Answer:
[tex]\large \boxed{\text{-486 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.
2H₂ + O₂ ⟶ 2H-O-H
Bonds: 2H-H 1O=O 4H-O
D/kJ·mol⁻¹: 436 498 464
[tex]\begin{array}{rcl}\Delta H & = & \sum{mD_{\text{reactants}}} - \sum{nD_{\text{products}}}\\& = & 2 \times 436 +1 \times 498 - 4 \times 464\\&=& 1370 - 1856\\&=&\textbf{-486 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{-486 kJ}}$}.[/tex]
