Chemistry - Thermochemistry Question with Solution | TestHub

Reaction${\text{N}}_{\text{2}}\text{+}\frac{\text{1}}{\text{2}}{\text{O}}_{\text{2}}\to {\text{N}}_{\text{2}}\text{O;}\text{ΔH}\text{=}\text{82}\text{kJ}{\text{mol}}^{-\text{1}}$
The structure of${\mathrm{N}}_2\mathrm{O}$is$\mathrm{N}=\mathrm{N}=\mathrm{O}$
${\text{Δ}}_{\text{f}}\text{H}$(calculated) = Bond energy of reactant – Bond energy of product
$=\left[(\mathrm{N}\equiv \mathrm{N})+\frac{1}{2}(\mathrm{O}=\mathrm{O})\right]-\left[\right(\mathrm{N}=\mathrm{N})+(\mathrm{N}=\mathrm{O}\left)\right]$
$\text{=}\left[\text{946}\text{+}\frac{\text{1}}{\text{2}}\text{498}\right]-\left[\text{418}\text{+}\text{607}\right]\text{=}\text{170}\text{kJ}$
Resonance energy = experimental${\text{Δ}}_{\text{f}}\text{H}\left({\text{N}}_{\text{2}}\text{O}\right)-$calculated${\text{Δ}}_{\text{f}}\text{H}\left({\text{N}}_{\text{2}}\text{O}\right)$
$\text{=}\text{82}-\text{170}\text{=}-\text{88}\text{kJ}{\text{mol}}^{-\text{1}}$.