I learned that all reactions that yield hydrogen are endothermic (such has reforming) and reactions that use up hydrogen are exothermic (FCC cracking, hydrogenation, etc.). But why is this so? I know what exothermicity/endothermicity means, but I haven't found a specific explanation for the correlation with hydrogen generation or consumption yet.

  • 2
    $\begingroup$ In hydrogenation entropy is lowered so heat must be generated - otherwise change in enthalpy also wouldn't be favorable and reaction wouln't happen $\endgroup$
    – Mithoron
    Commented Jun 3, 2016 at 18:43
  • $\begingroup$ @Mithoron I like this line of reasoning. $\endgroup$
    – Zhe
    Commented May 3, 2017 at 15:55

2 Answers 2


Let's take a look at an example of a hydrogenation reaction, namely the hydrogenation of ethene to give ethane

$$\ce{CH2\bond{=}CH2\ (g) + H2 (g)-> CH3\bond{-}CH3 (g)}$$

And now let's determine the enthalpy change of this reaction. By Hess's Law, the enthalpy change of the reaction is essentially the same as the enthalpy change of breaking all the bonds of the reactants and forming all the bonds in the molecules of the product. We can see that

  • 1 $\ce{H\bond{-}H}$ bond has been broken
  • 1 $\ce{C\bond{=}C}$ bond has been broken
  • 1 $\ce{C\bond{-}C}$ bond has been formed
  • 2 $\ce{C\bond{-}H}$ bonds have been formed

The other bonds remain unchanged by the end of the reaction

The following is the list of bond enthalpies of the aforementioned bonds, obtained from the Internet

  • $\ce{H\bond{-}H}$ bond: +432 kJ/mol
  • $\ce{C\bond{=}C}$ bond: +614 kJ/mol
  • $\ce{C\bond{-}C}$ bond: +347 kJ/mol
  • $\ce{C\bond{-}H}$ bond: +413 kJ/mol

Breaking bonds gives a positive enthalpy change since it is an endothermic process, and vice versa for bond formation. We can now estimate the enthalpy change of the reaction

$\Delta$$H$ = 2(-413) + (-347) + 614 + 432 = -127 kJ/mol

As indicated by the negative value of the enthalpy change, the reaction is exothermic.

In fact, the enthalpy changes of hydrogenation reactions of other alkenes are also similar (though more substituted alkenes may have a lower magnitude of heat of hydrogenation, but still the reaction is exothermic). You may notice that for other alkenes, the way you calculate the enthalpy change is the same since they involve the same bond breaking and bond forming processes aforementioned (any substituents bonded to the carbons with the $\ce{C\bond{=}C}$ bond remain unaffected throughout the reaction and will still be present in the product being bonded to the carbon).

To simply put, the energy released from bond forming is larger than the energy required (absorbed) for bond breaking, and that is why you get net energy released from the reaction.


Only one bond is broken, as in H-H bond in hydrogen gas. But two bonds are made, that are C-H bonds. One extra bond is made. As we know bond making is exothermic, so more energy due to that one bond made will be released than absorbed.

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    $\begingroup$ You're breaking a C-C pi bond as well. $\endgroup$ Commented May 3, 2017 at 14:16

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