Question

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" ?`

Answer 1

` {:(,FeO(s),+,CO(g),hArr,Fe(s),+,CO_(2)) ,("Intial pressures",,,1*4 "atm",,,,0*80"atm"),(,,,,,,,):}`
` Q_(p) = p_(CO_(2))/ (p_(CO))= = (0*80)/1*4 = 0*571 `
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)) = (0*80- p) "atm" , p_(CO) = (1*4 + p)"atm "`
` K_(p) = p_(CO_(2))/(p_(CO)) :. 0*265 = (0*80 - p)/(1*4 +p) or 0*265 (1*4 +p)= 0*80 - p `
or ` 0* 371 + 0*265 p = 0*80 - p or 1 * 265 p = 0*49 or p = 0*339 "atm"`
` :. (p_(CO))_(eq) = 1*4 + 0*339 " atm" = 1*739 " atm" , (p_(CO_(2)))_(eq) = 0*80 - 0*339 "atm" = 0*46`