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?

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

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

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

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

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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

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).

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

Sample Problem

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