# Why does the energy required to break a bond decrease as the number of bonds increase?

I noticed that the energy to break a single C bond is $348\ \mathrm{kJ}$, to break a double C bond would be $612\ \mathrm{kJ}$, and to break a triple C bond would be $837\ \mathrm{kJ}$. How come the energy required per bond breakage decreases as the number of bonds decrease? It seems that for double bonds, the energy required to break a bond is $(612/2) = 306\ \mathrm{kJ}$. Why is this so?

Thank you.

• It guess its obvious why it takes more energy to break a double bond that a single one, ( more electrons in a double bond for example) but are you asking why the energy to break a double bond is not twice that of a single? Sep 26, 2016 at 7:43
• I thought the bond-breaking enthalpy is different for different compounds...
– DHMO
Sep 26, 2016 at 10:44
• You are making the false assumption that a double bond is composed of two single bonds, when in fact it is not so. The two bonds in a double bond are not identical. In simplistic terms a double bond is composed of a sigma bond and a pi bond. Sigma bonds are the strongest type of covalent bond and therefore require more energy to break than a pi bond. This is why a carbon-carbon double bond doesn't require twice the energy to break as a carbon-carbon single bond. Other factors come into play, but thats the answer in simplistic terms. Sep 26, 2016 at 16:21
• There is no simple answer other than "because chemistry and stuff". The bonds are different; one is $\sigma$, another is $\pi$. To make everything worse, think of nitrogen, which is the other way around: triple bond $\ce{N\equiv N}$ is stronger than three ordinary bonds $\ce{N-N}$. Oct 19, 2016 at 7:36
• @Keaton I'd like to encourage you to transform your comment into an answer. While it is indeed a very simple explanation and Ivan is also correct, it does provide a good starting point for the OP (and you might gain some reputation). Jan 17, 2017 at 12:27