I suppose the first question supporting the main question is, has tritium water ever been synthesized in sufficient quantity to test chemical properties?

If so, and apart from the obvious radioactive nature of the molecule, and possibly self-heating nature, what unusual properties does tritium water have over protium water?

For example, deuterium oxide is not (easily) metabolized by living things; it cannot support life in the same manner as protium oxide. Also protium water is blue and deuterium oxide is clear. What color is tritium oxide?


1 Answer 1


Yes, $\ce{T2O}$ has been prepared and is available in significant quantity. When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil. It must be transported in a shielded, cryogenic dewar.

A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds. The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest. We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a primary kinetic isotope effect. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows: $$\frac{K_H}{K_D} \sim 6-7$$ $$\frac{K_H}{K_T} \sim 13-14$$ In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.

These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds. The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon. Still, as you pointed out, the effects can be disastrous in biological systems. Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.

The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation


Here $\ce{\mu}$ is the reduced mass of the system (e.g. the bond involved in the vibration that is producing the light absorption) and is given as

$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$

where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond. We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is inversely related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration. Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (reference, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be colorless.

  • $\begingroup$ Great answer! And so if I had a long enough column of $T_2O$ and, and assuming it wouldn't boil away before I viewed light coming up through it, the light would be red? $\endgroup$
    – docscience
    May 17, 2015 at 17:16
  • $\begingroup$ see my edit, $\ce{D2O}$ is colorless. $\endgroup$
    – ron
    May 17, 2015 at 18:26
  • $\begingroup$ Can you take advantage of the tritium to helium decay to convert a tritium substituted centre to a radical ? Would be cool although not very useful. $\endgroup$
    – J. LS
    May 17, 2015 at 21:27
  • $\begingroup$ @J.LS Interesting, I think it would. That might be a good question to post on its own. The electron released carries about 6 keV of energy - that might do some local damage! $\endgroup$
    – ron
    May 17, 2015 at 21:56

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