Question 15.7: Solving an Equilibrium Problem (Involving a Linear Equation ...
Solving an Equilibrium Problem (Involving a Linear Equation in x)
The reaction
\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)
is used to increase the ratio of hydrogen in synthesis gas (mixtures of \mathrm{CO} and \mathrm{H}_{2} ). Suppose you start with 1.00 mol each of carbon monoxide and water in a 50.0-\mathrm{L} vessel. How many moles of each substance are in the equilibrium mixture at 1000^{\circ} \mathrm{C} ? The equilibrium constant K_{c} at this temperature is 0.58 .
PROBLEM STRATEGY
The solution of an equilibrium problem involves three steps. In Step 1, you use the given information to set up a table of starting, change, and equilibrium concentrations. In this problem, you are given starting amounts of reactants. You use these to obtain the starting concentrations. Express the change concentrations in terms of x, as we did in Example 15.1. In Step 2, you substitute the equilibrium concentrations in the table into the equilibrium-constant expression and equate this to the value of the equilibrium constant. In Step 3, you solve this equilibriumconstant equation. This example gives an equation that is a perfect square; taking the square root of it gives a linear equation, which is easy to solve. (More generally, an equilibrium equation involves solving an n th-order equation for its roots. For n=2, you can use the quadratic formula, as illustrated in Example 15.8. For n>2, it is time to think about using a computer.)
STEP 1 The starting concentrations of \mathrm{CO} and \mathrm{H}_{2} \mathrm{O} are
[\mathrm{CO}]=\left[\mathrm{H}_{2} \mathrm{O}\right]=\frac{1.00 \mathrm{~mol}}{50.0 \mathrm{~L}}=0.0200 \mathrm{~mol} / \mathrm{L}
The starting concentrations of the products, \mathrm{CO}_{2} and \mathrm{H}_{2}, are 0 . The changes in concentrations when the mixture goes to equilibrium are not given. However, you can write them all in terms of a single unknown. If you let x be the moles of \mathrm{CO}_{2} formed per liter, then the moles of \mathrm{H}_{2} formed per liter is also x. Similarly, x moles each of \mathrm{CO} and \mathrm{H}_{2} \mathrm{O} are consumed. You write the changes for \mathrm{CO} and \mathrm{H}_{2} \mathrm{O} as -x. You obtain the equilibrium concentrations by adding the change in concentrations to the starting concentrations, as shown in the following table:
\begin{array}{lc} \text { Concentrations }(M) & \mathrm{CO}(\mathrm{g})+&\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) & \rightleftharpoons \mathrm{CO}_{2}(\mathrm{~g})&+\mathrm{H}_{2}(\mathrm{~g}) \\ \text { Starting } & 0.0200 & 0.0200 & 0 & 0 \\ \text { Change } & -x & -x & +x & +x \\ \text { Equilibrium } & 0.0200-x & 0.0200-x & x & x \end{array}
STEP 2 You then substitute the values for the equilibrium concentrations into the equilibrium-constant equation,
K_{c}=\frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{[\mathrm{CO}]\left[\mathrm{H}_{2} \mathrm{O}\right]}
and you get
0.58=\frac{(x)(x)}{(0.0200-x)(0.0200-x)}
or
0.58=\frac{x^{2}}{(0.0200-x)^{2}}
STEP 3 You now solve this equilibrium equation for the value of x. Note that the right-hand side is a perfect square. If you take the square root of both sides, you get
\pm 0.76=\frac{x}{0.0200-x}
We have written ± to indicate that you should consider both positive and negative values, because both are mathematically possible. Rearranging the equation gives
x=\frac{0.0200 \times 0.76}{1.76}=0.0086 \text { and } x=\frac{-0.0200 \times 0.76}{0.24}=-0.063
You can dismiss the negative value as physically impossible ( x can only be positive; it represents the concentration of \mathrm{CO}_{2} formed). If you substitute for x in the last line of the table, the equilibrium concentrations are 0.0114 \mathrm{M CO}, 0.0114 \mathrm{M} \mathrm{H}_{2} \mathrm{O}, 0.0086 M \mathrm{CO}_{2}, and 0.0086 M \mathrm{H}_{2}. To find the moles of each substance in the 50.0 -L vessel, you multiply the concentrations by the volume of the vessel. For example, the amount of \mathrm{CO} is
0.0114 \mathrm{~mol} / \mathrm{L} \times 50.0 \mathrm{~L}=0.570 \mathrm{~mol}
The equilibrium composition of the reaction mixture is 0.570 \mathrm{~mol} \mathrm{CO}, 0.570 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}, 0.43 \mathrm{~mol} \mathrm{CO}_{2}, and 0.43 \mathrm{~mol} \mathrm{H}_{2}.
ANSWER CHECK
A rough estimate of the answer can sometimes save you from giving the wrong answer. If this reaction could go completely to products, 1 mol CO could give only 1 mol CO_2. (Is the molar amount that you obtained for CO_2 less than 1?) The magnitude of K_c suggests that the reaction will give a mixture of approximately similar molar amounts of reactants and products. (Does your answer agree with this?)