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)),("Initial pressure","1.4 atm","0.80 atm"),:}`
`Q_p=p_(CO_2)/p_(CO)=0.80/1.4`=0.571
Since `Q_p > K_p` , the reaction will move in the backward direction . Therefore , pressure of `CO_2` will decrease and that of CO will increase to attain equilibrium.
If p is the decrease in pressure of `CO_2` , then increase in pressure of CO= p
`therefore` At equilibrium `p_(CO_2)` =(0.80-p) atm
`p_(CO)` = (1.4+p) atm
`K_p=(0.80-p)/(1.4+p)`=0.265
or 0.265 (1.4+ p) =0.80 -p
or 0.371 + 0.265 p = 0.80 - p
or 1.265 p = 0.429
or `p=0.429/1.265` =0.339 atm
At equilibrium : `p_(CO)`= 1.4+0.339 =-1.739 atm
`p_(CO_2)` =0.80 - 0.339 =0.461 atm