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 atm at 1050K 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?

The initial partial pressures of CO and CO2 are 1.40 atm and 0.8. atm respectively. The expression for the reaction quotient is: Qp=PCO2 / PCO=0.80 / 1.4 = 0.571 As the value of the reaction quotient is greater than the value of the equilibrium constant, the reaction will move in the backward direction. To attain the equilibrium, the partial pressure of CO2 will decrease and the partial pressure of CO will increase. Let p atm be the decrease in the partial pressure of CO2. The increase in the partial pressure of CO will be p atm. The equilibrium partial pressure of CO2 is (0.80 – p) atm. The equilibrium partial pressure of CO is (1.4 + p) atm. The equilibrium constant is: Kp = PCO2 / PCO= (0.80p) / (1.4 + p) =0.265 0.265 (1.4+p) = (0.80 – p) 0.371+0.265p = (0.80 – p) p= 0.339 atm The equilibrium partial pressure of CO is 1.4+0.339= 1.739. The equilibrium partial pressure of CO2 is 0.80?0.339= 0.461 atm.

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); K_p = 0.265 atm at 1050K

What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: p_CO= 1.4 atm and p_CO2= 0.80 atm?

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Vishal Baghel | Contributor-Level 10

9 months ago

The initial partial pressures of CO and CO₂ are 1.40 atm and 0.8. atm respectively.
The expression for the reaction quotient is:
Q_p=P_CO2 / P_CO=0.80 / 1.4 = 0.571
As the value of the reaction quotient is greater than the value of the equilibrium constant, the reaction will move i