## Chapter 13

## Q. 13.12

**Calculating Equilibrium Partial Pressures from Initial Partial Pressures **

One reaction that occurs in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and carbon dioxide. The equilibrium constant K_p for the reaction at 1000 K is 0.259.

\mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_2(g) \quad K_{\mathrm{p}}=0.259 \text { at } 1000 \mathrm{~K}What are the equilibrium partial pressures of CO and CO_2 at 1000 K if the initial partial pressures are P_{CO} = 1.000 atm and P_{CO_2} = 0.500 atm?

**STRATEGY **

We can calculate equilibrium partial pressures from initial partial pressures and K_p in the same way that we calculate equilibrium concentrations from initial concentrations and K_c. Follow the steps in Figure 13.6, but substitute partial pressures for concentrations.

## Step-by-Step

## Verified Solution

**Step 1** The balanced equation is \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_2(g)

**Step 2** Define x as the partial pressure of CO that reacts. Set up a table of partial pressures of the gases under the balanced equation:

**Step 3** Substitute the equilibrium partial pressures into the equilibrium equation for K_p:

As usual for a heterogeneous equilibrium, we omit the pure solids from the equilibrium equation. Rearranging the equilibrium equation and solving for x give

\begin{aligned}& 0.259-0.259 x=0.500+x \\& x=\frac{-0.241}{1.259}=-0.191\end{aligned}**Step 4** Calculate the equilibrium partial pressures from the calculated value of x:

**Step 5** Check the results by substituting them into the equilibrium equation:

A negative value for x means that the reaction goes from products to reactants to reach equilibrium. Tis makes sense because the initial reaction quotient, Q_p = 0.500/1.000 = 0.500, is greater than the equilibrium constant, K_p = 0.259. When Q_p > K_p, the net reaction always goes from products to reactants (right to left).