Lesson 13.4 Calculating Equilibrium Concentrations
Suggested Reading

Zumdahl Chapter 13 Section 13.5
Essential Question

How can the equilibrium constant be used to calculate equilibrium concentrations?
Learning Objectives:

Calculate the concentrations of reactants and products at equilibrium.
Introduction
Once you have determined the equilibrium constant for a reaction, you can use it to calculate
the concentrations of substances in an equilibrium mixture. These calculations are best learned
by doing. I will show you the three typical problem types, but you will learn this material by
working problems. The following examples will provide a model for your practice, and should be
included in your notes.
Problem Type 1
Obtaining One Equilibrium Concentration Give the Others
A gaseous mixture contains 0.30 mol CO, 0.10 mol H2, and 0.020 mol H2O, plus an unknown
amount of CH4, in each liter. This mixture is at equilibrium at 1200 K.
CO(g) + H2(g) ⇌ CH4(g) + H2O(g)
What is the concentration of CH4 in this mixture? The equilibrium equals 3.92.
Solution:
Step 1: Write the equilibrium constant expression.
Step 2: Substitute the known concentrations and the value for K.
Step 3: Solve for the concentration of CH4.
Step 4: State the answer.
The concentration of CH4 in the mixture is 0.059 mol/L.
Problem Type 2
Usually you begin a reaction with known starting quantities of substances and want to calculate
the quantities at equilibrium. The next example illustrates the steps used to solve this type of
problem.
Solving an Equilibrium Problem (Involving a Linear Equation in x)
The reaction, CO(g) + H2O(g) ⇌ CO2(g) + H2(g), is used to increase the ratio of hydrogen in
synthesis gas (mixtures of CO and H2). Suppose you start with 1.00 mol of each reactant in a
50.0 L vessel. How many moles of each substance are in the equilibrium mixture at 1000∘C?
The equilibrium constant at this temperature is 0.58.
Solution:
The solution of an equilibrium problem of this type involves three steps.
1. Set up a table of concentrations showing the starting, change, and equilibrium
concentrations in x (ICE it!, just like we did in lesson 13.2).
2. Substitute the expressions in x for equilibrium concentrations into the equilibriumconstant equation.
3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.
The starting concentrations of CO and H2O are [CO] = [H2O] = 1.00 mol /50.0 L = 0.0200 mol/L
Step 1: Ice it!
Concentration
Initial
Change
Equilibrium
CO
0.0200
H2O
0.0200
CO2
0
H2
0
-x
0.0200 - x
-x
0.0200 - x
+x
x
+x
x
Step 2: Substitute the equilibrium values from the table into the equilibrium-constant
expression.
Step 3: Solve
The right-hand side above is a perfect square (both the numerator and denominator are
squared), so if you take the square root of both sides you get
We write ± because both positive and negative value are mathematically possible. Rearranging
gives
You can dismiss the negative value, because x can only be positive since it represents the
concentration of CO2 formed. If you substitute for x into the last line of the ICE table, the
equilibrium concentrations are 0.0114 M CO, 0.0114 M H2O, 0.0086 M CO2, and 0.0086 M H2.
To find the moles of each substance in the 50.0 L vessel, you multiply the concentration by the
volume of the vessel. For example, the amount of CO is
0.0114 mol/L x 50.0 L = 0.570 mol.
Step 4: State the answer.
The equilibrium composition of the reaction mixture is 0.570 mol CO, 0.570 mol H2O, 0.43 mol
CO2, & 0.43 mol H2.
Problem Type 3
In the previous example, if you had not started with the same number of moles or reactants, you
would not have gotten an equation with a perfect square. In that case you would have had to
solve a quadratic equation. The next example illustrates how to solve such a problem.
Recall that a quadratic equation is an equation that takes the form ax2 + bx + c = 0 and has the
solutions
This equation for x is called the quadratic formula.
Solving an Equilibrium Problem (Involving a Quadratic Equation in x)
Hydrogen an iodine react according to the equation
H2(g) + I2(g) ⇌ 2HI(g)
Suppose 1.00 mol H2 and 2.00 mole of I2 are placed in a 1.00 L vessel. How many moles of
substances are in the gaseous mixture when it comes to equilibrium at 458∘C? The equilibrium
constant Kc at this temperature is 49.7.
Solution:
You follow the four-step approach outlined in the previous example. The equilibrium equation
will be a quadratic equation which you can use the quadratic formula to solve.
Step 1: ICE it!
Concentration (M)
Initial
H2(g)
1.00
I2(g)
2.00
2HI(g)
0
Change
Equilibrium
-x
1.00 - x
-x
2.00 - x
+2x
2x
Step 2: Substitute into the equilibrium constant expression.
Step 3: Solve
Because the right-hand side is not a perfect square, you must use the quadratic formula to solve
for x. The equilibrium expression rearranges to give
There are two mathematical solutions to a quadratic equation. You obtain one solution by taking
the upper (positive) sign in ± and the other by taking the lower (negative) sign. Doing this gives
x = 2.33 and x = 0.93
However, x = 2.33 gives a negative value to 1.00 - x (the equilibrium concentration of H2), which
is physically impossible. Only x = 0.93 remains. You substitute this value of x into the last line of
the table in Step 1 to get the equilibrium concentrations, and then multiply these by the volume
of the vessel (1.00 L) to get the amounts of substances. The last line of the table rewritten is
Concentrations
(M)
Equilibrium
H2(g)
I2(g)
2HI(g)
1.00 - x = 0.07
2.00 - x = 1.07
2x = 1.86
Step 4: State the answer.
The equilibrium composition is 0.07 mol H2, 1.07 mol I2, and 1.86 mol HI.
Homework: Book questions pg. 615 questions 21, 23, 27, 33, 37,39,41,45