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GENERAL CHEMISTRY
Principles and Modern Applications
PETRUCCI
HERRING
MADURA
TENTH EDITION
BISSONNETTE
Principles of Chemical
Equilibrium
15
PHILIP DUTTON
UNIVERSITY OF WINDSOR
DEPARTMENT OF CHEMISTRY AND
BIOCHEMISTRY
Slide 1 of 32
General Chemistry: Chapter 15
Copyright © 2011 Pearson Canada Inc.
Principles of Chemical
Equilibrium
Slide 2 of 32
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
Copyright © 2011 Pearson Canada Inc.
15-1 Dynamic Equilibrium
Equilibrium –opposing
processes taking place at
equal rates.
H2O(l)
NaCl(s)
I2(H2O)
I2(CCl4)
H2O(g)
H2O
CO(g) + 2 H2(g)
NaCl(aq)
CH3OH(g)
FIGURE 15-1
Dynamic equilibrium in a physical process
Slide 3 of 32
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)
Slide 5 of 32
General Chemistry: Chapter 15
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FIGURE 15-3
Three approaches to equilibrium in the reaction
2 Cu2+(aq) + Sn2+(aq)
2 Cu+(aq) + Sn4+(aq)
Slide 6 of 32
General Chemistry: Chapter 15
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The Equilibrium Constant and Activities
k-1
Slide 7 of 32
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 8 of 32
General Chemistry: Chapter 15
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Activity
A similar expression applies to gases
BPB
aB =
= BPB
0
PB
Slide 9 of 32
PB0 is a standard reference state
= 1 bar (ideal conditions)
General Chemistry: Chapter 15
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Reconsider the equilibrium between Cu2+ and Sn2+
Slide 10 of 32
General Chemistry: Chapter 15
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Reconsider the equilibrium between Cu2+ and Sn2+
Slide 11 of 32
General Chemistry: Chapter 15
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A general expression for K
a A + b B …. → g G + h H ….
Equilibrium constant = Kc=
[G]g[H]h ….
[A]m[B]n ….
Thermodynamic Equilibrium constant
Keq=
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1
(aG)g(aH)h ….
≈
c°
(aA)a(aB)b ….
Δn
[G]g[H]h ….
1
=
m
n
….
c°
[A] [B]
General Chemistry: Chapter 15
Δn
Keq
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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 13 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)
Slide 14 of 32
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
2 SO2(g) + O2(g)
aSO3 =
PSO3
P
2 SO3(g)
aSO2 =
KP =
Slide 15 of 32
KP =
PSO2
(aSO2)2(aO2)
aSO3 =
P
(PSO3 )2
(aSO3)
PO2
P
P
(PSO2)2(PO2)
General Chemistry: Chapter 15
Copyright © 2011 Pearson Canada Inc.
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:
Slide 18 of 32
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
Slide 19 of 32
General Chemistry: Chapter 15
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Burnt Lime
CaCO3(s)
Kc = [CO2]
Slide 20 of 32
CaO(s) + CO2(g)
KP = PCO2(RT)
General Chemistry: Chapter 15
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15-4 The Significance of the Magnitude of the
Equilibrium Constant.
Slide 21 of 32
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. [G]tg[H]th
Qc =
Slide 22 of 32
[A]tm[B]tn
At equilibrium Qc = Kc
General Chemistry: Chapter 15
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Reaction Quotient
Slide 23 of 32
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?
Slide 24 of 32
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]
Slide 25 of 32
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
Slide 26 of 32
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.
Slide 27 of 32
General Chemistry: Chapter 15
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Effect of the Change of Volume
Slide 28 of 32
ntot =
1.16 mol gas
KP =
415
1.085 mol gas
338
General Chemistry: Chapter 15
Copyright © 2011 Pearson Canada Inc.
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.
Slide 29 of 32
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.
Slide 30 of 32
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.
Slide 31 of 32
General Chemistry: Chapter 15
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End of Chapter Questions
Problems can be solved by eliminating
errors from your approach.
There may be nothing wrong with your
strategy, but for some reason the problem
is not solving.
Be willing to make errors.
Be able to recognize them.
Slide 32 of 32
General Chemistry: Chapter 15
Copyright © 2011 Pearson Canada Inc.