Given the following reactions
Fe2O3(s) + 3CO (s) → 2Fe (s) + 3CO2(g)ΔH
=-28.0 kJ
3Fe (s) + 4CO2(s) → 4CO (g) +Fe3O4(s)
ΔH = +12.5 kJ
the enthalpy of the reaction of
Fe2O3 with CO
3Fe2O3(s) + CO (g) → CO2(g) + 2Fe3O4(s)
is __________ kJ
3 answers
jlkmf;slda,
33.33KJ
The enthalpy of the reaction can be determined by adding the enthalpies of the two given reactions. However, to make the reaction match, we need to reverse the second reaction and multiply it by 3.
Reversed reaction 2:
4CO(g) + Fe3O4(s) → 3Fe(s) + 4CO2(g)
Multiplying by 3:
12CO(g) + 3Fe3O4(s) → 9Fe(s) + 12CO2(g)
Adding with reaction 1:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -28.0 kJ
+ 12CO(g) + 3Fe3O4(s) → 9Fe(s) + 12CO2(g) ΔH = -3(12.5 kJ) = -37.5 kJ
----------------------------------
3Fe2O3(s) + 15CO(g) → 6Fe3O4(s) + 15CO2(g) ΔH = -28.0 kJ - 37.5 kJ = -65.5 kJ
Therefore, the enthalpy of the reaction of Fe2O3 with CO is -65.5 kJ.
Reversed reaction 2:
4CO(g) + Fe3O4(s) → 3Fe(s) + 4CO2(g)
Multiplying by 3:
12CO(g) + 3Fe3O4(s) → 9Fe(s) + 12CO2(g)
Adding with reaction 1:
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) ΔH = -28.0 kJ
+ 12CO(g) + 3Fe3O4(s) → 9Fe(s) + 12CO2(g) ΔH = -3(12.5 kJ) = -37.5 kJ
----------------------------------
3Fe2O3(s) + 15CO(g) → 6Fe3O4(s) + 15CO2(g) ΔH = -28.0 kJ - 37.5 kJ = -65.5 kJ
Therefore, the enthalpy of the reaction of Fe2O3 with CO is -65.5 kJ.
3 answers