Question
The enthalpy of atomisation of $$C{H_4}$$ and $${C_2}{H_6}$$ are $$360$$ and $$620\,kcal\,mo{l^{ - 1}}$$ respectively. The $$C-C$$ bond energy is expected to be
A.
$$210\,kcal\,mo{l^{ - 1}}$$
B.
$$80\,kcal\,mo{l^{ - 1}}$$
C.
$$130\,kcal\,mo{l^{ - 1}}$$
D.
$$180\,kcal\,mo{l^{ - 1}}$$
Answer :
$$80\,kcal\,mo{l^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Atomisation of methane}} \cr
& C{H_4}\left( g \right) \to C\left( g \right) + 4H\left( g \right); \cr
& \Delta H = 360\,kcal \cr
& \therefore \,\,C - H\,{\text{bond energy}} = \frac{{360}}{4} \cr
& = 90\,kcal\,mo{l^{ - 1}} \cr
& {C_2}{H_6}\left( g \right) \to 2C\left( g \right) + 6H\left( g \right); \cr
& \Delta H = 620\,kcal \cr
& {\text{or}}\,\,{H_{C - C}} + 6\,{H_{C - H}} = 620 \cr
& \therefore \,\,{H_{C - C}} = 620 - 6\,{H_{C - H}} \cr
& = 620 - 6 \times 90 = 80\,kcal\,mo{l^{ - 1}} \cr} $$