Using the equations H₂ (g) + F₂ (g) → 2 HF (g) ∆H° = -79.2 kJ/mol
C (s) + 2 F₂ (g) → CF₄ (g) ∆H° = 141.3 kJ/mol
2 C(s) + 2 H₂ (g) → C₂H₄ (g) ∆H° = -97.6 kJ/mol

Determine the molar enthalpy (in kJ/mol) for the reaction
C₂H₄ (g) + 6 F₂ (g) → 2 CF₄ (g) + 4 HF (g).