Q.
Given that:
(l)$\Delta_{f} H^{\circ} \, of \, N_{2} O \,is\, 82 KJ mol^{-1}$
(ii) Bond energies of $N = N, N=N, O=O$ and $N = O$ are $946,418,498$ and $607\, kJ\, mol^{-1}$respectively,
The resonance energy of $N_{2} O$ is :
Solution:
$N_{2} \left(g\right)+\frac{1}{2} O_{2} \to N_{2} O \left(g\right)$
$N=N \left(g\right)+\frac{1}{2} \left(O=O \to\right)$
$\Delta H_{f}^{0} =$ [Energy required for breaking of bonds]
-[Energy released for forming of bonds]
$= (\Delta H_{N=N} +\frac{1}{2} \Delta H_{o=o} - \left(\Delta H_{N=N}+\Delta H_{N=O}\right)$
$= \left(946+\frac{1}{2}\times498\right)-\left(418+607\right)=170 kJ mol^{-1}$
Resonance energy = 170-82 = 88 kJ $ mol^{-1}$
