Using the data given below (all values are kilories per mol at `25^(@)C)`, calculate the bond energy of `C -C` and `C -H` bonds.
`C(s) rarr C(g), DeltaH = 172`
`H_(2) rarr 2H, DeltaH = 104`
`H_(2)+(1)/(2)O_(2) rarr H_(2)O(l), DeltaH =- 68.0`
`C(s) +O_(2) rarr CO_(2), DeltaH =- 94.0`
Heta of combustion of `C_(2)H_(6) =- 372.0`
Heat of combustion of `C_(3)H_(8) =- 530.0`

Explanation

Let x kcal be the C - C bond energy and y kcal be the C - H bond energy per mole.
`implies 2C(gr) + 3H_(2) (g) rarr C_(2)H_(6)(g),`
`Delta H^(@) = - 2 xx 94 - 3 xx 68 + 372 = -20 kcal`
`implies - 20 kcal = 2 xx 172 + 3 xx 104 - BE (C_(2)H_(16)) implies BE(C_(6)H_(6)) = 676 kcal`
Similarly, `3C (gr) + 4H_(2) (g) rarr C_(3)H_(8) (g) , Delta H^(@) = -3xx 94 - 4 xx 68 + 530 = -24 kcal`
`implies -24 kcal = 3 xx 172 + 4 xx 104 - BE(C_(3)H_(8)) implies BE (C_(3)H_(8)) = 956 kcal`
Also, `BE (C_(2)H_(6)) = 676 kcal = x + 6y " ".....(i)`
`BE (C_(2)H_(8)) = 956 kcal = 2x + 8y" ".....(ii)`
Solving eqs. (i) and (ii) gives `y = 99 kcal (C - H) BE, x = 82 kcal (C - C) BE`.