Chemical Equilibrium
Chapter 16
General Chemistry - Principles and Modern Applications
Petrucci • Harwood • Herring, 8th Edition; Prentice-Hall © 2002
An Equilibrium is reached when two opposing
processes take place at equal rates.
Dynamic Equilibria of
“Physical” Processes
Equilibrium – two opposing processes taking place at equal rates.
Saturated aqueous solution of I2
in contact with CCl4
Phase Transfer (Chapter 14)
I2(H2O)
I2(CCl4)
Evaporation (Chapter 14)
H2O(l)
H 2O
H 2O
CCl4
CCl4
Time = 0
H2O(g)
Dissolution of a solid in a solvent (water)
(Chapter 19)
NaCl(s)
Time = Equilibrium
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Na+ (aq) + Cl-(aq)
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Dynamic Equilibrium of a
Chemical Process
CO(g) + 2 H2(g)
CH3OH(g)
Forward:
k1
CO(g) + 2 H2(g) → CH3OH(g)
Rfwrd = k1[CO][H2]2
Reverse:
k-1
CH3OH(g) → CO(g) + 2 H2(g)
Rrvrs = k-1[CH3OH]
CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = Rrvrs
k1
k-1

[CH3OH]
=
[CO][H2]2
k1
k-1
CH3OH(g)
k1[CO][H2]2 = k-1[CH3OH]
= Kc (Equilibium Constant)
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
3 Trial & Error Approaches to the
Equilibrium Constant of the Methanol
Reaction (Data from Table 16.1)
CO(g) + 2 H2(g)
k1
k-1
CH3OH(g)
[CH3OH]
[CH3OH]
[CH3OH]
[CO][H2]
[CO](2[H2])
[CO][H2]2
Kc(1) =
1.19 M-1
0.596 M-1
14.2 M-2
Kc(2) =
2.17 M-1
1.09 M-1
14.2 M-2
Kc(3) =
2.55 M-1
1.28 M-1
14.2 M-2
Kc =
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Relationship between Kc and the
Balanced Chemical Equation
a A + b B ….
g G + h H ….
[G]g[H]h ….
Equilibrium Constant (concentrations) = Kc =
[A]m[B]n ….
A more accurate equilibrium constant is Keq, which will be discussed in
detail in Physical Chemistry I (59-240).
Thermodynamic Equilibrium constant = Keq =
aB =
[B]
cB0
= B[B]
(aG)g(aH)h ….
(aA)a(aB)b ….
cB0 is a standard reference state
= 1 mol L-1 (ideal conditions)
aB is dimensionless and so is Keq!
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Important Relationships
• When we reverse a chemical equation, the value of Kc
inverses;
[CH3OH]
CO(g) + 2 H2(g)
CH3OH(g)
= Kc = 14.5
2
[CO][H2]
CH3OH(g)
CO(g) + 2 H2(g)
[CO][H2]2
[CH3OH]
= Kc-1 = 0.069
• When we multiply the coefficients in a balanced equation
by a common factor, we raise the equilibrium constant to
the corresponding power;
• When we divide the coefficients in a balanced equation
by a common factor, we take the corresponding root of
the equilibrium constant;
[CH3OH]
2 = 2.10102
2 CO(g) + 4 H2(g)
2 CH3OH(g)
=
(K
)
c
[CO][H2]2
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Chemical Equilibrium Expressions Might
Be Combined to Find an Unknown KC
N2O(g) + ½O2
2 NO(g) Kc= ?
Known:
N2(g) + ½O2
N2(g) + O2
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)
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Working with Gases – KP
• Mixtures of gases are solutions as are mixtures with
liquids.
• With gases it is more convenient to use KP, which is
based upon partial pressures of gases (Chapter 6), than
the concentration based KC.
2 SO2(g) + O2(g)
2 SO3(g)
[SO3]2
Kc =
[SO2]2[O2]
Use the ideal gas law (Chapter 6):
[SO3] =
nSO3
V
=
PSO3
RT
[SO2] =
[O2] =
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
nSO2
V
nO2
V
=
=
PSO2
RT
PO2
RT
PSO3
2
2
PSO3
]2
RT
[SO3
=
=
Kc =
RT
2
2
2
[SO2] [O2]
PSO2 PO2
PSO2 PO2
RT
RT
KP =
2
PSO3
2
PSO2 PO2
 Kc = KP(RT) and KP = Kc(RT)-1
For a general balanced chemical equation
a A + b B ….
g G + h H ….
KP = Kc(RT)Δn(gas)
Dn = the sum of the stoichiometric coefficients of gaseous
products minus the sum of the stoichiometric coefficients
of gaseous reactants;
Dn = (g + h + …) – (a + b + …)
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Do Examples 16-2 and 16-3
Pure Liquids and Solids
Concentration terms for solid or liquid phases of a
single component (that is, pure solids or liquids)
are, by definition, set to 1 and, therefore, do not
appear in the equilibrium constant expressions!
C(s) + H2O(g)
CO(g) + H2(g)
PCOPH2
[CO][H2]
(RT)-1
Kc =
=
[H2O]
PH2O
CaCO3(s)
CaCO3
CaO
Kc = [CO2]
CaO(s) + CO2(g)
KP = PCO2
KP = Kc(RT)
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Wide Range of KP Values
Reversible Reactions (KC or P = > 10-10 and < 1010);
Quasi Non-Reversible Reactions (KC or P = < 10-10 and > 1010);
The large value of K does not imply that the reaction proceeds at
a given temperature. Why?
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
The Reaction Quotient, Q: Predicting
the Direction of Net Change
• An equilibrium might be approached in various ways;
• The calculation of Q allows us to qualitatively determine
the change of an initial condition as the equilibrium is
approached;
[G]tg[H]th
Qc =
[A]ta[B]tb
At equilibrium Qc = Kc
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Example 16-5
CO(g) + 2 H2(g)
k1
k-1
CH3OH(g)
Direction of Net Chemical Change in Established Equilibrium
KC = 1 at 1100 ºC; Starting amounts are 1 mol each of CO and
H2O as well as 2 mols each of CO2 and H2.
Which substance will be present in greater amounts and which
in lesser amounts once the equilibrium has been established?
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Altering Equilibrium Conditions:
Le Châtellier’s Principle
When a system in equilibrium 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.
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Effect of Changing the Amounts of
Reacting Species on Equilibrium
k1
2 SO2(g) + O2(g)
k-1
2 SO3(g)
Kc = 2.8102 at 1000K
Q > Kc
[SO3]2
Q=
= Kc
2
[SO2] [O2]
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
new equilibrium
concentrations
Effect of Changes in Pressure or
Volume on Equilibrium
• Addition or removal of a gaseous reactant or product
changes Pgases at equilibrium as discussed before.
• Addition of an inert gas changes the total pressure.
– Relative partial pressures and, therefore K, remain
unchanged!
• Change of the volume of the system causes a change in
the equilibrium position.
nSO3
Kc =
2
2
nSO3
]2
[SO3
V
2
=
=

V
=
2.8

10
2
2
[SO2]2[O2]
nSO2 nO2
nSO2 nO2
V
V
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Kc =
[G]g[H]h
[A]a[B]b
[A]a = (nA / VA)a
= nAa / VAa
g
=
=
h
nG nH
nAa nBb
g
nG
nAa
 V(a+b)-(g+h)
h
nH
nBa
 V-Δn
Dn = (g + h + …) – (a + b + …)
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.
Do not confuse with Kc = KP(RT) and KP = Kc(RT)-1
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Effect of Temperature on
Equilibrium
• Raising the temperature of a reaction
mixture at equilibrium shifts the equilibrium
condition in the direction of the endothermic
reaction.
• Lowering the temperature causes a shift in
the direction of the exothermic reaction.
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
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 equilibrium
concentrations and constant.
– But does affect the rate at which
equilibrium is attained!
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Synthesis of ammonia (NH3) from
H2 and N2
The optimum conditions
are only for the
equilibrium position and
do not take into account
the rate at which
equilibrium is attained.
low temp. and
high pressure
high temp. and
lower pressure
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005
Examples 16-10 to -12
and
Feature Problem 91 (p. 664)
Do it!
S. Holger Eichhorn; University of Windsor; General Chemistry II 59-141; © 2005