Why do lighter isotopes evaporate faster than heavier isotopes?

I have just read that we are able to estimate the temperature of the earth thousands of years ago by measuring the ratio of certain isotopes present in ice cores that froze thousands of years ago. My understanding of this idea is that $$\mathrm{H_2O}$$ molecules containing lighter isotopes (i.e. with $$\mathrm{^{16}O}$$) evaporate quicker than those molecules that contain heavier isotopes (i.e. with $$\mathrm{^{18}O}$$) and that this difference in evaporation rate exhibits a temperature-dependent relationship.

My problem is that the equipartition theorem says that each molecule should contain an energy of $$\frac{1}{2}kT$$ per degree of freedom regardless of the molecular mass. If we assume that all the liquid $$\mathrm{H_2O}$$ molecules roughly obey a Maxwell-Boltzmann distribution, then the heavier isotopes will be moving slower than the lighter isotopes (on average at a given temperature) but their greater mass will mean that all isotopes will still possess the same average kinetic energy (in accordance with the equipartition theorem). So if all the water molecules possess the same kinetic energy on average (regardless of the isotopes they contain), why do those molecules with lighter mass preferentially evaporate? I understand that they are moving quicker than their heavier counterparts but the determination of whether a particle escapes a potential well and evaporates is its kinetic energy not its velocity.

In effect, I am asking why the speed of the molecules determines their proclivity to evaporate (to escape the potential well they inhabit due to intermolecular attraction) when it should be their kinetic energy that determines whether they escape their intermolecular potential well or not. If the evaporation rate is determined by kinetic energy (which I think it should be?) as opposed to velocity, then there shouldn't be any difference in the evaporation rates of different water molecules containing different isotopes because they all contain the same kinetic energy on average.

Why is hydrogen bonding stronger in heavy water? To get some intuition, let's consider the quantum harmonic oscillator. If you solve the Schrödinger equation with a potential energy function of $$V(x) = \frac12 kx^2$$, you get solutions with energy $$\hbar \omega (n + \frac12)$$ where $$n$$ is an integer and $$\omega = \sqrt{\frac{k}{m_r}}$$. Here $$m_r$$ is the reduced mass $$\frac{m_1m_2}{m_1+m_2}$$. Thus, a quantum harmonic oscillator with higher masses at the endpoints (heavy water molecules) will have lower $$\omega$$ and a lower ground state energy $$\hbar \omega/2$$. This means that more energy is needed to excite the bonding electrons from the ground state to the dissociation energy, so the bond should be slightly stronger.
I should mention that this explanation is rather oversimplified in a lot of important ways. In fact, surprisingly many important questions about hydrogen bonding in water are still not well understood. It is known that the hydrogen bond in a heavy water dimer is stronger than the one in a light water dimer (as we'd expect based on the argument above). There is some theoretical evidence that this conclusion should generalize to similar small systems. However, it's possible that the relevant physics may be different (and unfortunately, less elegant) for larger amounts of water. One intriguing experimental study found that the hydrogen bond in light water is actually shorter than the hydrogen bond in heavy water, which would imply a stronger bond. The study also reported that the number of hydrogen bonds per water molecule drops from $$3.76 \pm 0.1$$ in heavy water to $$3.62 \pm 0.1$$ in light water because the geometric packing of the molecules is slightly different. Perhaps this effect may be more important than the strength of the individual bonds.
• This is a truly fantastic answer! I think your explanation using the QHO provides more than enough intuition for me. Just to be clear on some specifics before I accept your answer though: I usually see heavy water defined as $2^H_20$ specifically as opposed to the set of all heavier-than-usual water molecules ("Heavy water"={$H_2^{18}0$, $^2H_2^{18}0$, ...etc}). Is my thinking correct that the QHO argument should in principle apply equally well to all of these other heavier molecules(not just $^2H_20$) since in each of these cases, the H bond viewed as a QHO will always have a greater mass? Commented Jul 31, 2021 at 9:44
• @SalahTheGoat Yes, the same argument should apply to other heavier-than-usual water molecules. For example, the vapor pressure of $^{18}$O enriched water is lower than the vapor pressure of normal water. Here's some experimental evidence: sciencedirect.com/science/article/abs/pii/S0021961498903810 Commented Jul 31, 2021 at 9:52