chemistry question

Question

Given that

~H2(g)+F2(g) 2H2(g)+O2(g) ⟶ 2HF(g). ~Δ𝐻∘rxn=−546.6 kJ⋅mol−1

~2H2(g)+O2(g)⟶ 2H2O(l)

~Δ𝐻∘rxn=−571.6 kJ⋅mol−1

calculate the value of Δ𝐻∘rxn for

2F2(g)+2H2O(l) ⟶ 4HF(g)+O2(g)

Δ𝐻∘rxn= kJ⋅mol−1

Answer 1

You kinda messed up the formatting, so I'll assume the following is what you meant...

Given:

(1) H2(g) + F2(g) ==> 2HF(g) ... ∆H = -546.6 kJ/mol

(2) 2H2(g) + O2(g) ==> 2H2O(l) ... ∆H = -571.6 kJ/mol

Calculate ∆H for:

2F2(g) + 2H2O(l) ==> 4HF(g) + O2(g) ... ∆H = ?

This is an example of using Hess' Law. Rearrange the 2 given equations to obtain the unknown:

Procedure:

copy (1) x2: 2H2(g) + 2F2(g) ==> 4HF(g) ... ∆H = -546.6 x 2 = -1093.2 kJ

reverse (2): 2H2O(l) ==> 2H2(g) + O2(g) ... ∆H = +571.6 kJ (change sign)

____________________________________________add the 2 reactions and combine or cancel like terms

2F2(g) + 2H2O(l) ==> 4HF(g) + O2(g) ∆H = -1093.2 + 571.6 = -521.6 kJ