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

Bond energies are calculated from heat of formation of a compound. Now from the data given for heats of combustion for ethane and propane, we can calculate the heats of formation of two compounds `(C_2H_6 and C_3H_8)` as follows:
For ethane: The equation for combustion of ethane:
`C_(2)H_(6)(g) + 7//2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)Delta_("comb")H^(Theta) = -372.0`
From definitaion of `Delta H^(Theta)` of reaction : `DeltaH^(Theta) = Delta H_(p)^(Theta) - DeltaH_(R )^(Theta)`
The enthalpy of a compound is the enthalpy of formation of that compound at standard conditions (i.e.`Delta_(f)H^(-)`).
`Delta_("comb") H^(Theta) = [2Delta_(f)H^(Theta)(CO_2) + 3Delta_(f) H^(Theta) (H_2O)] - [Delta_(f)H^(Theta)(C_(2)H_(6)) - 7//2Delta_(f)H^(Theta)(O_2)]`
`[Delta_(f)H^(Theta)(O_2) = 0` (as enthalpy of formation of an element in standard state is taken as zero).
`implies -372 = 2 xx (-94) + 3 xx (-68) - Delta_(f)H^(theta)(C_2H_6) implies Delta_(f)H^(Theta)(C_2H_6)= -20 kcal`
For propane : The equation for combustion of propane.
`C_(3)H_(8)(g) + 5O_(2_(g) to 3CO_(2)(g) + 4H_(2)O (l) Delta_("comb") H^(Theta) = -530.0`
From definition of `DeltaH` of a reaction : `Delta H = H_P - H_R`
`Delta_("comb") H^(theta) = [3Delta_fH^@(CO_2)]+4Delta_fH^(theta)(H_2O) - [Delta_fH^(theta)(C_3H_8)) = Delta_(f)H^(Theta)(O_(2))]`
`implies -530 = 3 xx (-94) + 4 xx (-68) - Delta_(f) H^(Theta)(C_3H_8) implies Delta_(f)H^(Theta) (C_3H_8) = -24 kcal`.