What is the overall enthalpy of reaction for the equation shown below?

ΔH₁₂ = -867.2 Kj

Explanation:

Find enthalpy for 3H₂ + O₃ => 3H₂O given ...

2H₂ + O₂ => 2H₂O      ΔH₁ = -483.6 Kj

        3O₂ => 2O₃        ΔH₂ = + 284.6 Kj

3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O       (multiply by 3 to cancel O₂)

6H₂ + 3O₂ => 6H₂O        ΔH₁ = 3(-483.6 Kj) = -1450.6Kj

          2O₃ => 3O₂           ΔH₂ = -284.6Kj              (reverse rxn to cancel O₂)

6H₂ + 2O₃ => 6H₂O         ΔH₁₂ = -1735.2 Kj       (Net Reaction - not reduced)

divide by 2 => target equation (Net Reaction - reduced)

3H₂ + O₃ => 3H₂O            ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj    

the answer would be a