Calculate the resonance enegry of `N_(2)O` form the following data
`Delta_(f)H^(Theta) of N_(2)O = 82 kJ mol^(-1)`
Bond enegry of `N-=N, N=N, O=O,` and `N=O` bond is `946, 418, 498`, and `607 kJ mol^(-1)`, respectively.

Explanation

`C_(2)H_(6)(g) +(7)/(2)O_(2) rarr 2CO_(2)(g) +3H_(2)O(l), DeltaH^(Theta) =- 372.0`
`Delta_(f)H_((C_(2)H_(6)))^(Theta) = 2 xx (-94.0) +3 xx (-68.0) + 372.0`
`=- 20 kcal`
`C_(3)H_(8)(g) +5O_(2) rarr 3CO_(2)(g) +4H_(2)O (l), DeltaH^(Theta) =- 530.0`
`Delta_(f)H_((C_(3)H_(8)))^(Theta)=2xx(-94.0) +4xx (-68.0) +530.0`
`=- 24 kcal`
`2C(s) +3H_(2)(g) rarr C_(2)H_(6)(g), DeltaH^(Theta) =- 20.0`
`2C(g)rarr 2C(s),DeltaH^(Theta) =- 344.0`
`4H(g) rarr 3H_(2)(g), DeltaH^(Theta) =- 312.0`
`"Adding"ulbar(2C(g)+6H(g)rarrC_(2)H_(6)(g),DeltaH=-676kcal)`
So, the enthalpy of formation of six `(C-H)` and one `(C-C)` bonds is `-676.0 kcal`.
`3C(s) +4H_(2)(g) rarr C_(3)H_(8)(g), DeltaH^(Theta)=- 24.0`
`3C(g) rarr 3C(s),DeltaH^(Theta) =- 516.0`
`8H (g)rarr 4H_(2)(g),DeltaH^(Theta) =- 416.0`
`"Adding"ulbar(3C(g)+8H(g)rarrC_(3)H_(8)(g),DeltaH=-956.0kcal)`
So, the enthalpy of formation of eight `(C-H)` and two `(C-C)` bond is `-965kcal`.
Let the bond enegry of `C-C` be `x` and of `C-H` be `y` kcal.
In ethane, `x+6y = 676`
In propane, `2 x +8y = 956`
On solving, `x = 82` and`y =99`
Thus, bond energy of `C-C =82` kcal and bond enegry of `C-H = 99 kcal`.