How does thermochemistry treat the making and breaking of chemical bonds?

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How does thermochemistry treat the making and breaking of chemical bonds?

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Answer 1 · 2015-09-26T19:25:54

Bond breaking requires energy; bond making releases energy. See if you can remember that mantra.

Explanation:

Whether a reaction is exothermic or endothermic depends of course on the balance between bond-breaking, and bond-making. When we look at combustion reactions, a typical reaction is the combustion of hydrocarbon (the which drives our industrial society). For combustion of hexane, we could write,

$C_{6} H_{14} + \frac{19}{2} O_{2} \to 6 C O_{2} + 7 H_{2} O$ $C_6H_14 + 19/2O_2 rarr 6CO_2 + 7H_2O$

The reaction requires the breaking of bonds: viz. the breaking of strong $C - H$ $C-H$ , and $O = O$ $O=O$ bonds, but it also involves the formation of stronger $C = O$ $C=O$ and $O - H$ $O-H$ bonds, in the form of energetically stable $C O_{2}$ $CO_2$ and $H_{2} O$ $H_2O$ molecules. If you look up the bond enthalpies (and you should!), you will find that bond formation is indeed more energetic than bond cleavage. Of course it is, otherwise the reaction would not take place.

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How does thermochemistry treat the making and breaking of chemical bonds?

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