Given that H2(g)+F2(g)⟶2HF(g)Δ𝐻∘rxn=−546.6 kJ 2H2(g)+O2(g)⟶2H2O(l)Δ𝐻∘rxn=−571.6 kJ calculate the value of Δ𝐻∘rxn for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g)

Answer 1

Hess' Law

2H2 + 2F2 ==> 4HF ... ∆H = -546.6 x 2 = -1093 kJ

2H2O ===> 2H2 + O2 ... ∆H = +571.6 kJ

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2H₂ + 2F₂ + 2H₂O ==> 4HF + 2H₂ + O₂ ... ∆H = -521 kJ