Answer:
[tex]\large \boxed{\text{761 kJ}}[/tex]
Explanation:
You calculate the energy required to break all the bonds in the reactants.
Then you subtract the energy needed to break all the bonds in the products.
N₂ + O₂ ⟶ 2NO
N≡N + O=O ⟶ 2O-N=O
Bonds: 2N≡N 1O=O 2N-O + 2N=O
D/kJ·mol⁻¹: 941 495 201 607
[tex]\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.[/tex]
