Calculate the ΔH for the following reaction: Fe2O3 + 3 CO → 2 Fe + 3 CO2
based on the reactions below.
3Fe2O3 + CO→2Fe3O4 + CO2 ΔH=−48.5kJ
Fe+CO2 →FeO+CO. ΔH=−11.0kJ
Fe3O4 +CO→3FeO+CO2 ΔH=+22.0kJ
3Fe2O3 + CO --> 2Fe3O4 + CO2 , dH = - 48.5 KJ .............(1)
Fe + CO2 --> FeO + CO , dH = -11 KJ .................(2)
Fe3O4 + CO --> 3FeO + CO2 , dH = 22 KJ ...............(3)
eq(1) +2eq(3) - 6eq(2) gives us eq (4) i.e required eq
3Fe2O3 + CO --> 2Fe3O4 + CO2 , dH = - 48.5 KJ
2Fe3O4 + 2CO --> 6FeO + 2CO2 , dH = 2 x 22 KJ = 44 KJ
6FeO + 6CO --> 6Fe + 6CO2 , dH = - 6 x ( -11) = 66 KJ
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3Fe2O3 + 9CO --> 6Fe + 9CO2 , dH = 61.5 KJ
now dividing whole eq by 3 we get
Fe2O3 + CO --> 2Fe + 3CO2 , dH = ( 61.5/3) = 20.5 KJ
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