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.

  • $\begingroup$ 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? $\endgroup$ – porphyrin Sep 26 '16 at 7:43
  • $\begingroup$ I thought the bond-breaking enthalpy is different for different compounds... $\endgroup$ – DHMO Sep 26 '16 at 10:44
  • 7
    $\begingroup$ 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. $\endgroup$ – KeatonB Sep 26 '16 at 16:21
  • 3
    $\begingroup$ 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 $\ce{N-N}$. $\endgroup$ – Ivan Neretin Oct 19 '16 at 7:36
  • 4
    $\begingroup$ @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). $\endgroup$ – Martin - マーチン Jan 17 '17 at 12:27

You are considering all the bonds equivalent and that's where you are making mistake.The energy required to break second bond will depend upon several factors such as repulsion, configuration, also note that there are different types of bond formed (sigma and pi bonds) and we know that energy required to break a sigma bond is greater than that of pi bond.

| improve this answer | |
  • $\begingroup$ I would be interested to know why a sigma bond takes more energy to break than a pi bond. $\endgroup$ – AMT Feb 16 '17 at 13:05
  • 3
    $\begingroup$ The reason to that is that a sigma bond is formed by the head on overlap of orbitals,whereas a pi bond is formed by sideways overlap,resulting in a less stronger bond $\endgroup$ – SubZero Mar 18 '17 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.