Question

Use the information provided to determine Δ_rH° for the following reaction:

   Δ_fH° (kJ mol^-1)  3Fe₂O₃(s) + CO(g) → 2Fe₃O₄(s) + CO₂(g)  Δ_rH° = ?
Fe₂O₃(s)  -824
Fe₃O₄(s)  -1118
CO(g)  -111
CO₂(g)  -394

Answers:

-577 kJ mol^-1

+144 kJ mol^-1

+277 kJ mol^-1

-47 kJ mol^-1

-111 kJ mol^-1

Answer 1

we have:

Hof(Fe2O3(s)) = -824.0 KJ/mol

Hof(CO(g)) = -111.0 KJ/mol

Hof(Fe3O4(s)) = -1118.0 KJ/mol

Hof(CO2(g)) = -394.0 KJ/mol

we have the Balanced chemical equation as:

3 Fe2O3(s) + CO(g) ---> 2 Fe3O4(s) + CO2(g)

deltaHo rxn = 2*Hof(Fe3O4(s)) + 1*Hof(CO2(g)) - 3*Hof( Fe2O3(s)) - 1*Hof(CO(g))

deltaHo rxn = 2*(-1118.0) + 1*(-394.0) - 3*(-824.0) - 1*(-111.0)

deltaHo rxn = -47 KJ/mol

Answer: -47 KJ/mol