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`?

Explanation

`{:(FeO(s),+,CO(g),hArr,Fe(s),+,CO_(2)(g)),(,,1.4 "atm",,,,0.80 "atm"),(,,,,,,"Initial pressure"):}`
`Q_(p)=p_(CO_(2))/p_(CO)=0.80/1.4=0.571`
As `Q_(p)gt K_(p)` reaction will move in the backward direction, i.e., pressure of `CO_(2)` will decrease and that of `CO` will increase to attain equilibrium. Hence, if `p` is the decrease in pressure of `CO_(2)` increase in pressure of `CO=p`
`:.` At equilibrium, `p_(CO_(2))=(0.80-p) atm`,
`p_(CO)=(1.4+p) atm`
`K_(p)=p_(CO_(2))/p_(CO)`
`:. 0.265=(0.80-p)/(1.4+p)`
or `0.265(1.4+p)=0.80-p`
or `0.371+0.265p=0.80-p`
or `1.265p=0.429`
or `p=0.339 atm`
`:. (p_(CO))_(eq)=1.4+0.339 atm =1.739 atm`
`(p_(CO_(2)))_(eq)=0.80-0.339 atm =0.461 atm`