Introduction

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

Example: N₂O_{4 (g)} 2 NO_{2 (g)}    K = 11 (at 25^oC)

Sample equilibrium conditions: P_N2O4 = 0.027 atm, P_NO2 = 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: P_NO2 = 0.546 atm x 4 = 2.18 atm
P_N2O4 = 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. (More on Q in the next section, II-4.)

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: P_NO2 = 2.18 atm
and P_N2O4 = 0.108 atm

Q tells us the reaction will shift in the reverse direction, N₂O_{4 (g)} 2 NO_{2 (g)}. The problem is set up in the following:

	N2O4	NO2
Po	0.108 atm	2.18 atm
P	+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 P_NO2 and P_N2O4

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:
N₂O_{4 (g)} 2 NO_{2 (g)} H^o = + 57.2 kJ/mol

What happens if T increases?

The equilibrium conditions will shift to absorb heat.

Since H^o, this reaction shifts to the right, N₂O_{4 (g)} 2 NO_{2 (g)}, until equilibrium partial pressures are reached.