One of the reactions that takes place in producing steel from iron ore is the reduction of iron $(II)$ oxide by carbon monoxide to give iron metal and $CO_{2}$.
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g);$ $K_{p} = 0.265$ at $1050 \, K$
What are the equilibrium partial pressures of $CO$ and $CO_{2}$ at $1050 \, K$ if the initial partial pressures are: $p_{CO} = 1.4 \, atm$ and $p_{CO_{2}} = 0.80 \, atm$?

(N/A) For the given reaction,
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g)$
Initially: $p_{CO} = 1.4 \, atm$,$p_{CO_{2}} = 0.80 \, atm$
$Q_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80}{1.4} \approx 0.571$
Since $Q_{p} > K_{p}$ $(0.571 > 0.265)$,the reaction proceeds in the backward direction.
Let $x$ be the change in pressure at equilibrium.
$K_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80 - x}{1.4 + x} = 0.265$
$0.80 - x = 0.265(1.4 + x)$
$0.80 - x = 0.371 + 0.265x$
$1.265x = 0.429$
$x = 0.339 \, atm$
Equilibrium partial pressures:
$p_{CO_{2}} = 0.80 - 0.339 = 0.461 \, atm$
$p_{CO} = 1.4 + 0.339 = 1.739 \, atm$