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


Hydrogen oxidation.
$$\ce{H2 + 0.5 O2 -&gt; H2O(g)}$$
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 $K=1.2\times10^{40}$.
Deuterium isotope exchange between water and hydrogen.

A 1976 paper gives the value for this reaction.

$$\ce{H2O(g) + HD -&gt; HDO(g) + H2}$$
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.
Add the above reactions. 
$$\ce{HD + 0.5 O2 -&gt; HDO(g)}$$
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 &permil; (i.e. one per mille).