Le Châtelier`s Principle

Le Châtelier's Principle
“If a chemical system at equilibrium experiences a change in
concentration, temperature, volume, or total pressure, then
the equilibrium shifts to partially counteract the imposed change
change.”
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Changes in Concentration
PCl5 PCl3 + Cl2 Kc = 0.030
At equilibrium:
 Adding reactants shifts the reaction toward products
 Adding products shifts the reaction toward reactants
 Removing
g reactants … shifts the reaction toward reactants
 Removing products … shifts the reaction toward products
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Le Châtelier Sample Problem
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Changes in Pressure and Volume
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Pre-solved Example
N2 (g) + 3 H2 (g)
2 NH3 (g)
A system with 2.5 atm of N2 and 7.5 atm of H2 is allowed to equilibrate at 500°C,
then the overall pressure is increased by a factor of 10. What happens? 2.0% of total pressure
Before compression
After
compression
PNH3 = 0.2 atm
PNH3 = 8.4 atm
PN2 = 2.4 atm
PN2 = 21 atm
PH2 = 7.2 atm
PH2 = 62 atm
9.2% of total pressure
 The reaction shifts toward products because this
minimizes the total number of molecules in the system
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Further Examples
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Changes in Temperature
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Sample Question
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Adding a Catalyst
A catalyst is a substance that increases the rate of a reaction
without itself being consumed in the reaction
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Question
Which of the following equilibria would not be affected by changes in overall pressure? (a) 2 NO(g) + O2(g) 2 NO2(g) (b) N2O4(g) 2 NO2(g) (c) 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) (d) CaCO3(s) CaO(s) + CO2(g) (e) CO(g) + H2O(g) CO2(g) + H2(g) 72
New Question
What would happen if O2 were removed from the following system at equilibrium at 25°C? 2 NO2(g) 2 NO(g) + O2(g) Kc = 7.4 x 10–1 (a) The NO2 and NO concentrations would increase. (b) The NO2 and NO concentrations would decrease. (c) The NO2 concentration would increase and the NO concentration would decrease. (d) The NO2 concentration would decrease and the (d) The NO
concentration would decrease and the
NO concentration would increase. (e) none of the above 73
Yet Another Question
The following chemical reaction has reached equilibrium. Which of the changes listed below would cause the equilibrium to shift back toward the reactants? N
N2(g) + 3 H
( ) 3 H2(g) 
( )  2 NH
2 NH3(g) H = 92.2
( ) H
92 2 kJ/mol
kJ/ lrxn
(a) increasing the pressure (b) increasing the concentration of N2
(b) increasing the concentration of N
(c) increasing the temperature (d) decreasing the concentration of NH3 (e) none of these
(e) none of these
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Le Châtelier’s Principle Summary
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The Haber-Bosch Process
 Artificial nitrogen fixation process to produce ammonia
N2 2 (g) (g) + 3 H
3 H2 2 (g) (g)
2 NH3 3 (g) ΔH
2 NH
(g)
ΔH° = ‐92.2
92.2 kJ/mol
kJ/mol
‐ Fertilizer for 1/3 of world population, chemicals, explosives
Fritz Haber, 1918
Yield and rate
°C
C
Keq
300
4.34 x 10–3
400
1.64 x 10–4
 But increasing the temperature lowers Kc (yield)
450
4.51 x 10–5
What should we do to achieve the b t
best compromise of K
i
f Kc and rate?
d t ?
500
1.45 x 10–5
550
5.38 x 10
5
38 x 10–6
600
2.25 x 10–6
 Low temperature results in a very slow reaction
Low temperature results in a very slow reaction
(low rate)
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The Haber-Bosch Process
Le Châtelier says:
1) Use high pressure to increase Kc (typically 150‐250 atmospheres)
N2 (g) + 3 H2 (g) → 2 NH3 (g)
2) Use a moderate temperature (300‐550°C) to get a good reaction rate g c too much
without lowering K
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The Haber-Bosch Process
3) Use a catalyst to speed up the reaction
‐ Catalysts do not change K
Catalysts do not change Kc
‐ Catalysts lower the activation energy of a reaction, increasing its rate
 Most Haber reactors use a porous iron catalyst made from Fe
b
l
d f
3O4
4) Periodically remove NH3 from the reactor
N2 + 3 H2 → 2 NH3 ‐ forces the reaction toward products
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Adding a Catalyst
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A Haber-Bosch Plant
An American ammonia plant, c. 1970 80
Equilibrium Thermodynamics
∆G° measures the difference between the free energies of the reactants and
products when all species are present at 1 atm (gases) and 1 M (solutes)
Reactants
Products
∆G° < 0 → spontaneous → products favored → K > 1
∆G° > 0 → not spontaneous
p
→ reactants favored → K < 1
∆G° and K are related:
exponential
ti l relationship
l ti
hi
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Standard Free Energy and Equilibrium
∆G°
∆G
signifies standard conditions:  1 atm for gases
 1 M for solutes
at the standard state, Qc = Qp = 1
∆G° tells us how far the standard state is from equilibrium
and which direction the reaction must shift to reach equilibrium
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∆G°
K
large +
small +
0
small
ll large -
<< 1
<1
=1
>1
>> 1
Free Energy at Non-standard Conditions
What is the value of the free energy change when we are not at
standard conditions?
(i.e., ≠ 1 atm for all gases and 1 M for all solutes)
∆G at any Q:
For a rxn with
∆G° = -100 kJ/mol:
∆Gdelta
(kJ
J/mol)
aG
 when Q = 1, ∆G = ∆G°
 when Q = K,
K ∆G = 0
100
50
0
-50
50
equilibrium
-100
-150
-200
-40
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spontaneous
not
spontaneous
∆G = ∆G°
-20
0
ln Q
20
40
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Temperature Dependence of Keq
Equilibrium constants change with temperature
- Predictable based on ∆H° of the reaction
We know:
and
So:
∆H°
∆H
+ (endo)
- (exo)
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K
exp. increases with T
exp. decreases with T
Temperature Dependence of Keq
• ∆H° and ∆S° are essentially constant
for moderate changes in temperature
Exothermic
ln K
Endothermic
1/T
higher temp
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Temperature Take-Homes
The value of the equilibrium constant changes with temperature for
a given reaction.
Master Equations:
 Kc will increase with temperature if the reaction is endothermic
(takes up heat).
A + B + heat
A + B + heat
C+D
C + D
 Kc will decrease with temperature if the reaction is exothermic
(releases heat)
heat).
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A + B C + D + heat
Temperature Dependence Example
The dimerization of NO2 is exothermic (b/c of bond making): 2 NO2
∆H°
- (exo)
Temperature
°C
N2O4
∆H° = -57.2
∆H
57 2 kJ/mol
K
decreases with T
Kc
Kp
‐78
400,000,000
25,000,000
0
1400
62.5
25
170
6.95
100
2.1
0.069
 The dimer
Th di
i f
is favored at low temperatures
d tl t
t
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Sample Problem
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