Calculating
Equilibrium
Concentrations
from Initial
Conditions
AP Chemistry
Ms. Grobsky
Types of Equilibrium Problems
A
typical equilibrium problem involves
finding the equilibrium concentrations (or
pressures) of reactants and products


Initial concentrations (or pressures) are
given
Equilibrium constant is given
General Procedure to Solve
For Equilibrium Concentrations
1.
Write the balanced chemical equation for
the reaction

2.
3.
4.
The chemical equation, once written out,
represents a chemical equilibrium
Write the equilibrium expression
List or calculate initial concentrations (or
pressures) with information given
If there are products present initially,
calculate Q to make sure system is not
already at equilibrium

If Q ≠ K, set up an ICE table!
The ICE Table – A Useful Tool


To set up the ICE table, first write the equilibrium
equation on your paper
Then at the beginning of the first line under the
equilibrium equation, write the letter “I”


On the second line below the equilibrium reaction,
write the letter “C” under the letter “I”


“I” stands for initial concentrations (or pressures) for
each species in the reaction mixture
“C” represents the change in the concentrations
(or pressures) for each species as the system
moves towards equilibrium
And on the third line below the equilibrium
reaction, write the letter “E” under the letter “C”

“E” represents the equilibrium concentrations (or
pressures) of each species when the system is in a
state of equilibrium
The ICE Table
 After
this part of the process, the table should look
like the example table below using the general
equilibrium reaction:
aA + bB ⇌ dD + eD
aA
+
bB
⇌
dD
+
eE
I
C
E


In the chemical equilibrium, a, b, d, and e represent
the stoichiometric coefficients of the reaction and can
be 1, 3, 4, 2, ½, etc.
A, B, D and E represent the reactants and products
The ICE Table


Now, fill the table with all known concentrations
 These are the concentrations that are given in the problem
Example - Suppose for the equilibrium reaction below, the
initial concentrations of A and B were given as 0.750 M in A
and 1.500 M in B
 The table will look as follows:
aA
I
C
E
0.750
+
bB
1.500
⇌
dD
0
+
eE
0
The ICE Table


Now, the rest of the table should be filled out
This is done by first noting that the equation will shift in
only one direction


Indicate an amount, x, that it shifts


In this case, it can only shift right since there are no products
present
Always write the stoichiometric coefficient in front of x
Put a minus sign in front of any species for which the
concentration decreases and use a plus sign in front of x
of any species for which the concentration increases
aA
+
bB
⇌
dD
+
eE
I
0.750
1.500
0
0
C
-ax
-bx
+dx
+ex
E
The ICE Table

Fill the last line of the table

Each concentration on the last line is the sum of the
concentrations for that species as shown on the “I” line and
the “C” line. The table should now read:
aA


+
bB
⇌
dD
+
eE
I
0.750
1.500
0
0
C
-ax
-bx
+dx
+ex
E
0.750-ax
1.500-bx
dx
ex
The expressions for each species on the “E” line represent the
concentrations of each species at equilibrium
These expressions should now be substituted in the equilibrium
expression

Solve for unknown “x”
Assumptions for ICE Table
 If
K is small, assume x is negligible
 In other words:
[A]int – x = [A]eq ≈ [A]int
 Remember,
a small K means there are mostly
reactants at equilibrium so initial quantities vary
very little from final quantities
 Differences
are negligible
Assumptions for the ICE Table
 The
5% Rule of Thumb
 Once you solve for x, check to see if
assumption is justified
 Divide
“x” by initial concentration
 If
error is less than 5%, then assumption
was justified!
 If
K is not small or error was greater than
5%, you must use quadratic formula to
solve for x!
−b ± b 2 − 4ac
2a
Practice!

A sample of phosgene gas, COCl2, is allowed to
decompose:
COCl2 (g)


CO (g) + Cl2 (g)
The value of Kc for the equilibrium is 2.2 x 10-10 at
100 oC
If the initial concentration of COCl2 is 0.095M, what
will be the equilibrium concentrations for each of
the species involved?
Practice!
COCl2 (g) ↔ CO (g) +
Cl2 (g)
I
0.095
0.000
0.000
C
-X
+X
+X
E
(0.095 -X)
X
X
Kc =
[ CO ] [Cl2 ]
[ COCl2 ]
=
X2
(0.095 - X)
Practice!
 Since
Kc<<1, assume the change in
[COCl2] is negligible and solve for X:
Kc
[ CO ] [Cl2 ]
= [ COCl ]
2
=
X = 4.5 x 10-6 M = [CO] = [Cl2]
[COCl2] = 0.095 M
X2
(0.095)