Question

What is the enthalpy of the following reactions? Use the appendix in your text book using heat formatino values.

1. SO3+ H2O-----H2SO4

2. 3K2O2+3H2O----6KOH+O3

3. Fe3O4+8HCL----2FeCl3+FeCl2+4H2O

Answer 1

Solution :-

Formula to calculate the enthalpy change of the reaction is

Delta H rxn = sum of Delta Hf product - sun of delta Hf reactant.

1. SO3+ H2O-----H2SO4

Solution :-

Delta H rxn = [H2SO4 * 1] - [ (SO3*1)+(H2O*1)]

                 = [-907.51*1]-[(-396*1)+(-285.8*1)]

                = -225.71 kJ

2. 3K2O2+3H2O---- > 6KOH+O3

solution :-

Delta H rxn = [(KOH * 6)+(O3*1)] - [(K2O2*1)+(H2O*3)]

                  = [(-424.8*6)+(143*1)]-[(-496*1)+(-285.8*3)]

                  = -152.4 kJ

3. Fe3O4+8HCL----2FeCl3+FeCl2+4H2O

Solution :-

Delta H rxn = [(FeCl3*2)+(FeCl2*1) +(H2O*4)] - [(Fe3O4*1)+(HCl*8)]

                  = [(-399.5*2)+(-341.8*1)+(-285.8*4)] - [ (-825.5*1) +(-167.46*8)]

                = -118.82 kJ