Calculate standerd enthalpy of the reaction
Fe₂O₃ (s) + 3 CO (g) → 2Fe (s) + 3CO₂ (g)
from the following data:
ΔfH° (Fe₂O₃) = -824.2 kJ mol⁻¹ , ΔfH° (CO) =-110.5 kJ mol⁻¹, ΔfH° (CO₂) = -393.5 kJ mol⁻¹
(-24.8kJ)
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This what I come up with, i get 2/3 Fe3O4 on
one side and 1/3 Fe3O4 on the other side
giving me a 1/3 of Fe3O4 that dose not cancel
each other out. Fe2O3(s) + 3 CO(g) → 2 Fe(s) +
3 CO2(g) ∆H= ? Fe2O3(s) + 1/3CO(g) →
1/3Fe3O4(s) + 1/3CO2(g) ∆Ho = -16.17kJ 2FeO(s)
+ 2CO(g) → 2 Fe(s) + 2CO2(g) ∆Ho = +22.0 kJ
2/3Fe3O4(s) + 2/3CO(g) → 2FeO(s) + 2/3CO2(g)
∆Ho = +14.67 kJ
one side and 1/3 Fe3O4 on the other side
giving me a 1/3 of Fe3O4 that dose not cancel
each other out. Fe2O3(s) + 3 CO(g) → 2 Fe(s) +
3 CO2(g) ∆H= ? Fe2O3(s) + 1/3CO(g) →
1/3Fe3O4(s) + 1/3CO2(g) ∆Ho = -16.17kJ 2FeO(s)
+ 2CO(g) → 2 Fe(s) + 2CO2(g) ∆Ho = +22.0 kJ
2/3Fe3O4(s) + 2/3CO(g) → 2FeO(s) + 2/3CO2(g)
∆Ho = +14.67 kJ
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