Question

One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2.
$FeO (s) + CO (g) ⇋ Fe (s) + CO_2 (g);\ K_p = 0.265\ \text {atm}$ at 1050 K
What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: $P_{CO}$ = 1.4 atm and $P_{CO_2}$ = 0.80 atm?

Answer 1

For the given reaction,
$FeO(g), + CO(g) ↔ Fe(s) + CO_2(g)$
Initialy, $1.4\ atm$ $0.80\ atm$

$Q_P = \frac {P_{CO_2}}{P_{CO}}$

$Q_P = \frac {0.80}{1.4}$
$Q_P = 0.571$
It is given that $K_P = 0.265$
Since $Q_P > K_P$, the reaction will proceed in the backward direction.
Therefore, we can say that the pressure of CO will increase while the pressure of CO2 will decrease.
Now, let the increase in pressure of CO = decrease in pressure of CO2 be p.
Then, we can write,
$K_P = \frac {P_{CO_2}}{P_{CO}}$

⇒ $0.265 =\frac { 0.80 - p}{1.4 + p }$
⇒ $0.371 + 0.265 p = 0.80 - p$
⇒ $1.265 p = 0.429$
⇒ $p = 0.339$ atm
Therefore, equilibrium partial of CO2, Pco2 = 0.80 - 0.339 = 0.461 atm.
And, equilibrium partial pressure of Co, Pco = 1.4 + 0.339 = 1.739 atm.