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.
