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2 years ago

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

Chemistry

1 answer:

Rudiy272 years ago

4 0

Answer:

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

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