Question

Calcualate the resonance energy of `N_(2)O` from the following data `:`
`Delta_(f)H^(@)` of `N_(2)O= 82kJ mol^(-1)` , Bond energies of `N -= N , N = N , O=O` and `N=O` bonds are 946, 418, 498 and `607 kJ mol^(-1)` respectively.

Answer 1

The equation for the formation of one mole of `N_(2)O` will be
`N -= N(g) + (1)/(2)O= O(g) rarr N = N = O`
Calculate value of `Delta_(f)H^(@)` for this reaction will be
`Delta_(f)H^(@) = `[ B.E.`(N-=N) + (1)/(2) `B.E. ( `O=O )] - [B.E(N=N) + B.E. ( N=O)]`
`= [ 946 +(1)/(2) ( 498 ) ] - [ 418 + 607] kJ mol^(-1) = 170 kJ mol^(-1)`
Resonance energy `=` Observed `Delta_(f)H^(@)` - Calculated `Delta_(f)H^(@) = 82- 170 = -88kJ mol^(-1)`