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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
Solution
∆Hf (Fe2O3) = −741 kJ mol−1
∆Hf (CO) = −137 kJ mol−1
∆Hf (CO2) = −394.5 kJ mol−1
\[\ce{Fe2O3 + 3CO -> 2Fe + 3CO2}\]
ΔHr = ?
ΔHr = `Σ (Δ"H"_"f")_"products" − Σ (Δ"H"_"f")_"reactants"`
ΔHr = [2 ∆Hf (Fe) + 3 ∆Hf (CO2)] – [∆Hf (Fe2O3) + 3∆Hf (CO)]
ΔHr = [0 + 3 (−394.5)] − [−741 + 3 (−137)]
ΔHr = [−1183.5] − [−1152]
ΔHr = −1183.5 + 1152
ΔHr = −31.5 kJ mol−1
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