If the enthalpy of formation of N_2O is 82 , | vedclass

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(A) The enthalpy of formation of $N_2O$ (experimental) is $\Delta_f H = 82 \, kJ \, mol^{-1}$.The reaction is: $N_2 + \frac{1}{2} O_2 \longrightarrow

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(A) The enthalpy of formation of $N_2O$ (experimental) is $\Delta_f H = 82 \, kJ \, mol^{-1}$.The reaction is: $N_2 + \frac{1}{2} O_2 \longrightarrow N_2O$.Using bond energies: $(N \equiv N) + \frac{1}{2}(O = O) \longrightarrow (N = N) + (N = O)$.$\Delta_f H^{\circ} = \Sigma B.E._{	ext{reactants}} - \Sigma B.E._{	ext{products}}$.$\Delta_f H^{\circ} = [946 + \frac{1}{2}(498)] - [418 + 607] = [946 + 249] - [1025] = 1195 - 1025 = 170 \, kJ \, mol^{-1}$.Resonance energy = $\Delta_f H_{	ext{experimental}} - \Delta_f H_{	ext{calculated}}$.Resonance energy = $82 - 170 = -88 \, kJ \, mol^{-1}$.

If the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.
$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$
$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$

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If the enthalpy of formation of $N_2O$ is $82 \, kJ \, mol^{-1}$,calculate the resonance energy of $N_2O$ in $kJ \, mol^{-1}$.$N \equiv N \, (946 \, kJ \, mol^{-1}); \, N = N \, (418 \, kJ \, mol^{-1})$$O = O \, (498 \, kJ \, mol^{-1}); \, N = O \, (607 \, kJ \, mol^{-1})$