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AP Chemistry
Equilibrium Notes
Consider some reaction with stoichiometric coefficients as follows:
A + 2B
↔
2C + D
The reaction will proceed until there is no longer any net reactants or products being formed.
This tells us
nothing about the rate of the reaction (kinetics).
Back and forth reactions continue to occur at the microscopic level but macroscopically the system is in
equilibrium.
Equilibrium Law:
K = Product of the concentration of products/Product of the concentrations of reactants
where K is the equilibrium constant at a particular temperature.
For concentrations expressed in molarity (M) units, the symbol is K
c
Given
aA
+ bB
↔
cC + dD then (small case letters being the coefficients of the balanced equation):
K
c
=
[C]
c
[D]
d
[A]
a
[B]
b
Values in brackets represent the concentration of that particular substance expressed in molarity units.
Note
: Substances with constant concentrations (i.e. solids, pure liquids and solvents when solute < 1M)
are not included in the formula since their concentrations change very little.
Example:
Write the equilibrium expressions for the following reversible reactions.
a)
4NH
3(g)
+5O
2(g)
↔
4NO
(g)
+ 6H
2
O
(g)
b)
BaO
(s)
+ CO
2(g)
↔
BaCO
3(s)
Modifying the equilibrium constant
1) Reversing the reaction
K
c
= 1/K
c
’
=K
c
’
1
Reactions that are spontaneous in one direction (K
c
>1) are nonspontaneous in the reverse direction.
2) Multiplying coefficients in a reaction equation by some number, z.
K
c
’ = K
z
This is useful for finding K
c
values for unknown reactions using known reaction values, as we shall see.
3)
Adding chemical reactions together
K
overall
= K
1
x K
2
x …
Ex.
aA + bB
↔
cC
K
1
cC + dD
↔
bB + fF
K
2
aA + dD
↔
fF
K
overall
= K
1
x K
2
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View Full DocumentExample:
Given the two reactions below, what is the equation for the addition of these two reactions, and
what is the final Kc value?
NO
2(g)
↔
NO
(g)
+ ½ O
2(g)
K
2
= 0.012
SO
2(g)
+ ½ O
2(g)
↔
SO
3(g)
K
1
= 20
The
Reaction Quotient (or massaction expression)
, Q
The reaction quotient identifies the current state of the concentrations of a reaction.
When the
system is in equilibrium the Q = K
c
.
The equation is set up in exactly the same manner.
Reaction progress:
Comparing the Values of Q and K
If Q<K then reactant
→
product
If Q>K then product
→
reactant
If Q=K the equilibrium is established
In order to solve an
equilibrium expression
, you must determine 1) The initial conditions of the system
(M), 2) What concentration changes will take place 3) and the concentrations of the system at equilibrium.
Often it is easiest to determine these values with an
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 Fall '10
 Feebeck
 Equilibrium

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