Given that:(i) Delta_f H^o of N_2O is 82 kJ | vedclass

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(A) The reaction for the formation of $N_2O$ is: $N_{2(g)}   \frac{1}{2} O_{2(g)} 	o N_2O_{(g)}$.The theoretical enthalpy of formation $(\Delta H_{ca

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$-88$

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(A) The reaction for the formation of $N_2O$ is: $N_{2(g)}   \frac{1}{2} O_{2(g)} 	o N_2O_{(g)}$.The theoretical enthalpy of formation $(\Delta H_{calc})$ using bond energies is calculated as:$\Delta H_{calc} = [	ext{Energy required to break bonds}] - [	ext{Energy released in forming bonds}]$$\Delta H_{calc} = (BE_{N \equiv N}   \frac{1}{2} BE_{O = O}) - (BE_{N = N}   BE_{N = O})$$\Delta H_{calc} = (946   \frac{1}{2} 	imes 498) - (418   607) = (946   249) - 1025 = 1195 - 1025 = 170 \ kJ \ mol^{-1}$.Resonance energy is defined as the difference between the experimental enthalpy of formation and the theoretical enthalpy of formation:$	ext{Resonance Energy} = \Delta H_{calc} - \Delta H_{exp} = 170 - 82 = 88 \ kJ \ mol^{-1}$.Note: The resonance energy is typically expressed as a negative value representing stability,so the answer is $-88 \ kJ \ mol^{-1}$.

Given that:
$(i) \Delta_f H^o$ of $N_2O$ is $82 \ kJ \ mol^{-1}$
$(ii)$ Bond energies of $N \equiv N, N = N, O = O$ and $N = O$ are $946, 418, 498$ and $607 \ kJ \ mol^{-1}$ respectively.
The resonance energy of $N_2O$ is $...... \ kJ \ mol^{-1}$.

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Given that:$(i) \Delta_f H^o$ of $N_2O$ is $82 \ kJ \ mol^{-1}$$(ii)$ Bond energies of $N \equiv N, N = N, O = O$ and $N = O$ are $946, 418, 498$ and $607 \ kJ \ mol^{-1}$ respectively.The resonance energy of $N_2O$ is $...... \ kJ \ mol^{-1}$.