What will be $$\Delta H$$  for the reaction, $$C{H_2}C{l_2} \to C + 2H + 2Cl?$$
( $$B.E.$$  of $$C- H$$   and $$C- Cl$$   bonds are $$416\,kJ\,mo{l^{ - 1}}$$   and $$325\,kJ\,mo{l^{ - 1}}$$   respectively )

Solution :

$$\eqalign{ & \Delta H = \sum {B.E{._R}} - \sum {B.E{._P}} \cr & \,\,\,\,\,\,\,\,\,\,\, = 2 \times B.E.\left( {C - H} \right) + 2 \times B.E.\left( {C - Cl} \right) \cr & \,\,\,\,\,\,\,\,\,\,\, = 2 \times 416 + 2 \times 325 \cr & \,\,\,\,\,\,\,\,\,\,\, = 1482\,kJ \cr} $$