• # question_answer 28) Calculate the enthalpy change for the process: $CC{{l}_{4}}(g)\to C(g)+4Cl(g)$ Calculate bond enthalpy by $C-Cl$in $CC{{l}_{4}}(g)$ Given:$\Delta H_{vap}^{{}^\circ }CC{{l}_{4}}=30.5kJ\,mo{{l}^{-1}}$ ${{\Delta }_{a}}{{H}^{{}^\circ }}(C)=715kJ\,mo{{l}^{-1}}$ (Enthalpy of atomisation) ${{\Delta }_{a}}{{H}^{{}^\circ }}({{l}_{2}})=715kJ\,mo{{l}^{-1}}$ (Enthalpy of atomisation)

Given: (i) $\text{CC}{{\text{l}}_{4}}(l)\to CC{{l}_{4}}(g)$$\Delta {{H}^{{}^\circ }}=+30.5kJmo{{l}^{-1}}$ (ii) $\Delta {{H}^{{}^\circ }}=-135kJ\,mo{{l}^{-1}}$ (iii) $C(s)\to C(g)$$\Delta {{H}^{{}^\circ }}=+715kJ\,mo{{l}^{-1}}$ (iii) $C{{l}_{2}}(g)\to 2Cl(g)$$\Delta {{H}^{{}^\circ }}=+242kJ\,mo{{l}^{-1}}$ (iv)$C{{l}_{2}}(g)\to 2Cl(g)$$\Delta {{H}^{{}^\circ }}=+242kJ\,mo{{l}^{-1}}$ Required equation $CC{{l}_{4}}(g)\to C(g)+4Cl(g)$$\Delta H=?$ Equation (iii) + 2 x Equation (iv) -Equation (ii) Equation (i) gives the required equation $\Delta H=715\text{ }+\text{ }2\left( 242 \right)\text{ }-30.5\text{ }-\left( -135.5 \right)$ $=1304\text{ }kJmo{{l}^{-1}}$ Bond enthalpy of $(C-Cl)$bond $=\frac{1304}{4}=326kJ\,mo{{l}^{-1}}$