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](https://en.wikipedia.org/wiki/Bond-dissociation_energy#BDE_versus_bond_energy) 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](http://www.engineeringtoolbox.com/water-vapor-d_979.html). 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](https://en.wikipedia.org/wiki/Water_splitting#Thermal_decomposition_of_water) 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!


 4