II. General Solution to Equilibria Problems | ||||
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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: N_{2}O_{4 (g)} 2 NO_{2 (g)} K = 11 (at 25^{o}C)
Sample equilibrium conditions: P_{N2O4} = 0.027 atm, P_{NO2} = 0.546 atm in a 4.0 L container at 25^{o}C.
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: 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, N_{2}O_{4 (g)} 2 NO_{2 (g)}. The problem is set up in the following:
N_{2}O_{4} | NO_{2} | |
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P_{o} | 0.108 atm | 2.18 atm |
P | +0.5x atm | -x atm |
P_{eq} | (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 the temperature of a chemical system changes K. Changing concentrations or pressures shifts the equilibrium conditions, but does not change K.
Example:
N_{2}O_{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, N_{2}O_{4 (g)} 2 NO_{2 (g)}, until equilibrium partial pressures are reached.
Equilibrium Practice Problems | |
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