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Chapter 14
Ch 14
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2
Dynamic Processes
Irreversible process:
A
B
A is converted to B until
there is no A and only B.
Reversible process:
A
B
B
A
A
B
Number
A
A is in equilibrium with B
Neither A nor B are completely
consumed during the reaction.
B
3
Equilibrium
• The change is reversible
A
B
• The system is “closed”—no substance can enter or leave
• The system is dynamic - At the macroscopic level, it appears as if nothing is happening,
but at the particulate level, reversible changes are occurring continuously.
• Can be at physical equilibrium or chemical equilibrium
Physical Equilibrium- forward and reverse processes occur
at the same rate but there is no chemical change.
Solubility
Vapor Pressure
Liquid
At time = 0
At time > 0
rate of
dissolution
=
At time = ∞
rate of
precipitation
At time = 0
At time > 0
rate of
=
vaporization
At time = ∞
rate of
condensation
4
Equilibrium
• The change is reversible
A
B
• The system is “closed”—no substance can enter or leave
• The system is dynamic - At the macroscopic level, it appears as if nothing is happening,
but at the particulate level, reversible changes are occurring continuously
• Can be at physical equilibrium or chemical equilibrium
Chemical Equilibrium• the rates of the forward and reverse reactions are equal .
• a chemical change is occurring (intramolecular bonds broken/formed)
• concentrations of the reactants and products remain constant.
• this does not mean [conc] of reactants and products are equal!
NO2
N2O4 (g)
2NO2 (g)
5
Equilibrium and Rate
A
B
rateAB = kA [A]
B
A
rateBA = kB [B]
Number
B
A
0 1 2
4
n
time
At equilibrium!
At time ~0:
[B] = 0, [A] > 0
B formation = kA[A]
At time 1
[B] > 0, [A] > 0
rateAB >> RateBA
At time 2
[B] = [A]
kA/rateAB = kB / rateBA
At time 4
[B] > 0, [A] > 0
rateAB > RateBA
At time n  ∞
[B] unchanged, [A] unchanged
rateAB = RateBA
6
Equilibrium and Rate
N2O4 (g)
2NO2 (g)
Ratefwd = kfwd[NO2]2
Raterev = krev[N2O4]
At equilibrium:
Each compound is
produced and consumed
at the same rate.
Ratefwd = Raterev
7
Equilibrium and Rate
N2O4 (g)
equilibrium
2NO2 (g)
equilibrium
equilibrium
Start with NO2
Start with N2O4
Start with NO2 & N2O4
No matter what concentrations you start with, at some point:
Ratefwd = Raterev
The reaction will reach equilibrium!
8
Three Species Equilibrium
2SO2(g)  O2(g)
2SO3(g)
No matter what concentrations you start with, at some point:
Ratefwd = Raterev
The reaction will reach equilibrium!
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Chapter 14
Ch 14
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Equilibrium Concentrations
A
B
• Equilibrium does not mean that concentrations are all equal!!
• However, concentrations at equilibrium are related to the fwd/rev rate constants.
• Every equilibrium has its own equilibrium constant.
rateAB = kA [A]
rateBA = kB [B]
At equilibrium:
rateAB = rateBA
kA [A] = kB [B]
Rate constant AB
Rate constant BA
kA
kB
=
[B]
[A]
=
K
11
Reaction Diagram
A
Reverse
Reaction
B
Forward
Reaction
A
B
At equilibrium, the number of balls rolling over the
hill in each direction is equal.
12
Equilibrium Constant
For a general reversible reaction such as:
aA + bB
cC + dD
the equilibrium constant is written as
m
c
d
[products ]
[C] [D]
K

n
a
b
[reactants ]
[A] [B]
Law of Mass Action
The equilibrium constant for the overall reaction includes
stoichiometric coefficients.
General practice is not to include units for the equilibrium constant.
13
Equilibrium Constant
N2O4 (g)
2NO2 (g)
[NO2]2
K =
[N2O4]
K = 4.6 x 10-3
14
Three Species Equilibrium
No matter how you set up the reaction, the value of the equilibrium
constant will be the same if the temperature is the same.
15
Equilibrium Constant
aA + bB
cC + dD
K=
[C]c[D]d
[A]a[B]b
K >> 1
Favor product formation
kfwd > krev
aA + bB
cC + dD
K << 1
Favor reactant formation
kfwd < krev
aA + bB
cC + dD
16
Equilibrium Constant
Which reaction most favors the formation of products?
Which reaction most favors the formation of reactants?
[products ]m
K
[reactants ]n
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Chapter 14
Ch 14
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18
Equilibrium Constant
For a general reversible reaction such as:
aA + bB
cC + dD
• Equilibrium constants can be expressed using Kc or Kp.
• Kc uses the concentration of reactants and products.
[C]c [D]d
Kc =
[A]a [B]b
• Kp uses the pressure of the gaseous reactants and products.
PCc PDd
Kp = a b
PA PB
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14.2
The following equilibrium process has been studied at 230°C:
2NO(g) + O2(g)
2NO2(g)
In one experiment, the concentrations of the reacting species at equilibrium are found to
be [NO] = 0.0542 M, [O2] = 0.127 M, and [NO2] = 15.5 M. Calculate the equilibrium constant
(Kc) of the reaction at this temperature.
Solution The equilibrium constant is given by
[NO2 ]2
Kc =
[NO]2[O2 ]
If you are given
pressures, you
would use Kp
Substituting the concentrations, we find that
(15.5) 2
5
Kc =
=
6.44
×
10
(0.0542) 2 (0.127)
20
14.3
The equilibrium constant KP for the decomposition of phosphorus pentachloride to
phosphorus trichloride and molecular chlorine
PCl5(g)
PCl3(g) + Cl2(g)
is found to be 1.05 at 250°C. If the equilibrium partial pressures of PCl5 and PCl3 are
0.875 atm and 0.463 atm, respectively, what is the equilibrium partial pressure of
Cl2 at 250°C?
Kp =
1.05 =
PCl2 =
PPCl3 PCl2
Given Kp, PPCl5, and PCl3
PPCl5
Find PCl2
(0.463)(PCl2 )
(0.875)
(1.05)(0.875)
= 1.98 atm
(0.463)
21
Equilibrium
N2O4 (g)
In terms of pressure (KP)
In terms of concentration (Kc)
Kc =
2NO2 (g)
[NO2]2
PNO2
Kp = P 2
NO
[N2O4]
2 4
Kc  Kp
…at least not always!
22
Kc vs Kp
aA(g)
bB(g)
Substitution
23
Kc vs Kp
aA + bB
[C]c [D]d
Kc =
[A]a [B]b
cC + dD
PCc PDd
Kp = a b
PA PB
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
Dn = (c + d) – (a + b)
When does Kp equal Kc?
24
14.4
Methanol (CH3OH) is manufactured industrially by the reaction
CO(g) + 2H2(g)
CH3OH(g)
The equilibrium constant (Kc) for the reaction is 10.5 at 220°C. What is the value of
KP at this temperature?
Kp = Kc(RT)Dn
R = 0.0821 L atm/K mol
T = 273 + 220 = 493 K
Dn = = 1 - 3 = -2
KP = (10.5) (0.0821 x 493)-2
KP = 6.41 x 10-3
25
Equilibrium Constants Example
The equilibrium concentrations for the reaction between carbon monoxide and
molecular chlorine to form COCl2 (g) at 74°C are [CO] = 0.012 M, [Cl2] = 0.054 M,
and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g)
Kc =
[COCl2]
[CO][Cl2]
=
COCl2 (g)
0.14
0.012 x 0.054
= 220
Kp = Kc(RT)Dn
R = 0.0821 L atm/K mol
T = 273 + 74 = 347 K
Dn = 1 – 2 = -1
Kp = 220 x (0.0821 x 347)-1 = 7.7
26
Equilibria: Phases Matter
Homogenous equilibriumall reacting species are in the
same phase.
gas – gas
liquid-liquid
liquid-aqueous
N2O4 (g)
2NO2 (g)
Heterogeneous equilibriumreacting species are in
different phases.
gas – liquid
liquid-solid
solid-gas
CaCO3 (s)
CaO (s) + CO2
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Homogenous Equilibria
-all reacting species are in the same phase.
gas-gas equilibria
N2(g) + 3H2(g) ⇄ 2NH3(g)
Kc =
H2O is often used as a
solvent(l). However, if H2O
is written as a gas (g), then
its concentration must be
considered in Kc and Kp.
[NH2]2
[N2][H2]3
PNH32
Kp =
PN2 PH23
CH4(g) + 3H2O(g) ⇄ CO(g) + 3H2(g)
Kc =
[CO][H2]2
[CH4][H2O]3
PCO PH2
Kp =
PCH4 PH2O3
28
Homogenous Equilibria
-all reacting species are in the same phase.
liquid equilibrium
CH3COOH (aq) + H2O (l)
[CH3COO-][H3O+]
Kc‘ =
[CH3COOH][H2O]
CH3COO- (aq) + H3O+ (aq)
[H2O] = solvent
[H2O] = constant
The concentration of pure liquids are not included in the
expression for the equilibrium constant.
[CH3COO-][H3O+]
Kc =
[CH3COOH]
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14.1
Write expressions for Kc, and KP if applicable, for the following reversible reactions at
equilibrium:
H3O+(aq) + F-(aq)
(a) HF(aq) + H2O(l)
[H 3O + ][F - ]
Kc =
[HF]
(b) 2NO(g) + O2(g)
2NO2(g)
[NO2 ]2
Kc =
[NO]2 [O 2 ]
(c) CH3COOH(aq) + C2H5OH(aq)
Kc =
Kp =
2
PNO
2
2
PNO
PO2
CH3COOC2H5(aq) + H2O(l)
[CH 3COOC2 H5 ]
[CH 3COOH][C2 H5OH]
30
Heterogenous Equilibria
-reacting species are in different phases.
• Can include liquids, gases and solids as either
reactants or products.
• Equilibrium expression is the same as that for a
homogeneous equilibrium.
CaCO3 (s)
CaO (s) + CO2 (g)
Note CaO will
not form a neat
separated pile!
time
31
Heterogenous Equilibria
CaCO3 (s)
Kc’
CaO (s) + CO2 (g)
=
[CaO][CO2]
[CaCO3]
PCO2 or [CO2] does not depend on the amount of CaCO3 or CaO
32
Heterogenous Equilibria
CaCO3 (s)
CaO (s) + CO2 (g)
Kc’
=
[CaO][CO2]
[CaCO3]
Activity- is a measure of the “effective concentration” of a species in a mixture.
Activity of a solid = 1
Kc’
=
Kc’ = [CO2]
[1][CO2]
[1]
Kp = PCO 2
The concentration of solids are not included in the expression for the
equilibrium constant.
33
14.6
Consider the following heterogeneous equilibrium:
CaCO3(s)
CaO(s) + CO2(g)
At 800°C, the pressure of CO2 is 0.236 atm. Calculate:
(a) KP
Kp = PCO 2
KP = 0.236
(b) Kc
Kp = Kc(RT)Dn
R = 0.0821 L atm/K mol
T = 273 + 800 = 1073 K
Dn = 1 – 0 = 1
0.236 = Kc(0.0821 x 1073)1
Kc = 2.68 x 10-3
34
14.5
Write the equilibrium constant expression Kc, and KP if applicable, for
each of the following heterogeneous systems:
(a) (NH4)2Se(s)
2NH3(g) + H2Se(g)
Kc = [NH3]2[H2Se]
2
K p = PNH
PH 2Se
3
Ag+(aq) + Cl-(aq)
(b) AgCl(s)
Kc = [Ag+][Cl-]
(c) P4(s) + 6Cl2(g)
4PCl3(l)
1
Kc =
[Cl 2 ]6
1
Kp = 6
PCl 2
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Chapter 14
Ch 14
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36
Equilibrium Concentrations
A
B
• Equilibrium does not mean that concentrations are all equal!!
• However, concentrations at equilibrium are related to the fwd/rev rate constants.
• Every equilibrium has its own equilibrium constant.
rateAB = kA [A]
rateBA = kB [B]
At equilibrium:
rateAB = rateBA
kA [A] = kB [B]
Rate constant AB
Rate constant BA
kA
kB
=
[B]
[A]
=
K
37
Equilibrium Constant
For a general reversible reaction such as:
aA + bB
cC + dD
the equilibrium constant is written as
m
c
d
[products ]
[C] [D]
K

n
a
b
[reactants ]
[A] [B]
Law of Mass Action
The concentration of pure liquids or solids are not included in
the expression for the equilibrium constant.
38
Equilibrium Constant
For a general reversible reaction such as:
aA + bB
cC + dD
• Equilibrium constants can be expressed using Kc or Kp.
• Kc uses the concentration of reactants and products.
[C]c [D]d
Kc =
[A]a [B]b
• Kp uses the pressure of the gaseous reactants and products.
PCc PDd
Kp = a b
PA PB
39
Kc vs Kp
aA + bB
[C]c [D]d
Kc =
[A]a [B]b
cC + dD
PCc PDd
Kp = a b
PA PB
Kp = Kc(RT)Dn
Dn = moles of gaseous products – moles of gaseous reactants
Dn = (c + d) – (a + b)
40
Writing Equilibrium Constant Expressions
1. The concentrations of the reacting species in the condensed
phase are expressed in M. In the gaseous phase, the
concentrations can be expressed in M or in atm.
2. The concentrations of pure solids, pure liquids and solvents do
not appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must specify
the balanced equation and the temperature.
41
Multiple Equilibria
– A reaction can be an individual reaction step or a
multistep reaction
– If the overall reaction is the sum of two or more
reactions, the overall reaction Equilibrium
Constant is the product of the Equilibrium
Constants for the steps
K overall = K1 × K 2 × K 3 × ....
42
Multiple Equilibria
Product molecules of one equilibrium constant are involved in a
second equilibrium process.
A+B
C+D
A+B
Kc = Kc‘ x Kc‘‘
C+D
E+F
E+F
Kc‘
Kc‘‘
Kc
[C][D] x [E][F]
=
[A][B] [C][D]
[C][D]
Kc‘ =
[A][B]
[E][F]
Kc‘‘=
[C][D]
Kc =
[E][F]
[A][B]
If a reaction can be expressed as the sum of two or more reactions, the
equilibrium constant for the overall reaction is given by the product of the
equilibrium constants of the individual reactions.
43
Practice Problem
Determine the overall equilibrium constant for the reaction
between Nitrogen & Oxygen to form the toxic gas Nitrogen
Dioxide – a component of atmospheric smog
N2(g) +
2NO(g) +
Overall:
K c1
O2(g) ⇆ 2NO(g)
O2(g) ⇆ 2NO2(g)
N2(g) + 2O2(g) ⇆
[NO]2
=
[N 2 ][O 2 ]
Kcoverall = K c1 × K c2
K c2
Kc1
Kc2
2NO2(g)
[NO 2 ]2
=
[NO]2 ][O 2 ]
[NO 2 ]2
[NO 2 ]2
[NO]2
=
×
=
2
[N 2 ][O 2 ] [NO] [O 2 ]
[N 2 ][O 2 ]2
44
Multiple Equilibria
Overall Equilibrium:
Equilibrium 1:
Equilibrium 2:
x
45
Rules for Manipulating K
• The overall reaction equilibrium constant (K) is the product
of the equilibrium constants for the steps.
K overall = K1 ×K 2 ×K 3 ×....
• If the equation is reversed, the equilibrium constant is
inverted.
3
[ C]

[A ][ B]2
A  2B  3C
K forward
3C  A  2B
[A ][ B]2
1
K reverse 

3
[ C]
K forward
• If the equation is multiplied by a factor, the equilibrium
constant is raised to the same factor.
14.7
Write the equilibrium constant expression for each formulation :
(a) N2(g) + 3H2(g)
1
3
(b) N2(g) + H2(g)
2
2
1
(c) N2(g) + H2(g)
3
[NH 3 ]2
Ka =
[N 2 ][H 2 ]3
2NH3(g)
Kb =
NH3(g)
[NH3 ]
[N 2
2
NH3(g)
3
Kc =
1
]2 [H
[NH 3
[N 2
2
3
]2
2
]3
1
]3 [H
2]
(d) How are the equilibrium constants related to one another?
K a =K b2
K a =K c3
K b2  K c3
or
Kb 
3
Kc2
47
Writing Equilibrium Constant Expressions
1. The concentrations of the reacting species in the condensed
phase are expressed in M. In the gaseous phase, the
concentrations can be expressed in M or in atm.
2. The concentrations of pure solids, pure liquids and solvents do
not appear in the equilibrium constant expressions.
3. The equilibrium constant is a dimensionless quantity.
4. In quoting a value for the equilibrium constant, you must specify
the balanced equation and the temperature.
5. If a reaction can be expressed as a sum of two or more reactions,
the equilibrium constant for the overall reaction is given by the
product of the equilibrium constants of the individual reactions.
48
Chapter 14
Ch 14
Page 623
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