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

`FeO(s)+CO(g)hArr Fe(s)+CO_(2)(g)`
`" Initial pressure : " " "1.4 atm " "0.8atm `
`Q_(p)=(Pco_(2))/(Pco) = ((0.8atm))/((1.4 atm)) =0.571`
Since `Q_(p) gt K_(p) (0.265),` this means that the reaction will move in the backward direction to attain the equilibrium. Therefore , partial pressure of `CO_(2)` will decrease while that of CO will increase so that the equilibrium may be attained again. Let P atm be teh decrease in the partial pressure of `CO_(2)`. Therefore, the partial pressure of CO will increase by the same magnitude i.e.,p atm.
`Pco_(2) =(0.8 -p) " atm " , Pco(g) =(1.4 +p ) " atm "`
`"At equilibrium "" "K_(p) =(Pco_(2))/(Pco)= ((0.8-p) atm)/((1.4 +p) atm) = ((0.8 -p))/((1.4+p))`
`" or "" "0.265 =((0.8 -p))/((1.4 +p))`
`0.371 +0.265p= 0.8 -P or 1.265p = 0.8 -0.371 =0.429`
`p=0.429//1.265 =0.339 " atm "`
`(Pco)_(eq) = (1.4+0.339)=1.739 " atm "`
`(Pco_(2))_(eq)=(0.8-0.339)=0.461` atm