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.

Options

A. `-88KJmol^(-1)`
B. `-44KJmol^(-1)`
C. `-22KJmol^(-1)`
D. None of these

Answer

A

Explanation

`N_(2)(g)+(1)/(2)O_(2)(g)rarrN_(2)O`
`N-=N+(1)/(0)0=0rarrN=N=0`
Calculated
`DeltaH_(f)^(0)(N_(2)O)=[B.E._((N-=N))+(1)/(2)B.E.(0=0)]`
`-[B.E._(N=N))+B.E._(N=0)]`
`=[946+(498)/(2)]-[418+607]=+170KJ//mol`
Resonance energy=observed `DeltaH_(f)^(0)` -calculated
`DeltaH^(0)_(f)=82-170=-88KJmol^(-1)`