In: Chemistry
10. Given the data below, what is ∆Hr for the reaction:
FeO + CO → CO2 + Fe
∆Hr
Fe2O3 + 3CO → 2Fe + 3CO2 –35kJ
3Fe2O3 + CO → CO2 + 2Fe2O4 –70 kJ
Fe3O4 + CO → 3FeO + CO2 +45 kJ
Fe2O3 + 3CO → 2Fe + 3CO2...........(1) ∆Hr = –35kJ
3Fe2O3 + CO → CO2 + 2Fe3O4.......(2) ∆Hr = –70 kJ
Fe3O4 + CO → 3FeO + CO2............(3) ∆Hr = +45 kJ
Multiplying (3) 2
2Fe3O4 + 2CO → 6FeO + 2CO2............(4) ∆Hr = +90 kJ
Writing (2) as it is
3Fe2O3 + CO → CO2 + 2Fe3O4.......(2) ∆Hr = –70 kJ
Multiplying (1) 3 and reversing
6Fe + 9CO2 → 3Fe2O3 + 9CO...........(5) ∆Hr = +105 kJ
Adding (4) + (2) + (5)
2Fe3O4 + 2CO + 3Fe2O3 + CO + 6Fe + 9CO2 → 6FeO + 2CO2 + CO2 + 2Fe3O4 + 3Fe2O3 + 9CO......∆Hr = +125 kJ
2Fe3O4 + 3CO + 3Fe2O3 + 6Fe + 9CO2 → 6FeO + 3CO2 + 2Fe3O4 + 3Fe2O3 + 9CO......∆Hr = +125 kJ
6Fe + 6CO2 → 6FeO + 6CO......∆Hr = +125 kJ
Reversing the equation
6FeO + 6CO → 6Fe + 6CO2......∆Hr = -125 kJ
Dividing by 6
FeO + CO → Fe + CO2......∆Hr = -20.8 kJ
So answer is -20.8 kJ