## 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. ## 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:

$\begin{array}{lcccc} & \mathrm{FeO}(s)+\mathrm{CO}(g) & \rightleftharpoons & \mathrm{Fe}(s)+\mathrm{CO}_2(g) \\\hline \text { Initial pressure }(\mathrm{atm}) & 1.000 & & 0.500 \\\text { Change }(\mathrm{atm}) & -x & & +x \\\text { Equilibrium pressure }(\mathrm{atm}) & (1.000 – x) & & (0.500+x)\end{array}$

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

$K_{\mathrm{p}}=0.259=\frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}}=\frac{0.500+x}{1.000-x}$

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:

\begin{aligned}& P_{\mathrm{CO}}=1.000-x=1.000-(-0.191)=1.191 \mathrm{~atm} \\& P_{\mathrm{CO}_2}=0.500+x=0.500+(-0.191)=0.309 \mathrm{~atm}\end{aligned}

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

$K_{\mathrm{p}}=0.259=\frac{P_{\mathrm{CO}_2}}{P_{\mathrm{CO}}}=\frac{0.309}{1.191}=0.259$

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).