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Dr. Donald VanDerveer
Office: Boggs 2-5
Office Hours: 8:00-9:30 MWF
e-mail: don.vanderveer@
chemistry.gatech.edu
Phone/voice-mail: 404-894-8517
Location of Boggs 2-5
B6A
Equilibrium
Equilibrium
In order to have equilibrium for a chemical
reaction, the process must be able to move
both forwards and backwards.
2 H2 + O2
spark
2 H2O
2 H2 + O2
2 H2O
spark
2 H2O
2 H2 + O2
Equilibrium processes include both
reactions which have chemically
different reactants and products and
phase changes.
Equilibrium
H2O(l)
H2O(g)
25oC
H2O(l)
H2O(g)
After a given period of time, the pressure
does not increase nor does it decrease.
H2O(l)
H2O(g)
H2O(g)
H2O(l)
H2O(g)
H2O(l)
As long as there is liquid water present,
there will be a liquid-gas equilibrium.
As long as there is liquid water present,
there will be a liquid-gas equilibrium.
At any given temperature, the equilibrium
pressure will be the same.
As long as there is liquid water present,
there will be a liquid-gas equilibrium.
At any given temperature, the equilibrium
pressure will be the same.
This pressure is independent of the
volume of the container.
Equilibrium constant:
Equilibrium constant:
Equilibrium constants are calculated
based on the phases of the reactants
and products.
Equilibrium constant:
If reactants and products are gasses, the
equilibrium constant is calculated from
partial pressures.
Equilibrium constant:
If reactants and products in solution, the
equilibrium constant is calculated from
concentrations.
Equilibrium constant:
How the value of an equilibrium
constant is determined varies,
depending on the type of
reaction.
Equilibrium constant:
For a liquid - gas equilibrium, the
constant, K, is equal to the vapor
pressure of the gas at equilibrium.
Equilibrium constant:
For a liquid - gas equilibrium, the
constant, K, is equal to the vapor
pressure of the gas at equilibrium.
Px
K=
1 atm
H2O(g)
H2O(l)
K = 0.03126
25o C
H2O(g)
H2O(l)
30o C
K = 0.04187
Chemical equilibrium
2 NO2(g)
N2O4(g)
2
17 e-
Place different amounts of pure NO2
in a container, equilibrate at 25oC and
measure the partial pressures of both
gasses.
N2O4(g)
2 NO2(g)
PNO
2
Know PN O and PNO .
2 4
2
Know PN O and PNO .
2 4
2
Make equilibrium constant
from an expression of
partial pressures.
Partial pressures
P1 + P2 + … Pn = Ptotal
PN O
2 4
= K(PNO )2
2
PNO
2
2 NO2(g)
N2O4(g)
PN O
2 4
= K(PNO )2
2
PNO
2
2 NO2(g)
N2O4(g)
PN O
2 4
K=
= K(PNO )2
2
PN O
2 4
(PNO )2
2
PNO
2
2 NO2(g)
N2O4(g)
PN O
2 4
K=
= K(PNO )2
2
PN O
2 4
(PNO )2
2
PNO
2
K = 8.8 @ 25oC
Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.
Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.
3. Dynamic balance of forward
and reverse processes.
Characteristics of equilibrium states
1. No changes in macroscopic states
(pressure, volume, etc.)
2. Equilibrium is reached through
spontaneous processes.
3. Dynamic balance of forward
and reverse processes.
4. Are the same regardless of
direction of approach.
General form of equilibrium constant
when all species are low-pressure gasses
General form of equilibrium constant
when all species are low-pressure gasses
Balanced chemical equation:
aA + bB
c d
PCPD
PAa PBb
cC
=K
+ dD
c d
PCPD
PAa PBb
=K
2 NOCl
2
PNOPCl
2
PNOCl
2 NO + Cl2
=K
Calculating equilibrium constants
Calculating equilibrium constants
Data
PNO
2
Calculating equilibrium constants
Data
K=
PN O
2 4
(PNO )2
2
PNO
2
Calculating equilibrium constants
K=
PN O
2 4
(PNO )2
2
PNO
2
Calculating equilibrium constants
PN O
K=
2 4
(PNO )2
2
PN O = 5 atm
2 4
PNO
2
Calculating equilibrium constants
PN O
K=
2 2
(PNO )2
2
PN O = 5 atm
2 4
PNO
PNO
2
2
= 0.75 atm
Calculating equilibrium constants
5 atm
K=
(0.75 atm)2
PN O = 5 atm
2 4
PNO
PNO
2
2
= 0.75 atm
Calculating equilibrium constants
5 atm
K=
=
(0.75 atm)2
K = 8.9
PNO
2
Calculating equilibrium constants
5 atm
K=
=
(0.75 atm)2
K = 8.9
PNO
2
K = 8.8 @ 25oC
Calculating equilibrium constants
without all partial pressures.
Calculating equilibrium constants
without all partial pressures.
4 NO2
2 N2O + 3 O2
Calculating equilibrium constants
without all partial pressures.
4 NO2
2 N2O + 3 O2
Given: starting partial pressures,
equilibrium partial pressure for NO2.
4 NO2
2 N2O + 3 O2
Initial (atm)
3.6
5.1
8.0
change (atm)
?
?
?
Equilibrium
2.4
?
?
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
?
?
Equilibrium
?
?
2.4
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
?
?
Equilibrium
?
?
2.4
For every 4 molecules of NO2 that react,
2 molecules of N2O and 3 molecules of O2
are formed.
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -4x
+2x
+3x
?
?
Equilibrium
2.4
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -4x
+2x
+3x
5.1 + 2x
8.0 + 3x
Equilibrium
2.4
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -4x
+2x
+3x
5.1 + 2x
8.0 + 3x
Equilibrium
2.4
3.6 - 4x = 2.4
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -4x
+2x
+3x
5.1 + 2x
8.0 + 3x
Equilibrium
2.4
3.6 - 4x = 2.4
x=
2.4 - 3.6
-4
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -4x
+2x
+3x
5.1 + 2x
8.0 + 3x
Equilibrium
2.4
3.6 - 4x = 2.4
x=
2.4 - 3.6
-4
= 0.3 atm
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
+0.6
+0.9
5.7
8.9
Equilibrium
2.4
3.6 - 4x = 2.4
x=
2.4 - 3.6
-4
= 0.3 atm
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
+0.6
+0.9
5.7
8.9
Equilibrium
2.4
(PN O)2(PO)3
2
2
(PNO )4
2
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
+0.6
+0.9
5.7
8.9
Equilibrium
2.4
(PN O)2(PO)3
2
(PNO )4
2
2
(5.7)2(8.9)3
=
(2.4)4
Volume changes are  coefficients
4 NO2
Initial (atm)
2 N2O + 3 O2
3.6
5.1
8.0
change (atm) -1.2
+0.6
+0.9
5.7
8.9
Equilibrium
2.4
(PN O)2(PO)3
2
2
(PNO )4
2
=
(5.7)2(8.9)3
(2.4)4
= 6.9 x 102
Relationships among equilibrium
expressions.
Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.
Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.
2 NO2
K=
N2O4
PN O
2 4
(PNO )2
2
Relationships among equilibrium
expressions.
1. The equilibrium constant of a
reverse reaction is the reciprocal
of the forward reaction.
2 NO2
K1 =
N2O4
PN O
2 4
(PNO )2
2
N2O4
K2 =
2 NO2
(PNO2)2
PN O
2 4
2 NO2
K1 =
N2O4
PN O
2 4
N2O4
K2 =
(PNO )2
2 NO2
(PNO2)2
PN O
2 4
2
K1K2 =1
2 NO2
N2O4
PN O
K1 =
N2O4
K2 =
2 4
(PNO )2
K1K2 =1
K1
K1
(PNO2)2
PN O
2 4
2
1
2 NO2
=1
2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.
2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.
2 NO2
K1 =
N 2O 4
4 NO2
PN O
2 4
(PNO )2
2
K2 =
2 N2O4
(PN O )2
2 4
(PNO )4
2
2. If coefficients in a balanced equation
are all multiplied by a single factor, the
equilibrium constant is raised to a power
equal to that factor.
2 NO2
K1 =
N 2O 2
4 NO2
PN O
2 4
K2 =
(PNO )2
2
K2 = K12
2 N2O2
(PN O )2
2 4
(PNO )4
2
3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.
3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.
Exercise page 290
CF4 + 2 H2O
2 CO2 + 4 HF
K1 = 5.9 x 1023
3. Operations of addition and subtraction,
applied to chemical equations, lead to
operations of multiplication and division
of their corresponding equilibrium expressions
and constants.
Exercise page 290
CF4 + 2 H2O
CO2 + 4 HF
K1 = 5.9 x 1023
CO + 1/2 O2
CO2
K2 = 1.3 x 1035
CF4 + 2 H2O
CO + 1/2 O2
2 CF4 + 4 H2O
CO2 + 4 HF
K1 = 5.9 x 1023
CO2
K2 = 1.3 x 1035
2 CO + 8 HF + O2
CF4 + 2 H2O
CO + 1/2 O2
2 CF4 + 4 H2O
CO2 + 4 HF
K1 = 5.9 x 1023
CO2
K2 = 1.3 x 1035
2 CO + 8 HF + O2
CF4 + 2 H2O
CO + 1/2 O2
2 CF4 + 4 H2O
CO2 + 4 HF
K1 = 5.9 x 1023
CO2
K2 = 1.3 x 1035
2 CO + 8 HF + O2
CF4 + 2 H2O
2 CO + O2
CO2 + 4 HF
K1 = 5.9 x 1023
2 CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O
2 CO + 8 HF + O2
2 CF4 + 4 H2O
2 CO + O2
2 CO2 + 8 HF
K1 = 34.8 x 1046
2 CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O
2 CO + 8 HF + O2
2 CF4 + 4 H2O
2 CO2 + 8 HF
2 CO + O2
2 CF4 + 4 H2O
2 CO + 8 HF + O2
2 CF4 + 4 H2O
2 CO + O2
2 CO2 + 8 HF
K1 = 34.8 x 1046
2 CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O
K3 =
2 CO + 8 HF + O2
K1 x K2-1
2 CF4 + 4 H2O
2 CO + O2
2 CO2 + 8 HF
K1 = 34.8 x 1046
2 CO2
K2 = 1.7 x 1070
2 CF4 + 4 H2O
K3 =
2 CO + 8 HF + O2
K1 x K2-1
34.8 x 1046 = 20.5 x 10-24
K3 =
1.7 x 1070
Reaction quotient
Q =
c d
PCPD
PAa PBb
Reaction quotient
Q =
c d
PCPD
PAa PBb
PA , PB, etc. not at equilibrium
Reaction quotient
Q =
c d
PCPD
PAa PBb
Q will go to the value of K as
the partial pressures go to
equilibrium
Q vs K can
predict where
the reaction is
in respect to
equilibrium.
K =
Q =
c d
PCPD
Q<K
PAa PBb
aA + bB
c d
PCPD
PAa PBb
cC + dD
K =
Q =
c d
PCPD
Q<K
PAa PBb
aA + bB
c d
PCPD
PAa PBb
cC + dD
Too much reactant,
not enough product.
K =
Q =
c d
PCPD
Q<K
PAa PBb
aA + bB
c d
PCPD
PAa PBb
cC + dD
Q>K
aA + bB
cC + dD
K =
Q =
c d
PCPD
Q<K
PAa PBb
aA + bB
c d
PCPD
PAa PBb
cC + dD
Q>K
aA + bB
cC + dD
Too much product, not enough
reactant.
K =
Q =
c d
PCPD
Q<K
PAa PBb
aA + bB
c d
PCPD
PAa PBb
Q<K
cC + dD
Q>K
aA + bB
PAa PBb
cC + dD
Large vs
c d
PCPD
Q =
c d
PCPD
PAa PBb
Exercise page 291
P4
2 P2
K = 1.39 @ 400oC
1.40 mol P4
1.25 mol P2
Volume = 25.0 L
P4
Q =
2 P2
(PP 2)2
PP 4
P4
Q =
2 P2
(PP 2)2
PP 4
P =
nRT
T = 673 K
V
V = 25.0 L
R = 0.0820578 L atm mol-1 K-1
P4
Q =
2 P2
(PP 2)2
PP
4
nRT = (1.4)(0.0820578)(673)
PP =
= 3.09 atm
4
25.0
V
P4
Q =
2 P2
(PP 2)2
PP
4
nRT = (1.4)(0.0820578)(673)
PP =
= 3.09 atm
4
25.0
V
PP =
2
nRT
V
(1.25)(0.0820578)(673)
=
= 2.76 atm
25.0
P4
Q =
PP = 2.76 atm
2
PP = 3.09 atm
4
2 P2
(PP 2)2
PP
=
(2.76)2
3.09
4
=
P4
Q =
PP = 2.76 atm
2 P2
(PP 2)2
PP
=
3.09
4
2
PP = 3.09 atm
4
Q = 2.46
K=
(2.76)2
1.39
=
P4
Q =
2 P2
(PP 2)2
PP
PP = 2.76 atm
=
3.09
4
2
PP = 3.09 atm
Q = 2.46
4
K=
P4
(2.76)2
1.39
2 P2
=
Exercise page 292
CO + H2O
CO2 + H2
Exercise page 292
CO + H2O
PCO = 2.00 atm
PCO 2 = 0.80 atm
PH 2 = 0.48 atm
K = 0.64 @ 900 K
CO2 + H2
Exercise page 292
CO + H2O
PCO = 2.00 atm
PCO 2 = 0.80 atm
PH 2 = 0.48 atm
K = 0.64 @ 900 K
CO2 + H2
K=
PCO 2 PH 2
PCO PH O
2
Exercise page 292
CO + H2O
PCO = 2.00 atm
CO2 + H2
K=
PCO 2 PH 2
PCO PH O
PCO 2 = 0.80 atm
PH 2 = 0.48 atm
K = 0.64 @ 900 K
2
PH O =
2
PCO 2 PH 2
PCO K
Exercise page 292
CO + H2O
CO2 + H2
PCO = 2.00 atm
PCO 2 = 0.80 atm
PH 2
= 0.48 atm
K = 0.64 @ 900 K
PH O =
2
PCO 2 PH 2
=
PCO K
(0.80)(0.48)
= 0.30 atm
(2.00) (0.64)
Converting between partial
pressures and concentrations.
Converting between partial
pressures and concentrations.
Concentration =
moles
V
Converting between partial
pressures and concentrations.
Concentration =
moles
V
PV = nRT
Converting between partial
pressures and concentrations.
Concentration =
moles
V
For gas ‘A’
PV = nRT
[A] =
nA
V
Converting between partial
pressures and concentrations.
Concentration =
moles
V
For gas ‘A’
PV = nRT
[A] =
nA
V
=
PA
RT
Converting between partial
pressures and concentrations.
Concentration =
moles
V
For gas ‘A’
PV = nRT
[A] =
nA
V
PA = RT[A]
=
PA
RT
PA = RT[A]
2 NO2(g)
N2O4(g)
PA = RT[A]
2 NO2(g)
Pref = 1 atm
N2O4(g)
PA = RT[A]
N2O4(g)
2 NO2(g)
Pref = 1 atm
PN O /Pref
2 4
(PNO
2
=K
/Pref)2
PA = RT[A]
N2O4(g)
2 NO2(g)
PN O = RT[N2O4]
2 4
Pref = 1 atm
PN O /Pref
2 4
(PNO
2
=K
/Pref)2
PA = RT[A]
2 NO2(g)
PN O = RT[N2O4]
2 4
Pref = 1 atm
PNO = RT[NO2]
PN O /Pref
2 4
(PNO
2
2
=K
/Pref)2
N2O4(g)
N2O4(g)
2 NO2(g)
PN O = RT[N2O4]
2 4
PNO = RT[NO2]
PN O /Pref
2 4
(PNO
2
=K
/Pref)2
2
K=
[N2O4](RT/Pref)
[NO2](RT/Pref)
=
[N2O4]
[NO2]
x
(
RT
Pref
-1
)
N2O4(g)
2 NO2(g)
K=
[N2O4](RT/Pref)
=
[NO2](RT/Pref)
[N2O4]
[NO2]
=K
[N2O4]
[NO2]
(
RT
Pref
)
x
(
RT
Pref
-1
)
[N2O4]
[NO2]
=K
(
aA + bB
c d
PCPD
PAa PBb
RT
Pref
)
cC
[C]c[D]d
=K
[A]a[B]b
=K
+ dD
(
RT
Pref
a+b-c-d
)
[C]c[D]d
[A]a[B]b
=K
(
RT
Pref
a+b-c-d
)
Exercise page 297
CH4 + H2O
CO + 3 H2
[H2]=[CO]=[H2O]= 0.00642 mol L-1
K = 0.172
[C]c[D]d
=K
[A]a[B]b
CH4 + H2O
(
RT
Pref
a+b-c-d
)
CO + 3 H2
[H2]=[CO]=[H2O]= 0.00642 mol L-1
K = 0.172
(0.00642)4
[CH4](0.00642)
= 0.172
(
RT
Pref
-2
)
CH4 + H2O
(0.00642)3
CO + 3 H2
= 0.172
[CH4]
(
RT
Pref
-2
)
K(RT)-2 = 3.153 x 10-5
3
(0.00642)
[CH4] =
= 8.39 x 10-2 mol L-1
3.153 x 10-5