# Question #a896f

Sep 25, 2015

The energy stored in the bonds of ${H}_{2}$ and ${O}_{2}$ molecules is higher than it is in ${H}_{2} O$ molecule.

#### Explanation:

Exothermic reactions are characterized by their negative enthalpy $\Delta H$.

$\Delta H$ can be calculated using the following expression:
$\Delta H = \sum n \times {D}_{\text{bonds broken")-sumnxxD_("bonds formed}}$

where ${D}_{\text{bonds broken}}$ is the bond energy of broken bonds and ${D}_{\text{bonds formed}}$ is the bond energy of formed bonds.

For the following reaction:
${H}_{2} \left(g\right) + \frac{1}{2} {O}_{2} \left(g\right) \to {H}_{2} O \left(g\right)$

$\Delta H = \left({D}_{H - H} + \frac{1}{2} {D}_{O = O}\right) - 2 {D}_{O - H}$

$\Delta H = \left(1 \cancel{m o l} \times 432 \frac{k J}{\cancel{m o l}} + \frac{1}{2} \cancel{m o l} \times 498 \frac{k J}{\cancel{m o l}}\right) - \left(2 \cancel{m o l} \times 467 \frac{k J}{\cancel{m o l}}\right) = - 253 k J$

The calculated $\Delta H = - 253 k J$ is negative and therefore the reaction is exothermic.

In words, we can state that the energy stored in the bonds of ${H}_{2}$ and ${O}_{2}$ molecules is higher than it is in ${H}_{2} O$ molecule.