The progression of the combustion is shown for a temperature of $T=1900$\;K in Fig.~\ref{fig:t1900a}. It is visible how first the intermediate substances are produced and then nearly all oxygen is converted to water.

For a temperature of $T=2900$\;K, as shown in Fig.~\ref{fig:t2900a}, the reaction follows the same pattern and only is speeded up by approximately of 100.

For a temperature of $T=900$\;K, as shown in Fig.~\ref{fig:t900a}, the reaction sets in later. However the set in of the reaction here seem to be an error in the integration. Switching the adaptive time stepping off (Fig.~\ref{fig:t900}) reveals no set in of the reaction. It is expected that the integration sets in eventually at 900\;K. However, it was not feasible to calculate much longer intervalls using the presented code.

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T1900ap2t7.png}
 \caption{Reaction at $T=1900$\;K. Calculated using adaptive timestepping. Plotted are concentrations (unitless) versus time (in seconds).}
 \label{fig:t1900a}
\end{figure}

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T2900ap2t9.png}
 \caption{Reaction at $T=2900$\;K. Calculated using adaptive timestepping.}
 \label{fig:t2900a}
\end{figure}

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T900ap2t7.png}
 \caption{Reaction at $T=900$\;K. Calculated using adaptive timestepping.}
 \label{fig:t900a}
\end{figure}

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T900d14t7.png}
 \caption{Reaction at $T=900$\;K. Calculated using fixed timestepping.}
 \label{fig:t900}
\end{figure}

In Fig.~\ref{fig:t1900ad}, the reaction progression is shown for a non-isothermic system, i. e. the temperature was allowed to change during the process. Unfortunately, this leads to the adaptive timesteps to become very small when the temperature, so this calculation was canceled at $t=(7.65\cdot10^{-10})$\;s. The temperature behavior is shown in Fig.~\ref{fig:t1900adT}. 

In these two plots the typical feedback reaction of combustion processes can be seen: The reaction rise the temperature, which then leeds to an acceleration of the reaction, this lets the temperature rise even faster and so on.

Because of the hight temperature, more free radicals can be found in the compound. The result can probably be considered to be in the state of a plasma.

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T1900adiabaticz.png}
 \caption{Reaction at varable temperature, starting at $T=2900$\;K. Calculated using adaptive timestepping and adiabatic.}
 \label{fig:t1900ad}
\end{figure}

\begin{figure}
 \centering
 \includegraphics[width=0.8\textwidth]{../T1900adiabaticTz.png}
 \caption{Reaction temperature. Calculated using adaptive timestepping and adiabatic.}
 \label{fig:t1900adT}
\end{figure}