Question

One of the reaction that takes plece 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) hArr Fe(s)+CO_(2)(g), K_(p)=0.265` atm at `1050 K`
What are the equilibrium partial pressure of `CO` and `CO_(2)` at `1050 K` if the partical pressure are: `p_(CO)=1.4 atm` and `p_(CO_(2))=0.80 atm`?

Answer 1

Correct Answer - `[P_(CO)] = 1.739`atm, `[P_(CO2)] = 0.461` atm.
For the given reaction ,
`{:(FeO_((g)),+,CO_((g)),harr,Fe_((s)),+,CO_(2(g))),("Initially",,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) gt 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 `CO_(2)` will decrease.
Now, let the increase in pressure of CO = decrease in pressure of `CO_(2)` be p.
Then, we can write,
`K_(P) = (p_(CO_(2)))/(p_(CO))`
`rArr 0.265 = (0.80 - p)/(1.4 + p)`
`rArr 0.37 + 0.265 p = 0.80 - p`
`rArr 1.265 p = 0.429`
`rArr p = 0.339 "atm"`
Therefore, equilibrium partial of
`CO_(2), p_(CO_(2)) = 0.80-0.339 = 0.461 "atm"`
And, equilibrium partial pressure of
`CO, p_(CO) = 1.4 + 0.339 = 1.739 "atm"`.