# Difference between dissociation energy and dissociation temperature for water?

I was trying to find the temperature at which water dissociates into hydrogen and oxygen, and so I came across this Wikipedia page. It states that it takes $$\pu{493.4 kJ/mol} + \pu{424.4 kJ/mol}$$ to dissociate the bonds in a water molecule which gives a total of about $$\pu{920 kJ/mol}$$ or about $$\pu{51 MJ/kg}$$. So I tried to calculate the temperature that would give that huge amount of energy using the heat capacity of water vapor. I took $$\pu{3.3 kJ/kgK}$$ for water vapor as an average specific heat, and so by simply dividing $$\pu{51 MJ}$$ by $$\pu{3.3 kJ}$$, I ended up with about $$\pu{15,000 K}$$.

But on this other Wikipedia page, I found: "at $$\pu{2200 ^\circ C}$$ about three percent of all $$\ce{H2O}$$ molecules are dissociated into various combinations of hydrogen and oxygen atoms. At the very high temperature of $$\pu{3000 ^\circ C}$$ more than half of the water molecules are decomposed".

So now I am very confused. If $$\pu{3000 ^\circ C}$$ is enough to dissociate more than half of the water molecules, then it should be expected that almost all the molecules should dissociate at around $$\pu{4000 ^\circ C}$$ or so and not the $$\pu{15,000 K}$$ I calculated earlier.

So am I missing something here? Or is there an explanation to this?

One key is an assumption about various degrees of dissociation.

It states that it takes 493.4 kJ/mol + 424.4 kJ/mol to dissociate the bonds in a water molecule which gives a total of about 920 kJ/mol or about 51 MJ/kg.

That sounds like the energy required for $$\ce{H2O <=> 2H + O}$$.

But on this other Wikipedia page, I found: "at 2200 °C about three percent of all H2O molecules are dissociated into various combinations of hydrogen and oxygen atoms.

Here it sounds like they are considering less extreme dissociations than complete dispersal to individual atoms. 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.

In effect, if just one OH bond breaks in water, that only requires 493.4 kJ/mol and leads to the formation of species such as $$\ce{OH}$$ and $$\ce{H}$$ atoms. Those species have a different heat capacity than $$\ce{H2O}$$ vapor. The heat capacity page you linked to says:

The values above apply to undissociated states. At high temperatures above 1500 K dissociation becomes appreciable and pressure is a significant variable.

In the calculation you are doing you are not including a specific pressure variable either. Heat capacity could also be a function of pressure. Presumably the data you linked to is all for 1 atm of pressure, although it didn't really say. Likewise, the claim that at 2200 °C, 3% of water is dissociated must also be pressure-dependent. For which pressure is it true?

The final complexity is the entropic term. The three dissociated atoms have 9 translational degrees of freedom but intact water only has three.

These pressure and entropy complexities matter because $$\Delta G$$ is what determines equilibrium, and $$\Delta G = \Delta U + \Delta (PV) - T \Delta S$$. The bond dissociation energies you quote represent the $$\Delta U$$ term, so the calculation you are doing is implicitly assuming that $$\Delta U \gg \Delta (PV) - T \Delta S$$.

I think that your calculation is admirable in that it comes so close to the right answer while still neglecting all these complexities!

• Quick caveat: the bond dissociation energies are closer to $\Delta H$ than $\Delta U$. So I mis-spoke a tad above. But they are "standard" enthalpies, meaning they probably assume that water and its dissociation products are ideal gases. Thus a real-gas correction is still required. That said, the $T \Delta S$ term is probably driving most of the difference between your calculated temperature and the observed temperature for equilibration. – Curt F. Sep 24 '15 at 19:21