Question

Calculate the enthalpy change for the reaction \[\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\] from the following data.

\[\ce{2Fe + 3/2O2 -> Fe2O3}\]; ΔH = −741 kJ

\[\ce{C + 1/2O2 -> CO}\]; ΔH = −137 kJ

\[\ce{C + O2-> CO2}\]; ΔH = −394.5 kJ

Answer 1

∆H_f (Fe₂O₃) = −741 kJ mol⁻¹

∆H_f (CO) = −137 kJ mol⁻¹

∆H_f (CO₂) = −394.5 kJ mol⁻¹

\[\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\]

ΔH_r = ?

ΔH_r = `Σ (Δ"H"_"f")_"products" − Σ (Δ"H"_"f")_"reactants"`

ΔH_r = [2 ∆H_f (Fe) + 3 ∆H_f (CO₂)] – [∆H_f (Fe₂O₃) + 3∆H_f (CO)]

ΔH_r = [0 + 3 (−394.5)] − [−741 + 3 (−137)]

ΔH_r = [−1183.5] − [−1152]

ΔH_r = −1183.5 + 1152

ΔH_r = −31.5 kJ mol⁻¹