Question

Calculate the standard enthalpy change for the following reaction: Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g) given the following standard enthalpy of formation values: ΔHf°[Fe2O3(s)] = -824.2 kJ/mol ΔHf°[CO(g)] = -110.5 kJ/mol ΔHf°[Fe(s)] = 0 kJ/mol ΔHf°[CO2(g)] = -393.5 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H reaction  =  Hf products  -  Hf reactants For the products, we have 2 moles of Fe s  and 3 moles of CO2 g : Hf products  =  2  0 kJ/mol  +  3  -393.5 kJ/mol  = -1180.5 kJ/molFor the reactants, we have 1 mole of Fe2O3 s  and 3 moles of CO g : Hf reactants  =  -824.2 kJ/mol  +  3  -110.5 kJ/mol  = -1155.7 kJ/molNow, we can calculate the standard enthalpy change for the reaction:H reaction  = -1180.5 kJ/mol -  -1155.7 kJ/mol  = -24.8 kJ/molThe standard enthalpy change for the reaction is -24.8 kJ/mol.