Question

Determine the standard enthalpy change for the following reaction: 2 Fe (s) + 3/2 O2 (g) → Fe2O3 (s) Given the following data: - ΔHf° of Fe2O3 (s) = -826 kJ/mol - ΔHf° of O2 (g) = 0 kJ/mol - The heat capacity of the calorimeter is 135 J/°C Assume that the reaction takes place at constant pressure and that the heat absorbed by the calorimeter is negligible.

Answer 1

To determine the standard enthalpy change  H  for the given reaction, we can use the following equation:H =  Hf products  -  Hf reactants In this case, the products are Fe2O3  s  and the reactants are Fe  s  and O2  g . We are given the standard enthalpy of formation  Hf  for Fe2O3  s  and O2  g , but not for Fe  s . However, since Fe  s  is in its elemental form, its Hf is 0 kJ/mol.Now we can plug in the values into the equation:H = [1   -826 kJ/mol ] - [2  0 kJ/mol +  3/2   0 kJ/mol]H = -826 kJ/molTherefore, the standard enthalpy change for the given reaction is -826 kJ/mol.