Question

At 25°C, the following heats of reaction are known: 2ClF(g) + O2(g) → Cl2O(g) + F2O(g) ΔH°rxn = 167.4 kJ/mol 2ClF3(g) + 2O2(g) → Cl2O(g) + 3F2O(g) ΔH°rxn = 341.4 kJ/mol 2F2(g) + O2(g) → 2F2O(g) ΔH°rxn = –43.4 kJ/mol At the same temperature, use Hess's law to calculate ΔH°rxn for the reaction: ClF(g) + F2(g) → ClF3(g)

Answer 1

Answer:

-108.7 KJ/mol

Explanation:

Hess's law state that in a multiple stage reaction, the enthalpy change is the sum of all the enthalpy changes.

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn =167.4  KJ/mol

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

flip the above reaction

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

2CIF (g) + O₂ (g) → Cl₂O(g) + F₂O (g)        ΔH°rxn = 167.4  KJ/mol

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

add all the enthalpy change = -341.4 kJ/mol + 167.4  KJ/mol – 43.4 kJ/mol = -217.4

2CIF (g) + 2F₂(g)  → 2ClF₃(g)          ΔH°rxn =  -217.4 KJ/mol

ClF(g) + F2(g) → ClF3(g) =         (ΔH°rxn =  -217.4 KJ/mol) / 2 = -108.7 KJ/mol