Question

Calculate the standard enthalpy change of the following reduction reaction if the standard enthalpy of formation of FeO is -272.0 kJ/mol and that of Fe is 0 kJ/mol.FeO(s) + H2(g) → Fe(s) + H2O(l)

Answer 1

To calculate the standard enthalpy change of the reaction, we need to use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change of the reaction, and Hf is the standard enthalpy of formation of the respective substances.We are given the standard enthalpy of formation of FeO  -272.0 kJ/mol  and Fe  0 kJ/mol . We also need the standard enthalpy of formation of H2O l , which is -285.8 kJ/mol, and H2 g , which is 0 kJ/mol.Now we can plug these values into the equation:H = [Hf Fe  + Hf H2O ] - [Hf FeO  + Hf H2 ]H = [ 0 kJ/mol  +  -285.8 kJ/mol ] - [ -272.0 kJ/mol  +  0 kJ/mol ]H = -285.8 kJ/mol + 272.0 kJ/molH = -13.8 kJ/molThe standard enthalpy change of the reduction reaction is -13.8 kJ/mol.