Principles of Chemical
Equilibrium
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CONTENTS
15-1
Dynamic Equilibrium
15-2
The Equilibrium Constant
Expression
15-3
Relationships Involving
Equilibrium Constants
15-4
The Magnitude of an
Equilibrium Constant
15-5
The Reaction Quotient A:
Predicting the Direction of Net
Change
15-6
Altering Equilibrium
Conditions: Le Châtelier’s
Principle
15-7
Equilibrium Calculations:
Some Illustrative Examples
General Chemistry: Chapter 15
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15-1 Dynamic Equilibrium
Equilibrium –opposing
processes taking place at
equal rates.
H2O(l)
NaCl(s)
I2(H2O)
H2O(g)
H2O
NaCl(aq)
I2(CCl4)
FIGURE 15-1
Dynamic equilibrium in a physical process
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General Chemistry: Chapter 15
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15-2 The Equilibrium Constant
Expression
The oxidation-reduction reaction of copper(II)
and tin(II) in aqueous solution is reversible.
2 Cu2+(aq) + Sn2+(aq)
2
Cu+(aq)
2
Cu2+(aq)
+
Sn4+(aq)
+
Sn2+(aq)
k1
k-1
k1
k-1
2 Cu+(aq) + Sn4+(aq)
2 Cu2+(aq) + Sn2+(aq)
2 Cu+(aq) + Sn4+(aq)
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General Chemistry: Chapter 15
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The Equilibrium Constant and Activities
k-1
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General Chemistry: Chapter 15
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A general expression for K
a A + b B …. → g G + h H ….
Equilibrium constant = Kc=
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[G]g[H]h ….
[A]a[B]b ….
General Chemistry: Chapter 15
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Activity
Thermodynamic concept introduced by Lewis.
Dimensionless ratio referred to a chosen reference
state.
B[B]
aB =
= B[B]
0
cB
cB0 is a standard reference state
= 1 mol L-1 (ideal conditions)
Accounts for non-ideal behaviour in solutions and gases.
An effective concentration.
Slide 7 of 32
General Chemistry: Chapter 15
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Activity
A similar expression applies to gases
BPB
aB =
= BPB
0
PB
Slide 8 of 32
PB0 is a standard reference state
= 1 bar (ideal conditions)
General Chemistry: Chapter 15
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15-3 Relationships Involving the Equilibrium
Constant
Relationship of K to the Balanced Chemical Equation
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.
Slide 9 of 32
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)
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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(aSO3)2
KP =
(aSO2)2(aO2)
PSO2
aSO3 =
P
(PSO3 )2
PO2
P
P
(PSO2)2(PO2)
General Chemistry: Chapter 15
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Pure Liquids and Solids (heterogeneous reactions)
solids and liquids in contact with a gas or solution phase
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)0
Kc =
=
[H2O]
PH2O
The conc. of a component in pure liquid or solid phase cannot change.
15-4 The Significance of the Magnitude of the
Equilibrium Constant.
H2O is much more thermodynamically stable than a mixture of
O2 and H2 as it lies at a lower energy state.
However, this reaction is very slow at 298 K (Ea is very high)
so we must control the kinetics of the reaction by adding a catalyst
or by increasing temp.
If Kc is greater than 1010 we can tell that the reaction goes to
completion.
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What Does the Value of K Mean?
If K>>1, the reaction is
product-favored; product
predominates at
equilibrium.
• If K<<1, the reaction is
reactant-favored; reactant
predominates at
equilibrium.
Le Châtelier’s Principle
“If a system at equilibrium is disturbed by a
change in temperature, pressure, or the
concentration of one of the components, the
system responds by attaining a new
equilibrium that partially offsets the impact
of the change.
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
What happens if we add SO3 to this equilibrium?
Slide 17 of 32
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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]
Kc = 2.8102 at 1000K
Q > Kc
A net change occurs in the direction that reduces [SO3]
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
Slide 19 of 32
V
General Chemistry: Chapter 15
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Effect of Change in Volume
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.
Slide 20 of 32
General Chemistry: Chapter 15
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Effect of the Change of Volume
ntot =
Slide 21 of 32
1.16 mol gas
1.085 mol gas
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.
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; equilibrium is achieved more
rapidly.
So, an equilibrium condition is
independent of the reaction
mechanism.
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 =
[G]ig[H]ih
[A]im[B]in
At equilibrium Qc = Kc
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General Chemistry: Chapter 15
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Reaction Quotient
Gases: we can express The Equilibrium Constant in terms
of concentrations KC or partial pressures KP
2 SO2(g) + O2(g)
[SO3]=
nSO3
V
=
PSO3
[SO2]=
RT
2 SO3(g)
PSO2
RT
[O2] =
PO2
RT
PX
(aX ) =
[X]
=
c◦
RT
c◦
PX = [X] RT
Gases: The Equilibrium Constant, KC
2 SO2(g) + O2(g)
KP =
(aSO3)2
(aSO2)2(aO2)
= P
=
= P
RT
2 SO3(g)
(PSO3 )2
PX = [X] RT
(PSO2)2(PO2 )
([SO3] RT)2
([SO2] RT)2([O2] RT)
([SO3])2
([SO2])2([O2])
=
KC
RT
P
Where P = 1 bar
An Alternative Derivation
2 SO2(g) + O2(g)
PSO3
Kc =
2 SO3(g)
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:
RT
KP = Kc(RT)-1
KP = Kc(RT)Δn