Question

Calculate the resonance energy of $N_{2}O$ from the following data: $\Delta H_f$ of $N_{2}O$ is $82\text{kJ/mol}$
$N\equiv N946\text{kJ/mol}$
$N=N418\text{kJ/mol}$
$O=O498\text{kJ/mol}$
$N=O607\text{kJ/mol}$

• A. $-88\text{kJ/mol}$
• B. $88\text{kJ/mol}$
• C. $-252\text{kJ/mol}$
• D. $252\text{kJ/mol}$

Answer 1

The correct option is A $-88\text{kJ/mol}$

The required equation is
$N_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to N_{2}O\left(g\right),\Delta H_f=82\text{kJ/mol}$
The structure of $N_{2}O$ is $\stackrel{-}{N}=\stackrel{+}{N}=O$
Now, we know that
$\Delta H_f\text{(calculated)}=$ Bond energy of reactant - Bond energy of product
$=\left[\right(N\equiv N)+\frac{1}{2}(O=O\left)\right]-\left[\right(N=N)+(N=O\left)\right]$
or, $\Delta H_f=[946+\frac{498}{2}]-[418+607]=+170\text{kJ/mol}$
$\therefore$ Resonance energy = Experimental $\Delta H_f\left(N_{2}O\right)$ - Calculated $\Delta H_f\left(N_{2}O\right)$
$=82-170=-88\text{kJ/mol}$