Calculate the enthalpy of formation of ammonia from the following bond energy data:
`(N-H) bond = 389 kJ mol^(-1), (H-H) bond = 435 kJ mol^(-1)`, and `(N-=N)bond = 945.36kJ mol^(-1)`.

Explanation

To compute resonance enegry, we compare the calaculated value of `Delta_(f)H^(Theta)` (benzene, g) with the given one. To calculate `Delta_(f)H^(Theta)` (benzene, g) we add the following reactions.
(a). `6C(g) +6H(g) rarr C_(6)H_(6)(g), DeltaH^(Theta) =- (3BE_(C-C) +3BE_(C=C) +6BE_(C-H))`
(b) `6C ("graphite") rarr 6C(g) DeltaH^(Theta) = 6 xx 718.4 kJ mol^(-1)`
(c) `3H_(2)(g) rarr 6H(g) DeltaH^(Theta) = 3 xx 435.9 kJ mol^(-1)`
`ulbar(6C("graphite")+3H_(2)(g)rarr C_(6)H_(6)(g))`
The corresponding enthalpy change is
`Delta_(f)H^(Theta) =- (3BE_(C-C)+ 3BE_(C=C) +6BE_(C-H)) +[6 xx 718.4 +3 xx 435.9] kJ mol^(-1)`
`=[-(3xx331.4 xx 591.1 +6 xx 718.4 +3 xx 435.9] kJ mol^(-1)`
The given `Delta_(f)H^(Theta)` is `Delta_(f)H^(Theta)` (benzene, g) `= 82.9 kJ mol^(-1)`
This means benzene become more stable by `(352.8 -82.9) kJ mol^(-1)`, i.e., `269.7 kJ mol^(-1)`
This is its resonance enegry.