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Chia sẻ tài liệu: hoa dai cuong_chuong 16 thuộc Bài giảng khác
Nội dung tài liệu:
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 16: Principles of Chemical Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 2 of 27
Contents
16-1 Dynamic Equilibrium
16-2 The Equilibrium Constant Expression
16-3 Relationships Involving Equilibrium Constants
16-4 The Significance of the Magnitude of an Equilibrium Constant
16-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 3 of 27
Contents
16-6 Altering Equilibrium Conditions:
Le Châtelliers Principle
16-7 Equilibrium Calculations:
Some Illustrative Examples
Focus On The Nitrogen Cycle and the
Synthesis of Nitrogen Compounds
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 4 of 27
16-1 Dynamic Equilibrium
Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
I2(H2O) I2(CCl4)
CO(g) + 2 H2(g) CH3OH(g)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 5 of 27
16-2 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]
= Kc
k1
k-1
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 6 of 27
Three Approaches to Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 7 of 27
Three Approaches to the Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 8 of 27
Three Approaches to Equilibrium
[CH3OH]
[CO][H2]2
Kc(1) =
14.2 M-2
Kc(2) =
14.2 M-2
Kc(3) =
14.2 M-2
[CH3OH]
[CO][H2]
Kc =
[CH3OH]
[CO](2[H2])
0.596 M-1
1.09 M-1
1.28 M-1
1.19 M-1
2.17 M-1
2.55 M-1
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 9 of 27
General Expressions
a A + b B …. → g G + h H ….
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 10 of 27
16-3 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 11 of 27
Combining Equilibrium Constant Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
N2(g) + ½O2 N2O(g) Kc(2)= 2.710+18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31
= 1.710-13
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 12 of 27
Gases: The Equilibrium Constant, KP
Mixtures of gases are solutions just as liquids are.
Use KP, based upon partial pressures of gases.
2 SO2(g) + O2(g) 2 SO3(g)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 13 of 27
The Equilibrium Constant, KP
2 SO2(g) + O2(g) 2 SO3(g)
Kc = KP(RT)
KP = Kc(RT)-1
KP = Kc(RT)Δn
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 14 of 27
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)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 15 of 27
Burnt Lime
CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2]
KP = PCO2(RT)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 16 of 27
16-4 The Significance of the Magnitude of the Equilibrium Constant.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 17 of 27
16-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.
Equilibrium can be approached various ways.
Qualitative determination of change of initial conditions as equilibrium is approached is needed.
At equilibrium Qc = Kc
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 18 of 27
Reaction Quotient
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 19 of 27
16-6 Altering Equilibrium Conditions:
Le Châtellier’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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 20 of 27
Le Châtellier’s Principle
Q > Kc
Kc = 2.8102 at 1000K
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 21 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 22 of 27
Effect of change in volume
Kc =
[C]c[D]d
[G]g[H]h
=
V(a+b)-(g+h)
nG
a
nA
nB
g
nH
h
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 23 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 24 of 27
Effect of a Catalyst on Equilibrium
A catalyist 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 25 of 27
16-7 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 26 of 27
Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds.
KP = 4.7 10-31 at 298K and 1.3 x 10-4 at 1800K
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 27 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 28 of 27
Chapter 16 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 16: Principles of Chemical Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 2 of 27
Contents
16-1 Dynamic Equilibrium
16-2 The Equilibrium Constant Expression
16-3 Relationships Involving Equilibrium Constants
16-4 The Significance of the Magnitude of an Equilibrium Constant
16-5 The Reaction Quotient, Q: Predicting the Direction of a Net Change
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 3 of 27
Contents
16-6 Altering Equilibrium Conditions:
Le Châtelliers Principle
16-7 Equilibrium Calculations:
Some Illustrative Examples
Focus On The Nitrogen Cycle and the
Synthesis of Nitrogen Compounds
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 4 of 27
16-1 Dynamic Equilibrium
Equilibrium – two opposing processes taking place at equal rates.
H2O(l) H2O(g)
I2(H2O) I2(CCl4)
CO(g) + 2 H2(g) CH3OH(g)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 5 of 27
16-2 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]
= Kc
k1
k-1
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 6 of 27
Three Approaches to Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 7 of 27
Three Approaches to the Equilibrium
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 8 of 27
Three Approaches to Equilibrium
[CH3OH]
[CO][H2]2
Kc(1) =
14.2 M-2
Kc(2) =
14.2 M-2
Kc(3) =
14.2 M-2
[CH3OH]
[CO][H2]
Kc =
[CH3OH]
[CO](2[H2])
0.596 M-1
1.09 M-1
1.28 M-1
1.19 M-1
2.17 M-1
2.55 M-1
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 9 of 27
General Expressions
a A + b B …. → g G + h H ….
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 10 of 27
16-3 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 11 of 27
Combining Equilibrium Constant Expressions
N2O(g) + ½O2 2 NO(g) Kc= ?
N2(g) + ½O2 N2O(g) Kc(2)= 2.710+18
N2(g) + O2 2 NO(g) Kc(3)= 4.710-31
= 1.710-13
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 12 of 27
Gases: The Equilibrium Constant, KP
Mixtures of gases are solutions just as liquids are.
Use KP, based upon partial pressures of gases.
2 SO2(g) + O2(g) 2 SO3(g)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 13 of 27
The Equilibrium Constant, KP
2 SO2(g) + O2(g) 2 SO3(g)
Kc = KP(RT)
KP = Kc(RT)-1
KP = Kc(RT)Δn
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 14 of 27
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)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 15 of 27
Burnt Lime
CaCO3(s) CaO(s) + CO2(g)
Kc = [CO2]
KP = PCO2(RT)
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 16 of 27
16-4 The Significance of the Magnitude of the Equilibrium Constant.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 17 of 27
16-5 The Reaction Quotient, Q: Predicting the Direction of Net Change.
Equilibrium can be approached various ways.
Qualitative determination of change of initial conditions as equilibrium is approached is needed.
At equilibrium Qc = Kc
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 18 of 27
Reaction Quotient
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 19 of 27
16-6 Altering Equilibrium Conditions:
Le Châtellier’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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 20 of 27
Le Châtellier’s Principle
Q > Kc
Kc = 2.8102 at 1000K
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 21 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 22 of 27
Effect of change in volume
Kc =
[C]c[D]d
[G]g[H]h
=
V(a+b)-(g+h)
nG
a
nA
nB
g
nH
h
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 23 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 24 of 27
Effect of a Catalyst on Equilibrium
A catalyist 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 25 of 27
16-7 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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 26 of 27
Focus on the Nitrogen Cycle and the Synthesis of Nitrogen Compounds.
KP = 4.7 10-31 at 298K and 1.3 x 10-4 at 1800K
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 27 of 27
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.
Prentice-Hall © 2002
General Chemistry: Chapter 16
Slide 28 of 27
Chapter 16 Questions
Develop problem solving skills and base your strategy not on solutions to specific problems but on understanding.
Choose a variety of problems from the text as examples.
Practice good techniques and get coaching from people who have been here before.
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