Name
Chem 163 Section: ______ Team Number: ______
ALE 8. Using ICE Tables to Solve Equilibrium Problems
(Reference: 17.5 Silberberg 5th edition)
How can an equilibrium constant, K, be used to determine the equilibrium
concentrations of the reactants and products?
The Model: ICE Tables
When reactants (and sometimes even products) are combined in a reaction vessel, the system will nearly
always not start off at equilibrium. The reactants are consumed, so their amounts (either measured through
concentrations or partial pressures) go down over time. Consequently, the amounts of products increase.
Eventually the system reaches a state of dynamic equilibrium, in which the amounts of all species no longer
change with the passage of time. The equilibrium amount and the initial amount of at least one species can
be monitored so as to determine how much the amount of that species has changed over the course of the
reaction. How much the amounts of the other species in the reaction change is dependent on the
stoichiometric coefficients.
Example 1
At 25 C, 0.200 mol of dinitrogen monoxide and 0.560 mol of oxygen gas were placed in a 10.0-L
reaction vessel and allowed to react:
2 N2O(g) + 3 O2(g)
4 NO2(g)
When the system reached equilibrium at 25 C, the concentration of nitrogen dioxide was found to be
0.0200 M. What is the value of Kc for the oxidation of dinitrogen monoxide to nitrogen dioxide at 25
C?
Solution: We set up an “ICE table” (i.e., a table showing the Initial concentrations, the Changes in
concentrations, and the Equilibrium concentrations for the species in the reaction) and enter
whatever values we know. For each species, the entry in the row labeled “C” is a product
between an algebraic variable and the species’ stoichiometric coefficient.
[N2O]
[O2]
[NO2]
I
0.0200
0.0560
0
C
-2x
-3x
+0.0200
E
0.0200 – 2x
0.0560 – 3x
0.0200
Key Questions
1. In the row labeled “C”, why is the entry negative for [N2O] and [O2] but positive for [NO2]?
Page 1 of 4
2. If 0.200 mol of N2O is the starting amount of dinitrogen monoxide, why in the table is its “Initial” value
equal to 0.0200?
3. What allows us to enter a “0” for the initial concentration of nitrogen dioxide?
4. What is the relationship between I, C and E in the ICE table.
5. a. The changes in concentrations for N2O and O2 are “-2x” and “-3x”, respectively. Explain the
coefficients that appear in front of the x for each.
b. In terms of x, how can the change in the concentration of NO2 be expressed?
Exercises
6. Given the information in the Model for the equilibrium:
2 N2O(g) + 3 O2(g)
4 NO2(g),
calculate the value of Kc for the reaction. Hints:  Use your answer to Question 5b and the entries in
the “[NO2]” column of the ICE table to solve for x.  Use the ICE table to solve for [N2O]eq and
[O2]eq.  Set up a law of mass action for the reaction and solve for Kc.
Page 2 of 4
7. 0.0487 mol of nitrosyl chloride was injected into an otherwise empty 1.00-L reaction vessel at 500. K. The
nitrosyl chloride decomposed into nitrogen monoxide and chlorine:
2 NOCl(g)
2 NO(g) + Cl2(g)
When the system reached equilibrium at 500. K, the total pressure within the reaction vessel was found to be
2.63 atm. What is Kp for the above reaction at 500. K? Hints:  What are the initial pressures of NOCl, NO,
and Cl2?  Set up an ICE table and express the equilibrium partial pressures of the gases in terms of an
algebraic variable.  Use Dalton’s law of partial pressures to determine the value of that algebraic variable as
well as the values of the equilibrium partial pressures for the three gases.  Write the law of mass action for the
reaction, and substitute the equilibrium partial pressures to solve for the equilibrium constant.
8. If the equilibrium constant for a reaction is already known, whenever initial amounts are provided, one can use
an ICE table and the law of mass action to determine the equilibrium concentrations of the reactants and
products.
PCl5(g)
PCl3(g) + Cl2(g)
Kp = 0.49 at 500. K
Pure phosphorous pentachloride is injected into an otherwise empty reaction vessel so that it has an initial
pressure of 1.64 atm at 500. K. If the system remains at 500. K, what are the equilibrium partial pressures of
the three gases?
Hints:  Set up an ICE table. For each gas, determine its initial pressure, write its change in pressure in terms
of an algebraic variable, and express its equilibrium partial pressure in terms of the variable.
 Write the law of mass action.  Substitute the given value of Kp and the algebraic expressions for the
equilibrium partial pressures into the law of mass action and solve for the algebraic variable. To do this, you
will need to use the Quadratic Formula.
Given the equation: ax2 + bx + c = 0,
the solutions are given by the quadratic formula: x 
 b  b 2  4ac
2a
 Don’t forget to substitute the solved value of the variable into the algebraic expressions
for the equilibrium partial pressures of the three gases!
Page 3 of 4
The Model: Making Assumptions to Simplify the Algebra
There will come times when you will encounter either a quadratic or a higher-order polynomial, but it is not always
necessary to solve it by “brute force”. We can often find ways to make the solution process easier. One such way is
to make an assumption which simplifies an algebraic expression so that we do not have to solve a high-order
polynomial.
Example 2
The chemical equation for the decomposition of phosgene into carbon monoxide and chlorine is…
COCl2(g)
Kp = 6.7  10-9 at 100. C
CO(g) + Cl2(g)
If a reaction vessel is initially filled with only phosgene at 2.3 atm and 100. C, what will be the equilibrium
partial pressures of all three gases?
Solution: We set up an ICE table:
PCOCl2
PCO
PCl2
I
2.3
0
0
C
-x
+x
+x
E
2.3 – x
x
x
We now set up the law of mass action and substitute the values of Kp and the equilibrium partial pressures:
x2
6.7  10-9 = 
2.3 – x
This certainly could be solved using the quadratic formula. But we reason: “Since the equilibrium constant is so
small, very little phosgene will decompose. Therefore, the equilibrium partial pressure of phosgene should be
virtually the same as its initial value.” The approximation we make is: 2.3 – x  2.3 (i.e., x is negligible compared
to the initial value of 2.3 atm). The problem simplifies to:
x2
6.7  10-9 =   x2 = 2.3(6.7  10-9) = 1.5  10-8
 x = 1.2  10-4
2.3
And then we must confirm that our assumption is valid. Is 2.3 – 0.00012  2.3? Indeed, it is. Therefore, the
equilibrium partial pressures are: PCOCl2 = 2.3 atm and PCO = PCl2 = 1.2  10-4 atm.
Exercise
9. Suppose the initial concentrations of H2S, H2, and S2 are 1.00 M, 0 M, and 0 M, respectively:
2 H2S(g)
2 H2(g) + S2(g)
Kc = 1.00  10-6
What would be the equilibrium concentrations of the reactants and products at the temperature for which the
above Kc is valid? Hint: After setting up the law of mass action, you will encounter a cubic equation. Do NOT
solve it by “brute force”. Instead, make an assumption. Carefully state the assumption and make the appropriate
simplification to the law of mass action. You should end up being able to simply take a cube root instead of
solving the cubic equation. After you’ve solved for the algebraic variable, verify that the assumption is valid.
Then don’t forget to solve for the equilibrium concentrations!
Page 4 of 4