I understand that forming a bond between $\ce{A}$ and $\ce{B}$ is an exothermic process, with negative enthalpy, but won't there be energy required to get $\ce{A}$ and $\ce{B}$ close enough together for the bond to form? So we would get an energy hill with activation energy on one side and heat of formation on the other side.

Is it ever possible in a chemical reaction to have more energy required to get the reactants $\ce{A}$ and $\ce{B}$ close enough together than the energy released in forming the bond? In terms of atoms I am thinking that the repulsion between two atoms may reach a peak at some distance greater than the bond length, so we might get an energy hill diagram where:

energy hill

  1. What about a situation like breaking and forming bonds on elemental crystals like graphite and diamond? If graphite pieces $\ce{(A)}$ are sloughed off of a large piece of graphite $\ce{(B)}$ this obviously requires energy to break the bonds, but the graphite flakes don't spontaneously re-bond to the large piece upon light contact.
    Likewise, a fractured diamond does not re-bond on contact. Also, in formation of a diamond, why is there a need for a continual input of heat? If the energy of formation is greater than the energy needed to activate the reaction, then wouldn't the released heat of formation cause other diamond bonds to form without more input of energy. This indicates to me that some energy is required to begin the process of combining $\ce{A + B}$ to form $\ce{AB}$, even if the actual bond formation is exothermic - that there is a two sided energy hill in reactions of $\ce{A + B -> AB}$, though the activation portion may be smaller than the formation portion.

  2. What about an analogous $\ce{A + B -> AB}$ nuclear fusion reaction with heavy species above iron? Clearly in this case there is a large input of energy needed to fuse elements above iron, and a small release as the nucleons reach a distance that allows the strong nuclear force to dominate. In this case the energy hill has a greater required activation energy than an energy of formation. How is this fundamentally different from chemical bonds?
    I understand that a critical component is that nuclear attraction becomes stronger than electrostatic repulsion at a critical distance, but might there be some instances of chemical reactions where $\ce{A}$ and $\ce{B}$ have to pass through a less stable conformation to get to the more stable bonded state?

So there are three possible answers:

  1. $\ce{A + B -> AB}$ reactions can be described energetically solely by the heat of formation of the bonds between $\ce{AB}$.

  2. $\ce{A + B -> AB}$ can be energetically described by an energy hill, but the heat of formation will always exceed the activation energy, or the energy required to move $\ce{A}$ and $\ce{B}$ through any intermediate states that are less stable than $\ce{A + B}$ (far apart).

  3. $\ce{A + B -> AB}$ can sometimes be exothermic where the energy required to move $\ce{A}$ and $\ce{B}$ into an arrangement that will allow bonding can exceed the heat of formation of the bond between them.

  • $\begingroup$ If anyone can explain how I can take an image that I draw and save on Google docs and turn it into an image that can be directly formatted into a question, I would appreciate an explanation. $\endgroup$ Dec 7, 2016 at 2:51
  • $\begingroup$ Easiest way I know is to take a screenshot, paste into Paint (or whatever image editor), crop, and save; then insert wherever into the post. $\endgroup$
    – hBy2Py
    Dec 7, 2016 at 3:39
  • $\begingroup$ Well, I draw them in paint. How do I insert them into the post? I tried to copy and paste. $\endgroup$ Dec 7, 2016 at 4:11
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  • $\begingroup$ $\ce{He + He -> He2}$ would be endothermic, I guess. That's why it just doesn't occur. $\endgroup$ Dec 7, 2016 at 7:20

2 Answers 2


A bond does not simply form because two atoms are close enough. You also have to bring them into the right electronic state to do so. Which takes various amounts of energy. So yes, an A+B -> AB reaction can be endothermic.

Also beware of mistaking the abscissa in those energy diagrams as a distance: It's usually just a symbolic "reaction progress", or it is a line on a multidimensional hypersurface describing the energy potential with respect to all angles, distances, electronic parameters etc. in the reaction.

  • $\begingroup$ I know it's not a distance, though distances could be part of it given variable strength of static electric field. $\endgroup$ Dec 7, 2016 at 6:13

Yes it is possible for a $\ce{A + B -> AB}$ reaction to be endothermic but it will generally not occur.

Whether the reaction proceeds spontaneously depends on the change in Gibbs free energy $\Delta G = \Delta H - T\Delta S$. ${\Delta H}$ is the enthalpy (negative for exothermic reactions), ${T}$ the temperature and ${\Delta S}$ the change in entropy (a measure of disorder in the system).

If ${\Delta G \le 0}$ then the reaction will proceed.

When you combine two molecules, such as $\ce{A + B -> AB}$ you are forcing order on the system (molecules must now stay close to each other), so the entropy goes down. So in an endothermic combination reaction you have ${\Delta H > 0}$ (endothermic) and ${-\Delta S > 0}$ (decreasing disorder). This means that ${\Delta G > 0}$ and the reaction will not happen.

The exception is high pressures (e.g. heavy nuclear fusion) where a decrease in volume caused by the fusion can provide the necessary driving force (It's included in the ${H}$ term)

A decomposition reaction $\ce{AB -> A + B}$ however will increase the disorder (molecules may now move about freely), allowing endothermic reactions to proceed provided that ${\Delta H < T\Delta S}$.

  • $\begingroup$ While you are correct, as far as I tried to follow, you have also completely missed the point of the original question. ;-) When you think about formation of single molecules, thermodynamics is hardly an issue. $\endgroup$
    – Karl
    Dec 7, 2016 at 23:29

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