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

Sep 26, 2015

${C}_{6} {H}_{14} + \frac{19}{2} {O}_{2} \rightarrow 6 C {O}_{2} + 7 {H}_{2} O$
The reaction requires the breaking of bonds: viz. the breaking of strong $C - H$, and $O = O$ bonds, but it also involves the formation of stronger $C = O$ and $O - H$ bonds, in the form of energetically stable $C {O}_{2}$ and ${H}_{2} O$ 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.