Kc =
[C]c[D]d
g
=
h
nG nH
nAa nBa
=
g
nG
V(a+b)-(g+h)
h
nH
nAa nBa
V-Δn
 When the volume of an equilibrium mixture of
gases is reduced, a net change occurs in the
direction that produces fewer moles of gas. When
the volume is increased, a net change occurs in the
direction that produces more moles of gas.
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General Chemistry: Chapter 15
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Effect of the Change of Volume
ntot =
1.16 mol gas
KP =
415
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1.085 mol gas
338
General Chemistry: Chapter 15
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Effect of Temperature on Equilibrium
 Raising the temperature of an equilibrium
mixture shifts the equilibrium condition in the
direction of the endothermic reaction.
 Lowering the temperature causes a shift in the
direction of the exothermic reaction.
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General Chemistry: Chapter 15
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Effect of a Catalyst on Equilibrium
 A catalyst changes the mechanism of a
reaction to one with a lower activation energy.
 A catalyst has no effect on the condition of
equilibrium.
 But does affect the rate at which equilibrium is
attained.
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General Chemistry: Chapter 15
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15-7 Equilibrium Calculations:
Some Illustrative Examples.
 Five numerical examples are given in the text
that illustrate ideas that have been presented in
this chapter.
 Refer to the “comments” which describe the
methodology. These will help in subsequent
chapters.
 Exercise your understanding by working
through the examples with a pencil and paper.
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General Chemistry: Chapter 15
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