# Burning different isotope combinations of hydrogen and oxygen, how much does the energy vary?

In this comment I suggested that there would be a (very) small difference in energy released from burning a stoichiometric mixture of hydrogen and oxygen, depending on the isotopes, due to a slight difference in bond energy. The differences can be simply related to different masses of the nuclei, or to other quantum mechanical effects due to the large differences in nuclear spin and moments.

I'd like to repeat the calculation in this answer just to see the sensitivity to the isotopic shifts.

Approximately how large are these small differences in bond energies? Are the differences mostly due to the mass difference or due to the other nuclear properties?

Isotopes in question: $\ce{{}^1H, {}^2H, {}^3H$, ${}^{16}O$, ${}^{17}O$, ${}^{18}O}$

Either the bond energies themselves, or just a link to a source where I can look it up myself would be fine.

In the Born-Oppenheimer Approximation the nuclei are infinitely heavy point particles that move in the effective field of the electrons. Since the mass is considered to be infinitely heavy, within the BO approximation there is no change in the potential energy as a function of the internuclear distance. Effects of the finite nuclear mass can be incorporated and are referred to as adiabatic corrections to the BO energies. These corrections are very small (the difference in the adiabatic corrections for H$_2$ and D$_2$ is 3 cm$^{-1}$ or 0.0085 kcal/mol [Pachucki and Komasa, J. Chem. Phys. 141, 223103 (2014)]). Finite nuclear size effects are much smaller than that and can only be observed in high-precision spectroscopy of the simplest molecules.

A more important effect of the different mass for the different isotopes is the effect on the vibrational energy structure. When solving the Schrodinger Equation for the nuclear potential, one obtains the rovibrational energies (and wave functions). As you probably know, the lowest energy solution is not located at the zero of the potential energy curve, but has an offset which is typically referred to as the zero point energy. The zero point energy is roughly equal to halve of the harmonic frequency of the potential and since the harmonic frequency scales as $\omega_e\sim \sqrt{\frac{1}{\mu}}$, with $\mu$ the reduced mass of the system, the dissociation energy (i.e. bond energy) with respect to the ground state of the potential becomes larger with increasing $\mu$ as the ZPE becomes smaller.

For H$_2$, $\omega_e=4401$ cm$^{-1}$ using the scaling with $\mu$, we find for D$_2$ a value for $\omega_e(\text{D}_2)=\omega_e(\text{H}_2)/\sqrt(2)=3112$ cm$^{-1}$ and thus a difference of 1289 cm$^{-1}$. The difference in the dissociation energy $D_0$ is half of the difference in the harmonic frequency, so $\Delta D_0=644.5$ cm$^{-1}$. In good agreement with the experimental result of 630 cm$^{-1}$ (Liu et al. J. Chem. Phys. 130, 174306 (2009) and Sprecher et al., J. Chem. Phys. 132, 154301 (2010)).

When looking at the energy released in a reaction, you should also consider the change in $D_0$ for the products of course.

• Thank you for the very helpful answer! I hadn't even through of the vibration/rotation energies. Thanks also for the references. At say 20C the molecules with have some distribution of excited (rotational/vibrational) states. In addition to the shift in zero-point energy, the level spacing may also shift in some cases at least. But that doesn't actually matter so much because that would just slightly shift the population fractions so that the average energy would still be $k_B T$? – uhoh Dec 14 '16 at 12:55
• Yes, the different mass also reduces the rotational spacing and therefore the peak of the distribution might be at a different rotational level, but overall it will still be a Boltzmann distribution. – Paul Dec 15 '16 at 13:55

Let me add to Paul's excellent answer with a simple thermodynamic calculation.

1. Hydrogen oxidation. $$\ce{H2 + 0.5 O2 -> H2O(g)}$$ This reaction has a standard free energy change of about -229 kJ/mol. Using $\Delta G^\circ = -RT \ln K$, I get the equilibrium constant for this reaction of $K=1.2\times10^{40}$.

2. Deuterium isotope exchange between water and hydrogen.

A 1976 paper gives the value for this reaction.

$$\ce{H2O(g) + HD -> HDO(g) + H2}$$ The value given is $K = 3.5$, which by the standards of isotope exchange reactions is very large. Using $\Delta G^\circ = - RT \ln K$, we can calculate that the standard free energy change of this reaction is $\Delta G ^\circ =$ -3.1 kJ/mol.

3. Add the above reactions. $$\ce{HD + 0.5 O2 -> HDO(g)}$$ We can add the free energies of the reactions to get that $\Delta G =$-231 kJ / mol.

Thus, burning $\ce{HD}$ gas provides about 1.4% more free energy than burning $\ce{H2}$ gas. And this is by far the largest difference that we would observe for any common stable isotope. For nearly any isotopic exchange reaction that involves other isotopes, such as $\ce{^{13}C}$ or $\ce{^{18}O}$, the relevant equilibrium constants will be $0.9 \lt K \lt 1.1$ or so. Only deuterium vs. protium (i.e. $\ce{D}$ or $\ce{^2H}$ vs $\ce{^1H}$) would often have values outside that range. Thus for isotopes like $\ce{^{13}C}$ or $\ce{^{18}O}$, the difference will be closer to 0.1%, or, as isotopic geochemists like to say, 1 ‰ (i.e. one per mille).

• That you for working the problem through. I'm really surprised the difference is so large, but it makes sense it would be mostly from the large H/D mass difference. Does this have implications for the question I've linked to? Would exposure of typical water to HD gas lead to an exothermic 'reaction' or 'exchange'? - not sure what to call it. if there's a relevant answer maybe post there. – uhoh Dec 16 '16 at 23:07
• This calculation here is a thermodynamic calculation, but the question you are asking now is about kinetics. "Is an exchange reaction energetically favorable" is different than "will an exchange reaction occur". Similarly, when you visit the Grand Canyon, "is me falling all the way to the bottom energetically favorable" has a (hopefully) different answer than "will I spontaneously fall all the way to the bottom". – Curt F. Dec 19 '16 at 19:19
• Thank's for the visuals. I'll ask a new question once my vertigo subsides. :) – uhoh Dec 20 '16 at 1:10