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Question
Answer the following question.
Calculate standard enthalpy of reaction,
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g), from the following data.
Δf H°(Fe2O3) = - 824 kJ/mol,
Δf H°(CO) = - 110 kJ/mol,
Δf H°(CO2) = - 393 kJ/mol
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Solution
Given:
Δf H°(Fe2O3) = - 824 kJ/mol,
Δf H°(CO) = - 110 kJ/mol,
Δf H°(CO2) = - 393 kJ/mol
To find: Standard enthalpy of the given reaction (ΔrH°)
Formula: Δ H° = ∑ Δf H° (products) - ∑ Δf H° (reactants)
Calculation:
The reaction is
Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g)
ΔrH° = ∑ Δf H° (products) - ∑ Δf H° (reactants)
= [2 Δf H° (Fe) + 3 Δf H° (CO2)] - [Δf H° (Fe2O3) + 3 Δf H° (CO)]
= [0 + 3 mol × (- 393 kJ mol-1)] - [1 mol × (- 824 kJ mol-1) + 3 mol × (- 110 kJ mol-1)]
= - 1179 + 824 + 330
= - 25 kJ
The standard enthalpy of the given reaction is –25 kJ.
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