The bond enthalpy of C-Cl in CCl₄(g) is to be calculated given: Δ Hf°(CCl₄,g) = -95.7 kJ/mol , Δ Ha(C,g) = 715 kJ/mol ,
For CCl₄(g) → C(g) + 4Cl(g) , Δ H = Δ Ha(C) + 2 × Δ Ha(Cl₂) - Δ Hf°(CCl₄,g) = 715 + 2 × 242 - (-95.7) = 715 + 484 + 95.7 = 1294.7 kJ/mol . Bond enthalpy = 1294.7 / 4 = 323.7 kJ/mol .
Ref: NCERT Class 11 Chemistry > Chapter 5: Thermodynamics > Topic: Bond Enthalpy and Entropy and Spontaneity
