Question

If the resonance energy of $N{O}_2(:O--N==O:)$ is X kJ The measured enthalpy formation of $N{O}_2\left({\Delta }_f{H}^{\ominus }\right)$ is $34kJmo{l}^{-1}$. The bond energies given are:
$N--O\Rightarrow 222kJmo{l}^{-1}$
$N\equiv N\Rightarrow 946kJmo{l}^{-1}$
$O==O\Rightarrow 498kJmo{l}^{-1}$
$N==O\Rightarrow 607kJmo{l}^{-1}$
Find out the value of X

• A. $X=-108$
• B. $X=+108$
• C. $X=-59$
• D. None of these

Answer 1

The correct option is A $X=-108$

Sol. $R.E.={\Delta }_{f}H\left(observed\right)-{\Delta }_{f}H\left(calculated\right)$
The balanced chemical reaction for the formation of nitrogen dioxide is as shown.
$\frac{1}{2}{N}_2\left(g\right)+O_2\left(g\right)⟶N{O}_2\left(g\right)$ ${\Delta }_f{H}_{cal}=\text{(BE of reactants - BE of products)}$ $=\left(\begin{array}{c}\frac{1}{2}\text{BE of}{N}_2+\text{BE of}{O}_2\end{array}\right)-\left(\begin{array}{c}\frac{1}{2}\text{BE of}N==O+\text{BE of}N--O\end{array}\right)$

$=\left(\begin{array}{c}\frac{1}{2}\times 946+498\end{array}\right)-(607+222)$

$=971-829$

$=142kJ$ $\Delta H_{\left(observed\right)}$

$=34kJ$

$\therefore$ Resonacne .Energy $=34-142$

$=-108kJ$
Hence, the value of $X$ is $-108$.