Question

N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol

Answer 1

`\large \boxed{\text{761 kJ}}`

Step-by-step 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

`\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 * 941 +1 * 495 - (2 * 201 + 2* 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.`