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)=0.80atm ?

One of the reaction that takes place in producing steel from iron ore

1.	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)=0.80atm ?
Answer» Solution :`{:("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