Given the bond dissociation energies: $ ext{N}\equiv ext{N}$ is $946 ext{ kJ/mol}$, $ ext{O=O}$

Question

Given the bond dissociation energies: $	ext{N}\equiv	ext{N}$ is $946 	ext{ kJ/mol}$, $	ext{O=O}$ is $498 	ext{ kJ/mol}$, and $	ext{N=O}$ is $631 	ext{ kJ/mol}$. The standard enthalpy change ($\Delta H^\circ$) for the gas-phase synthesis reaction $\mathrm{N}_{2(g)} + \mathrm{O}_{2(g)} ightarrow 2\mathrm{NO}_{(g)}$ is

Professor PC Thomas Classes · Answer

(a) : The standard enthalpy of reaction is calculated by subtracting the total energy released upon bond formation from the total energy consumed to break reactant bonds: $\Delta H^\circ = \sum 	ext{B.E.}(	ext{reactants}) - \sum 	ext{B.E.}(	ext{products})$. For this reaction: $\Delta H^\circ = [	ext{B.E.}(	ext{N}\equiv	ext{N}) + 	ext{B.E.}(	ext{O=O})] - [2 	imes 	ext{B.E.}(	ext{N=O})]$. Substituting the provided numerical values: $\Delta H^\circ = [946 + 498] - [2 	imes 631] = 1444 - 1262 = +182 	ext{ kJ}$.