Question

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.

Answer 1

`N-= N(g) +(1)/(2)(O=O) rarr N=N =O (g)`
`Delta_(f)H_((N_(2)O))^(Theta)=[DeltaH_((N^-=N))^(Theta) +(1)/(2)DeltaH_((O=O))^(Theta)]`
`-[DeltaH_((N=O))^(Theta)+DeltaH_((N=N))^(Theta)]`
`=(946 +(1)/(2)xx498) - (607+418)`
`=946 +249 - 607 - 418`
`= 1195 -1025 = 170 kJ mol^(-1)`
Resonance energy = Observed heat of formation -Calculated heat of formation
`=82 - 170 =- 88 kJ mol^(-1)`