# Sum of Bond Enthalpies in an Exothermic Reaction

The following reaction is exothermic.

In an exothermic reaction, heat is given out to the surroundings so the enthalpy of the reaction mixture should decrease.

Hence $\sum$ Reactant Bond Enthalpies $> \sum$ Product Bond Enthalpies

Yet I have just read that:

$\sum$ Reactant Bond Enthalpies $< \sum$ Product Bond Enthalpies

which does not make sense to me.

Am I right or wrong, please explain?

EDIT

I see where I made a mistake.

The products have a lower enthalpy so they are in a more stable state. Therefore more energy is required to break the bonds in the products so the sum of the bond enthalpies of the products is greater than that of the reactants.

I think this is right.

$\sum$ Reactant Bond Enthalpies $< \sum$ Product Bond Enthalpies
\begin{align} D(\ce{I-I})+D(\ce{H-H})&=(152.25\pm0.57)\ \mathrm{kJ\ mol^{-1}}+(435.7799\pm0.0001)\ \mathrm{kJ\ mol^{-1}}\\ &=(588.03\pm0.57)\ \mathrm{kJ\ mol^{-1}} \end{align}
\begin{align} 2\times D(\ce{H-I})&=2\times(298.26\pm0.10)\ \mathrm{kJ\ mol^{-1}}\\ &=(596.52\pm0.20)\ \mathrm{kJ\ mol^{-1}} \end{align}