1. Chapter 15: Principles of Chemical Equilibrium
Chemistry 140 Fall 2002
CHEMISTRY
Ninth
Edition
GENERAL
Principles and Modern Applications
Petrucci • Harwood • Herring • Madura
Chapter 15: Principles of Chemical Equilibrium
Philip Dutton
University of Windsor, Canada
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General Chemistry: Chapter 15
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2. Contents 15-1 Dynamic Equilibrium
Chemistry 140 Fall 2002
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
Focus On The Nitrogen Cycle and the Synthesis of Nitrogen Compounds
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3. General Chemistry: Chapter 15
15-1 Dynamic Equilibrium
Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
NaCl(s) NaCl(aq)
H2O
I2(H2O) I2(CCl4)
CO(g) + 2 H2(g) CH3OH(g)
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4. General Chemistry: Chapter 15
Dynamic Equilibrium
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5. 15-2 The Equilibrium Constant Expression
Methanol synthesis is a reversible reaction. k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
CH3OH(g) CO(g) + 2 H2(g) k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
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6. Three Approaches to the Equilibrium
Chemistry 140 Fall 2002
Three Approaches to the Equilibrium
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7. Three Approaches to Equilibrium
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8. Three Approaches to Equilibriumk1
CO(g) + 2 H2(g) CH3OH(g)
k-1
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9. The Equilibrium Constant Expressionk1
Forward:
CO(g) + 2 H2(g) → CH3OH(g)
Rfwrd = k1[CO][H2]2
k-1
Reverse:
CH3OH(g) → CO(g) + 2 H2(g)
Rrvrs = k-1[CH3OH]
At Equilibrium:
CO(g) + 2 H2(g) CH3OH(g) k1
k-1
Rfwrd = Rrvrs
k1[CO][H2]2 = k-1[CH3OH]
[CH3OH]
[CO][H2]2 = k1
k-1
= Kc
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10. The Equilibrium Constant and Activities
Activity
Thermodynamic concept introduced by Lewis.
Dimensionless ratio referred to a chosen reference state.
aB =
[B]
cB0
cB0 is a standard reference state
= 1 mol L-1 (ideal conditions)
= B[B]
Accounts for non-ideal behaviour in solutions and gases.
An effective concentration.
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11. General Expressions a A + b B …. → g G + h H …. [G]g[H]h ….
Chemistry 140 Fall 2002
General Expressions
a A + b B …. → g G + h H ….
[G]g[H]h ….
Equilibrium constant = Kc=
[A]m[B]n ….
Thermodynamic
Equilibrium constant = Keq=
(aG)g(aH)h ….
(aA)a(aB)b ….
[G]g[H]h ….
(G)g(H)h …. = 
[A]m[B]n ….
(A)a(B)b …. 1
under ideal conditions
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12. 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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13. Combining Equilibrium Constant Expressions
N2O(g) + ½O NO(g) Kc= ?
[N2][O2]½
[N2O] =
N2(g) + ½O N2O(g) Kc(2)= 2.710+18
[N2][O2]
[NO]2 =
N2(g) + O NO(g) Kc(3)= 4.710-31
Kc=
[N2O][O2]½
[NO]2 =
[N2][O2]½
[N2O]
[N2][O2]
[NO]2
Kc(2) 1
Kc(3) =
= 1.710-13
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14. 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 )2 3 2
2 SO2(g) + O2(g) SO3(g) = P
PSO3
aSO 3 = P
PSO2
aSO 2 = P
PO2
aSO 3
KP =
(PSO )2(PO )
(PSO )2 3 2 P
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15. Gases: The Equilibrium Constant, KC
In concentration we can do another substitution
2 SO2(g) + O2(g) SO3(g)
[SO3]= V
nSO3 = RT
PSO3
PSO2
PO2
[SO2]=
[O2] = RT RT PX
[X] RT PX =
[X] RT
(aX ) = = c◦ c◦
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16. Gases: The Equilibrium Constant, KC
2 SO2(g) + O2(g) SO3(g)
(aSO )2
(PSO )2 PX =
[X] RT
KP = = P 3 3
(aSO )2(aO )
(PSO )2(PO ) 2 2 2 2
([SO3] RT)2 = P
([SO2] RT)2([O2] RT)
([SO3])2 KC P = =
Where P = 1 bar RT
([SO2])2([O2]) RT
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17. An Alternative Derivation
2 SO2(g) + O2(g) SO3(g)
Where P = 1 bar RT
PSO3 2
PSO2
PO2 =
Kc =
[SO2]2[O2]
[SO3] = RT
PSO3 2
PSO2
PO2
Kc = KP(RT)
KP = Kc(RT)-1
In general terms:
KP = Kc(RT)Δn
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18. 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)
Kc =
[H2O]2
[CO][H2] =
PH2O2
PCOPH2
(RT)1
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19. General Chemistry: Chapter 15
Chemistry 140 Fall 2002
Burnt Lime
CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2]
KP = PCO2(RT)
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20. 15-4 The Significance of the Magnitude of the Equilibrium Constant.
Chemistry 140 Fall 2002
15-4 The Significance of the Magnitude of the Equilibrium Constant.
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21. 15-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.k1
CO(g) + 2 H2(g) CH3OH(g)
k-1
Equilibrium can be approached various ways.
Qualitative determination of change of initial conditions as equilibrium is approached is needed.
Qc =
[A]tm[B]tn
[G]tg[H]th
At equilibrium Qc = Kc
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22. General Chemistry: Chapter 15
Reaction Quotient
General Chemistry: Chapter 15
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23. 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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24. Le Châtelier’s Principle
2 SO2(g) + O2(g) SO3(g) k1
k-1
Kc = 2.8102 at 1000K
Q =
= Kc
[SO2]2[O2]
[SO3]2
Q > Kc
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25. 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. V
nSO3 2
nSO2
nO2 =
Kc =
[SO2]2[O2]
[SO3] = V
nSO3 2
nSO2
nO2
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26. Effect of Change in Volume
[G]g[H]h nG g nH h
Kc = =
V(a+b)-(g+h)
[C]c[D]d nA a nB a
V-Δn nG a nA nB g nH h =
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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27. Effect of the Change of Volume
ntot =
1.16 mol gas
1.085 mol gas
KP =
415
338
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28. 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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29. 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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30. 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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31. Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds.
N2(g) + O2(g) NO(g) k1
k-1
KP = at 298K and 1.3 x 10-4 at 1800K
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32. General Chemistry: Chapter 15
Synthesis of Ammonia
The optimum conditions are only for the equilibrium position and do not take into account the rate at which equilibrium is attained.
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33. 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.
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