We know $\ce{H2}$ is formed with covalent bonds (just as $\ce{O2}$ is formed via covalent bonding). Hence, these molecules gain some stability. In spite of this stability, why do they react with each other to form $\ce{H2O}$ or $\ce{H2O2}$?

  • 2
    $\begingroup$ Burning hydrogen in oxygen is a hugely exothermic reaction. Hence, they gain even more stability by combusting. $\endgroup$
    – Jon Custer
    May 15, 2017 at 19:58
  • $\begingroup$ That is, they gain even more stability by breaking those bonds and forming other bonds instead. $\endgroup$ May 15, 2017 at 20:04

1 Answer 1


Indeed, much of the chemistry that takes place in the universe involves the breaking of covalent bonds and the formation of new covalent bonds. The key is that the energy of the products needs to be in a lower state than that of the reactants. This is the thermodynamic requirement anyway. Lets consider one of your examples:

$$\ce{2H2(g) + O2(g) -> 2H2O(g) + \pu{572kJ}}$$

Although we had to break the covalent bonds of hydrogen and oxygen, the new bonds that were formed are even stronger, thus the reaction is thermodynamically favored.

The complicating point of the above example is the fact that there is a large kinetic barrier to the reaction. This basically means that there is an intermediate product in the steps of this reaction that is energetically very high (unfavorable), so some catalyst is required (say, a spark) for this reaction to get over that energetic barrier and proceed on any reasonable time scale.

Summary, TL;DR:
The breaking and creating of covalent bonds plays a major role in chemistry. The criteria for covalently bound compounds to participate in a chemical reaction is that the products have a lower energy state than than the reactants. If this is not the case, the reaction will not happen. Even if it is the case (that the products are lower energy / more stable than the reactants), there may be kinetic limitations requiring a catalyst in order for the reaction to proceed. But, no catalyst can make a reaction proceed if the above criteria is not met, that the products be more stable than the reactants.

  • $\begingroup$ You suggest in your summary paragraph that endothermic reactions do not happen, 'if this is not the case, the reaction will not happen'. $\endgroup$
    – porphyrin
    May 15, 2017 at 22:04
  • $\begingroup$ @porphyrin , thx for the input. What I'm shooting for is a statement referring to the Gibbs free energy of the reaction needing to decrease for the reaction to have a chance. I hope the way I've re-worded gets that across. $\endgroup$
    – airhuff
    May 15, 2017 at 22:16
  • $\begingroup$ I suppose it depends on what you mean by 'stable' in your last line. The point is that the Gibbs energy must be negative for a spontaneous reaction, rather than just the enthalpy, so entropy change is crucial also. The rate does not matter as far as the thermodynamics goes, and, as you write, the catalyst does not change the reactants or products but only the mechanism of how one gets from one to the other. $\endgroup$
    – porphyrin
    May 16, 2017 at 17:13

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