In comments below this answer in Space Exploration SE I've said:

(Wikipedia) says that dissociating water costs about 4.4 eV per bond. If 𝑘𝐵𝑇= 4.4 eV and $k_B$ is 8.617E-05 eV/K then that corresponds to a characteristic temperature of something like 50,000 K. Does a nuclear rocket's exhaust get close to that? (I have no idea)

and the reply was:

but en.wikipedia.org/wiki/Water_splitting says 3000 K. Can anyone clarify?

That Wikipedia article actually says:

At the very high temperature of 3000 °C more than half of the water molecules are decomposed, but at ambient temperatures only one molecule in 100 trillion dissociates by the effect of heat.

Question: If it takes 4.4 eV to break one hydrogen-oxygen bond, how could half of the water be dissociated at only 3273 K where $k_B T$ only about 0.28 eV?

$$\exp \left( -\frac{0.28 \ \text{eV}}{4.4 \ \text{eV}} \right) \approx 0.94$$

so I'd naively expect only about 6% of the water to be dissociated at 3000 °C.

update: I've just realized that in my naivety I've forgotten to consider that forming H2 and O2 from free atomic H and O is probably exothermic, so my 4.4 eV is an overestimate of the energy difference between initial and final states. But I don't know how to take this into account.

  • 1
    $\begingroup$ chemistry.stackexchange.com/questions/37562/… $\endgroup$
    – Mithoron
    Commented Dec 1, 2019 at 19:29
  • $\begingroup$ @Mithoron Thanks for the link! Yep, answers there certainly looks helpful and consistent with my update to the question, but an answer to this question addressing "...and how to calculate?" would need to go further than "For example, $\ce{H2O <=> H + OH}$ and perhaps even $\ce{H2O <=> H2 + O}$ or $\ce{H2O <=> 2H + \frac{1}{2}O2}$ and related equilibria may be happening." $\endgroup$
    – uhoh
    Commented Dec 2, 2019 at 0:43


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