# Why is the combustion of methanol an exothermic reaction?

Mar 2, 2017

The breaking of bonds REQUIRES energy; the making of bonds RELEASES energy.

#### Explanation:

Now we represent the combustion of methanol as:

$C {H}_{3} O H \left(l\right) + \frac{3}{2} {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right)$

Now, clearly, we have to break $C - H$, $O - H$, and $O = O$ bonds. But the reaction necessarily makes $C = O$ and $O - H$ bonds. The difference between bond making and bond breaking is the energy associated with the reaction, and, in this instance, the energy released on bond-making is greater than energy required for bond-breaking, and thus the reaction is EXOTHERMIC.

Certainly, we can look up standard tables of bond enthalpies (which can be derived experimentally, and also theoretically from first principles), and certainly, the value we get from these tables are identical to the actual values, i.e. the values obtained by experiment, i.e.

$\sum \text{bonds broken"-sum"bonds formed"=DeltaH_"rxn}$