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![Page 1: Prentice-Hall © 2007 General Chemistry: Chapter 15 Slide 1 of 33 Philip Dutton University of Windsor, Canada Prentice-Hall © 2007 CHEMISTRY Ninth Edition](https://reader035.vdocument.in/reader035/viewer/2022081421/56649d605503460f94a4108a/html5/thumbnails/1.jpg)
Prentice-Hall © 2007General Chemistry: Chapter 15Slide 1 of 33
Philip DuttonUniversity of Windsor, Canada
Prentice-Hall © 2007
CHEMISTRYNinth
Edition GENERAL
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
Chapter 15: Principles of Chemical Equilibrium
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 2 of 33
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, Q: Predicting the Direction of a Net Change
15-6 Altering Equilibrium Conditions:
Le Châtelier’s Principle
15-7 Equilibrium Calculations: Some Illustrative Examples
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 3 of 33
15-1 Dynamic Equilibrium
Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
NaCl(s) NaCl(aq)H2O
CO(g) + 2 H2(g) CH3OH(g)
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 4 of 33
Dynamic Equilibrium
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 5 of 33
15-2 The Equilibrium Constant Expression
Methanol synthesis is a reversible reaction.
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
CH3OH(g) CO(g) + 2 H2(g)
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 6 of 33
Three Approaches to the Equilibrium
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 7 of 33
Three Approaches to Equilibrium
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 8 of 33
Three Approaches to Equilibrium
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 9 of 33
The Equilibrium Constant Expression
Forward: CO(g) + 2 H2(g) → CH3OH(g)
Reverse: CH3OH(g) → CO(g) + 2 H2(g)
At Equilibrium:
Rfwrd = k1[CO][H2]2
Rrvrs = k-1[CH3OH]
Rfwrd = Rrvrs
k1[CO][H2]2 = k-1[CH3OH]
[CH3OH]
[CO][H2]2=
k1
k-1
= Kc
CO(g) + 2 H2(g) CH3OH(g) k1
k-1
k1
k-1
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 10 of 33
General Expressions
a A + b B …. → g G + h H ….
Equilibrium constant = Kc= [A]m[B]n ….
[G]g[H]h ….
Thermodynamic Equilibrium constant = Keq= (aG)g(aH)h ….
(aA)a(aB)b ….
[A]m[B]n ….
[G]g[H]h …. (G)g(H)h ….
(A)a(B)b ….=
1 under ideal conditions
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 11 of 33
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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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 12 of 33
Combining Equilibrium Constant Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
Kc= [N2O][O2]½
[NO]2
=[N2][O2]½
[N2O][N2][O2]
[NO]2
Kc(2)
1Kc(3)= = 1.710-13
[N2][O2]
[NO]2
=
[N2][O2]½
[N2O]=N2(g) + ½O2 N2O(g) Kc(2)= 2.710+18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 13 of 33
Gases: The Equilibrium Constant, KP
Mixtures of gases are solutions just as liquids are. Use KP, based upon activities of gases.
KP = (aSO )2(aO )
(aSO )23
22
=P
PSO3 aSO 3=
PPSO2aSO 2 =
PPO2aSO 3
KP = (PSO )2(PO )
(PSO )23
22
P
2 SO2(g) + O2(g) 2 SO3(g)
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 14 of 33
Gases: The Equilibrium Constant, KC
In concentration we can do another substitution
2 SO2(g) + O2(g) 2 SO3(g)
[SO3]=V
nSO3 = RT
PSO3 [SO2]=RT
PSO2 [O2] = RT
PO2
(aX ) = [X]c◦
RT
PX
c◦= PX = [X] RT
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 15 of 33
Gases: The Equilibrium Constant, KC
2 SO2(g) + O2(g) 2 SO3(g)
KP = (aSO )2(aO )
(aSO )23
22
=(PSO )2(PO )
(PSO )23
22
P PX = [X] RT
=([SO2] RT)2([O2] RT)
([SO3] RT)2
P
=([SO2])2([O2])
([SO3])2P
RT=
RT
KC Where P = 1 bar
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 16 of 33
An Alternative Derivation
2 SO2(g) + O2(g) 2 SO3(g)
Kc = [SO2]2[O2]
[SO3] RT
PSO3
2
RT
PSO2
RT
PO2
=
2
= RTPSO3
2PSO2PO2
2
Kc = KP(RT) KP = Kc(RT)-1
KP = Kc(RT)ΔnIn general terms:
Where P = 1 bar
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 17 of 33
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).
Kc = [H2O]2
[CO][H2] =PH2O
2
PCOPH2 (RT)1
C(s) + H2O(g) CO(g) + H2(g)
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 18 of 33
Burnt Lime
CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2] KP = PCO2(RT)
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 19 of 33
15-4 The Significance of the Magnitude of the Equilibrium Constant.
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 20 of 33
15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.
CO(g) + 2 H2(g) CH3OH(g)k1
k-1
Equilibrium can be approached various ways. Qualitative determination of change of initial
conditions as equilibrium is approached is needed.
At equilibrium Qc = Kc Qc = [A]t
m[B]tn
[G]tg[H]t
h
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 21 of 33
Reaction Quotient
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 22 of 33
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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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 23 of 33
Le Châtelier’s Principle
Q = = Kc [SO2]2[O2]
[SO3]2
Q > Kc
Kc = 2.8102 at 1000K2 SO2(g) + O2(g) 2 SO3(g)k1
k-1
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 24 of 33
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.
Kc = [SO2]2[O2]
[SO3] V
nSO3
2
V
nSO2
V
nO2
=
2
= VnSO3
2nSO2nO2
2
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 25 of 33
Effect of Change in Volume
Kc = [C]c[D]d
[G]g[H]h
= V(a+b)-(g+h)nG
anA nB
g nHh
a
V-ΔnnG
anA nB
g nHh
a
=
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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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 26 of 33
Effect of the Change of Volume
1.085 mol gas1.16 mol gasntot =
KP = 338415
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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 27 of 33
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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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 28 of 33
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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Prentice-Hall © 2007General Chemistry: Chapter 15Slide 29 of 33
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.