Calculate the resonance enegry of `NO_(2) ( :O-N=O: )`
The measured enthalpy formation of `NO_(2)(Delta_(f)H^(Theta))` is `34 kJ mol^(-1)`. The bond energies given are:
`N -O rArr 222 kJ mol^(-1)`
`N -= N rArr 946 kJ mol^(-1)`
`O = O rArr 498 kJ mol^(-1)`
`N = O rArr 607 kJ mol^(-1)`

Explanation

`R.E. =Delta_(f)H("observed") -Delta_(f)H ("calculated")`
`(1)/(2)N_(2)(g) +O_(2)(g) rarr NO_(2)(g)`
`Delta_(f)H_(cal) = (BE "of Reactant" - BE "of Products")`
`= ((1)/(2) BE of N_(2) +BE of O_(2))`
` -((1)/(2)BE of N=O +BE of N-O)`
`= ((1)/(2) xx 946 +498) - (607 +222)`
`=971 - 829 = 142 kJ`
`DeltaH_(("observed")) = 34 kJ`
`:. R.E. = 34 - 142 =- 108 kJ`