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

ΔH = -135.05 kJ

Step-by-step 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

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

ΔH = -135.05 kJ/mol