Adiabation Flame Temperature Overestimation (Hydrogen Gas)

I'm working on some code that computes adiabatic flame temperature given a balanced equation and the relevant thermodynamic properties and I'm starting with the simplest combustion reaction I can think of: stoichiometric combustion of pure hydrogen with pure oxygen.

$2H_2+O_2 \Rightarrow 2H_2O$

I'm using these thermodynamic values:

$h_{f,H_2} = 0.0 \text{ kJ/kmol}$

$h_{f,O_2} = 0.0 \text{ kJ/kmol}$

$h_{f,H_2O} = -241820 \text{ kJ/kmol}$

$c_p = 1.864$ kJ/kg-K at 300K for $H_2O_{(g)}$

$c_p = 3.217$ kJ/kg-K at 4000K for $H_2O_{(g)}$

If the combustion equation I wrote is on a kmol basis, I get a total change in enthalpy of 483640 kJ with the combustion. If that heat goes into heating the product (water vapor), I get temperature increases ranging from 4173K to 7201K (final temperatures of 4471K and 7499K) depending on the specific heat value you use.

$\frac{483640\text{ kJ}}{(3.217 \text{ kJ/kg-K})(36.03\text{ kg})} = 4173\text{ K}$

$\frac{483640\text{ kJ}}{(1.864 \text{ kJ/kg-K})(36.03\text{ kg})} = 7201\text{ K}$

When I run the code that actually considers changes in the specific heat throughout the whole heating process (interpolating based on a table at each step), I get 5024K. Even my lowest value, assuming constant specific heat at the highest value, gives a temperature increase higher than the number listed on Wikipedia, ~3500K. Am I making some kind of error or is there some kind of factor that makes the real thing different from the idealization? Thanks.

• You assume complete combustion but even in ideal conditions Le Chatelier's principle prohibits that. en.wikipedia.org/wiki/… .High temperatures affect the reaction equilibrium constants.
– Gert
Jul 10 '16 at 1:43

The assumed reaction $2H_2+O_2 \Rightarrow 2H_2O$ is not true at high temperature. The molecules will dissociate to $H_2, H, O_2, O, OH, HO_2$. The dissociation reactions (e.g. $H_2O \rightarrow OH + H + O$) are endothermic and thus reduce the flame temperature.