# If C-C and C-H bonds are strong, why are hydrocarbons used as fuels?

##### 1 Answer
Sep 10, 2017

Well, the $\text{BREAKING of BONDS REQUIRES ENERGY.....}$

#### Explanation:

$\text{And the making of bonds RELEASES ENERGY.}$

And the goto example is the combustion of hydrocarbons...and we could represent this by say the combustion of propane.....

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right)$

Given this reaction, CLEARLY, we have to break strong $H - C$, and $C - C$, and $O = O$ bonds.......but WE MAKE, even stronger $C = O$ and $H - O$ bonds, and we may quantitatively measure this energy transfer, i.e........

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + {\Delta}_{1}$

Here ${\Delta}_{1} \equiv - 2220 \cdot k J \cdot m o {l}^{-} 1$ (the negative sign means that energy is released to the surroundings, i.e. the reaction is exothermic); and this seems to agree with our conception of bond energies inasmuch as we may take values of bond enthalpy from standard tables, and we can calculate enthalpies that agree with the experimental values.

For butane, we could write.....

${C}_{4} {H}_{10} \left(g\right) + \frac{13}{2} {O}_{2} \left(g\right) \rightarrow 4 C {O}_{2} \left(g\right) + 5 {H}_{2} O \left(l\right) + {\Delta}_{2}$

$\text{A priori}$, without looking at experimental data, would you predict that ${\Delta}_{1} > {\Delta}_{2}$ or the converse? Why or why not?

And as to how we use this energy? Well, we use it to drive our cars, and to light and heat our homes, and cook our dinners....