Chemical Equilibrium
The big one!! This is your first required FRQ
on the AP test.
Generally it is considered that all chemical reactions go to
completion of the products and then stop.
This is only true under a certain set of conditions. Most of the
time, chemical reactions can go in the opposite direction as
well if the condition changes made to the system are favorable.
H2 + O2 --> H2O (synthesis)
and
H2O --> H2 + O2
(decomposition)
become
H2 + O 2 
H2O
(both)
The equilibrium constant is a mathematical value that is constant for a
specific chemical reaction. It, like the “k” value for kinetics is
temperature dependent only. Concentrations may vary as much as
necessary.
The equilibrium constant, “k”, is determined from the
equilibrium expression that can be determined by the stoichiometric
equation:
jA + kB  lC + mD
leads to:
[C]l [D]m
[A]j [B]k
That is:
Products coefficeints
Reactantscoefficients
Write the equilibrium expressions for the following equilibrium
reactions:
4NH3 (g) + 7O2 (g)  4NO2 (g) + 6H2O
N2 (g) + O2 (g)  2NO (g)
N2O4 (g)  2NO2 (g)
(g)
[NO2]4[H2O]6
[NH3]4[O2]7
[NO]2
[N2][O2]
[NO2]2
[N2O4]
2PBr3 (g) + 3Cl2 (g)  2PCl3 (g) + 3Br2 (g)
[PCl3]2[Br2]3
[PBr3]2[Cl2]3
Once the equilibrium expression has been written, you can
input equilibrium concentration values into the formula and
find the value of the equilibrium constant “k”.
The following equilibrium concentration were
observed for the synthesis of ammonia from its elements:
N2
(g)
+ H2 (g)  NH3
(g)
a. Write the equilibrium expression for the reaction
after balancing the equation.
b. plug in the values for the given concentrations:
[NH3] = 3.1 x 10 -2 M
[N2] = 8.5 x 10 -1 M
[H2] = 3.1 x 10 -3 M
Find the value of “k” for the reverse reaction
2NH3  N2 + 3H2
10 -2 M
[NH3] = 3.1 x
[N2] = 8.5 x 10 -1 M
[H2]
= 3.1 x 10 -3 M
Calculate the value of “k” for the reaction given below
1/2 N2 + 3/2 H2
 NH3
Now compare the equilibrium expressions in example
b and c to a:
What do you find is true?
For the reaction H2 + Br2 --> 2HBr
K = 3.5 x 10 4 at 1495 K. What is the value of K under the following
circumstances?
invert and square root
a. HBr  1/2 H2 + 1/2 Br2
invert only
b. 2HBr  H2
+ Br2
square
c. 1/2H2 + 1/2
Br2 root
HBronly
You can determine if the reaction will shift to the product
side or the reactant side in a similar process:
Q = [products]coefficients
[reactants]coefficients
Instead of using equlibrium concentrations, you will use
beginning of reaction concentrations.
If k = Q, No shift in equilibrium will occur
If Q > k the system will shift to the left, consuming
products and turning them into reactants.
If Q < k the system will shift to the right, consuming
reactants and turning them into products
Equilibrium expressions that involve pressure instead of
concentration.
Because PV = nRT can be written P = (n/v)RT or P=CRT
we can replace concentration with pressure.
R is always a constant and for “k” to remain
constant T must also be a constant for equilibrium
expressions.
Now you can put partial pressures of gases into your
equilibrium expressions to calculate for a “k” value.
Since “K” or Kc are representative of concentration
equilibrium constants, we will use Kp when we solve
with partial pressures.
Solve for Kp of the reaction:
2NO (g) + Cl2 (g)  2NOCl (g)
when the partial pressures of each gas found at equlibrium are equal to:
PNOCl = 1.2 atm
PNO = 5.0 x 10 -2 atm
PCl2 = 3.0 x 10 -1 atm
Kp = PNOCl 2
(PNO)2 (PCl2)
=
(1.2)2
(5.0 x 10 -2)2 (3.0 x 10 -1)
= 1.9 x 10 3
.
Now because there is a difference in measurements between partial pressure
measurements and concentration measurements of the same system at the same
time Kc ≠Kp, but there is a relationship to convert from one to another.
Kp = Kc (RT) Dn
(The derivation for this relationship
can be found on page 619)
K is equilibrium constant for concentration
R is the ideal gas constant
T is Temperature in Kelvin
Dn is sum of product coefficients - sum reactant coefficients
Convert the Kp calculated earlier to a K value using the
balanced equation and the mathematical relationship between
Kp and Kc:
2NO (g) + Cl2 (g)  2NOCl (g)
Kp = K(RT) Dn
The reaction is taking place at 25° C.
The states of matter become very important to calclulating Kp
when there are more than gas particles present in a system. If
this is the case, only the gas particles in the system should be
included in the equilibrium expression equation.
2H2O (l)
2H2O (g)
 2H2 (g) + O2 (g)
 2H2 (g) + O2 (g)
Write Kc and Kp expressions for the following:
a. The decomposition of solid phosphous pentachloride to
liquid phosphorous trichloride and chlorine gas.
b. copper (II) oxide and carbon dioxide gas are combined
to form solid copper (II) carbonate
Now since we can calculate Kp, if we are given Kp and have a
missing concentration or pressure we can rearrange the
equation to solve for that unknown if given K.
N2O4 (g) 
2NO2 (g)
If the Kp of the above reaction is 0.133 when the
equilibrium pressure of N2O4 is found to be 2.71 atm, find the
pressure of NO2 at equilibrium.
Now sometimes we will be asked to convert from original
concentrations to equilibrium concentrations while being given a
known k value.
H2 (g) + F2 (g)  2HF (g)
K = 1.15 x 10 2
The system starts with 3.000 mols of each component in a
1.5 L flask.
Our job is find the equilibrium concentrations of each
component (3variables!!!) We can do this by writing the change in
every system as a function of the same variable.
2NO(g) + Br2 (g) <-> 2NOBr (g)
I
C
E
If 1.0 M and 0.8 M of NO and Br2
gases were mixed, one quarter of the
Br2 would be consumed. Find the
equilibrium concentrations and solve
for the value of the equilibrium
constant.
We call this an ICE table
H2
Initial
I
Change
C
Equlibrium
E
F2
2HF
Equilibrium values, with variables included can now be
plugged into the equation and you will solve for “x”.