At 25°c, the following heats of reaction are known: 2 clf(g) + o2(g) cl2o(g) + f2o(g)δh°rxn=167.4 kj/mol2 clf3(g) + 2 o2(g) cl2o(g) + 3 f2o(g)δh°rxn=341.4 kj/mol2 f2(g)+ o2(g)2 f2o(g)δh°rxn=-43.4 kj/molat the same temperature, use hess’s law to calculate δh°rxnfor the following reaction: clf(g) + f2(g) clf3(g)
a) -217.5 kj/mol
b) -130.2 kj/mol
c) 217.5 kj/mol
d) -108.7 kj/mol
e) 465.4 kj/mol

D) -108.7 kJ/mol

Explanation:

The Hess's law states that overall ΔH° of a reaction could be obtained from the ΔH°rxn in which the overall reaction could be divided.

For the reactions:

2 ClF(g) + O₂(g) → Cl₂O(g) + F₂O(g) ΔH°rxn = 167.4 kJ/mol (1)

2 ClF₃(g) + 2O₂(g) → Cl₂O(g) + 3F₂O(g) ΔH°rxn =341.4 kJ/mol (2)

2 F₂(g)+ O₂(g) → 2F₂O(g) ΔH°rxn=-43.4 kJ/mol (3)

The sum of the reactions (1) + (3) minus (2)  will give the reaction:

2ClF(g) + 2F₂(g) → 2ClF3(g) ΔH°rxn = 167,4 kH/mol +(-43,4kJ/mol) -341,4kJ/mol = -217,4 kJ/mol

Dividing the reaction in 2, ΔH°rxn of ClF(g) + F₂(g) → ClF3(g) is -217,4 kJ/mol/2

ΔH°rxn = -108,7 kJ/mol

Right answer is:

D) -108.7 kJ/mol

I hope it helps!

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the product of the pressure and volume is also constant