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Le Chatelier's Principle

Introduction

Le Chatelier's Principle: When a system at equilibrium is disturbed, the equilibrium conditions shift to counteract the disturbance.

Example: N2O4 (g) <--> 2 NO2 (g)    K = 11 (25 oC)

Sample equilibrium conditions: PN2O4 = 0.027 atm, PNO2 = 0.546 atm in a 4.0 L container at 25 oC.

What would disturb this equilibrium?

  1. Changing the amounts of the reactants or products in the container.
  2. Changing pressure.
  3. Changing temperature.


Changing Pressure

Example: Decrease the volume of the container in the above example from 4.0 L to 1.0 L (Remember PV = nRT).

New partial pressures: PNO2 = 0.546 atm x 4 = 2.18 atm
PN2O4 = 0.027 atm x 4 = 0.108 atm

In which direction will the reaction proceed to reestablish equilibrium?
Calculate the reaction quotient, Q, to find out.

Q = (2.18 atm)2/0.108 atm = 44 atm
Q > K so the reaction will proceed in the reverse direction.

How do we find the new equilibrium partial pressures?

Start with the new initial conditions: PNO2 = 2.18 atm
and PN2O4 = 0.108 atm

Q tells us the reaction will shift in the reverse direction, N2O4 (g) ---> 2 NO2 (g). The problem is set up in the following:

N2O4NO2
Po0.108 atm2.18 atm
DeltaP+0.5x atm-x atm
Peq(0.108+0.5x) atm(2.18-x) atm

K = 11 = (2.18-x)2 / (0.108+0.5x)

Find x, then find PNO2 and PN2O4


Changing Temperature

Changing the temperature of a chemical system changes K. Changing concentrations or pressures shifts the equilibrium conditions, but does not change K.

Example:
N2O4 (g) <--> 2 NO2 (g) DeltaHo = + 57.2 kJ/mol

What happens if T increases?

The equilibrium conditions will shift to absorb heat.

Since DeltaHo, this reaction shifts to the right, N2O4 (g) ---> 2 NO2 (g), until equilibrium partial pressures are reached.


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