kinetics_computations.tex

\subsubsection{Kinetics models}
The kinetics models are all given in Tab.~\ref{antioch::kinMod}.
The kinetics model will render the temperature evolution of
the rate constant, thus the only differential that is needed
at that level is the temperature differentiation.

The typical interpretation of the kinetics model comes from
Arrhenius (See Fig.~\ref{Arrhenius_interp}). The activation
energy is the wall the reactants have to overcome to reassemble
into the products. This quantity is a purely kinetics one, meaning
it is related only to the rate of the reaction. The enthalpy of reaction
is the difference of enthalpy between the reactants and the products.
This is, contrary of the activation energy, a purely thermodynamics
quantity, therefore containing information upon the equilibrium
state (the relative quantity of products/reactants).

This interpretation is accurate only in an elementary reaction, when
indeed a reaction consists of the reactants forming a transitional
state and going to the products (Fig.~\ref{Arrhenius_interp}, blue curve).
A reaction, more generally, will have one, or several, reaction intermediate(s), 
and the Arrhenius interpretation will be an approximation (Fig.~\ref{Arrhenius_interp},
red curve). This explains
why the Arrhenius equation typically needs some modifications (see Tab.~\ref{antioch::kinMod}).

\begin{figure}
\centering
\includegraphics{Arrhenius_interpretation}
\caption{\label{Arrhenius_interp}The activation energy (\AcEn) is the energy wall 
(in most chemical cases the enthalpy) between the reactants and the
transitional form. The enthalpy of reaction (\Denth[r]) is the difference
of enthalpy between the reactants state and the products state. The activation
energy is a purely kinetics quantity while the enthalpy of reaction is purely
thermodynamics.}
\end{figure}

\begin{table}
\centering\renewcommand{\arraystretch}{1.5}
\begin{tabular}{clr}\toprule
\multirow{2}{*}{Kinetics model}
                & Expression                     & Parameters \\
                & $\frac{\dd}{\dd T}$ Expression &\\\midrule
\multirow{2}{*}{Hercourt-Hessen} 
                & $\kinMod = \PreExp \left(\frac{\Temp}{\Tref}\right)^\Power$ 
                        & \multirow{2}{*}{\PreExp, \Power}\\
                & $\ddoverdT{\kinMod} = \kinMod \frac{\Power}{\Temp}$ \\[10pt]
\multirow{2}{*}{Berthelot}
                & $\kinMod = \PreExp \exp\left(\BerthExp\Temp\right)$ 
                        & \multirow{2}{*}{\PreExp, \BerthExp}\\
                & $\ddoverdT{\kinMod} = \kinMod \BerthExp$ \\[10pt]
\multirow{2}{*}{Arrhenius}
                & $\kinMod = \PreExp \exp\left(-\frac{\AcEn}{\Rg\Temp}\right)$ 
                        & \multirow{2}{*}{\PreExp, \AcEn}\\
                & $\ddoverdT{\kinMod} = \kinMod \frac{\AcEn}{\Rg\Temp^2}$ \\[10pt]
\multirow{2}{*}{Kooij}
                & $\kinMod = \PreExp \left(\frac{\Temp}{\Tref}\right)^\Power\exp\left(-\frac{\AcEn}{\Rg\Temp}\right)$
                        & \multirow{2}{*}{\PreExp, \Power, \AcEn}\\
                & $\ddoverdT{\kinMod} = \kinMod \left(\frac{\Power}{\Temp} + \frac{\AcEn}{\Rg\Temp^2}\right)$ \\[10pt]
\multirow{2}{*}{Berthelot Hercourt-Essen}
                & $\kinMod = \PreExp \left(\frac{\Temp}{\Tref}\right)^\Power\exp\left(\BerthExp\Temp\right)$
                        & \multirow{2}{*}{\PreExp, \Power, \BerthExp}\\
                & $\ddoverdT{\kinMod} = \kinMod \left(\frac{\Power}{\Temp} + \BerthExp\right)$ \\[10pt]
\multirow{2}{*}{Van't Hoff}
                & $\kinMod = \PreExp \left(\frac{\Temp}{\Tref}\right)^\Power\exp\left(\BerthExp\Temp-\frac{\AcEn}{\Rg\Temp}\right)$
                        & \multirow{2}{*}{\PreExp, \Power, \BerthExp, \AcEn}\\
                & $\ddoverdT{\kinMod} = \kinMod \left(\frac{\Power}{\Temp} + \BerthExp +  \frac{\AcEn}{\Rg\Temp^2}\right)$ \\
\multirow{2}{*}{Photochemistry}
                & $\kinMod = \int_0^\infty f(\wavelength)\crosssection\dd\wavelength$
                        & \multirow{2}{*}{\wavelength, \crosssection}\\
                & $\ddoverdT{\kinMod} = 0$\\
\bottomrule
\end{tabular}
\caption[Kinetics models]{\label{antioch::kinMod}Kinetics models available in \Antioch. In the photochemical
kinetics model, $f$ is the photon flux, which is user-defined, \wavelength\ and \crosssection\
are vectors.}
\end{table}

\subsubsection{Chemical processes}
\label{subsec:chem_proc}
The chemical processes is how this kinetics model behave with respect to
the densities of the molecules in the medium. Either they're all equivalent,
and only the total density matters (falloff type), or each species can 
influence the rate constant in a specific manner (three-body type).
The chemical processes are given in Tab.~\ref{antioch::chemProc}. This is where
chemical modeling begins to be stretched a little: as seen on section~\ref{kinetics_gen},
we assume that every reaction modeled is an elementary step, even in the case of
non-elementary reactions. Therefore some corrections need to be made. This is where
the chemical process adds, if needed, more suppleness to the rate constant
function, from adding degrees of freedom to the temperature dependency (duplicate
chemical process) to adding a pressure dependency over the temperature dependency
(falloff).
\begin{table}
\centering\renewcommand{\arraystretch}{2}
\begin{tabular}{ccc}\toprule
\multirow{2}{*}{Chemical process}
                  & \multirow{2}{*}{Expression}
                                           & $\doverdT{\text{Expression}}$ \\
                  &                        & $\doverdc{\text{Expression}}$ \\\midrule
\multirow{2}{*}{Elementary}
                  & \multirow{2}{*}{$\chemProc = \kinMod$}  
                                           & $\doverdT{\chemProc} = \ddoverdT{\kinMod}$ \\
                  &                        & $\doverdc{\chemProc} = 0$ \\[10pt]
\multirow{2}{*}{Duplicate \dag}
                  & \multirow{2}{*}{$\chemProc = \displaystyle\sum_i^{\mathrm{N_{proc}}}\kinMod_i$}
                                           & $\doverdT{\chemProc} = \displaystyle\sum_i^\mathrm{N_{proc}}\ddoverdT{\kinMod_i}$ \\
                  &                        & $\doverdc{\chemProc} = 0$ \\[10pt]
\multirow{2}{*}{Three-Body}
                  & \multirow{2}{*}{$\chemProc = \kinMod \threeBody$}
                                           & $\doverdT{\chemProc} = \ddoverdT{\kinMod}\threeBody$ \\
                  &                        & $\doverdc[I]{\chemProc} = \kinMod\epsilon_i$ \\[10pt]
\multirow{2}{*}{Falloff}
                  & \multirow{2}{*}{$\chemProc = \frac{\conc[M]\kinModZ}{1 + \conc[M]\frac{\kinModZ}{\kinModI}}\Falloff$}
                                          & \ref{Falloff:doverdT} with $F = \FLind\text{ or }\FTroe$\\
                  &                       & \ref{Falloff:doverdc} with $F = \FLind\text{ or }\FTroe$\\[10pt]
\multirow{2}{*}{Three-Body falloff}
                  & \multirow{2}{*}{$\chemProc = \frac{\kinModZ\threeBody}{1 + \frac{\kinModZ}{\kinModI}\threeBody}\Falloff$}
                                          & \parbox{5cm}{\ref{Falloff:doverdT} with $F = \FTroe$ or $\FLind$ 
                                                         and $[M] = \threeBody$}\\
                  &                       & \ref{Falloff_Three:doverdc} with $F = \FTroe\text{ or }\FLind$\\[10pt]
\bottomrule
\end{tabular}
\caption[Chemical processes]{\label{antioch::chemProc}Chemical processes available in \Antioch.
\dag: the duplicate chemical process do not permit several kinetics
models to be mixed. The functions $F$ for the falloff are described in section~\ref{subsub:falloff}}
\end{table}

\paragraph{Elementary process}

An elementary process is the smallest unit of reaction. Within a chemical
network, elementary processes are the elementary steps, when the collision
of the reactants produces the reactants without any intermediate. All
reactions approximated to an elementary process are therefore coded with
a simple kinetics model.

\paragraph{Duplicate process}

In the case the elementary approximation breaks down, and one kinetics
process is not sufficient to properly render measurements, it is possible
to add several kinetics model to obtain the necessary temperature
dependency.

\paragraph{Three body process}
\label{subsub:three_body}

A third body may have an influence over the reaction, therefore
changing a generic chemical equation \ce{A + B -> C + D} into
\ce{A + B + M -> C + D + M}. It acts like an homogeneous catalyst.
Thus the rate of the reaction will be written
\[
\chemProc = \kinMod[ABM]\conc[A]\conc[B]\conc[M]
\]
instead of
\[
\chemProc = \kinMod[AB]\conc[A]\conc[B]
\]
with typically $\kinMod[ABM] \propto \kinMod[AB]$.

Not all molecules have the same effect, meaning \rcons[ABM]
is dependant of \ce{M}. Therefore it is needed to
consider coefficients that are specific to third-bodies,
with a typical default of unity. Thus we derive a mixture
formul\ae:
\begin{equation}
\chemProc = \threeBody\kinMod
\label{kinproc:Three_body}
\end{equation}

\paragraph{Pressure dependency: falloffs}
\label{subsub:falloff}

The falloffs are reactions having two different regimes: a low-pressure (\kinModZ) and
a high-pressure (\kinModI) regime. Any pressure's rate constant can be expressed in
terms if those two regimes:
\begin{equation}
\chemProc = \frac{\conc[M]\kinModZ}{1 + \conc[M]\frac{\kinModZ}{\kinModI}} F
\label{falloff:k}
\end{equation}
with \conc[M] being the total molar density and $F$ the falloff model.

The differentiations are:
\begin{equation}
\begin{split}
\doverdT{\chemProc} & = \chemProc 
                        \Bigg[\frac{1}{F}\doverdT{F} \\
                    &      + \left(\frac{1}{\kinModZ} - \frac{\conc[M]}{\kinModI}\frac{1}{1 + \conc[M]\frac{\kinModZ}{\kinModI}}\right)\doverdT{\kinModZ} \\
                    &      - \conc[M]\frac{\kinModZ}{\kinModI^2}\frac{1}{1 + \conc[M]\frac{\kinModZ}{\kinModI}}\doverdT{\kinModI}
                        \Bigg]
\end{split}
\label{Falloff:doverdT}
\end{equation}
and
\begin{equation}
\doverdc[I]{\chemProc} = \chemProc
                          \left(\frac{1}{\conc[M]} \frac{1}{1 + \conc[M]\frac{\kinModZ}{\kinModI}}
                               + \doverdT{F}\frac{1}{F}
                          \right)
\label{Falloff:doverdc}
\end{equation}

\paragraph{A twist in the physics: three-body falloff}

As all models above are approximations of the ``truth'', it can
happen that some reactions do not fit well into any category.
In the case of a pressure-dependant reaction that behaves
differently depending of a third-body, the third-body model
and the falloff model can be combined. Note that this is not
very rigorous, we're mixing here two different hypothesises that
do not belong together. 

A pressure-dependance reaction is a collision-driven reaction, hence
the change of regime depending of the collisions rate regime. Whereas
a three-body reaction is the contraction of $N$ reactions, one per
third-body. An increase in pressure for each of these $N$ reactions
is not the increase of the \threeBody\ term.

The rate of this type of reaction is:
\begin{equation}
\chemProc = \frac{\threeBody\kinModZ}{1 + \threeBody\frac{\kinModZ}{\kinModI}} F
\label{Falloff_Three:chem_proc}
\end{equation}
%
with the derivative with respect to densities:
%
\begin{equation}
\doverdc[I]{\chemProc} = \chemProc\epsilon_i
                          \left(\frac{1}{\threeBody} \frac{1}{1 + \threeBody\frac{\kinModZ}{\kinModI}}
                               + \doverdT{F}\frac{1}{F}
                          \right)
\label{Falloff_Three:doverdc}
\end{equation}

\paragraph{Falloff models}
\subparagraph{Lindemann}
The Lindemann falloff is the simplest falloff, it is given by:
\begin{equation}
\FLind = 1,
\label{Falloff:Lindeman}
\end{equation}
which gives:
\begin{equation}
\begin{split}
\doverdT{\FLind} & = 0 \\
\doverdc{\FLind} & = 0 \\
\end{split}
\label{Falloff:Lindemann:diff}
\end{equation}

\subparagraph{Troe}
The Troe model is more elaborate and rely on the definition of three or
four parameters:
\begin{itemize}
\item \Troealpha, with \nounit,
\item \TroeTone\ in \unit{K},
\item \TroeTtwo\ in \unit{K},
\item \TroeTthree\ in \unit{K}.
\end{itemize}
Sometimes, the parameter \TroeTtwo\ is not provided, as its contribution
may be negligible.
The falloff calculations need the definition of several parameters:
\begin{equation}
\begin{split}
\TroeFcent & = \left(1 - \Troealpha\right)\exp\left(-\frac{\Temp}{\TroeTthree}\right)
              +          \Troealpha       \exp\left(-\frac{\Temp}{\TroeTone}\right)
              +                           \exp\left(-\frac{\TroeTtwo}{\Temp}\right) \\[5pt]
\Troen     & =   \numprint{0.75} - \numprint{1.27}\log_{10}\left(\TroeFcent\right)  \\[5pt]
\Troec     & = - \numprint{0.4}  - \numprint{0.64}\log_{10}\left(\TroeFcent\right)  \\[5pt]
\Troed     & =   \numprint{0.14}
\end{split}
\label{Troe:parameter}
\end{equation}
to finally have
\newcommand{\pr}{\ensuremath{\log_{10}\left(\frac{\conc[M]\kinMod_0}{\kinMod_\infty}\right)}}
\begin{equation}
\log_{10}\left(\FTroe\right) 
        = \frac{\log_{10}\left(\TroeFcent\right)}
              {1 + \left[
                        \log_{10}\left(
                                        \frac{\pr + \Troec}
                                             {\Troen - \Troed\left(
                                                                \pr + \Troec
                                                            \right)}
                                 \right) 
                   \right]^2}
\label{Troe:F}
\end{equation}
Thus the differentials will be:
\begin{equation}
\begin{split}
\doverdT{\TroeFcent} & = - \frac{1 - \Troealpha}{\TroeTthree}\exp\left(-\frac{\Temp}{\TroeTthree}\right)
                         - \frac{\Troealpha}{\TroeTone}\exp\left(-\frac{\Temp}{\TroeTone}\right)
                         + \frac{\TroeTtwo}{\Temp^2}\exp\left(-\frac{\TroeTtwo}{\Temp}\right)\\[5pt]
\doverdT{\Troen}     & = - \frac{\numprint{1.27}}{\ln(10)\TroeFcent}\doverdT{\TroeFcent} \\[5pt]
\doverdT{\Troec}     & = - \frac{\numprint{0.64}}{\ln(10)\TroeFcent}\doverdT{\TroeFcent} \\[5pt]
\doverdT{\Troed}     & = 0
\end{split}
\label{Troe:dparameterdT}
\end{equation}
and of course
\begin{equation}
\doverdc{\TroeFcent}  = 0,\qquad
\doverdc{\Troen}      = 0,\qquad
\doverdc{\Troec}      = 0,\qquad
\doverdc{\Troed}      = 0
\label{Troe:dparameterdc}
\end{equation}
For simplicity of writing, we define:
\begin{equation}
\begin{split}
g(\kinMod_0,\kinMod_\infty) = & \conc[M] \frac{\kinMod_0}{\kinMod_\infty} \\[5pt]
h(\kinMod_0,\kinMod_\infty) = & \log_{10}(g) + \Troec \\[5pt]
l(\kinMod_0,\kinMod_\infty) = & \log_{10}\left(\frac{h}{\Troen - \Troed h}\right) \\
\end{split}
\end{equation}
Therefore:
\begin{equation}
\FTroe = \frac{\log_{10}\left(\TroeFcent\right)}{1 + l^2}
\end{equation}
When we derive, we obtain:
\begin{equation}
\begin{split}
\doverdT{g} & = g \left(\frac{1}{\kinMod_0}\doverdT{\kinMod_0} - \frac{1}{\kinMod_\infty}\doverdT{\kinMod_\infty} \right) \\[5pt]
\doverdc{g} & = g \frac{1}{\conc[M]}\doverdc{\conc[M]}
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\doverdT{h} & = \frac{1}{g \ln(10)}\doverdT{g} + \doverdT{\Troec} \\[5pt]
\doverdc{h} & = \frac{1}{g\ln(10)}\doverdc{g}
\end{split}
\end{equation}
\begin{equation}
\begin{split}
\doverdT{l} & = \frac{1}{h \ln(10)}\left[\left(1 + \frac{\Troed h}{\Troen - \Troed h}\right) \doverdT{h} - \frac{h}{\Troen - \Troed h}\doverdT{\Troen} \right]  \\[5pt]
\doverdc{l} & = \frac{1}{h \ln(10)}\left(1 + \frac{h\Troed}{\Troen - \Troed h} \right)\doverdT{h}
\end{split}
\end{equation}
Thus, we have:
\begin{equation}
\doverdT{\log_{10}\left(\FTroe\right)} = \log_{10}\left(\FTroe\right) 
                           \left[
                                  \frac{1}{\ln(10)\log_{10}(\TroeFcent)}\doverdT{\TroeFcent}
                                - \frac{2 l}{1 + l^2} \doverdT{l}
                          \right]
\end{equation}
and
\begin{equation}
\doverdc{\log_{10}\left(\FTroe\right)} = \log_{10}\left(\FTroe\right)
                           \left[
                                  \frac{1}{\ln(10)\log_{10}(\TroeFcent)}\doverdc{\TroeFcent}
                                - \frac{2 l}{1 + l^2} \doverdc{l}
                          \right]
\end{equation}
the final values are given by:
\begin{equation}
\doverdpar{\FTroe} = \ln(10)\FTroe\doverdpar{\log_{10}\left(\FTroe\right)}
                           \left[
                                  \frac{1}{\ln(10)\log_{10}(\TroeFcent)}\doverdpar{\TroeFcent}
                                - \frac{2 l}{1 + l^2} \doverdpar{l}
                          \right]
\end{equation}
with $x = \Temp$ or $x = \conc$.
\begin{figure}
\centering
\includegraphics{falloffs}
\caption[Example of falloff reaction]{\label{kinetics::falloffs}%
Example of falloff for the reaction \ce{CH3 + CH3 ( + M) -> C2H6 ( + M)}.
The rate constants are calculated for a temperature of $T = 1000~\unit{K}$. The kinetics model
used is a Kooij model. Parameters are:
$\PreExp_0      = \numprint{1.135}\,10^{36}~\unit{mol^{-2}\,cm^6\,s^{-1}}$,
$\Power_0       = \numprint{-5.245}$,
$\AcEn_0        = \numprint{1704.8}~\unit{cal\,mol^{-1}}$,
$\PreExp_\infty = \numprint{6.22}\,10^{16}~\unit{mol^-2\,cm^6\,s^{-1}}$,
$\Power_\infty  = \numprint{-1.174}$,
$\AcEn_\infty   = \numprint{653.8}~\unit{cal\,mol^{-1}}$,
$\Troealpha     = \numprint{0.405}$,
$\TroeTone      = \numprint{69.6}~\unit{K}$ and
$\TroeTthree    = \numprint{1120.0}~\unit{K}$.}
\end{figure}

\subsection{Thermodynamics}
\subsubsection{Macroscopic description}
\label{phys:NASA_therm}

\Antioch\ uses the NASA polynoms.
These relations are used in the standard state, \textit{i.e.} $\Press = \pz = 10^5~\unit{Pa}$.
Thus what is means is that, for any quantity $A$ (\enth, \entr, \dots), $A^0$ is $A$ at
\pz.
The relations are:
\begin{equation}
\frac{\enthZ(T)}{\Rg T} = - \tc{0} T^{-2} 
                       + \tc{1} T^{-1} \ln(T)
                       + \tc{2} 
                       + \tc{3} \frac{T}{2}
                       + \tc{4} \frac{T^2}{3}
                       + \tc{5} \frac{T^3}{4}
                       + \tc{6} \frac{T^4}{5}
                       + \tc{8} T^{-1}
\label{therm:DH}
\end{equation}
and
\begin{equation}
\frac{\entrZ(T)}{\Rg}    = - \tc{0} \frac{T^{-2}}{2}
                       - \tc{1} T^{-1}
                       + \tc{2} \ln(T)
                       + \tc{3} T
                       + \tc{4} \frac{T^2}{2}
                       + \tc{5} \frac{T^3}{3}
                       + \tc{6} \frac{T^4}{4}
                       + \tc{9}
\label{therm:DS}
\end{equation}
%
Table~\ref{therm:unit} gives the units of the parameters.
\begin{table}
\centering
\begin{tabular}{cccc}\toprule
Parameter & homegeneous SI unit &
Parameter & homegeneous SI unit \\\midrule
\tc{0} & \unit{K^2}         &
\tc{1} & \unit{K}           \\
\tc{2} & \nounit            &
\tc{3} & \unit{K^{-1}}      \\
\tc{4} & \unit{K^{-2}}      &
\tc{5} & \unit{K^{-3}}      \\
\tc{6} & \unit{K^{-4}}      &
\tc{7} & not used, always 0 \\
\tc{8} & \unit{K}           &
\tc{9} & \nounit            \\
\bottomrule
\end{tabular}
\caption[Units of NASA polynoms parameters]{\label{therm:unit}Units of the parameters of the NASA polynoms, as defined in
Eq.~\ref{therm:DH} and \ref{therm:DS}.}
\end{table}
%
These equations are easily differentiables:
\begin{equation}
\doverdT{\left(\frac{\enthZ(T)}{\Rg T}\right)} 
                     =   2 \tc{0} T^{-3} 
                       - \tc{1} T^{-2} \ln(T)
                       + \tc{1} T^{-2}
                       + \tc{3} \frac{1}{2}
                       + \tc{4} \frac{2}{3} T
                       + \tc{5} \frac{3}{4} T^2
                       + \tc{6} \frac{4}{5} T^3
                       - \tc{8} T^{-2}
\label{therm:dDH_dT}
\end{equation}
and
\begin{equation}
\doverdT{\left(\frac{\entrZ(T)}{\Rg}\right)} 
                     =   \tc{0} T^{-3}
                       + \tc{1} T^{-2}
                       + \tc{2} T^{-1}
                       + \tc{3}
                       + \tc{4} T
                       + \tc{5} T^2
                       + \tc{6} T^3
\label{therm:dDS_dT}
\end{equation}

\subsubsection{Microscopic description}