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If you add or subtract a species from a system at equilibrium, the system
will move to a new equilibrium. According to LeChatlier's
Principle, the system will move so as to partially compensate for the
change. If practice, this means that
- If you remove a species, the system will move to form more of that
species.
- If you add a species, the system will move so as to use up some of the
added species
If you compute the reaction
quotient Q for the new concentrations, you will find that LeChatlier's
Principle and the rules for which direction a reaction will run both agree.
You can compute the changes in concentrations or pressures using the same
methods that you use to compute the concentrations
of species at equilibrium.
Example: In the reaction below, you find that at a certain temperature
the equilibrium partial pressures of the gases are PNO2 =
0.856 atm and PN2O4 = 0.231. K for the reaction
is thus 3.17 atm.
- N2O4 < = > 2NO2(g)
If you
add enough N2O4 to bring the pressure up to
PN2O4 = 0.500 atm and let the system move to a
new equilibrium, what are the final equilibrium pressures?
Solution: First, compute Q for the reaction
- Q =
(PNO2)2/PN2O4
= (0.856 atm)2/0.500 atm = 1.46
Since Q is less than K, the
reaction will move to the right, as predicted by LeChatlier's Principle.
For every molecule of N2O4 that decomposes, two
molecules of NO2 form. Thus, if x is the drop in pressure for
N2O4 as the reaction adjusts to a new equilibrium, then
the pressure of NO2 should go up by 2x.
|
N2O4 |
NO2 |
| Initial pressure (atm) |
0.500 |
0.856 |
| Change in pressure (atm) |
-x |
+2x |
| Equilibrium pressure (atm) |
0.500-x |
0.856+2x |
Plug these back into the equation
for K: since only the pressures changed and not the temperature, K stays the
same.
- K =
(PNO2)2/PN2O4
= 3.17 = (0.856+2x)2/(0.500-x)
- 1.59 - 3.17x = 0.733 + 3.42x + 4x2
- 4x2 + 6.59x -0.857
Use the quadratic equation to solve
this. We get out two answers for x: 0.121 and -1.77. Choose x=0.121- we know
that x must be positive. (If not, then the pressure of N2O4
would go up and the pressure of NO2 would decrease- i.e., the
reaction would run to the left. The value of the reaction quotient Q tells us
that this is wrong.) Therefore, the final pressures are
- PN2O4 = 0.500-x = 0.379 atm
- PNO2 = 0.856 + 2x = 1.10 atm
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