Question

Calculate the resonance energy of $N_{2}O$ from the following data:

$\Delta H_f^{\circ }$ of $N_{2}O=82kJmol{e}^{-1}$

Bond energies of $N\equiv N$, $N=N$, $O=O$ and $N=O$ bonds are 946, 418, 498 and 607 kJ $mol{e}^{-1}$ respectively.

Answer 1

$N_2+\frac{1}{2}{O}_2\to N_{2}O(N=N=O)$

$\Delta H_{W.O.R}=(B.E_{N\equiv N}+\frac{1}{2}B.E_{O=O})-(B.E_{N=N}+B.E_{N=O})$

$\Delta H_{W.O.R}=(946+\frac{1}{2}\times 498)-(418+607)=1195-1025=170$ kJ/mole

also

$\Delta H_{W.R}=\Delta H_{W.O.R}-R.E_{Reactant}+R.E_{product}$

$82=170-0+R.E_{product}$

$R.E_{product}=82-170=-88kJ$

$R.E_{N_{2}O}=82-170=-88kJ$