|II. General Solution to Equilibria Problems|
|II-1. Introduction||II-2. Equilibrium Constant||II-3. Le Chatelier's Principle||II-4. Reaction Quotient||II-5. General Solution|
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 (at 25oC)
Sample equilibrium conditions: PN2O4 = 0.027 atm, PNO2 = 0.546 atm in a 4.0 L container at 25oC.
What would disturb this equilibrium?
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. (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: 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:
|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 PNO2 and PN2O4
Changing the temperature of a chemical system changes K. Changing concentrations or pressures shifts the equilibrium conditions, but does not change K.
N2O4 (g) 2 NO2 (g) Ho = + 57.2 kJ/mol
What happens if T increases?
The equilibrium conditions will shift to absorb heat.
Since Ho, this reaction shifts to the right, N2O4 (g) 2 NO2 (g), until equilibrium partial pressures are reached.