One of the reaction that takes place in producing steel from iron are 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 pressures of `CO` and `CO_(2)` and `1050` K if the initial partial pressures are : `p_(CO) = 1.4` atm and `= 0.80` atm?

Explanation

` {:(,Fe O(s) , +, CO(g) ,hArr, Fe(s) + CO_2(g) ),( " Initial pressure " , 1.4 " atm " ,, -,, 0.80 " atm " ):}`
` " " Q_ (rho ) =( rho _(CO_2))/(rho _(CO)) = (0.80 )/(1.4) = 0.571 `
Since ` Q_p gt K_rho ` , the reactions will proceed in the reverse directions , On account of this partial pressure of `CO_2` will decrease while that of CO will increase . Suppose , the decrease in the partial pressure of `CO_2 ` is ` rho ` atm.
` therefore ` At equilibrium
` rho _(CO_2) = (0.80 - rho ) " atm " and rho _(CO) = (1.4+ rho ) " atm " `

` therefore " " K_(rho ) = (rho _(CO_2))/(rho _(CO))` , we have
` " " = 0.265= ( 0 . 80 - rho )/( 1.4 + rho ) `
or ` 0.265 xx (1.4+rho ) = 0.80 - rho `
or ` 0.371+ 0.265 rho = 0.80 - rho`
or ` " " 1.265 rho = 0.429`
` or " " rho =( 0.429)/(1.265) = 0.339 ` atm
` therefore rho _(CO) ` at equilibrium ` = 1.4 + 0.339 = 1.739 ` atm .
and ` rho _(CO_2)` at equilibrium = 0.80 - 0.339 = 0.461 atm .