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

Answer:

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

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

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