Question

Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ΔH o rxn for the production of ClF3:
(1) 2 ClF(g) + O2(g) → Cl2O(g) + OF2(g) ΔHo = 167.5 kJ
(2) 2 F2(g) + O2(g) → 2 OF2(g) ΔHo = −43.5 kJ
(3) 2 ClF3(l) + 2 O2(g) → Cl2O(g) + 3 OF2(g) ΔHo = 394.1 kJ
calculate ΔH rxn in KJ

Answer 1

Answer:

ΔH = -135.05 kJ

Explanation:

(1) 2 ClF(g) + O₂(g) → Cl₂O(g) + OF₂(g)           ΔHo = 167.5 kJ

(2) 2 F₂(g) + O₂(g) → 2 OF₂(g)                         ΔHo = −43.5 kJ

(3) 2 ClF₃(l) + 2 O₂(g) → Cl₂O(g) + 3 OF₂(g)    ΔHo = 394.1 kJ

We can use Hess' Law to calculate the  ΔH of reaction:

ClF(g) + 1/2O₂(g) → 1/2Cl₂O(g) + 1/2OF₂(g)  ΔHo = 167.5  x 1/2 = 83.75 kJ

F₂(g) + 1/2O₂(g) →  OF₂(g)                             ΔHo = −43.5   x 1/2 = -21.75 kJ

1/2Cl₂O(g) + 3/2 OF₂(g) →  ClF₃(l) +  O₂(g)   ΔHo = 394.1  x -1 x 1/2 = -197.05kJ

______________________________________________

(4) ClF (g) + F₂ (g) → ClF₃ (l)      ΔH = 83.75 kJ-21.75 kJ-197.05kJ

ΔH = -135.05 kJ/mol