Oxidation of gaseous ClF by F2 yields liquid ClF3 an importa

Oxidation of gaseous ClF by F2 yields liquid ClF3, an important fluorinating agent. Use the following thermochemical equations to calculate ?H o rxn for the production of ClF3:

(1) 2 ClF(g) + O₂(g) ?Cl₂O(g) + OF₂(g) ?H^o = 167.5 kJ

(2) 2 F₂(g) + O₂(g) ? 2 OF₂(g) ?H^o = ?43.5 kJ

(3) 2 ClF₃(l) + 2 O₂(g) ? Cl₂O(g) + 3 OF₂(g) ?H^o = 394.1 kJ

Solution

The target equation is

ClF(g) + F₂(g) ---> ClF₃(l) ...............(4)

Thermochemical given equations are

(1) 2 ClF(g) + O₂(g) ?Cl₂O(g) + OF₂(g) ?H^o = 167.5 kJ

(2) 2 F₂(g) + O₂(g) ? 2 OF₂(g) ?H^o = ?43.5 kJ

(3) 2 ClF₃(l) + 2 O₂(g) ? Cl₂O(g) + 3 OF₂(g) ?H^o = 394.1 kJ

So we can obtain equation (4) from equation (1),(2) and (3) as

Equation (4) = 0.5 (1) + 0.5 (2) -0.5(3) = 1/2( equation (1) + equation (2) - equation (3))

So enthalpy of equation (4) = 1/2 [167.5 + (-43.5) -391.4] = -133.7 kJ