Question

One of the reaction that takes place in producing steel from iron ore is the reduction of $\text{iron(II)}$ oxide by carbon monoxide to give iron metal and ${\text{CO}}_{\text{2}}.$ $\text{FeO}\left(\text{s}\right)\text{+CO}\left(\text{g}\right)\rightleftharpoons \text{ Fe}\left(\text{s}\right){\left(\text{+CO}\right)}_{\text{2}}\left(\text{g}\right)\text{;}$ ${\text{K}}_{\text{P}}\text{= 0}\text{.265}\text{atm}$ at $\text{1050}\text{K}\text{.}$ What are the equilibrium partial pressures of $\text{CO}$ and $C{O}_2$ respectively at $\text{1050}\text{K}$ if the initial partial pressures are: $P_{co}=1.4$ atm and $P_{C{O}_2}=0.80atm$ ?

• A. $\left[P_{CO}\right]=1.739atmand{P}_{C{O}_2}=0.461atm$
• B. $\left[P_{CO}\right]=17.39atmand{P}_{C{O}_2}=0.461atm$
• C. $\left[P_{CO}\right]=1.79atmand{P}_{C{O}_2}=0.46atm$
• D. $\left[P_{CO}\right]=2.739atmand{P}_{C{O}_2}=1.461atm$

Answer 1

Answer: (A)

$K_P=0.265=\frac{0.8-x}{1.4+x}$
On solving, $\text{x}\text{=}\text{0}\text{.339}$
$\therefore \left[P_{CO}\right]=1.739$ atm and $P_{C{O}_2}=0.461atm$