Applying Stoichiometry to Equilibrium Mixtures

You can think of this section as a prep school for equilibrium. Here we are not really going to deal with equilibrium, just stoichiometry. So the only thing different in this and our earlier treatment of  stoichiometry is that we will be applying it to an equilibrium reaction instead of a plain reaction. Let's look at some examples now.

Example 1

A 1.355-mol sample of phosphorus pentachloride, PCl5, dissociates according to the following equation to give 0.188 mol of chlorine, Cl2, at equilibrium.  What is the composition of the final reaction mixture?

PCl5(g) <==> PCl3(g) + Cl2(g) 

The procedure we will use to solve this problem, and for that matter, most equilibrium problems, is to set up a table in which you write the starting, change, and equilibrium values of each substance under the balanced equation.

Amount(mol) PCl5(g) <==> PCl3(g)
+
 Cl2(g)
Starting 1.355 0 0
Change -x +x +x
Equilibrium 1.355 - x x x = 0.188 (from problem)

The problem statement gives the equilibrium amount for Cl2.  This tells you that x = 0.188 mol.  You can calculate the equilibrium amounts for the other substances from the expressions in the table, using this value of x.

Equilibrium amount of PCl5 = (1.355 - x) mol = (1.355 - 0.188) mol = 1.167 mol

Equilibrium amount of PCl3 =  x mol = 0.188 mol

Equilibrium amount of Cl2 =  x mol = 0.188 mol

Therefore, the amounts of substances in the equilibrium mixture are 1.167 mol PCl5,  0.188 mol of PCl3 and 0.188 mol of Cl2. 

Example 2

A reaction mixture initially consists of 0.805 mol of nitrogen, N2, and 2.615 mol of hydrogen, H2, which react according to the equation below. If 0.062 mol of NH3 is found at equilibrium, what is the composition of the equilibrium mixture?

N2(g) + 3 H2(g) <==> 2 NH3(g)

Again we will set up a table with the starting, change, and equilibrium values of each substance under the balanced equation.

Amount (mol) N2(g) + 3 H2(g) <==>
 2 NH3(g)
Starting 0.805 2.615 0
Change -x -3x +2x
Equilibrium 0.805 - x 2.615 - 3x 2x = 0.062 (from problem)

From the table, we see that 2x = 0.062 or x = 0.031 mol. Using this value of x and the expressions from the table we obtain.

Equilibrium amount of N2 = (0.805 - x) mol = (0.805 - 0.031) mol = 0.774 mol

Equilibrium amount of H2 = (2.615 - 3x) mol = (2.615 - 0.093) mol = 2.522 mol

Equilibrium amount of NH3 (given in problem) = 0.062 mol

Therefore, in the equilibrium mixture we have 0.774 mol of N2, 2.522 mol of H2 and 0.062 mol of NH3.

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