# 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? – J. Murray Apr 16 at 2:50
• my chemistry might be a bit rusty – asdfasdfasdf Apr 16 at 3:22

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.