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)  

Answer:

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 \[=\frac{1304}{4}=326kJ\,mo{{l}^{-1}}\]