Equilibrium Thermodynamics
Reversibility and Chemical Change
• Equilibrium vapor pressure
– Evaporation and condensation
– Triple point conditions
• Chemical Reactions:
– CaCO
3
(s) ⇔ CaO(s) + CO
2
(g)
– CaCO
3
(s) + 2NaCl(s) ⇔ CaCl
2
(s) + Na
2
CO
3
(s)
Chemical equilibrium
• Reversibility is a general property of chemical change.
• Macroscopic reversibility depends on law of mass action:
– Rate of a reaction is a function of how much material
is reacting (concentration or partial pressure).
– Chemical equilibrium is achieved when the rate of the
forward reaction equals the rate of the reverse.
– Phase changes often accompany chemical change.
• Le Chatelier’s Principle:
– Systems at equilibrium try to stay in equilibrium and
respond to external stresses accordingly.
Systems at Equilibrium
• Systems move spontaneously toward equilibrium.
• Equilibrium is a dynamic state.
• Approach to equilibrium is independent of
direction.
• Trade-off between organization and
randomization.
Simple System
H
2
(g) ⇔ 2H(g)
– Drive toward maximum entropy:
• Favors bond dissociation, converting H
2
molecules to free H atoms.
• Energy is required.
• Equilibrium shifts to the right.
– Drive to achieve minimum energy
• favors bond formation and H
2
molecules
over free H atoms.
• Equilibrium shifts to the left..
Hydrogen Iodide
Synthesis and Decomposition
The Equilibrium Constant
• p = partial pressure,
usually measured in
units of torr or atm.
• [conc] = [ mol/L]
∆n = difference in
moles (n) of products
and reactants:
∆n= n
p
- n
r
• For a general reaction:
aA + bB⇔cC + dD
•
K
p
= K
c
(RT)
∆
n
K
c =
[
C ]
c [
D]
d
[
A]
a [
B]
b
K
p =
pc p d
pa pb
The Equilibrium Constant
• 2HI(g) ⇔ H
2
(g) + I
2
(g)
K = [H
2
][I
2
]/[HI]2
• H
2
(g) + I
2
(g) ⇔ 2HI(g)
K’ = 1/K = [HI]2/[H
2
][I
2
]
• K
p
= K
c
because ∆n = 0
Ammonium Chloride
Synthesis and Decomposition
• Chemical equilibrium is
achieved from either
direction
• Equilibrium depends
on…
– Temperature
– Pressure
– Moles of reactants
and products
The Equilibrium Constant
• NH
4
Cl(s) ⇔ NH
3
(g) + HCl(g)
K
c
= [NH
3
][HCl]
K
p
= p
NH3
p
HCl
• NH
3
(g) + HCl(g) ⇔ NH
4
Cl(s)
K` = 1/K = [NH
3
][HCl]
K` = 1/K = 1/p
NH3
p
HCl
• K
p
≠ K
c
because ∆n ≠ 0
The Equilibrium Constant
• 3H
2
(g) + N
2
(g) ⇔ 2NH
3
(g)
K = [NH
3
]2/[H
2
]3[N
2
]
• 2NH
3
(g) ⇔ 3H
2
(g) + N
2
(g)
K` = 1/K = [H
2
]3[N
2
]/ [NH
3
]2
• K
p
≠ K
c
because ∆n ≠ 0
Le Chatelier’s Principle
• Systems in equilibrium tend to stay in
equilibrium unless acted upon by an
external stress such as…..
– changes in concentration
– changes in temperature
– changes in pressure/volume
• Catalysts alter only the rate at which
equilibrium is achieved.
Ammonia Synthesis
Le Chatelier’s Principle
– 3H
2
(g) + N
2
(g) ⇔ 2NH
3
(g)
∆H = -93 kJ
– CO
2
(g) + H
2
(g) ⇔ CO(g) + H
2
O(g)
∆H = +41 kJ
– 4HCl(g) + O
2
(g) ⇔ 2Cl
2
(g) + H
2
O(g)
∆H = +118 kJ
Examples
• Decomposition of nitrosyl bromide (NOBr)
– NO(g) + Br
2
(g) ⇔ NOBr(g)
• Carbon monoxide shift reaction
– CO(g) + H
2
O(g) ⇔ CO
2
(g) + H
2
(g)
• Hydrogen iodide formation
–
H
2
(g) + I
2
(g) ⇔ 2HI(g)
2NO
2
(colorless) ⇔ N
2
O
4
(red)
2NO
2
(colorless) ⇔ N
2
O
4
(red)
2NO
2
(colorless) ⇔ N
2
O
4
(red)
Soluble Salts in Water
• KI and K
2
CrO
4
:
– Potassium iodide and
chromate are soluble
– Lead chromate and
silver iodide are
insoluble…. sparingly
soluble:
• Ksp(PbCro4)
An agricultural scientist, Norman Borlaug was recognized
By the Nobel Peace Prize in 1970 for his work on food
and agriculture.
He often speculates that if Alfred Nobel had written
his will to establish the various prizes and endowed them
fifty years earlier, the first prize established would have
been for food and agriculture. However, by the time Nobel
wrote his will in 1895, there was no serious food production
problem haunting Europe like the widespread potato famine
in 1845-51, that took the lives of untold millions.
http://www.nobel.se/peace/laureates/1970/
The Equilibrium Constant
• p = partial pressure,
usually measured in
units of torr or atm.
• [conc] = [ mol/L]
∆n = difference in
moles (n) of products
and reactants:
∆n= n
p
- n
r
• For a general reaction:
aA + bB⇔cC + dD
•
K
p
= K
c
(RT)
∆
n
K
c =
[
C ]
c [
D]
d
[
A]
a [
B]
b
K
p =
pc p d
pa pb
The Equilibrium Constant
• 2HI(g) ⇔ H
2
(g) + I
2
(g)
K = [H
2
][I
2
]/[HI]2
• H
2
(g) + I
2
(g) ⇔ 2HI(g)
K’ = 1/K = [HI]2/[H
2
][I
2
]
• K
p
= K
c
because ∆n = 0
The Equilibrium Constant
• NH
4
Cl(s) ⇔ NH
3
(g) + HCl(g)
K
c
= [NH
3
][HCl]
K
p
= p
NH3
p
HCl
• NH
3
(g) + HCl(g) ⇔ NH
4
Cl(s)
K` = 1/K = [NH
3
][HCl]
K` = 1/K = 1/p
NH3
p
HCl
• K
p
≠ K
c
because ∆n ≠ 0
The Equilibrium Constant
• 3H
2
(g) + N
2
(g) ⇔ 2NH
3
(g)
K = [NH
3
]2/[H
2
]3[N
2
]
• 2NH
3
(g) ⇔ 3H
2
(g) + N
2
(g)
K` = 1/K = [H
2
]3[N
2
]/ [NH
3
]2
• K
p
≠ K
c
because ∆n ≠ 0
Le Chatelier’s Principle
• Systems in equilibrium tend to stay in
equilibrium unless acted upon by an
external stress such as…..
– changes in concentration
– changes in temperature
– changes in pressure/volume
• Catalysts alter only the rate at which
equilibrium is achieved.
Le Chatelier’s Principle
Enthalpy Change - Heat of Reaction
– 3H
2
(g) + N
2
(g) ⇔ 2NH
3
(g)
∆H = -93 kJ
– CO
2
(g) + H
2
(g) ⇔ CO(g) + H
2
O(g)
∆H = +41 kJ
Examples
• Decomposition of nitrosyl bromide (NOBr)
– NO(g) + Br
2
(g) ⇔ NOBr(g)
• Carbon monoxide shift reaction
– CO(g) + H
2
O(g) ⇔ CO
2
(g) + H
2
(g)
• Hydrogen iodide formation
–
H
2
(g) + I
2
(g) ⇔ 2HI(g)
Haber Ammonia
• Fertilizers/Explosives
– Ammonium salts
– Nitrates
– Nitric acid
• Refrigerant
• Drugs-Dyes-Fibers
• Photography
• Household
Haber Ammonia
C. Bosch
F. Bergius
Haber Ammonia
Haber Ammonia and
War Reparations
• 33 billion dollars = 50,000 tons of gold
– Could not resort to…
• Synthetic ammonia
• Dye industry
• German colonies
– Estimated total gold content of the oceans:
• 8 billion tons
• Based on estimates of 5-10 mg/metric ton
Gold from seawater (1923)
– Chemistry:
• Add lead acetate or mercuric nitrate,
followed by ammonium sulfide, precipitating
the sulfide (Au
2
S)
• Separate silver by dissolving in nitric acid
– Alchemy
N
2
O
4
(g,red) ⇔ 2NO
2
(g,colorless)
Kp=
p
NO
2
2
p
N
2
O
4
=
2α
(1+α)
P
T
2
(1−α)
(1+α)
P
T
P
T
= 4α2
1−α2
P
T
N
2
O
4
(g,red) ⇔ 2NO
2
(g,colorless)
• Sample problem:
– Consider a mixture of N
2
O
4
and NO
2
at a
total pressure of 1.5 atm… resulting
from the dissociation of N
2
O
4
.
• If Kp = 0.14 at the temperature of the
experiment, what fraction of the N
2
O
4
originally present dissociated?
– What happens if P
T
falls to 1.0 atm?
Phosgene Decomposition
•
COCl
2
(g) ⇔
CO(g) + Cl
2
(g)
– Write a general expression in terms of
• the fraction α decomposed
• the total pressure P
T
• the equilibrium constant K
p
– Demonstrates the pressure-dependency
for an equilibrium system where ∆n≠0
NH
4
HS(s) ⇔ NH
3
(g) + H
2
S(g)
• If Kp = 0.11 at the temperature of
the experiment, what is the the
partial pressure of NH
3
? Of H
2
S?
• Add solid NH
4
HS into a reactor
containing 0.50 atm of NH
3
and
calculate the partial pressures of
both gases at equilibrium.
