One of the reaction that takes place 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 pressures of CO and `CO_2` at 1050 K if the initial partial pressures are: `p_(CO)` = 1.4 atm and `2p_(CO_2)`=0.80 atm ?

Explanation

`{:("Equillbrium reaction :", FeO_((s))+ CO_((g)) hArr , Fe_((s))+CO_(2(g))),("Initial partial pressure :", "1.4 atm","0.80 atm"),("Change in pressure :", "-x atm", "+x atm"),("Partial pressure at equilibrium:",(1.4-x),(0.80-x)):}`
where , x=1 pressure change in forward reaction.
`K_p=(p_(CO_2))/(p_(CO))`
`therefore 0.265=(0.80+x)/(1.4-x)`
`therefore` 0.265(1.4)-0.265x=0.80 +x
`therefore` 0.371 -0.80 =x + 0.265x
`therefore` 1.265x=-0.429
`therefore x=(-0.429)/1.265`=-0.3391 atm
So, `p_(CO)`=1.4 - (-0.3391)=1.7391
`p_(CO_2)`= 0.80+x =0.80-0.3391
=0.4609 = 0.461 atm