Given that :
(i) $${\Delta _f}{H^ \circ }\,{\text{of}}\,{N_2}O\,\,{\text{is}}\,\,82\,kJ\,mo{l^{ - 1}}$$
(ii) Bond energies of $$N \equiv N,\,N = N,O = O$$      and $$N=O$$   are $$946,418,498$$    and $$607\,kJ\,mo{l^{ - 1}}$$   respectively,
The resonance energy of $${N_2}O$$  is :

Solution :
\[\begin{align} & {{N}_{2}}\left( g \right)+\frac{1}{2}{{O}_{2}}\to {{N}_{2}}O\left( g \right) \\ & N\equiv N\left( g \right)+\frac{1}{2}\left( O=O \right)\to \underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{N}}=\overset{+}{\mathop{N}}\,=\underset{\scriptscriptstyle\centerdot\centerdot}{\ddot{O}}\left( g \right) \\ \end{align}\]
$$\Delta {H_f}^ \circ = $$   [ Energy required for breaking of bonds ] – [ Energy released for forming of bonds ]
$$\eqalign{ & = (\Delta {H_{N \equiv N}} + \frac{1}{2}\Delta {H_{O = O}} - (\Delta {H_{N = N}} + \Delta {H_{N = O}}) \cr & = (946 + \frac{1}{2} \times 498) - (418 + 607) \cr & = 170\,kJ\,mo{l^{ - 1}} \cr & {\text{Resonance energy}} = 82 - 170 \cr & = - 88\,kJ\,mo{l^{ - 1}} \cr} $$