The following reaction is exothermic.

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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?


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.


The statement

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

is correct, since bond energies or enthalpies are usually defined as energy or enthalpy required to break a bond. See for example the IUPAC Gold Book entries on bond energy and bond-dissociation energy.

Values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL. for the example that is given in the question are as follows.

$$\begin{align} D(\ce{I-I})+D(\ce{H-H})&=(152.25\pm0.57)\ \mathrm{kJ\ mol^{-1}}+(435.7799\pm0.0001)\ \mathrm{kJ\ mol^{-1}}\\ &=(588.03\pm0.57)\ \mathrm{kJ\ mol^{-1}} \end{align}$$

$$\begin{align} 2\times D(\ce{H-I})&=2\times(298.26\pm0.10)\ \mathrm{kJ\ mol^{-1}}\\ &=(596.52\pm0.20)\ \mathrm{kJ\ mol^{-1}} \end{align}$$

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