# Question #538c9

Feb 2, 2017

$\text{Bond formation is exothermic} \ldots \ldots \ldots \ldots .$

#### Explanation:

$\text{And bond breaking (fission) is endotherimic.........}$

When we combust a hydrocarbon to get energy (to warm the house for instance), we have to break strong $C - H$, $C - C$, and $O = O$ bonds; such a process requires energy.

Upon complete combustion, we MAKE STRONGER $C = O$ and $O - H$ bonds in the combustion products, $C {O}_{2}$ and ${H}_{2} O$; such a process RELEASES energy, which we can use to heat our homes, or drive a motor. And the enthalpy or energy change of a reaction can be precisely measured:

$C {H}_{4} \left(g\right) + 2 {O}_{2} \left(g\right) \rightarrow C {O}_{2} \left(g\right) + 2 {H}_{2} O \left(l\right) + \Delta$

In the given reaction, we have to break STRONG $C - H$ and $O = O$ bonds. Such a process requires energy. But we make STRONGER $C = O$ and $O - H$ bonds. The balance, the energy difference is the enthalpy of reaction represented as $\Delta$ (and quoted as a NEGATIVE value if heat is released!).