Question

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 CO₂.

\[\ce{FeO (s) + CO (g) ↔ Fe (s) + CO2 (g)}\]; K_p= 0.265 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) +	CO2(g)
Initally		1.4 atm			0.80 atm

`"Q"_"P" = "p"_("CO"_2)/"p"_("CO")`

`= 0.80/1.4`

= 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" = ("p"_("CO"_2))/"p"_("CO")`

`=>0.265 = (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 `"CO"_2, "p"_("CO"_2)` = 0.80 - 0.399 = 0.461 atm

And, equilibrium partial pressure of  CO, `"p"_("CO")` = 1.4 + 0.339 = 1.739 atm