Given that H2(g)+F2(g)⟶2HF(g)Δ𝐻∘rxn=−546.6 kJ

Question

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's Law

2F₂ + 2H₂O ==> 4HF + O₂ ... TARGET EQUATION

Given:

eq.1: H₂ + F₂ ==> 2HF ... ∆H = -546.6 kJ

eq.2: 2H₂ + O₂ ==> 2H₂O ... ∆H = -571.6 kJ

Procedure:

multiply eq.1 by 2: 2H₂ + 2F₂ ==> 4HF ... ∆H = -1093.2 kJ (note multiplied by 2)

reverse eq.2 to get: 2H₂O ==> 2H₂ + O₂ ... ∆H = +571.6 kJ (note reversal of sign)

adding them we get...

2H₂ + 2F₂ + 2H₂O ==> 4HF + 2H₂ + O₂ ∑∆H = -1093.2 kJ + 571.6 kJ = -521.6 kJ

Canceling like terms we end up with ...

2F₂ + 2H₂O ==> 4HF + O₂ ... TARGET EQUATION

∆Hrxn = -521.6 kJ