Question

Question 7.20 One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO2.

FeO (s) + CO (g) <---> Fe (s) + CO2 (g); Kp= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO = 1.4 atm and

pCO2 = 0.80 atm?

Class XI Equilibrium Page 226

Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: FeO (s) + CO(g) == Fe(s) + CO2(g) ; Kp = 0.265 \: \: at \: \: 1050K \\ at \: t = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 1.4 \: atm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 0.8 \: atm$
Qp = Pco2/Pco
= 0.8/1.4 = 0.571
you can see that , Kp < Qp , the reaction will go in reverse direction . due to this pressure of CO2 will decreases and that of CO will increase to attain equilibrium .
Hence, Pco2 = (0.8 - P) and Pco = (1.4 + p)
Kp = Pco2/Pco
0.265 = (0.8 - p)/(1.4 + p)
0.265(1.4 + p) = 0.8 - p
0.265p + 0.265 × 1.4 = 0.8 - p
1.265p = 0.8 - 0.371 = 0.429
p = 0.429/1.265 = 0.339 atm

Hence, at equilibrium ,
pressure of CO2 ( Pco2) = 0.8 - 0.339 = 0.461 atm.
pressure of CO ( Pco ) = 1.4 + 0.339 = 1.739 atm