Question

Calculate the standard enthalpy change for the following reaction, given the standard enthalpies of formation:2Fe (s) + 3Cl2 (g) → 2FeCl3 (s)ΔH°f(FeCl3) = -399.4 kJ/molΔH°f(Fe) = 0 kJ/molΔH°f(Cl2) = 0 kJ/mol

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =  [Hf products ] -  [Hf reactants ]For the given reaction:2Fe  s  + 3Cl2  g   2FeCl3  s The products are 2 moles of FeCl3, and the reactants are 2 moles of Fe and 3 moles of Cl2.Now, we can plug the values into the equation:H = [2  Hf FeCl3 ] - [2  Hf Fe  + 3  Hf Cl2 ]H = [2   -399.4 kJ/mol ] - [2   0 kJ/mol  + 3   0 kJ/mol ]H =  -798.8 kJ  -  0 kJ H = -798.8 kJThe standard enthalpy change for the reaction is -798.8 kJ.