Overview for Solving Problems Using the Equilibrium Constant.
1. Write chemical equation and set up the ICE table.
2. Begin filling in the ICE table by writing in the initial concentration of each species given in the
problem, reactants and products. (Fill in the “I” line—the first line under the chemical equation).
3. Fill in line C of the ICE table, the change in concentrations of all species.
4. Fill in line E the ICE table, the concentration of species when the system is at equilibrium.
5. Write the equilibrium expression (this is the mathematical equation) and plug in the equilibrium
concentrations as they are represented in the “E” line in the ICE table.
6. Make any assumptions if convenient.
7. Determine the numerical value of x.
8. Check any assumptions that were made in step 6. If the assumption was not valid, use the
quadratic equation or the successive guess method to determine x.
9. Use the value of x to determine the equilibrium concentrations of all species. The equations
written in line E of the ICE table should be used to do this.
10. Reread the problem to make sure you answered all parts of the question because by this time you
may have forgotten the whole purpose of the problem!
The ICE Table and Solving Equilibrium Problems.
The ICE table is a useful tool for solving equilibrium problems. The letters in ICE stand for “Initial
Concentration”, “Change in Concentration” and “Equilibrium Concentration”.
When considering an equilibrium problem, there is usually some indication of the reactants and
products of a chemical equation. The chemical equation may or may not be given. The chemical
equation, once written out, represents a chemical equilibrium. The problem will usually also indicate
initial or equilibrium concentrations of reactants and/or products. Last, the problem usually gives Kc
or Kp. After reading the problem, begin by setting up the ICE table.
To set up the ICE table, first write the equilibrium equation on your paper. Then write at the
beginning of the first line under the equilibrium equation the letter I (see example below). The letter
I stands for the Initial Concentration or it can stand for the Initial Partial Pressure if working in
pressures instead of concentrations. On the second line below the equilibrium reaction, write the
letter C under the letter I. And on the third line write the letter E. After this part of the process the
table should look like the example table below using the general equilibrium reaction,
aA + bB ⇌ dD + eD.
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.
Table:
aA
+
bB
⇌
dD
+
eE
I
C
E
Now all the known concentrations should be filled in. These are the concentrations that are given in
the problem. Many equilibrium problems give the initial concentrations of reactants and the
equilibrium constant. Then the problem will ask for the final concentrations of reactants and
products. Suppose for the equilibrium above the initial concentrations of A and B were given as
0.750 M in A and 1.500 M in B, and the equilibrium constant was given. These initial concentrations
should be entered.
I
C
E
aA
+
0.750
bB
⇌
1.500
dD
0
+
eE
0
Notice that 0’s were placed in the places in the table for the initial concentrations of D and E.
Generally, if a problem in a text does not indicate there is an initial concentration of a species, it
means the initial concentration is zero. 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 (it can only shift right for this problem
since there are no products present). Then indicate an amount, x, that it shifts. Always write the
stoichiometric coefficient in front of x and 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. This will complete the second line. The table should now look like the example below.
I
C
E
aA +
0.750
-ax
bB
⇌
1.500
-bx
dD
0
+dx
+
eE
0
+ex
Now the last line should be filled in. 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:
I
C
E
aA +
0.750
-ax
0.750-ax
bB
⇌
1.500
-bx
1.500-bx
dD
0
+dx
dx
+
eE
0
+ex
ex
The expressions for each species on the “E” line represent the concentrations of each species at
equilibrium. These expressions should now be used in the mathematical form of the equilibrium
expression.
Kc 
[ D] d [ E ] e
becomes: (numberforK c ) 
[ A] a [ B]b
[dx] d [ex]e
[0.750  ax] a [1.500  bx]b
In the equation above, there will be a numeric value for Kc and the letters a, b, c and d will be
numbers (the stoichiometric coefficients)—oftentimes a, b, c, and/or d will be 1.
After determining x, the problem is not done. The equilibrium concentrations need to be calculated
using the mathematical relations on the “E” line. This is usually fairly easy compared to the rest of
the problem.
If a problem gives the initial concentrations of products, for example [D] = 0.500 M and
[E] = 0.850 M, the table will look like:
aA
I
C
E
0
+ax
ax
+
bB
0
+bx
bx
⇌
dD
+
0.500
-dx
0.500-dx
eE
0.850
-ex
0.850-ex
Many times assumptions can be made that will simplify the math of the problems (these will be
covered in class). Any time that an assumption is made, the validity of the assumption must be
checked. If it is not valid, the quadratic equation should be used (the quadratic equation will be
shown on the exam). The quadratic equation is used to determine the value of x when x is in a
mathematical equation that must be in the form:
ax2 + bx +c = 0
where a, b and c are numbers (they can be 1 or even 0). The quadratic equation is then set up to read:
 b  b 2  4ac
x
2a
After using the quadratic equation, there will be two possible values of x, one from using the + sign
up top and one from using the – sign up top. Only one of these values will be correct. The incorrect
value is usually obvious because it is usually negative or way too big.