# Hooke's law for atom-to-atom energy well can't be right

Hooke's law says that the potential energy between two atoms is, $$V(x)$$, for two atoms with distance $$r_1-r_0$$ is $$\frac{1}{2}kx^2$$.

At room temperature and at equilibrium, water would have an equilibrium constant $$k$$ of $$1$$. The covalent bond of water is $$497 \;kJ / mol$$. One $$mol$$ of water is $$0.055\; gm$$. According to Hook's law, it would take only $$27\; kJ$$ of energy to compress $$1\;gm$$ of water into almost nothing.

Even in the case of a non-singularity, a volume in the neighborhood of a singularity would not require more energy on compression, and would essentially allow the neutrons to pass right next to each other, overcoming $$1\; mol$$ of strong nuclear force. The limit as a diameter of a neighborhood approached zero compressing $$0.05\; gm$$ of $$H_2O$$ could not possibly go from nominal to $$\frac{1}{2}kx^2$$. It would take $$27\; kJ$$ to break all the bonds in $$1\;gm$$ of water.

This can't be right, right? Note that Hooke's law is used in molecular dynamics.

Even if this doesn't account for O-O bonds, the same can be applied to a solid crystal diamond lattice with C-C bond strength of $$347$$ and similar proportions.

And if I'm totally wrong, does that really mean it would only take $$\frac{1}{2}kx^2$$ of energy to push a nucleus directly up against another nucleus?

• "At room temperature, in equilibrium, water would have an equilibrium constant K of 1." What? Why? Commented Apr 16, 2021 at 2:50
• my chemistry might be a bit rusty Commented Apr 16, 2021 at 3:22

Hooke’s Law is a first-order linear approximation and is only valid over a small range of deformation. You cannot use it to conclude anything about the forces between atoms or molecules outside of this range - except that, as your example shows, these forces cannot be linear over their whole range.

• is there anywhere I can find how much this deviates from the actual value over temp/distance/etc or a more accurate approximation? would calculating such a thing require actually calculating QFT forces with the standard model and stuff? Commented Apr 16, 2021 at 3:30

Chemical bonds apply at the atomic level and the potential tying atoms and molecules up to liquids and solids cannot be approximated by the potential of a spring, it is too complicated for that. The atoms and molecules are in orbitals, which allow spaces where positive and negative electromagnetic forces allow for bonding,( see for links this answer of mine for example,) and the potentials are very complicated.

Any symmetric in x potential can be expanded in a series and the first term will be a $$x^2$$, (that is why the harmonic oscillator is so useful in quantum mechanical studies) but the coefficients on the terms determine how many of them should be included in order for a calculation to be correct within errors.