Chapter Thirteen:
CHEMICAL
EQUILIBRIUM
講義
Assignment for Chapter 13
6, 13, 23, 37, 46, 56, 63, 71, 76, 85, 93
Chapter 13 | Slide 2
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OVERVIEW
•
Chemical equilibrium
•
˙When a reaction takes place in a closed system, it reaches a condition where the concentrations of the reactants and
products remain constant over time
•
˙Dynamic state: reactants and products are interconverted continually
•
˙Forward rate= reverse rate
jA+kB
mC+nD
•
˙The law of mass action: for the reaction
•
m
n
K
[C] [D]
 equilibrium constant
j
k
[A] [B]
pure liquid or solid is never included in the equilibrium expression
r a gas-phase reaction the reactants and products can be described in terms of their partial pressures and the equilibrium constant is called Kp:
Kp  K ( RT )n
•Where ∆n is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants
•Equilibrium position
•˙A set of reactant and product concentrations that satisfies the equilibrium constant expression
• ˙There is one value of K for a given system at a given temperature
• ˙There are an infinite number of equilibrium positions at a given temperature depending on the initial concentrations
•˙A small value of K means the equilibrium lies to the left; a large value of K means the equilibrium lies to the right
• ˙The size of K has no relationship to the speed at which equilibrium is achieved
•˙Q, the reaction quotient, applies the law of mass action to initial concentrations rather than equilibrium concentrations
• ˙If Q>K, the system will shift to the left to achieve equilibrium
jA+kB
mC+nD
• ˙If Q<K, the system will shift to the right to achieve equilibrium
•˙Finding the concentrations that characterize a given equilibrium position:
[C]tm [D]tn
Q

1.Start with the given initial concentrations (pressures)
[A]tj [B]tk
2.Define the change needed to reach equilibrium
3.Apply the change to the initial concentrations (pressures) and solve for the equilibrium concentrations (pressures)
•Le chatelier’s principle
•˙ Enables qualitative prediction of the effects of changes in concentration, pressure, and temperature on a system at equilibrium
•˙If a change in conditions is imposed on a system at equilibrium, the system will shift in a direction that compensates for the imposed change
• ˙In other words, when a stress is placed on a system at equilibrium, the system shifts in the direction that relieves the stress
Chapter 13 | Slide 3
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Chemical Equilibrium
• The state where the concentrations of all
reactants and products remain constant
with time.
• On the molecular level, there is frantic
activity. Equilibrium is not static, but is a
highly dynamic situation.
13.1
Chapter 13 | Slide 4
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Equilibrium Is:
• Macroscopically static
• Microscopically dynamic
13.1
Chapter 13 | Slide 5
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Figure 13.1 a-d A Molecular
Representation of the Reaction
2NO2(g)  N2O4(g)
Chapter 13 | Slide 6
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Concentration
H2O(g) + CO(g)  H2(g) + CO2(g)
Equilibrium
Time
Chapter 13 | Slide 7
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Figure 13.3 a-d Gaseous Equilibrium
Mixture of Reactants and Products
H2O(g) + CO(g)  H2(g) + CO2(g)
Chapter 13 | Slide 8
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Chemical Equilibrium
• Concentrations reach levels where the
rate of the forward reaction equals the
rate of the reverse reaction.
13.1
Chapter 13 | Slide 9
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The Changes with Time in the Rates of
Forward and Reverse Reactions
13.1
Chapter 13 | Slide 10
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Concept Check
• Consider an equilibrium mixture in a closed
vessel reacting according to the equation:
H2O(g) + CO(g)
H2(g) + CO2(g)
• You add more H2O to the flask. How does the
concentration of each chemical compare to its
original concentration after equilibrium is
reestablished? Justify your answer.
13.1
Chapter 13 | Slide 11
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Concept Check
• Consider an equilibrium mixture in a closed
vessel reacting according to the equation:
H2O(g) + CO(g)
H2(g) + CO2(g)
• You add more H2 to the flask. How does the
concentration of each chemical compare to its
original concentration after equilibrium is
reestablished? Justify your answer.
13.1
Chapter 13 | Slide 12
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Figure 13.5 A Concentration Profile
N2(g) + 3H2(g)  2NH3(g)
Chapter 13 | Slide 13
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The Equilibrium Constant
jA + kB
lC + mD
l
m
[C] [D]
K= j k
[A] [B]
13.2
Chapter 13 | Slide 14
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Equilibrium Constant
[Molar concentrations of products] Stoichiometric coefficients
Kc  [Molar concentrations of reactants] Stoichiometric coefficients
aA+bB+cC+…pP+qQ+rR+…
p
q
r
K c  [ A]a [ B ]b [C ]c ...
[ P ] [ Q ] [ R ] ...
Chapter 13 | Slide 15
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Mifflin Company.
All rights
reserved.
Cato Guldberg
& Peter
Waage
(1864)
Equilibrium Expression
• Equilibrium expression for a reaction is the
reciprocal of that for the reaction written in
reverse.
• When balanced equation for a reaction is
multiplied by a factor of n, the equilibrium
expression for the new reaction is the original
expression raised to the nth power; thus Knew =
(Koriginal)n.
• K values are usually written without units.
13.2
Chapter 13 | Slide 16
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Equilibrium Constant
• K always has the same value at a given
temperature regardless of the amounts of
reactants or products that are present
initially.
• For a reaction at a given temperature,
there are many equilibrium positions but
only one value for K.
– Equilibrium position is a set of equilibrium
concentrations
13.2
Chapter 13 | Slide 17
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Anhydrous
Ammonia is
Injected into
the Solid to
Act as a
Fertilizer
Chapter 13 | Slide 18
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Table 13.1 Results of Three
Experiments for the Reaction
N2(g) + 3H2(g) -- 2NH3(g)
Chapter 13 | Slide 19
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jA + kB
lC + mD
l
m
[C] [D]
K= j k
[A] [B]
PV=nRT
n /V= P/RT
[B] = nB/V= PB/RT
[A] = nA/V= PA/RT
[D] = nD/V= PD/RT
[C] = nC/V= PC/RT
Kp = K(RT)Δn
Δn = (l+m) – (j+k)
Chapter 13 | Slide 20
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Equilibrium Expressions with Pressures
• K involves concentrations.
• Kp involves pressures.
Kp = K(RT)Δn
• Δn = sum of the coefficients of the
gaseous products minus the sum of the
coefficients of the gaseous reactants.
13.3
Chapter 13 | Slide 21
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Homogeneous Equilibria
• Homogeneous equilibria – involve the
same phase:
N2(g) + 3H2(g)
2NH3(g)
13.4
Chapter 13 | Slide 22
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Heterogeneous Equilibria
• Heterogeneous equilibria – involve more
than one phase:
2KClO3(s)
2KCl(s) + 3O2(g)
13.4
Chapter 13 | Slide 23
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The Seven Sisters Chalk Cliffs in East Sussex, England
Chapter 13 | Slide 24
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The Position of the Equilibrium
CaCO3(s)  CaO(s) + CO2(g)
does not depend on the amount of CaCO3 and CaO present
Kp=pCO2
Chapter 13 | Slide 25
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Heterogeneous Equilibria
• The position of a heterogeneous
equilibrium does not depend on the
amounts of pure solids or liquids present.
– The concentrations of pure liquids and solids
are constant
13.4
Chapter 13 | Slide 26
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Hydrated Copper (II) Sulfate on the Left. Water Applied to Anhydrous
Copper (II) Sulfate, on the Right, Forms the Hydrated Compound
Kp=p5H2O
The Position of the Equilibrium
CuSO4 ．5H2O (s)  CuSO4 (s) + 5H2O (g)
does not depend on the amount of CuSO4 and its hydrate present
Chapter 13 | Slide 27
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The Position of the Equilibrium
PCl5 (s)  PCl3 (l) + Cl2 (g)
does not depend on the amount of PCl5 and PCl3 present
Kp=pCl2
Chapter 13 | Slide 28
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Reaction
Chapter 13 | Slide 29
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Two Types of Molecules are Mixed
Together in the Following Amounts
Chapter 13 | Slide 30
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Conditions of Equilibrium Reactions
Chapter 13 | Slide 31
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Equilibrium Mixture
Chapter 13 | Slide 32
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Figure 13.7 a-b
The Size of K
and the Time
Required to
Reach
Equilibrium are
not Directly
Related as
Shown Here
Chapter 13 | Slide 33
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Concept Check
Consider the reaction:
+
+
K=25
13.5
Chapter 13 | Slide 34
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Concept Check Cont’d
A set of initial conditions for the reaction on the
previous slide.
+
13.5
Chapter 13 | Slide 35
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Concept Check Cont’d
• Draw a quantitative molecular picture that
shows what this system looks like after the
reactants are mixed and the system reaches
equilibrium.
• Support your answer with calculations.
13.5
Chapter 13 | Slide 36
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Reaction Quotient, Q
• Apply the law of mass action using initial concentrations
(the concentrations when you make your observation!)
instead of equilibrium concentrations.
jA + kB
l
lC + mD
m
l
m
eq
eq
t
t
eq
eq
[C] [D]
[C]
[D]
K = j k Q= j k
[A] [B]
[A] [B]
Chapter 13 | Slide 37
t
t
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Reaction Quotient, Q
• Q = K; The system is at equilibrium. No
shift will occur.
• Q > K; The system shifts to the left.
• Q < K; The system shifts to the right.
13.5
Chapter 13 | Slide 38
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Photo 13.4 Apollo II Lunar Landing
Chapter 13 | Slide 39
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Exercise
• Consider the reaction represented by the
equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
• Trial #1
6.00 M Fe3+(aq) and 10.0 M SCN-(aq) are
mixed at a certain temperature and at
equilibrium the concentration of FeSCN2+(aq)
is 4.00 M.
What is the value for the equilibrium constant
for this reaction?
13.5
Chapter 13 | Slide 40
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Set up ICE Table
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
Initial
6.00
10.00
0.00
Change
-4.00
-4.00
+4.00
2.00
6.00
4.00
Equilibrium
13.5
Chapter 13 | Slide 41
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Exercise
• Consider the reaction represented by the
equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
• Trial #2:
Initial: 10.0 M Fe3+(aq) and 8.00 M SCN−(aq)
(same temperature as Trial #1)
Equilibrium:
? M FeSCN2+(aq)
13.5
Chapter 13 | Slide 42
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Exercise
• Consider the reaction represented by the
equation:
Fe3+(aq) + SCN-(aq)
FeSCN2+(aq)
• Trial #3:
Initial: 6.00 M Fe3+(aq) and 6.00 M SCN−(aq)
Equilibrium:
? M FeSCN2+(aq)
13.5
Chapter 13 | Slide 43
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Exercise
Consider the reaction represented by the equation:
Fe3+(aq) + SCN-(aq)
Fe3+
SCN-
FeSCN2+(aq)
FeSCN2+
Trial #1
9.00 M
5.00 M
1.00 M
Trial #2
3.00 M
2.00 M
5.00 M
Trial #3
2.00 M
9.00 M
6.00 M
Find the equilibrium concentrations for all species.
Chapter 13 | Slide 44
13.6
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Concept Check
A 2.0 mol sample of ammonia is introduced into a
1.00 L container. At a certain temperature, the
ammonia partially dissociates according to the
equation:
NH3(g)
N2(g) + H2(g)
At equilibrium 1.00 mol of ammonia remains.
Calculate the value for K.
13.6
Chapter 13 | Slide 45
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Concept Check
A 1.00 mol sample of N2O4(g) is placed in a
10.0 L vessel and allowed to reach equilibrium
according to the equation:
N2O4(g)
2NO2(g)
K = 4.00 x 10-4
Calculate the equilibrium
concentrations of: N2O4(g) and NO2(g).
13.6
Chapter 13 | Slide 46
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Le Châtelier’s Principle
• If a change is imposed on a system at
equilibrium, the position of the equilibrium
will shift in a direction that tends to
reduce that change.
13.7
Chapter 13 | Slide 47
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Table 13.2 The Percent by Mass of
NH3 at Equilibrium in a Mixture of N2, H2,
and NH3 as a Function of Temperature
and Total Pressure
Chapter 13 | Slide 48
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Figure 13.8 a-c (a) The Initial Equilibrium Mixture
of N2, H2, and NH3 (b) Addition of N2. (c.) The New
Equilibrium Position for the System Containing
More N2 (due to Less H2, and More NH3 than in (a)
Chapter 13 | Slide 49
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Effects of Changes on the System
1. Concentration: The system will shift
away from the added component. If a
component is removed, the opposite
effect occurs.
2. Temperature: K will change depending
upon the temperature (endothermic –
energy is a reactant; exothermic –
energy is a product).
13.7
Chapter 13 | Slide 50
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Effects of Changes on the System
3. Pressure:
a) The system will shift away from the added
gaseous component. If a component is
removed, the opposite effect occurs.
b) Addition of inert gas does not affect the
equilibrium position.
c) Decreasing the volume shifts the equilibrium
toward the side with fewer moles of gas.
13.7
Chapter 13 | Slide 51
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Figure 13.9 a-c (a) A Mixture of NH3(g), N2(g), and
H2(g) at Equilibrium (b) The Volume is Suddenly
Decreased (c) The New Equilibrium Position for the
System Containing More NH3 and Less N2 and H2
Chapter 13 | Slide 52
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Photo 13.5 a-b LeChatelier's Principle
II
Chapter 13 | Slide 53
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LeChâtelier’s Principle
13.7
Chapter 13 | Slide 54
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Equilibrium Decomposition of N2O4
13.7
Chapter 13 | Slide 55
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Table 13.3 Observed Value of K for the
Ammonia Synthesis Reaction as a Function
of Temperature
Chapter 13 | Slide 56
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Table 13.4
Shifts in the
Equilibrium
Position for
the Reaction
58 kJ +
N2O4(g) -2NO2(g)
Chapter 13 | Slide 57
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Calculating equilibrium concentration
• H2(g)+Cl2(g)2HCl(g)
17
17
[H 2 ]  1.0  10 mol/L; [Cl2 ]  1.0  10 mol/L;
K c  4.0  1031at300K. [HCl]  ?
Solution:
Kc 
[HCl]2
[H 2 ][Cl 2 ]
 [HCl]  K c [H 2 ][Cl2 ]  0.28
1016 times higher than [Cl 2 ] and [H 2 ]!
Chapter 13 | Slide 58
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Classroom Exercise
• Suppose that the equilibrium molar concentrations of H2
and Cl2 at 300 K are both 1.0×10-16 mol/L. What is the
equilibrium molar concentration of HCl, given Kc= 4.0×1031?
• H2(g)+Cl2(g)2HCl(g)
[H 2 ]  [Cl2 ]  1.0  1016 mol/L; K c  4.0  1031 at 300K. [HCl]  ?
Solution:
Kc 
[HCl]2
[H 2 ][Cl2 ]
 [HCl]  K c [H 2 ][Cl2 ]  4.0  1031  1.0  1032  0.63,
almost 1016 times higher than [Cl2 ] and [H 2 ]!
Chapter 13 | Slide 59
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Using K to determine a partial pressure
PCl5 (g)  PCl3 (g)  Cl2 (g) K p  25 at 298K.
PPCl5  0.0021 atm, PCl2  0.48 atm, PPCl3  ?
Solution :
Kp 
Chapter 13 | Slide 60
PPCl3 PCl2
PPCl5
 PPCl3  K p 
PPCl5
PCl2
 0.11
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Classroom Exercise:
Using K to determine a partial pressure
2NOCl(g)  2NO(g)  Cl2 (g) K p  0.018 at 500 K.
PNO  0.11 atm, PCl2  0.84 atm, PNOCl  ?
Solution :
Kp 
2
PNO
PCl2
P 2 NOCl
 PNOCl 
Chapter 13 | Slide 61
2
PNO
PCl2
Kp

0.112 atm 2 0.84 atm
0.018 atm
 0.75 atm
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Quiz
PCl5 (g)  PCl3 (g)  Cl2 (g) K p  25 at 298K.
At certain stage, it is measured that
PPCl5  0.0021 atm, PCl2  0.48 atm, PPCl3  0.08 atm
Will the reaction be moving to formation of more product or not?
Solution :
Qp 
PPCl3 PCl2
PPCl5

0.080.48
0.0021
 18.3  K p
The reaction will be moving to formation of more product.
Equilibrium Tables
A table that shows the initial concentrations, the changes
needed to reach equilibrium, and the final equilibrium
compositions.
Chapter 13 | Slide 63
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Problem: 0.15 mol PCl5 is placed in reaction vessel of
volume 500 mL and allowed to reach equilibrium with its
decomposition products at 250 o C and K c  1.8. What is the
composition of the equilibrium mixture?
PCl5 (g)  PCl3 (g)  Cl 2 (g) K c 
[PCl3 ][Cl2 ]
[PCl5 ]
The initial concentration [PCl5 ]  0.150 mol/0.5L  0.30 mol/L
Draw an equilirium table:
Species
PCl
Step 1: Initial concentration
0.30
Step 2: Change in the molar concentration
-x
Step 3: equilibrium concentration
0.30 - x
PCl
0
 x
x
3
Cl
2
0
 x
x
2
x x
 x  1.8 x  0.54  0  x  0.262(positive
Copyright ©root).
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0.30  x
K 13 | Slide 64
Chapter
c
5
Classroom Exercise
A mixture of 0.002 mol/L hydrogen gas and 0.002 mol/L iodine
is allolwed to form hydroegn iodide at 773 o C by the reaction
H 2 (g)+I 2 (g)  2HI(g). At equilibrium, the concentration of HI is
0.002 mol/L. Calculate the equilibrium constant at this temperature.
The initial concentration [H2 ]  0.002 mol/L, [I2 ]  0.002 mol/L
Draw an equilirium table:
Species
H
Step 1: Initial concentration
2
0.002
Step 2: Change in the molar concentration
Step 3: equilibrium concentration
-x
I
2
0.002
x
0.002 - x 0.002  x
HI
0
 2x
0.002
x  0.001
2
(2 x )2
0.002
K 

 4.0
c
(0.002  x )(0.002 x )
2
Chapter 13 | Slide 65
0.001
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OVERVIEW
•
Chemical equilibrium
•
˙When a reaction takes place in a closed system, it reaches a condition where the concentrations of the reactants and
products remain constant over time
•
˙Dynamic state: reactants and products are interconverted continually
•
˙Forward rate= reverse rate
jA+kB
mC+nD
•
˙The law of mass action: for the reaction
•
m
n
K
[C] [D]
 equilibrium constant
j
k
[A] [B]
A pure liquid or solid is never included in the equilibrium expression
For a gas-phase reaction the reactants and products can be described in terms of their partial pressures and the equilibrium constant is called Kp:
Kp  K ( RT )n
•Where ∆n is the sum of the coefficients of the gaseous products minus the sum of the coefficients of the gaseous reactants
•Equilibrium position
•˙A set of reactant and product concentrations that satisfies the equilibrium constant expression
• ˙There is one value of K for a given system at a given temperature
• ˙There are an infinite number of equilibrium positions at a given temperature depending on the initial concentrations
•˙A small value of K means the equilibrium lies to the left; a large value of K means the equilibrium lies to the right
• ˙The size of K has no relationship to the speed at which equilibrium is achieved
•˙Q, the reaction quotient, applies the law of mass action to initial concentrations rather than equilibrium concentrations
• ˙If Q>K, the system will shift to the left to achieve equilibrium
jA+kB
mC+nD
• ˙If Q<K, the system will shift to the right to achieve equilibrium
•˙Finding the concentrations that characterize a given equilibrium position:
[C]tm [D]tn
Q

1.Start with the given initial concentrations (pressures)
[A]tj [B]tk
2.Define the change needed to reach equilibrium
3.Apply the change to the initial concentrations (pressures) and solve for the equilibrium concentrations (pressures)
•Le chatelier’s principle
•˙ Enables qualitative prediction of the effects of changes in concentration, pressure, and temperature on a system at equilibrium
•˙If a change in conditions is imposed on a system at equilibrium, the system will shift in a direction that compensates for the imposed change
• ˙In other words, when a stress is placed on a system at equilibrium, the system shifts in the direction that relieves the stress
Chapter 13 | Slide 66
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Chapter Thirteen
Chemical Equilibrium
案例討論
Figure 13.1 a-d A Molecular
Representation of the Reaction 2NO2(g)
- N2O4(g)
Chapter 13 | Slide 68
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Figure 13.2 H2O(g) + CO(g) - H2(g) +
CO2(g)
Chapter 13 | Slide 69
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Figure 13.3 a-d Gaseous Equilibrium
Mixture of Reactants and Products
Chapter 13 | Slide 70
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Figure 13.4 As the Concentrations of the
Reactants Decrease, the Forward Reaction
Slows Down
Chapter 13 | Slide 71
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Figure 13.5 A Concentration Profile
Chapter 13 | Slide 72
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Figure 13.6 a-b The Position of the
Equilibrium
Chapter 13 | Slide 73
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Reaction
Chapter 13 | Slide 74
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Two Types of Molecules are Mixed
Together in the Following Amounts
Chapter 13 | Slide 75
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Conditions of Equilibrium Reactions
Chapter 13 | Slide 76
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Equilibrium Mixture
Chapter 13 | Slide 77
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Figure 13.7 a-b
The Size of K
and the Time
Required to
Reach
Equilibrium are
not Directly
Related as
Shown Here
Chapter 13 | Slide 78
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Figure 13.8 a-c (a) The Initial Equilibrium Mixture
of N2, H2, and NH3 (b) Addition of N2. (c.) The New
Equilibrium Position for the System Containing
More N2 (due to Less H2, and More NH3 than in (a)
Chapter 13 | Slide 79
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Figure 13.9 a-c (a) A Mixture of NH3(g), N2(g), and
H2(g) at Equilibrium (b) The Volume is Suddenly
Decreased (c) The New Equilibrium Position for the
System Containing More NH3 and Less N2 and H2
Chapter 13 | Slide 80
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Anhydrous
Ammonia is
Injected into
the Solid to
Act as a
Fertilizer
Chapter 13 | Slide 81
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The Seven Sisters Chalk Cliffs in East
Sussex, England
Chapter 13 | Slide 82
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Hydrated Copper (II) Sulfate on the Left. Water
Applied to Anhydrous Copper (II) Sulfate, on
the Right, Forms the Hydrated Compound
Chapter 13 | Slide 83
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Photo 13.4 Apollo II Lunar Landing
Chapter 13 | Slide 84
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Photo 13.5 a-b LeChatelier's Principle
II
Chapter 13 | Slide 85
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Table 13.1 Results of Three
Experiments for the Reaction N2(g) +
3H2(g) -- 2NH3(g)
Chapter 13 | Slide 86
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Table 13.2 The Percent by Mass of
NH3 at Equilibrium in a Mixture of N2, H2,
and NH3 as a Function of Temperature
and Total Pressure
Chapter 13 | Slide 87
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Table 13.3 Observed Value of K for the
Ammonia Synthesis Reaction as a Function
of Temperature
Chapter 13 | Slide 88
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Table 13.4
Shifts in the
Equilibrium
Position for
the Reaction
58 kJ +
N2O4(g) -2NO2(g)
Chapter 13 | Slide 89
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Chapter Thirteen
Chemical Equilibrium
問答
Question
•
•
Consider the following equation:
• CO(g) + H2O(g)
CO2(g) + H2(g)
Which of the following must be true at
equilibrium?
–
–
–
–
–
Chapter 13 | Slide 91
[CO2] = [H2] because they are in a 1:1 mole ratio in
the balanced equation.
The total concentration of reactants is equal to the
total concentration of the products.
The total concentration of the reactants is greater
than the total concentration of the products.
The total concentration of the products is greater
than the total concentration of the reactants.
None of these is true.
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Answer
•e) None of these is true.
•Section 13.1, The Equilibrium Condition
•None of the given choices is true. At equilibrium,
the rate of the forward reaction is equal to the rate
of the reverse reaction. The concentrations of the
reactants and products can vary depending on the
temperature and initial conditions.
Chapter 13 | Slide 92
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Question
• Which of the following is true about chemical
equilibrium?
– It is microscopically and macroscopically static.
– It is microscopically and macroscopically dynamic.
– It is microscopically static and macroscopically
dynamic.
– It is microscopically dynamic and macroscopically
static.
Chapter 13 | Slide 93
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Answer
•d) It is microscopically dynamic and
static.
macroscopically
•Section 13.1, The Equilibrium Condition
•When a reaction system reaches equilibrium, the forward
and reverse reactions both continue (microscopically
dynamic). Because the rates of the reactions are equal,
however, the concentrations do not change
(macroscopically static).
Chapter 13 | Slide 94
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Question
•Which of the following statements is false regarding
chemical equilibrium?
–
–
–
–
–
Chapter 13 | Slide 95
A system that is disturbed from an equilibrium condition
responds in a manner to restore equilibrium.
The value of the equilibrium constant for a given reaction
mixture at constant temperature is the same regardless
of the direction from which equilibrium is attained.
When two opposing processes are proceeding at
identical rates, the system is at equilibrium.
A system moves spontaneously toward a state of
equilibrium.
All of the above statements are true.
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Answer
•e) All of the above statements are true.
•Section 13.1, The Equilibrium Condition;
Section 13.2, The Equilibrium Constant
•All of the statements describe a system at
equilibrium.
Chapter 13 | Slide 96
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Question
• The equilibrium constant for A + 2B
3C
is 2.1  10–6. Determine the equilibrium
constant for 2A + 4B
6C.
–
–
–
–
–
4.2  10–6
4.4  10–12
2.3  1011
1.8  10-11
None of these
Chapter 13 | Slide 97
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Answer
•b) 4.4  10–12
•Section 13.2, The Equilibrium Constant
•Since the equilibrium constant for the first reaction is
[C]3/[A][B]2 and the equilibrium constant for the second
reaction is [C]6/[A]2[B]4, the two differ by one being the
square of the other. Consequently, the value for the
equilibrium constant for the second reaction will be
•(2.1  10–6)2 = 4.4  10–12
Chapter 13 | Slide 98
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Question
• For which of the following reactions
does K = Kp at the same temperature?
–
–
–
–
N2(g) + 3H2(g)
2NH3(g)
CaCO3(s)
CaO(s) + CO2(g)
CO(g) + H2O(g)
CO2(g) + H2(g)
3Fe(s) + H2O(g)
Fe3O4(s) + 4H2(g)
Chapter 13 | Slide 99
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Answer
•c) CO(g) + H2O(g)
CO2(g) + H2(g)
•Section 13.3, Equilibrium Expressions
Involving Pressures
•P = (n/V) × RT. Since the numbers of moles
of gas on each side of the equation are the
same, the RT factor will cancel. Thus K = Kp
for this reaction.
Chapter 13 | Slide 100
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Question
•The value of the equilibrium constant, K, is dependent
on
•
I. The temperature of the system.
•
II. The nature of the reactants and the products.
•
III. The concentrations of the reactants.
•
IV. The concentrations of the products.
– I, II
– II, III
– III, IV
– I, II, III
– I, II, III, IV
Chapter 13 | Slide 101
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Answer
•a)
I, II
•Section 13.2, The Equilibrium Constant;
Section 13.7, Le Châtelier’s Principle
•The value of the equilibrium constant is
influenced by the temperature and the
phases of the reactants and products.
Chapter 13 | Slide 102
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Question
• For a certain reaction at 25.0C, the value
of K is 1.2  10–3. At 50.0C, the value of
K is 3.4  10–1. This means that the
reaction is
– exothermic.
– endothermic.
– More information is needed to answer the
question.
Chapter 13 | Slide 103
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Answer
• b) endothermic.
• Section 13.7, Le Châtelier’s Principle
• When the magnitude of the equilibrium
constant increased, the equilibrium shifted
to the right. The reaction from left to right
is endothermic.
Chapter 13 | Slide 104
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Question
• Suppose we have the following reaction
at equilibrium:
• PCl5(g)
PCl3(g) + Cl2(g)
• Which of the following statements is false?
– Adding PCl3 to the container shifts the
equilibrium to form more PCl5.
– Decreasing the volume of the container shifts
the equilibrium to form more PCl5.
– Removing PCl5 from the container shifts the
equilibrium to form more PCl3.
Chapter 13 | Slide 105
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Answer
•c) Removing PCl5 from the container shifts the
equilibrium to form more PCl3.
•Section 13.7, Le Châtelier’s Principle
•Removing a reactant that is part of the mass
action expression that is equal to the equilibrium
constant will cause the equilibrium to shift toward
the reactants.
Chapter 13 | Slide 106
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Question
• When the following reaction is at
equilibrium and the volume of the
container is decreased,
•
COCl2(g)
CO(g) + Cl2(g)
–
–
–
–
the forward reaction rate increases.
the reverse reaction rate increases.
the forward reaction rate decreases.
the equilibrium remains unchanged.
Chapter 13 | Slide 107
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Answer
•b)
the reverse reaction rate increases.
•Section 13.7, Le Châtelier’s Principle
•When the volume of the container is
decreased for a reaction at equilibrium, the
equilibrium shifts toward the side with the
fewer moles of particles.
Chapter 13 | Slide 108
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Question
• Consider a chemical system at equilibrium. The reaction
is exothermic (releases energy as heat) as written and
the temperature of the system is raised. Which of the
following is true?
–
–
–
–
–
Equilibrium shifts to the left and the value of K increases.
Equilibrium shifts to the right and the value of K increases.
Equilibrium shifts to the right and the value of K decreases.
Equilibrium shifts to the left and the value of K decreases.
Equilibrium shifts, but the value of K stays constant.
Chapter 13 | Slide 109
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Answer
•d) Equilibrium shifts to the left and the value of K
decreases.
•Section 13.7, Le Châtelier’s Principle
•An increase in temperature is an increase in heat. For an
exothermic reaction, heat can be considered as a product
(although heat is not an object). In this way, the addition of
heat is like the addition of a product and will shift
equilibrium to the left. The concentrations of the reactants
will increase and the concentrations of the products will
decrease, so the value of K will decrease.
Chapter 13 | Slide 110
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