CHE
116: General
Chemistry
The Concept
of Equilibrium
At equilibrium, the rate at which products form 1
is EQUAL
to
the
rate
at
which
products
CHAPTER FIFTEEN
decompose.
Forward reaction: A --> B rate = kf[A]
Copyright © Tyna L. Heise 2002
Reverse reaction: B --> A rate = kr[B]
All Rights Reserved
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem. 116 Prof. T.L. Heise
Chem.
1
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
TheConcept
ConceptofofEquilibrium
Equilibrium
2
At
Atequilibrium,
equilibrium,the
therate
rateatatwhich
whichproducts
productsform
form
isisEQUAL
EQUALtotothe
therate
rateatatwhich
whichproducts
products
decompose.
decompose.
Forward
Forwardreaction:
reaction: AA-->
-->BB rate
rate==kkf[A]
f[A]
Reverse
Reversereaction:
reaction: BB-->
-->AA rate
rate==kkr[B]
r[B]
At
Atequilibrium,
equilibrium,kkf[A]
f[A]==kk
r[B]
r[B]
[B]
[B]== kkf f== aaconstant
constant(K
(Keqeq) )
[A]
[A] kkr r
Chap 15.1
Chem 15.1
Chem. 116 Prof. T.L. Heise
Chem.
2
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
TheConcept
ConceptofofEquilibrium
Equilibrium
3
Once
At equilibrium,
equilibriumthe
is established,
rate at which
the
products form
concentrations
is EQUAL to the
of A
rate
andatBwhich
do notproducts
change.
decompose.
The fact that the composition remains
constant
Forwardwith
reaction:
time does
A -->
not
B mean
rate that
= kf[A]
A and
B stop reacting
Reverse reaction: B --> A rate = kr[B]
Compound A is still converted into
At equilibrium, kf[A] = kr[B]
compound B, but both processes occur at the
same rate
[B] = kf = a constant (Keq)
[A] kr
Indicated by double arrow 
Chap 15.1
Chem 15.1
Chem. 116 Prof. T.L. Heise
Chem.
3
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
TheConcept
ConceptofofEquilibrium
Equilibrium
4
At equilibrium, the rate at which products form
is EQUAL to the rate at which products
decompose.
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.1
Chem. 116 Prof. T.L. Heise
Chem.
4
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
TheConcept
ConceptofofEquilibrium
Equilibrium
5
At equilibrium, the rate at which products form
One of the most important chemical systems is
isthe
EQUAL
to the
rate at which
synthesis
of ammonia
fromproducts
nitrogen and
decompose.
hydrogen (HABER process)
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.1
Chem. 116 Prof. T.L. Heise
Chem.
5
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
TheConcept
ConceptofofEquilibrium
Equilibrium
6
At equilibrium, the rate at which products form
HABER process: N2 + 3H2  2NH3(g)
is EQUAL to the rate at which products
put N2 + 3H2 in a high pressure tank at a
decompose.
total pressure of several hundred
Forward
reaction:
--> B rate
kf[A]
atmospheres,
in theApresence
of a=catalyst,
and at areaction:
temperature
of several
Reverse
B -->
A rate hundred
= kr[B]
degrees Celsius.
At equilibrium, kf[A] = kr[B]
Two gases react, but does not lead to
[B] =consumption
kf = a constant
(Keq)
complete
of gases
[A] kr
Chap 15.1
Chem 15.1
Chem. 116 Prof. T.L. Heise
Chem.
6
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
7
At equilibrium, the rate at which products form
HABER process:
is EQUAL to the rate at which
N2 +products
3H2 
decompose.
2NH3(g)
Forward reaction: A at
-->
B rate =the
kf[A]
equilibrium,
relative concentrations
Reverse reaction: B of-->
= kr[B]
H2A
, N2 rate
and NH
3
are the same, regardless of
At equilibrium, kf[A] = kthe
r[B]
starting mixture, and
note that(K
the )equilibrium is
[B] = kf = a constant
eq
reached from
either
[A] kr
direction!
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
7
ChemProf.
116: Prof.
T.L.Heise
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107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
8
General
Equation: aA
+ bB
+ qQproducts form
At equilibrium,
the
rate
atpP
which
is A,
EQUAL
toQ
the
which products
B, P and
arerate
the at
chemicals
involved
decompose.
a, b, p and q are the coefficients in the balanced
Forwardequilibrium
reaction: A --> B rate = kf[A]
chemical
Reverse
--> A at
rate
= kr[B] the
According
to thereaction:
law of massBaction,
equilibrium
relationship between products and reactants is
At equilibrium, kf[A] = kr[B]
proportionally the same
[B]p[Q]
= qkf = a constant (Keq)
Keq = [P]
[A]
[A]a[B]kbr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
8
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
9
equilibrium,
the rate expression
at which products
form
TheAt
equilibrium
constant
depends
EQUAL
to the rate at of
which
products not on
onlyison
the stoichiometry
the reaction,
decompose.
its mechanism.
Forward
B rate = kf[A]
The only
thingreaction:
can alter A
the-->proportionality
constant
of a balanced
is
Reverse
reaction: chemical
B --> A equation
rate = kr[B]
TEMPERATURE, altering anything else will
At equilibrium, kf[A] = kr[B]
cause the reaction to shift in order to get back to
[B] = kf = a constant (Keq)
the same proportionality...
[A]
Chap 15.1
Chem 15.2
kr
Chem. 116 Prof. T.L. Heise
Chem.
9
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
10
At equilibrium,
rateofatthe
which
products
…which
is why thethe
rates
forward
and form
is EQUAL
to the
at which
products
reverse
reactions
arerate
always
EQUAL
and the
decompose.of each chemical always remains
concentration
CONSTANT
Forward reaction: A --> B rate = k [A]
f
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
10
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
11
At equilibrium,
the rate
which products
form
Sample
exercise: Write
theatequilibrium
constant
is EQUAL
rate at which products
expression
fortoHthe
2(g) + I2(g)  2HI(g)
decompose.
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
11
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
12
At equilibrium,
the rate
which products
form
Sample
exercise: Write
theatequilibrium
constant
is EQUAL
rate at which products
expression
fortoHthe
2(g) + I2(g)  2HI(g)
decompose.
Forward reaction: A --> B rate = kf[A]
Keq = [HI]2
Reverse reaction:
B
-->
A
rate
=
k
[B]
r
[H2][I2]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
12
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
13
At equilibrium,
rate
at whichofproducts
form
When
the reactantsthe
and
products
an
is EQUALare
to the
ratethe
at partial
which products
equilibrium
gases,
pressures can
decompose.
be used…
Forward
q --> B rate = kf[A]
K =reaction:
(P )p(P )A
p
P
Q
a(P )b
(P
)
Reverse reaction:
A
B B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
Dn
[B]
=
k
=
a
constant
(Keq)
K
=
K
(RT)
f
p
eq
[A]
Chap 15.1
Chem 15.2
kr
Chem. 116 Prof. T.L. Heise
Chem.
13
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
14
At equilibrium,
thethe
rateequilibrium
at which products form
Sample
exercise: For
is EQUAL
to the
rate
at which products
2SO3(g) 
2SO
2(g) + O2(g)
decompose.
at a temperature of 1000K, Keq has the
value Forward
of 4.08 x reaction:
10-3. Calculate
value
A --> Btherate
= kfor
[A]Kp.
f
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
14
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
15
At equilibrium,
thethe
rateequilibrium
at which products form
Sample
exercise: For
is EQUAL
to the
rate
at which products
2SO3(g) 
2SO
2(g) + O2(g)
decompose.
at a temperature of 1000K, Keq has the
value Forward
of 4.08 x reaction:
10-3. Calculate
value
A --> Btherate
= kfor
[A]Kp.
f
Dn
K
=
K
(RT)
Reversepreaction:
B --> A rate = kr[B]
eq
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
15
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
16
At equilibrium,
thethe
rateequilibrium
at which products form
Sample
exercise: For
is EQUAL
to the
rate
at which products
2SO3(g) 
2SO
2(g) + O2(g)
decompose.
at a temperature of 1000K, Keq has the
value Forward
of 4.08 x reaction:
10-3. Calculate
value
A --> Btherate
= kfor
[A]Kp.
f
Dn
K
=
K
(RT)
Reversepreaction:
B --> A rate = kr[B]
eq
= 4.08 x 10-3((0.0821)(1000))1
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
16
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
17
At equilibrium,
thethe
rateequilibrium
at which products form
Sample
exercise: For
is EQUAL
to the
rate
at which products
2SO3(g) 
2SO
2(g) + O2(g)
decompose.
at a temperature of 1000K, Keq has the
value Forward
of 4.08 x reaction:
10-3. Calculate
value
A --> Btherate
= kfor
[A]Kp.
f
Dn
K
=
K
(RT)
Reversepreaction:
B --> A rate = kr[B]
eq
= 4.08 x 10-3((0.0821)(1000))1
At equilibrium, kf[A] = -3kr[B]
= 4.08 x 10 (82.1)
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
17
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
18
At equilibrium,
thethe
rateequilibrium
at which products form
Sample
exercise: For
is EQUAL
to the
rate
at which products
2SO3(g) 
2SO
2(g) + O2(g)
decompose.
at a temperature of 1000K, Keq has the
value Forward
of 4.08 x reaction:
10-3. Calculate
value
A --> Btherate
= kfor
[A]Kp.
f
Dn
K
=
K
(RT)
Reversepreaction:
B --> A rate = kr[B]
eq
= 4.08 x 10-3((0.0821)(1000))1
At equilibrium, kf[A] =-3kr[B]
= 4.08 x 10 (82.1)
[B] == 0.335
kf = a constant (Keq)
[A]
Chap 15.1
Chem 15.2
kr
Chem. 116 Prof. T.L. Heise
Chem.
18
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
19
At equilibrium,
the rate
which
Equilibrium
Constants
canatbe
veryproducts
large or form
very
is EQUAL
to the rateofatthe
which
products
small.
The magnitude
equilibrium
decompose.
constant
provides us with important information
aboutForward
the equilibrium
reaction:mixture.
A --> B rate = k [A]
f
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
19
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
20
At equilibrium,
the rate
at which products
Sample
Exercise: The
equilibrium
constantform
for
EQUALHto
the rate at which products
the is
reaction
2(g) + I2(g)  2HI(g) varies with
decompose.
temperature
in the following way:
Keq =
794 at 298
K B rate = k [A]
Forward
reaction:
A -->
f
Keq = 54 at 700 K
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
Is the formation of products favored more
at the
or lower
At higher
equilibrium,
kf[A]temperature?
= kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.2
Chem. 116 Prof. T.L. Heise
Chem.
20
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
The
Constant
TheEquilibrium
Concept of Equilibrium
21
At equilibrium,
the rate
at which products
Sample
Exercise: The
equilibrium
constantform
for
EQUALHto
the rate at which products
the is
reaction
2(g) + I2(g)  2HI(g) varies with
decompose.
temperature
in the following way:
Keq =
794 at 298
K B rate = k [A]
Forward
reaction:
A -->
f
Keq = 54 at 700 K
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
Is the formation of products favored more
at the
or lower
At higher
equilibrium,
kf[A]temperature?
= kr[B]
[B] = kbecause
Lower
Keq is (K
larger
f = a constant
eq)
[A]
Chap 15.1
Chem 15.2
kr
Chem. 116 Prof. T.L. Heise
Chem.
21
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
22
equilibrium,
the equilibrium
rate at whichare
products
form
TheAt
substances
in the
of
is EQUAL
to the rate at which products
different
phases:
decompose.
the concentration of a pure solid or liquid
equals
itsreaction:
density divided
its molar
Forward
A --> Bbyrate
= k [A]mass
f
3 = mol
D
=
g/cm
Reverse reaction: B --> A rate = kr[B]
M g/mol cm3
At equilibrium, kf[A] = kr[B]
the density is constant at any given
[B] = kf = a constant (Keq)
temperature
so...
[A] k
r
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
22
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
23
At3equilibrium,
the+rate
products form
CaCo
(s)  CaO(s)
CO2at(g)which
If a pure solid or
is EQUAL to the rate at whichliquid
products
is used it is
Keq = [CaO][CO2]
NOT included in the
decompose.
[CaCO3]
equilibrium constant
Forward reaction: A --> B rate = kf[A]
Keq = constant 1[CO2]
Reverse constant
reaction: 2 B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
Keq’ = Keq constant 1 = [CO2]
[B] = constant
kf = a constant
(Keq)
2
[A]
Chap 15.1
Chem 15.3
kr
Chem. 116 Prof. T.L. Heise
Chem.
23
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
24
At equilibrium,
the rate
at equilibrium
which products form
Sample
Exercise: Write
the
is EQUAL
toK
the
rate at which products
expressions
for
eq and Kp for the reaction
decompose.
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
24
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
25
At equilibrium,
the rate
at equilibrium
which products form
Sample
Exercise: Write
the
is EQUAL
toK
the
rate at which products
expressions
for
eq and Kp for the reaction
decompose.
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Forward
reaction: A --> B rate = kf[A]
K
=
products
eq
reactants
Reverse
reaction: B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
25
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
26
At equilibrium,
the rate
at equilibrium
which products form
Sample
Exercise: Write
the
is EQUAL
toK
the
rate at which products
expressions
for
eq and Kp for the reaction
decompose.
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Forward
reaction:
A -->
4 B rate = kf[A]
K
=
products
=
[H
]
eq
2
4
reactants
[H
O]
Reverse reaction: B 2--> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
26
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
27
At equilibrium,
the rate
at equilibrium
which products form
Sample
Exercise: Write
the
is EQUAL
toK
the
rate at which products
expressions
for
eq and Kp for the reaction
decompose.
3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Forward
reaction:
A -->
4 B rate = kf[A]
K
=
products
=
[H
]
eq
2
4
reactants
[H
O]
Reverse reaction: B 2--> A rate = kr[B]
4
=
P
AtKequilibrium,
kf[A] = kr[B]
p
(H )
2 4
P(H O)
[B]2= kf
[A]
Chap 15.1
Chem 15.3
= a constant (Keq)
kr
Chem. 116 Prof. T.L. Heise
Chem.
27
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
28
At equilibrium,
the rateof
atthe
which
products form
Sample
Exercise: Which
following
is EQUAL
theH
rate
at which products
substances
- Hto2(g),
2O(g), O2(g) - when added
decompose.
to Fe
3O4(s) in a closed container at high
temperature,
of equilibrium
Forwardpermits
reaction:attainment
A --> B rate
= kf[A]
in the rxn 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
28
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Heterogeneous
The Concept ofEquilibria
Equilibrium
29
At equilibrium,
the rateof
atthe
which
products form
Sample
Exercise: Which
following
is EQUAL
theH
rate
at which products
substances
- Hto2(g),
2O(g), O2(g) - when added
decompose.
to Fe
3O4(s) in a closed container at high
temperature,
of equilibrium
Forwardpermits
reaction:attainment
A --> B rate
= kf[A]
in the rxn 3Fe(s) + 4H2O(g)  Fe3O4(s) + 4H2(g)
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
Only hydrogen
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.3
Chem. 116 Prof. T.L. Heise
Chem.
29
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
30
At equilibrium,
the ratechloride,
at whichNO
products
Sample
Exercise: Nitryl
in
2Cl, isform
is EQUALwith
to the
rate
at which products
equilibrium
NO
2 and Cl2:
decompose.
2NO2Cl(g)  2NO2(g) + Cl2(g)
At
equilibrium
the concentrations
the
Forward
reaction:
A --> B rate = kof
[A]
f
substances are [NO2Cl] = 0.00106 M, [NO2] =
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
0.0108 M, and [Cl2] = 0.00538 M. From these
dataAtcalculate
the equilibrium
equilibrium,
kf[A] = kr[B] constant, Keq.
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
30
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
31
At2NO
equilibrium,
rate
at +
which
products form
2NO
Cl2(g)
2Cl(g) the
2(g)
is At
EQUAL
to the rate
which productsof the
equilibrium
the at
concentrations
decompose.
substances
are [NO2Cl] = 0.00106 M, [NO2] =
0.0108Forward
M, and reaction:
[Cl2] = 0.00538
From
A --> BM.rate
= kfthese
[A]
data calculate the equilibrium constant, Keq.
Reverse reaction:
B --> A rate = kr[B]
Keq = products
At equilibrium, kf[A] = kr[B]
reactants
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
31
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
32
At2NO
equilibrium,
rate
at +
which
products form
2NO
Cl2(g)
2Cl(g) the
2(g)
is At
EQUAL
to the rate
which productsof the
equilibrium
the at
concentrations
decompose.
substances
are [NO2Cl] = 0.00106 M, [NO2] =
0.0108Forward
M, and reaction:
[Cl2] = 0.00538
From
A --> BM.rate
= kfthese
[A]
data calculate the equilibrium constant, Keq.
Reverse reaction:
B --> A rate = kr[B]
Keq = products = [NO2]2[Cl2]
At equilibrium, kf[A] = kr[B] 2
reactants [NO2Cl]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
32
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
33
At equilibrium,
rate at which of
products
form
At equilibrium
thethe
concentrations
the
is EQUAL
the 2rate
at0.00106
which products
substances
areto[NO
Cl] =
M, [NO2] =
decompose.
0.0108
M, and [Cl2] = 0.00538 M. From these
data calculate
equilibrium
eq.
Forward the
reaction:
A --> Bconstant,
rate = k K
[A]
f
2(0.00538)
Keq =Reverse
[NO2]2[Cl
]
=
(0.0108)
reaction:
B --> A rate = kr[B]
2
[NO2Cl]2
(0.00106)2
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
33
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
34
At equilibrium,
rate at which of
products
form
At equilibrium
thethe
concentrations
the
is EQUAL
the 2rate
at0.00106
which products
substances
areto[NO
Cl] =
M, [NO2] =
decompose.
0.0108
M, and [Cl2] = 0.00538 M. From these
data calculate
equilibrium
eq.
Forward the
reaction:
A --> Bconstant,
rate = k K
[A]
f
2(0.00538)
Keq =Reverse
[NO2]2[Cl
]
=
(0.0108)
reaction:
B --> A rate = kr[B]
2
[NO2Cl]2
(0.00106)2
At equilibrium, kf[A] = kr[B]
= 0.558
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
34
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Calculating
Equilibrium
Const
The Concept
of Equilibrium
35
At equilibrium,
the
which products
Rarely
are we given
allrate
theatinformation
as inform
the
EQUAL problem,
to the ratenormally,
at which products
firstisexample
one
decompose.will be given and stoichiometric
concentration
calculations
will
give youAthe
Forward
reaction:
-->others.
B rate = k [A]
f
Reverse reaction:
B --> A rate = kr[B]
Exercise
AtSample
equilibrium,
kf[A]pg= 572
kr[B]on board.
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.4
Chem. 116 Prof. T.L. Heise
Chem.
35
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
36
the reaction
rate at which
products
If KAtis equilibrium,
very large, the
will tend
to form
is proceed
EQUAL far
to the
at which
products
to rate
the right
creating
a large
decompose.
amount of products.
reaction:
A --> Bwill
rate
kf[A]
If K isForward
very small,
the reaction
not= proceed,
staying
to the left,
leaving
a large
Reversefar
reaction:
B -->
A rate
= kr[B]
amount of reactants unused.
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
36
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
37
At equilibrium,
the rate at which products form
Predicting
the Direction
is EQUAL to the rate at which products
use
initial concentrations to determine
decompose.
original Keq
Forward reaction: A --> B rate = kf[A]
obtain Keq at temperature you want
Reverse reaction:
B --> A rate = kr[B]
compare calculated Keq to actual Keq and it
At equilibrium, kf[A] = kr[B]
will tell you how the reaction is going
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
37
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
38
At equilibrium,
the rate at which products form
Predicting
the Direction
is EQUAL to the rate at which products
2.00
mol H2, 1.00 mol of N2, and 2.00 mol of
decompose.
NH3 at 472°C
Q = (2)2/(1)(2)3 = 0.500
Forward reaction: A --> B rate = kf[A]
according to S.E. 15.7, Keq = 0.105
Reverse reaction:
B --> A rate = kr[B]
our actual is much smaller than the original
At equilibrium, kf[A] = kr[B]
which tells me that I have quite a bit less
kf = reactant
a constant
(Keqanticipated,
)
product [B]
and=more
than
[A]is proceeding
kr
so reaction
to the left
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
38
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
39
At equilibrium,
rateKatthe
which
products
Sample
exercise: Atthe
1000
value
of Keq form
for
EQUAL2SO
to the
rate at which products
the is
reaction
3(g)  2SO2(g) + O2(g) is 4.08 x
10-3decompose.
. Calculate the value for Q, and predict the
direction
in which
the reaction
proceed
Forward
reaction:
A --> Bwill
rate
= kf[A]
toward equilibrium if the initial concentrations
Reverse
reaction:
B
-->
A
rate
=
k
[B]
-3
-3
r
are: [SO3] = 2 x 10 M, [SO2] = 5 x 10 M, and
[O2]At= equilibrium,
3 x 10-2 M. kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
39
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
40
At equilibrium,
rateKatthe
which
products
Sample
exercise: Atthe
1000
value
of Keq form
for
EQUAL2SO
to the
rate at which products
the is
reaction
3(g)  2SO2(g) + O2(g) is 4.08 x
10-3decompose.
. Calculate the value for Q, and predict the
direction
in which
the reaction
proceed
Forward
reaction:
A --> Bwill
rate
= kf[A]
toward equilibrium if the initial concentrations
Reverse
reaction:
B
-->
A
rate
=
k
[B]
-3
-3
r
are: [SO3] = 2 x 10 M, [SO2] = 5 x 10 M, and
[O2]At= equilibrium,
3 x 10-2 M. kf[A] = kr[B]
Q = [SO2]2[O2]
[B] = k2f = a constant (Keq)
[SO
]
3
[A] k
r
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
40
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
41
At equilibrium,
rateKatthe
which
products
Sample
exercise: Atthe
1000
value
of Keq form
for
EQUAL2SO
to the
rate at which products
the is
reaction
3(g)  2SO2(g) + O2(g) is 4.08 x
10-3decompose.
. Calculate the value for Q, and predict the
direction
in which
the reaction
proceed
Forward
reaction:
A --> Bwill
rate
= kf[A]
toward equilibrium if the initial concentrations
Reverse
reaction:
B
-->
A
rate
=
k
[B]
-3
-3
r
are: [SO3] = 2 x 10 M, [SO2] = 5 x 10 M, and
[O2]At= equilibrium,
3 x 10-2 M. kf[A] = kr[B]
Q = [SO2]2[O2] = (5 x 10-3)2(3 x 10-2)
[B] = k2f = a constant (K-3eq)2
[SO
]
(2
x
10
)
3
[A] k
r
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
41
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
42
At equilibrium,
rateKatthe
which
products
Sample
exercise: Atthe
1000
value
of Keq form
for
EQUAL2SO
to the
rate at which products
the is
reaction
3(g)  2SO2(g) + O2(g) is 4.08 x
10-3decompose.
. Calculate the value for Q, and predict the
direction
in which
the reaction
proceed
Forward
reaction:
A --> Bwill
rate
= kf[A]
toward equilibrium if the initial concentrations
Reverse
reaction:
B
-->
A
rate
=
k
[B]
-3
-3
r
are: [SO3] = 2 x 10 M, [SO2] = 5 x 10 M, and
[O2]At= equilibrium,
3 x 10-2 M. kf[A] = kr[B]
Q = [SO2]2[O2] = (5 x 10-3)2(3 x 10-2) = 0.2
[B] = k2f = a constant (K-3eq)2
[SO
]
(2
x
10
)
3
[A] k
r
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
42
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
43
At equilibrium,
rateKatthe
which
products
Sample
exercise: Atthe
1000
value
of Keq form
for
EQUAL2SO
to the
rate at which products
the is
reaction
3(g)  2SO2(g) + O2(g) is 4.08 x
10-3decompose.
. Calculate the value for Q, and predict the
direction
in which
the reaction
proceed
Forward
reaction:
A --> Bwill
rate
= kf[A]
toward equilibrium if the initial concentrations
Reverse
reaction:
B
-->
A
rate
=
k
[B]
-3
-3
r
are: [SO3] = 2 x 10 M, [SO2] = 5 x 10 M, and
[O2]At= equilibrium,
3 x 10-2 M. kf[A] = kr[B]
Q = 0.2 ; 4.08 x 10-3 is much smaller, so I
[B] = kf = a constant (Keq)
need less products
to
form
and
more
reactants
[A] kr
which indicates a shift
to
the
left
Chem. 116 Prof. T.L. Heise
43
Chap 15.1
Chem 15.5
ChemProf.
116: Prof.
T.L.Heise
Heise
Chem. 107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
44
equilibrium,
thethe
rateKat
products
It isAt
possible
to know
and some
of theform
eq which
is EQUAL toand
the rate
concentrations
needattowhich
workproducts
backwards to
finddecompose.
a missing concentration.
Example:
For the Haber
Forward reaction:
A --> Bprocess,
rate = kf[A]
Kp = 1.45 x 10-5 at 500°C. Partial pressures
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
known are H2 0.928 atm and N2 0.432 atm.
What
the pressure
of NH
?
At is
equilibrium,
kf[A]
= kr3[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
44
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
45
equilibrium,
thethe
rateKat
products
It isAt
possible
to know
and some
of theform
eq which
is EQUAL toand
the rate
concentrations
needattowhich
workproducts
backwards to
finddecompose.
a missing concentration.
Example:
For the Haber
Forward reaction:
A --> Bprocess,
rate = kf[A]
Kp = 1.45 x 10-5 at 500°C. Partial pressures
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
known are H2 0.928 atm and N2 0.432 atm.
What
the pressure
of NH
?
At is
equilibrium,
kf[A]
= kr3[B]
Kp = 1.45 x 10-5 = [NH3]2 =
x2
[B] = kf = a constant
(Keq) 3
3
[H
]
[N
]
(0.928)
(0.432)
2
2
[A] k
r
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
45
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
46
equilibrium,
thethe
rateKat
products
It isAt
possible
to know
and some
of theform
eq which
is EQUAL toand
the rate
concentrations
needattowhich
workproducts
backwards to
finddecompose.
a missing concentration.
Example:
For the Haber
Forward reaction:
A --> Bprocess,
rate = kf[A]
Kp = 1.45 x 10-5 at 500°C. Partial pressures
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
known are H2 0.928 atm and N2 0.432 atm.
What
the pressure
of NH
?
At is
equilibrium,
kf[A]
= kr3[B]
Kp = 1.45 x 10-5 =
x2
x = 5.01 x
[B] = kf = a constant3 (Keq)
-6
10
(0.928)
(0.432)
[A] k
r
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
46
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
47
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.5
Chem. 116 Prof. T.L. Heise
Chem.
47
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
48
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
Kp = [PCl3][Cl2]
[PCl[B]
5] = kf = a constant (Keq)
[A]
Chap 15.1
Chem 15.5
kr
Chem. 116 Prof. T.L. Heise
Chem.
48
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
49
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
Kp = [PCl3][Cl2] 0.497 = (.350)(x)
(Keq)
[PCl[B]
5] = kf = a constant
(0.860)
[A]
Chap 15.1
Chem 15.5
kr
Chem. 116 Prof. T.L. Heise
Chem.
49
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
50
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
Kp = [PCl3][Cl2] 0.497(0.860) = (.350)(x)
[PCl[B]
5] = kf = a constant (Keq)
[A]
Chap 15.1
Chem 15.5
kr
Chem. 116 Prof. T.L. Heise
Chem.
50
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
51
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
Kp = [PCl3][Cl2] 0.497(0.860) = (x)
= a constant (Keq)
[PCl[B]
5] = kf (0.350)
[A]
Chap 15.1
Chem 15.5
kr
Chem. 116 Prof. T.L. Heise
Chem.
51
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Applications
The ConceptofofKEquilibrium
eq
52
At equilibrium,
rate
at which
products form
Sample
exercise:Atthe
500
K the
reaction
is PCl
EQUAL
to the rate at which products
5(g)  PCl3(g) + Cl2(g) has Kp = 0.497.
decompose.
In an
equilibrium mixture at 500 K, the partial
pressure
of PClreaction:
atm
that=of
PCl3 is
5 is 0.860 A
Forward
--> and
B rate
kf[A]
0.350 atm. What is the partial pressure of Cl2 in
Reverse
reaction:
B
-->
A
rate
=
k
[B]
r
the equilibrium mixture?
At equilibrium, kf[A] = kr[B]
Kp = [PCl3][Cl2] 1.22 = (x)
[PCl[B]
5] = kf = a constant (Keq)
[A]
Chap 15.1
Chem 15.5
kr
Chem. 116 Prof. T.L. Heise
Chem.
52
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
53
At equilibrium, the rate at which products form
If a system at equilibrium is disturbed by a
ischange
EQUAL
to the rate at pressure,
which products
in temperature,
or the
decompose.
concentration of one of the components, the
system
will reaction:
shift its equilibrium
position
so as
Forward
A --> B rate
= kf[A]
to counteract the effect of the disturbance.
Reverse
reaction:
-->equilibrium
A rate = kr[B]
3 ways
to changeBan
At equilibrium,
kf[A] = akrreactant
[B]
add or remove
or product
Chap 15.1
Chem 15.6
[B] = kfthe
= apressure
constant (Keq)
change
[A] kr
change theChem.
temperature
116 Prof. T.L. Heise
Chem.
53
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
54
At equilibrium, the rate at which products form
Change in reactant or product concentrations
is EQUAL to the rate at which products
adding a reactant or product forces a shift
decompose.
that will use up what has been added
Forward reaction: A --> B rate = kf[A]
removing a reactant or product forces a
Reverse
reaction:
--> A ofrate
= kwas
shift that
will createB more
what
r[B]
taken
At equilibrium,
k [A] = k [B]
f
r
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
54
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
55
At equilibrium, the rate at which products form
Change in volume or pressure
is EQUAL to the rate at which products
decreasing volume causes increasing
decompose.
pressure and vice versa
Forward reaction: A --> B rate = kf[A]
increasing pressure forces a shift in
Reverse
reaction:
B -->
rate = kr[B]
equilibrium
towards
the A
production
of
fewer moles of
gas= k [B]
At equilibrium,
k [A]
f
r
decreasing
forces(Ka shift
in
[B] = kf pressure
= a constant
)
eq
equilibrium
[A] krtowards the production of
more moles of Chem.
gas116 Prof. T.L. Heise
Chap 15.1
Chem 15.6
Chem.
55
ChemProf.
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T.L.Heise
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Le
Châtelier’s
Principle
The
Concept of
Equilibrium
56
At equilibrium, the rate at which products form
Change in temperature
is EQUAL to the rate at which products
increasing temperature forces a shift in
decompose.
the endothermic direction so increased
Forward
reaction:
energy can
be used A
up--> B rate = kf[A]
Reverse
reaction:
B --> Aforces
rate =
kr[B]in
decreasing
temperature
a shift
the exothermic
direction
At equilibrium,
kf[A]
= kr[B]so more energy
can be produced to replace lost energy
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
56
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
57
Concentration
of
At equilibrium, the rate at which
products form
is EQUAL to the rate at which H
products
2 is altered and
affects are
decompose.
graphed
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
57
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
58
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (a) Cl2(g) is added
Forward reaction: A --> B rate = kf[A]
(b) the temperature is increased
Reverse
(c ) the
reaction:
volume isBdecreased
--> A rate = kr[B]
(d) PCl5(g) is added
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
58
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
59
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (a) Cl2(g) is added
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) +
Cl
At2(g)
equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
59
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
60
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (a) Cl2(g) is added
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) + Cl2(g)
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
60
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
61
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (b) the temperature is increased
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) + Cl2(g)
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
61
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
62
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (b) the temperature is increased
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) + Cl2(g)
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
62
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
63
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (c ) the volume is decreased
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
63
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
64
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (c ) the volume is decreased
Forward reaction: A --> B rate = kf[A]
...so pressure is increased, favors less
moles
Reverse reaction: B --> A rate = kr[B]
At equilibrium,
= k3r(g)
[B]+ Cl2(g)
PCl
PCl
f[A]
5(g) + energyk
1 mole [B] = kf = a1 constant
mole + 1 (K
mole
eq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
64
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
65
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (c ) the volume is decreased
Forward reaction: A --> B rate = kf[A]
...so pressure is increased, favors less
moles
Reverse reaction: B --> A rate = kr[B]
At equilibrium,
= k3r(g)
[B]+ Cl2(g)
PCl
PCl
f[A]
5(g) + energyk
1 mole [B] = kf = a1 constant
mole + 1 (K
mole
eq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
65
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
66
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (d) PCl5(g) is added
Forward reaction: A --> B rate = kf[A]
Reverse reaction:
B --> A rate = kr[B]
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
66
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
67
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (b) the temperature is increased
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) + Cl2(g)
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
67
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
68
Sample
At equilibrium,
Exercise:the
For
rate
theatreaction
which products form
is EQUAL
to +the
rate at
products
PCl5(g)
energy
which
PCl3(g)
+ Cl2(g)
in
decompose.
what direction will the equilibrium shift
when (b) the temperature is increased
Forward reaction: A --> B rate = kf[A]
Reverse reaction: B --> A rate = kr[B]
PCl5(g) + energy  PCl3(g) + Cl2(g)
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
68
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
69
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g)  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
use this result to determine how the
equilibrium
Reverse reaction:
constant for
B -->
theAreaction
rate =should
kr[B]
change with temperature.
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
69
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
70
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g)  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
-542.2
-288.07 + 0
Reverse reaction: B --> A rate = kr[B]
DH = products - reactants
At equilibrium,
2(-288.07) k- f2(-542.2)
[A] = kr[B]
508.26 kJ
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
70
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
71
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g)  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
-542.2
-288.07 + 0
Reverse reaction: B --> A rate = kr[B]
DH = products - reactants
At equilibrium,
2(-288.07) k- f2(-542.2)
[A] = kr[B]
508.26 kJ
[B] = kf = a constant (Keq)
Endothermic
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
71
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
72
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g) + energy  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
use this result to determine how the equilibrium
constant
Reverse
forreaction:
the reaction
B -->
should
A rate
change
= krwith
[B]
temperature.
At equilibrium, kf[A] = kr[B]
[B] = kf = a constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
72
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
73
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g) + energy  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
use this result to determine how the equilibrium
constant
Reverse
forreaction:
the reaction
B -->
should
A rate
change
= krwith
[B]
temperature.
At equilibrium, k2f[A] = kr[B]
Keq = [PCl3] [O2]
[B] =[POCl
kf = 3a]2constant (Keq)
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
73
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
74
Sample
At equilibrium,
Exercise:the
Using
ratethe
at which
thermodynamic
products form
data
is EQUAL
in Appendix
to the rate
C, determine
at which the
products
enthalpy
change
decompose.
for the reaction
2POCl3(g) + energy  2PCl3(g) + O2(g)
Forward reaction: A --> B rate = kf[A]
use this result to determine how the equilibrium
constant
Reverse
forreaction:
the reaction
B -->
should
A rate
change
= krwith
[B]
temperature.
At equilibrium, k2f[A] = kr[B]
Keq = [PCl3] [O2]
if energy inc.,
[B] =[POCl
kf = 3a]2constant
equilibrium
(Keq)
shifts
[A] kr
right, products are
bigger,
KeqProf.
is larger
Chem. 116 Prof. T.L. Heise
74
Chap 15.1
Chem 116:
T.L. Heise
Chem 15.6
Chem. 107: Prof. T. L. Heise
Le
Châtelier’s
Principle
The
Concept of
Equilibrium
75
Effects
At equilibrium,
of a Catalyst:
the rate
a catalyst
at which
changes
products
the form
rate
of
is aEQUAL
reactiontoonly,
the rate
not the
at which
amounts
products
of the
compounds!!
decompose.
Forward reaction: A --> B rate = kf[A]
Remember
Reverse reaction:
- Rates are B
Equal
--> A rate = kr[B]
At equilibrium,
Concentrations
kf[A] = kr[B]are Constant
[B] =Only
kf =temp.
a constant
can change
(Keq) a constant
[A] kr
Chap 15.1
Chem 15.6
Chem. 116 Prof. T.L. Heise
Chem.
75
ChemProf.
116: Prof.
T.L.Heise
Heise
107:
T. L.