Using the data (all values are in kilocalorie per mole at `25^(@)C`) given below, calculate the bond enegry of `C-C` and `C-H` bonds.
`DeltaH^(Theta)` combustion of ethane `=- 372.0`
`DeltaH^(Theta)` combustion of propane `=- 530.0`
`DeltaH^(Theta)` for `C`( garphite) `rarr C(g) =+ 172.0`
Bond enegry of `H-H` bond `=+ 104.0`
`Delta_(f)H^(Theta) of H_(2)O(l) =- 68.0`
`Delta_(f)H^(Theta) of CO_(2)(g) =- 94.0`

Explanation

Bond energies are calculated form heat of formation of a compound. Now from the date given for heats of combustion for ethane and propane, we can calculate the heats of formation of two compounds `(C_(2)H_(6)` and `C_(3)H_(8))` are follows:
For ethane: The equation for combustion of ethane:
`C_(2)H_(6)(g) +7//2O_(2)(g) rarr 2CO_(2)(g) +3H_(2)O(l) Delta_(comb)H^(Theta) =- 372.0`
From definition of `DeltaH^(Theta)` of reaction:
`DeltaH^(Theta) = DeltaH_(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^(Theta))`.
`Delta_(comb)H^(Theta) = [2Delta_(f)H^(Theta)(CO_(2))+3Delta_(f)H^(Theta)(H_(2)O)]-[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).
`rArr -371 = 2 xx (-94) +3 xx (-68) - Delta_(f)H^(Theta) (C_(2)H_(6))`
`rArr Delta_(f)H^(Theta) (C_(2)H_(6)) =- 20 kcal`
For propane: The equation for combustion of propane.
`C_(3)H_(8)(g) +5O_(2)(g) rarr 3CO_(2)(g) + 4H_(2)O(l) Delta_(comb)H^(Theta) =- 530.0`
From definition of `DeltaH` of a reaction, `DeltaH = H_(P) - H_(R)`
`Delta_(comb)H^(Theta) =[3Delta_(f)H^(Theta)(CO_(2))+4Delta_(f)H^(Theta)(H_(2)O)] -[Delta_(f)H^(Theta)(C_(3)H_(8)) -Delta_(f)H^(Theta)(O_(2))]`
`rArr -530 = 3 xx (-94) +4 xx (-68) - Delta_(f)H^(Theta) (C_(3)H_(8))`
`rArr Delta_(f)H^(Theta) (C_(3)H_(8)) =- 24 kcal`
Calculation of bond energies:
Let the enegry of `C -C` bond `=x` kcal
and the bond enegry of `C -H` bond `=y kcal`
For ethane, heat of formation is given as:
`2C(g)+3H_(2)(g) rarr C_(2)H_(6), Delta_(f)H = - 20`
Bond breaking `(DeltaH_(1))`:
`2[C(s) rarr C(g), +172]`
and `3[H_(2)(g) rarr 2H(g), +104]`
`rArr DeltaH_(1) = 2 xx 172 +3 xx 104 = 654`
Bond formation `(DeltaH_(2))`:
`[C +C rarr C-C, -x]`
and `6[C+H rarr C-H, -y]`
`rArr DeltaH_(2) =- (x +6y)`
`rArr Delta_(f)H = H_(1) +H_(2)`
`rArr -20 = 654 -(x+6y)`
`rArr (x+6y) = 674 ....(i)`
For propane: heat of formation is given as
`3C(s) +4H_(2)(g) rarr C_(3)H_(8), Delta_(f)H =- 24`
Bond breaking `(H_(1)):`
`2[C +C rarr C-C, -x]`
and `8[C +H rarr C-H, -y]`
`rArr DeltaH_(2) =- (2x +8y)`
`rArr Delta_(f)H = DeltaH_(1)+ DeltaH_(2)`
`rArr -24 = 932 -(2x +8y)`
`rArr x +4y = 478....(ii)`
Solving equations (i) and (ii), we get `x = 86` and `y = 98` Therefore, Bond energy of `C -C` bond `-86 kcal` and `C -H` bond `=98 kcal`.