Question

Calculate the resonance energy of $N_{2}O$ from the following data; ${\text{Δ}}_{\text{f}}\text{H}$ of $N_{2}O$ is 82 KJ $mo{l}^{-1}N\equiv N,$ 946 KJ $mo{l}^{-1}$ , $N=N$ , 418 KJ $mo{l}^{-1}$ , 0 = 0, 498 KJ $mo{l}^{-1}$ , $N=0,607$ KJ $mo{l}^{-1}$ .

• A. $+8.85\text{kJ}{\text{mol}}^{-1}$
• B. $-88\text{kJ}{\text{mol}}^{-1}$
• C. $-8.8\text{kJ}{\text{mol}}^{-1}$
• D. $+88\text{kJ}{\text{mol}}^{-1}$

Answer 1

Answer: (B)

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