Question

Calculate the standard enthalpy of reduction for the following reaction: Fe2O3 (s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g)Given: ΔH°f for Fe2O3 (s) = -824.2 kJ/mol ΔH°f for H2O(g) = -241.8 kJ/molΔH°f for Fe(s) = 0 kJ/mol

Answer 1

To calculate the standard enthalpy of reduction for the given reaction, we can use the following formula:Hrxn =  Hf products  -  Hf reactants For the products, we have 2 moles of Fe s  and 3 moles of H2O g . For the reactants, we have 1 mole of Fe2O3 s  and 3 moles of H2 g . The Hf for H2 g  is 0 kJ/mol since it is in its standard state.Hrxn = [ 2 mol Fe  Hf Fe   +  3 mol H2O  Hf H2O  ] - [ 1 mol Fe2O3  Hf Fe2O3   +  3 mol H2  Hf H2  ]Hrxn = [ 2 mol  0 kJ/mol  +  3 mol  -241.8 kJ/mol ] - [ 1 mol  -824.2 kJ/mol  +  3 mol  0 kJ/mol ]Hrxn = [0 kJ +  -725.4 kJ ] - [-824.2 kJ]Hrxn = -725.4 kJ + 824.2 kJHrxn = 98.8 kJ/molThe standard enthalpy of reduction for the given reaction is 98.8 kJ/mol.