Double bond formation in alkenes during cracking

In thermal or catalytic cracking, energy is used to break a long chain alkane into smaller alkanes/alkenes. Since the energy goes into *breaking* the bonds, how is the C=C bond in alkenes formed?




I don't know if I have understood well your question.
During the process of cracking, homolysis of long chains occurs. Radical intermediates result from breaking the bonds. These intermediates are highly unstable and undergo rearrangements. As a consequence of the rearrangements, smaller alkanes and alkenes are formed. We have to point out that, due the the high reactivity of radicals, the control of radical reactions is very difficult and we can not readily predict what happens inside the chemical reactor. 
I hope I've answered your question.


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To add a small point to [Yomen's excellent answer](https://chemistry.stackexchange.com/a/38150/13184):


Alkyl radicals can be turned into alkenes through loss of a hydrogen radical. For example suppose during ethane cracking, an ethyl radical is produced. It can lose a hydrogen radical to form ethylene.


$$\ce{^{\cdot}CH2-CH3 <=> CH2=CH2 + H^{\cdot}}$$


Alkyl radicals and H radicals are both relatively unstable, so this reaction is an equilibrium. Many other reactions that produce, interconvert, and consume radicals also happen during cracking. Ultimately the driving force is the high temperature, which favors products which more entropy. The entropy of many small molecules is much higher than a few larger molecules because of the additional translational degrees of freedom of the small molecules. That is what favors production of smaller molecules such as ethylene and hydrogen, as compared to larger molecules such as ethane.


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