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

For `N_(2)O,N_(2)+(1)/(2)O_(2) rarr N_(2)O, DeltaH_(exp)=82kJ`
Also from bond energy data, using `N=N=O` structure for `N_(2)O`
`DeltaH_(f cal. ) =-[1(N=N)+1(N=O)]+[(N-=N)+(1)/(2)(O=O)]`
`-[418+607]+[946+(1)/(2)(498)]=+170kJ`
Now resonance energy
`=Exp .DeltaH_(f N_(2)O)-`Calcu lated `DeltaH_(f N_(2)O)`
`=82-170=-88kJ mol^(-1)`