One of the reactions that take place in producing steel from ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO_(2) Fe O(s) + CO(g)hArr Fe (s) + CO_(2) (g) , K_(p) = 0*265 " at " 1050 K What are the equilibrium partial pressures of CO and CO_(2) at 1050 K if the intial pressures are : p_(co) = 1*4 " atm " and p _(CO_(2))= 0*80" atm" ?

One of the reactions that take place in producing steel from ore is th

1.	One of the reactions that take place in producing steel from ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO_(2) Fe O(s) + CO(g)hArr Fe (s) + CO_(2) (g) , K_(p) = 0265 " at " 1050 K What are the equilibrium partial pressures of CO and CO_(2) at 1050 K if the intial pressures are : p_(co) = 14 " atm " and p _(CO_(2))= 0*80" atm" ?
Answer» SOLUTION :` {:(,FeO(s),+,CO(g),hArr,Fe(s),+,CO_(2)) ,("Intial pressures",,,14 "atm",,,,080"atm"),(,,,,,,,):}` ` Q_(p) = p_(CO_(2))/ (p_(CO))= = (080)/14 = 0571 ` 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 equilibrim . Hence , if p is the decrease in pressure of ` CO_(2)`increase in pressure of CO= p ` :. "At equilibrium " , p_(CO_(2)) = (080- p) "atm" , p_(CO) = (14 + p)"atm "` ` K_(p) = p_(CO_(2))/(p_(CO)) :.0265 = (080 - p)/(14 +p) or 0265 (14 +p)= 080 - p ` or ` 0 371 + 0265 p = 080 - por 1 * 265 p = 049 or p = 0339 "atm"` ` :. (p_(CO))_(eq) = 14 + 0339 " atm" = 1739 " atm" , (p_(CO_(2)))_(eq)= 080 - 0339 "atm" = 046`