# Adiabatic flame temperature of hydrogen peroxide calculation

I will first demonstrate my cursory method for evaluating the adiabatic flame temperature of H2O2 at atmospheric pressure.

$$\ce{H2O2 -> H2O + 0.5O2}$$

At $$\pu{1200 K}$$,

\begin{align} h_\ce{H2O}^\pu{1200 K} - h_\ce{H2O}^\pu{298 K} &= \pu{34.506 kJ/mol}\\ h_\ce{O2}^\pu{1200 K} - h_\ce{O2}^\pu{298 K} &= \pu{29.761 kJ/mol}\\ \Delta_f H^\pu{298 K} &= \pu{-241.826 kJ/mol} \end{align} Sources: $$\ce{H2O}$$: https://janaf.nist.gov/tables/H-064.html; $$\ce{O2}$$: https://janaf.nist.gov/tables/O-029.html

At this temperature,

$$\text{Energy released} = 241.826 - 34.506 - 0.5 \times 29.761 = \pu{192.44 kJ/mol}$$

For Adiabatic flame temperature, the energy released should be equal to zero which is not so in this case (I am aware that the process is iterative). However, multiple sources indicate the flame temperature to be around $$\pu{1000 K}$$. A screenshot from RPA (a rocket propulsion software) is one such source.

The software obviously takes into account other products but, even as a first estimate my calculation shouldn't have been so off from the software calculation. I would appreciate any help in pointing out the mistake in my calculation.

• I have tried to fix your markup, I hope I didn't screw it up; please check. If you want to know more, please have a look here and here. I personally don't really understand your question, maybe it is an option for you to explain why you did these calculations and how you arrived at these formulae. – Martin - マーチン Apr 12 '20 at 12:55