Question

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

Answer 1

D) -108.7 kJ/mol

Step-by-step 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!