Calculate the enthalpy change for the process $CCl_{4(g)} \to C_{(g)} + 4Cl_{(g)}$ and calculate the bond enthalpy of the $C-Cl$ bond in $CCl_{4(g)}$.
$\Delta_{vap} H^{\theta}(CCl_{4}) = 30.5 \, kJ \, mol^{-1}$
$\Delta_{f} H^{\theta}(CCl_{4}) = -135.5 \, kJ \, mol^{-1}$
$\Delta_{a} H^{\theta}(C) = 715.0 \, kJ \, mol^{-1}$ (where $\Delta_{a} H^{\theta}$ is enthalpy of atomisation)
$\Delta_{a} H^{\theta}(Cl_{2}) = 242 \, kJ \, mol^{-1}$

The chemical equations for the given enthalpy values are:
$(i) \quad CCl_{4(l)} \longrightarrow CCl_{4(g)} \quad \Delta_{vap}H^{\theta} = 30.5 \, kJ \, mol^{-1}$
$(ii) \quad C_{(s)} \longrightarrow C_{(g)} \quad \Delta_{a}H^{\theta} = 715.0 \, kJ \, mol^{-1}$
$(iii) \quad Cl_{2(g)} \longrightarrow 2Cl_{(g)} \quad \Delta_{a}H^{\theta} = 242 \, kJ \, mol^{-1}$
$(iv) \quad C_{(s)} + 2Cl_{2(g)} \longrightarrow CCl_{4(l)} \quad \Delta_{f}H^{\theta} = -135.5 \, kJ \, mol^{-1}$
To find the enthalpy change for $CCl_{4(g)} \longrightarrow C_{(g)} + 4Cl_{(g)}$,we rearrange the equations:
$\Delta H = \Delta_{a}H^{\theta}(C) + 2\Delta_{a}H^{\theta}(Cl_{2}) - \Delta_{vap}H^{\theta} - \Delta_{f}H^{\theta}$
$\Delta H = 715.0 + 2(242) - 30.5 - (-135.5)$
$\Delta H = 715.0 + 484 - 30.5 + 135.5 = 1304 \, kJ \, mol^{-1}$
Bond enthalpy of $C-Cl$ bond in $CCl_{4(g)} = \frac{1304}{4} = 326 \, kJ \, mol^{-1}$.