# Why does bond energy not vary with position?

Consider the formation of a hydrogen bond: $$\ce{H + H -> H2}$$ The $$\Delta H$$ is $$\pu{-436 kJ mol-1}$$, so we can equally write the equation as: $$\ce{H + H -> H2 + \pu{436 kJ}}$$

But depending on the differences in position between the hydrogen atoms, there will be a different force of attraction between each other by Coulomb's law. So the potential energy will be larger the farther away the hydrogen atoms are from each other; thus, more energy would need to be released to form a covalent bond. So how does it always release $$\pu{436 kJ}$$ of energy when forming an $$\ce{H2}$$ bond if the energy must vary depending on initial position?

• Of course while the strict definition of the bond energy is based on infinite separation, the shape of the potential energy surface gets very flat very quickly. So the difference in energy between say a separation of 10 bond lengths and 5 bond lengths as a starting point might be so small it is inside the error bars of the value. Sep 14, 2023 at 16:08

It's a good point, and precisely because of this, bond dissociation energies (in fact, any kind of reaction energy or enthalpy) are defined such that different reactants are infinitely separated from each other. This includes, for example, the two hydrogen atoms on the left-hand side of

$$\ce{H + H -> H2}.$$

This means that there are no Coulombic forces between the hydrogen atoms (or, to be precise, no forces between the constituent particles of different hydrogen atoms).

You may have seen a diagram such as the one in this answer by porphyrin: https://chemistry.stackexchange.com/a/57157/16683

The infinite separation corresponds to the far right end of the diagram (where the energy $$E = 0$$), and the bond corresponds to the bottom of the well. That is why the bond dissociation energy is equal to the depth of the well.*

In general, if you have something like

$$\ce{A + B -> C + D}$$

what this means is that $$\ce{A}$$ and $$\ce{B}$$ are at infinite separation, and then you bring them together, they react to form $$\ce{C}$$ and $$\ce{D}$$, and you infinitely separate the products again.

* (Neglecting zero-point energy.)