given : $\mathrm{CCl}_{4(g)}$ → $\mathrm{C}_{(g)} + 4\mathrm{Cl}_{(g)}$ , $\Delta_{\mathrm{vap}}H^\circ(\mathrm{CCl}_4)$ = 30.5 $\text{kJ mol}^{-1}$ ... (i) $\mathrm{C}_{(s)} + 2\mathrm{Cl}_{2(g)}$ → $\mathrm{CCl}_{4(l)}$ , $\Delta_{\mathrm{f}}H^\circ(\mathrm{CCl}_4)$ = –135.5 $\text{kJ mol}^{-1}$ ... (ii) $\mathrm{C}_{(s)}$ → $\mathrm{C}_{(g)}$ , $\Delta_{\mathrm{a}}H^\circ$ = 715.0 $\text{kJmol}^{-1}$ ... (iii) $\mathrm{Cl}_{2(g)}$ → $2\mathrm{Cl}_{(g)}$ , $\Delta_{\mathrm{a}}H^\circ$ = 242 $\text{jmol}^{-1}$ ... (iv) Required equation is : $\mathrm{CCl}_{4(g)}$ → $\mathrm{C}_{(g)} + 4\mathrm{Cl}_{(g)}$ ; ΔH = ? From Hess’s law, eqn.(iii) + 2 × eqn.(iv) – eqn.(i) – eqn.(ii) gives required equation : ∴ ΔH = 715.0 + 2(242) – 30.5 – (–135.5) = 1304 $\text{kJ mol}^{-1}$ Bond enthalpy of C – Cl in $\mathrm{CCl}_4$ (average value) = $\frac{1304}{4}$ = 326 $\text{kJmol}^{-1}$