Calculate the enthalpy change for the process

The chemical equations implying to the given values of enthalpies are:

(i) $\quad \mathrm{CCl}_{4(l)} \longrightarrow \mathrm{CCl}_{4(8)} \Delta_{\mathrm{vap}} \mathrm{H}^{\theta}=30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(ii) $\mathrm{C}_{(s)} \longrightarrow \mathrm{C}_{(s)} \Delta_{a} H^{\theta}=715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iii) $\mathrm{Cl}_{2(g)} \longrightarrow 2 \mathrm{Cl}_{(g)} \Delta_{a} H^{\theta}=242 \mathrm{~kJ} \mathrm{~mol}^{-1}$

(iv) $\mathrm{C}_{(g)}+4 \mathrm{Cl}_{(g)} \longrightarrow \mathrm{CCl}_{4(g)} \Delta_{f} H=-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Enthalpy change for the given process $\mathrm{CCl}_{4(g)} \longrightarrow \mathrm{C}_{(g)}+4 \mathrm{Cl}_{(g)}$, can be calculated using the following algebraic calculations as:

Equation (ii) + 2 × Equation (iii) – Equation (i) – Equation (iv)

$\Delta H=\Delta_{2} H^{\theta}(\mathrm{C})+2 \Delta_{a} H^{\theta}\left(\mathrm{Cl}_{2}\right)-\Delta_{\text {vap }} H^{\theta}-\Delta_{f} H$

$=\left(715.0 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)+2\left(242 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-\left(30.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)-\left(-135.5 \mathrm{~kJ} \mathrm{~mol}^{-1}\right)$

$\therefore \Delta H=1304 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Bond enthalpy of $\mathrm{C}-\mathrm{Cl}$ bond in $\mathrm{CCl}_{4(g)}$

$=\frac{1304}{4} \mathrm{~kJ} \mathrm{~mol}^{-1}$

$=326 \mathrm{~kJ} \mathrm{~mol}^{-1}$