concentration temperature volume total pressure -...
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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 ”the equilibrium shifts to partially counteract the imposed change.
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Changes in ConcentrationPCl5 PCl3 + Cl2 Kc = 0.030
At equilibrium:
Adding reactants shifts the reaction toward products
Adding products shifts the reaction toward reactants
Removing reactants … shifts the reaction toward reactants g
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 ExampleN2 (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?
Before compression
Aftercompression
PNH3 = 0.2 atm PNH3 = 8.4 atm2.0% of total pressure
9.2% of total pressure
PN2 = 2.4 atm PN2 = 21 atm
PH2 = 7.2 atm PH2 = 62 atm
pressure 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 CatalystA 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)
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New QuestionWhat 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 NO2 concentration would decrease and the NO concentration would increase.
(e) none of the above
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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 ( ) 3 H ( ) 2 NH ( ) H 92 2 kJ/ l N2(g) + 3 H2(g) 2 NH3(g) H = 92.2 kJ/molrxn
(a) increasing the pressure (b) increasing the concentration of N(b) increasing the concentration of N2
(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
N2 (g) + 3 H2 (g) 2 NH3 (g) ΔH° = ‐92.2 kJ/mol
Artificial nitrogen fixation process to produce ammonia
N2 (g) 3 H2 (g) 2 NH3 (g) ΔH 92.2 kJ/mol
‐ Fertilizer for 1/3 of world population, chemicals, explosives
Fritz Haber, 1918
°C K Low temperature results in a very slow reaction
Yield and rateC Keq
300 4.34 x 10–3
400 1.64 x 10–4
Low temperature results in a very slow reaction(low rate)
450 4.51 x 10–5
500 1.45 x 10–5
550 5 38 x 10–6
But increasing the temperature lowers Kc (yield)
What should we do to achieve the b t i f K d t ?
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550 5.38 x 10
600 2.25 x 10–6best compromise of Kc and rate?
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The Haber-Bosch Process
1) Use high pressure to increase Kc (typically 150‐250 atmospheres)
Le Châtelier says:
N2 (g) + 3 H2 (g)→ 2 NH3 (g)
2) Use a moderate temperature (300‐550°C) to get a good reaction rate without lowering Kc too much
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g c
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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
b l d f Most Haber reactors use a porous iron catalyst made from Fe3O4
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
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Equilibrium Thermodynamics
Reactants Products
∆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)
∆G° < 0 → spontaneous → products favored → K > 1 ∆G° > 0 → not spontaneous → reactants favored → K < 1 p
∆G° and K are related:
ti l l ti hi81
exponential relationship
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Standard Free Energy and Equilibrium
∆G°signifies standard conditions: 1 atm for gases
1 M for solutes∆G
at the standard state, Qc = Qp = 1
∆G° tells us how far the standard state is from equilibriumand which direction the reaction must shift to reach equilibrium
large + << 1∆G° Klarge +small +
0ll
<< 1< 1= 1> 1
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small -large -
> 1>> 1
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Free Energy at Non-standard ConditionsWhat 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:∆G at any Q:
when Q = 1, ∆G = ∆G° when Q = K ∆G = 0 100 spontaneous
not spontaneous when Q = K, ∆G = 0
50
0
50
100a
GJ/
mol
) spontaneous spontaneous
For a rxn with
-150
-100
-50
delta
∆G (k
J
∆G = ∆G°
equilibriumFor a rxn with
∆G° = -100 kJ/mol:
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-2006040200-20-40
ln Q
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Temperature Dependence of Keq
Equilibrium constants change with temperature
- Predictable based on ∆H° of the reaction
We know:
and
So:So:
∆H° K+-
∆H Kexp. increases with Texp. decreases with T
(endo)(exo)
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Temperature Dependence of Keq
• ∆H° and ∆S° are essentially constant for moderate changes in temperaturefor moderate changes in temperature
ln K
Exothermic
ln KEndothermic
1/Thigher temp
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Temperature Take-HomesThe 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 C + DA + B + heat C + D
Kc will decrease with temperature if the reaction is exothermic (releases heat)
86A + B C + D + heat
(releases heat).
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Temperature Dependence Example
2 NO N O
The dimerization of NO2 is exothermic (b/c of bond making):
∆H° = 57 2 kJ/mol2 NO2 N2O4 ∆H = -57.2 kJ/mol
∆H° K
Temperature Kc Kp
- decreases with T(exo)
°C
‐78 400,000,000 25,000,000
0 1400 62.5
25 170 6.95
100 2.1 0.069
Th di i f d t l t t
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The dimer is favored at low temperatures
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Sample Problem
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