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 $CO_{2}$ respectively at $\text{1050} \, \text{K}$ if the initial partial pressures are: $P_{c o}=1.4$ atm and $P_{C O_{2}}=0.80 \, atm$ ?

• A. $\left[P_{C O}\right]=1.739 \, atm \, and \, P_{C O_{2}}=0.461 \, atm$
• B. $\left[P_{C O}\right]=17.39 \, atm \, and \, P_{C O_{2}}=0.461 \, atm$
• C. $\left[P_{C O}\right]=1.79 \, atm \, and \, P_{C O_{2}}=0.46 \, atm$
• D. $\left[P_{C O}\right]=2.739 \, atm \, and \, P_{C O_{2}}=1.461 \, atm$

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_{C O}\right]=1.739 \, $ atm and $P_{C O_{2}}=0.461 \, atm$