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%%% Copyright (c) 2003 by Oliver Schneider                                  %%%
%%% Portions Copyright (c) Lehrstuhl Chemische Reaktionstechnik BTU Cottbus %%%
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%%%   My exam preparations for CRE. Most parts taken from the original      %%%
%%%   script, but slightly changed, corrected and commented.                %%%
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%%% Part: Summary                                                           %%%
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\special{pdf: docinfo << /Author   (Oliver Schneider [cre AT assarbad DOT net])
                         /Title    (My CRE exam preparation [Part: Summary] [v1.03])
                         /Subject  (Chemical Reaction Engineering)
                         /Keywords (BTU Cottbus, Assarbad, CRE, Chemical Reaction Engineering, Stoichiometry, Thermodynamics, Microkinetics, Reactor Design, PFR, CSTR, Batch Reactor, Excersises, Solutions)
                         /Creator  (LaTeX2e) >>}
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\begin{document}
%% The document was written using \LaTeXe
%% ---------------------------------------------------------------------------
  \section{Stoichiometry}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Definitions}
  \begin{enumerate}
    \item \textsc{Basics}:\\
      $N$ usually denotes the number of species/components in a reaction, whereas $M$
      denotes the number of reactions in a reaction system. We use the variables
      $i$ and $j$ as follows
      \begin{eqnarray*}
        i \in \left[ 1, \dots N \right] \\
        j \in \left[ 1, \dots M \right]
      \end{eqnarray*}
    \item \textsc{Element species matrix}:\\
      $\underline{B}$ contains the number of atoms of an element per molecule of a species.
      You have to set it up as follows for the four species $O_2$, $CO_2$, $H_O$ and $C_2H_5OH$:
      \begin{displaymath}
        \underline{B} =
        \left(\begin{array}{l|llll}
        {} & O_2            & CO_2            & H_2O            & C_2H_5OH \\ \hline
         C & \beta_{C, O_2} & \beta_{C, CO_2} & \beta_{C, H_2O} & \beta_{C, C_2H_5OH} \\
         O & \beta_{O, O_2} & \beta_{O, CO_2} & \beta_{O, H_2O} & \beta_{O, C_2H_5OH} \\
         H & \beta_{H, O_2} & \beta_{H, CO_2} & \beta_{H, H_2O} & \beta_{H, C_2H_5OH}
        \end{array}\right) =
        \left(\begin{array}{l|cccc}
        {} & O_2 & CO_2 & H_2O & C_2H_5OH \\ \hline
         C & 0   & 1    & 0    & 2 \\
         O & 2   & 2    & 1    & 1 \\
         H & 0   & 0    & 2    & 6
        \end{array}\right)
      \end{displaymath}
    \item \textsc{Amount of substance}:\\
      \begin{displaymath}
        \underline{\Delta n_i} =
        \left(\begin{array}{c}
          \Delta n_1 \\
          \vdots \\
          \Delta n_N
        \end{array}\right) =
        \left(\begin{array}{l}
          \Delta n_{O_2}  \\
          \Delta n_{CO_2} \\
          \Delta n_{H_2O} \\
          \Delta n_{C_2H_5OH}
        \end{array}\right)
      \end{displaymath}
    \item \textsc{Reactions need to be in balance}:\\
      The overall amount of elements is constant even though the amount of substance
      of particular species may change:
      \begin{eqnarray*}
        \underline{0} & = & \underline{B} \cdot \underline{\Delta n} \\
        \leadsto
        \left(\begin{array}{l}
          0 \\
          0 \\
          0 \\
          0
        \end{array}\right)
        & = & \left(\begin{array}{l|llll}
        {} & O_2            & CO_2            & H_2O            & C_2H_5OH \\ \hline
         C & \beta_{C, O_2} & \beta_{C, CO_2} & \beta_{C, H_2O} & \beta_{C, C_2H_5OH} \\
         O & \beta_{O, O_2} & \beta_{O, CO_2} & \beta_{O, H_2O} & \beta_{O, C_2H_5OH} \\
         H & \beta_{H, O_2} & \beta_{H, CO_2} & \beta_{H, H_2O} & \beta_{H, C_2H_5OH}
        \end{array}\right) \cdot
        \left(\begin{array}{l}
          \Delta n_{O_2}  \\
          \Delta n_{CO_2} \\
          \Delta n_{H_2O} \\
          \Delta n_{C_2H_5OH}
        \end{array}\right) \\
      \end{eqnarray*}
      One will usually search for the $\Delta n_i$ while $\underline{B}$ is known. To do this:
      First one should apply the \textsc{Gauss} elimination and then develop the linear
      equation system from it. The matrix form actually is already an equation system and
      it is possible to use it directly to get the needed values for the unknown $\Delta n_i$.\\
      Please note, that the key components/species will be given (or in a laboratory measured).
    \item \textsc{Extent of reaction} (dt.: Reaktionslaufzahl):\\
      The extent of reaction denotes the amount of substance which has already been
      consumed or produced. The used symbol is the small greek letter Xi ($\xi$; the
      capital letter Xi looks like this: $\Xi$).
      The extent of reaction remains the same throughout one reaction!
      \begin{displaymath}
        \xi = \frac{n_i - n_{i0}}{\nu_i} = \frac{\Delta n_i}{\nu_i} = \frac{\Delta n_1}{\nu_1} = \cdots \frac{\Delta n_N}{\nu_N}
      \end{displaymath}
    \item \textsc{Conversion} (dt.: Umsatzgrad):\\
      The conversion denotes the percentage of amount of substance which has already reacted.
      It is specific to a single species but can be transferred to be meaningful for other species, too
      (see \textsc{Stoichiometric Table}).
      \begin{displaymath}
        X_i = \frac{n_{i0} - n_i}{n_{i0}}
      \end{displaymath}

  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Changes in amount of substance and key components}
  Basic equation: $\underline{B} \cdot \underline{\Delta n} = \underline{0}$\\
  Approach:
  \begin{enumerate}
    \item Set up the element species matrix $\underline{B}$.
    \item Determine the rank of $\underline{B}$.\\
      $\leadsto$ Number of key components: $N_{kc} = N - Rg(\underline{B})$
    \item Measure $\Delta n$ of key components.\\
      $\leadsto$ $\Delta n$ of key components are known (or given).
    \item Plug in $\Delta n$ of key components into the basic equation.
    \item Solve the linear equation system (in form of a matrix) to get the $\Delta n$ for non-key components.
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Extent of reaction and key reactions}
  Basic equation: $\underline{N} \cdot \underline{\xi} = \underline{\Delta n}$\\
  Approach:
  \begin{enumerate}
    \item Set up the matrix of stoichiometric coefficients, $\underline{N}$.
    \item Determine the rank of $\underline{N}$.\\
      $\leadsto$ Number of key reactions: $N_{krc} = Rg(\underline{N})$
    \item Linearly independent rows describe the key reactions stoichiometrically (i.e. form the key reactions by taking the $\nu$'s from the rearranged matrix - rearranging was done using the \textsc{Gauss} elimination).
    \item Delete $\xi_j$ of the linearly dependent non-key reactions (i.e. cancel out the rows for the linearly dependent non-key reactions in the $\xi$-matrix/vector).
    \item Delete $\nu_{ij}$ and $\Delta n_i$ for the non-key reactions. (i.e. remove the respective rows from these matrices/vectors).
    \item Solve the equations for $\xi_j$ of key reactions.
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \newpage
  \section{Thermodynamics}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Important definitions}
  \begin{enumerate}
    \item Enthalpy change on reaction $\Delta H_R$ is the molar heat of formation
      of the products $\Delta H_{f,i}$ minus the molar heat of formation of educts.\\
      $\leadsto$ The heat tonality can be exo- ($\Delta H_R < 0$; i.e. heat is produced
      by the reaction), endo- ($\Delta H_R > 0$; i.e. the reaction consumes energy)
      or diathermic (i.e. neutral).
    \item \textsc{Hess}'s law of heat summation: $\Delta H_R$ is independent of the reaction
      path. (This allows to calculate the heat of reaction for any reaction, although
      it might not be possible to realize it experimentally!)
    \item \textsc{Le Chatelier}'s principle: A system in chemical equilibrium shifts in response
      to any imposed change of factors governing the equilibrium.\\
      \textsc{Note}: Catalysts \underline{never} have any influence on the equilibrium. They
      accellerate or inhibit a reaction, but they never shift the equilibrium!
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Important formulae}
  \begin{enumerate}
    \item Enthalpy change on reaction / heat of reaction
      \begin{eqnarray*}
        \Delta H_R^0   & = & \sum_{i=1}^N \nu_i \cdot \Delta H_{f,i}^0 \quad \mbox{this holds for STP} \\
        \Delta H_R (T) & = & \Delta H_R^0 + \int\limits_{T^{(0)}}^T \Delta c_p (T) dT \\
        \Delta c_p (T) & = & \sum_{i=1}^N \nu_i \cdot c_{pi} (T)
      \end{eqnarray*}
      Often the formula to calculate $\Delta c_{pi}$ appears similar to this
      \begin{displaymath}
        \Delta c_{pi} (T) = A + BT + CT^2 + DT^3,
      \end{displaymath}
      where A, B, C and D are coefficients one can find in standard tables.
    \item Gibbs free energy
      \begin{eqnarray*}
        \Delta G_R^0   & = & \sum_{i=1}^N \nu_i \cdot \Delta G_{f,i}^0 \quad \mbox{this holds for STP} \\
                       & = & - RT \cdot \ln K_p^0 \\
                       & = & - RT \cdot \ln K_a^0 \\
        \ln K_p (T)     & = & \ln K_p^0 + \int\limits_{T^{(0)}}^T \frac{\Delta H_R^0}{RT^2} dT \\
        \ln K_a (T)     & = & \ln K_a^0 + \int\limits_{T^{(0)}}^T \frac{\Delta H_R^0}{RT^2} dT \\
        K_a^0          & \sim & K_p^0
      \end{eqnarray*}
    \item Equilibria
      \begin{enumerate}
        \item $K_c\quad\longrightarrow$ concentration equilibrium
        \item $K_a\quad\longrightarrow$ activity equilibrium
        \item $K_p\quad\longrightarrow$ pressure equilibrium
        \item $K_x\quad\longrightarrow$ conversion equilibrium
      \end{enumerate}
      \begin{eqnarray*}
        K_c & = & \prod_{i=1}^N c_i^{\nu_i} = \prod_{i=1}^N \left( \frac{c_i}{c^{(0)}} \right)^{\nu_i} \quad\mbox{("normalized form")}\\
        K_a & = & \prod_{i=1}^N a_i^{\nu_i} \\
        K_p & = & \prod_{i=1}^N p_i^{\nu_i} = K_c \cdot \left( \frac{c^{(0)} \cdot RT}{p^{(0)}} \right)^{\sum\limits_{i=1}^N \nu_i} = \prod_{i=1}^N \left( \frac{p_i}{p^{(0)}} \right)^{\nu_i} \quad\mbox{("normalized form")}\\
        K_x & = & \prod_{i=1}^N x_i^{\nu_i} = K_p \cdot \left(\frac{p}{p^{(0)}} \right)^{- \sum\limits_{i=1}^N \nu_i} \quad \mbox{(where p is the overall pressure)}
      \end{eqnarray*}
      The equilibria values relate as follows:
      \begin{displaymath}
        K_a = \left(p^{(0)}\right)^{- \sum\limits_{i=1}^N \nu_i} \cdot K_p = \left( \frac{p}{p^{(0)}}\right)^{\sum\limits_{i=1}^N \nu_i} \cdot K_x = c^{- \sum\limits_{i=1}^N \nu_i} \cdot K_c
      \end{displaymath}
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Explanations}
  \begin{enumerate}
    \item STP means: \textbf{S}tandard \textbf{T}emperature and \textbf{P}ressure
    \item Standard temperature: $25^0C = 298K$
    \item Standard pressure: $p^{(0)} = 1,013 bar = 101,3 kPa = 1,013 \cdot 10^5 Pa$
    \item Standard concentration: $c^{(0)} = 1 \frac{mol}{l}$
    \item \textsc{Important:} $\Delta G_{f,i}^0 = \Delta H_{f,i}^0 = 0$ for elements!
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \newpage
  \section{Microkinetics (Reaction kinetics)}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Description of kinetics}
  \begin{enumerate}
    \item Reaction rate (N being the number of species, M being the number of reactions and i=1,..,N; M=1,..,M):
      \begin{displaymath}
        r_j = \frac{1}{\nu_{i,j}} \cdot \frac{1}{V} \cdot \frac{dn_i}{dt} = \frac{1}{\nu_{i,j}} \cdot \frac{dc_i}{dt} \quad\quad \mbox{(because }V=const. \mbox{ and } c = \frac{n}{V}\mbox{)}
      \end{displaymath}
    \item Power law:
      \begin{displaymath}
        r_j = k_j \cdot \prod_{i=1}^N c_i^{\kappa_{i,j}} \quad\quad
        \underbrace{
          \mbox{with } \kappa_{i,j} =
          \left\{
          \begin{array}{ll}
            \left| \nu_{i,j} \right| & \mbox{for educts} \\
            0 & \mbox{for products}
          \end{array}
          \right.
          \mbox{ \textbf{\underline{or}} }
          \begin{array}{ll}
            0 & \mbox{for educts} \\
            \left| \nu_{i,j} \right| & \mbox{for products}
          \end{array}}_{\mbox{ for \textbf{\underline{elementary reactions}} only!}}
      \end{displaymath}
    \item \textsc{Arrhenius}' law (dependency on temperature):
      \begin{displaymath}
        k_j (T) = k_{\infty, j} \cdot e^{- \frac{E_{a,j}}{RT}}
      \end{displaymath}
    \item Net rate of changes in amount of substance:
      \begin{displaymath}
        R_i = \frac{dc_i}{dt} = \sum_{j=1}^M \nu_{i,j} r_j
      \end{displaymath}
      For ideal gases we have $c_i = \frac{p_i}{R \cdot T}$, hence:
      \begin{displaymath}
        R_i = \frac{dc_i}{dt} = \frac{1}{RT} \cdot \frac{dp_i}{dt}
      \end{displaymath}
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Determination of kinetics}
  Measure the values for concentration at different points in time: $c(t); c(t + \Delta t)$
  \begin{enumerate}
    \item Differential method
      \begin{enumerate}
        \item Calculate approximation values for r:
          \begin{displaymath}
            r = \frac{1}{\nu_i} \cdot \frac{c_i}{dt} \approx \frac{1}{\nu_i} \cdot \underbrace{\frac{c_i(t + \Delta t) - c_i(t)}{\Delta t}}_{\mbox{similar to the definition of the differential}}
          \end{displaymath}
        \item Use the power law ($r = k \cdot \prod\limits_{i=1}^N c_i^{\kappa_{i}}$)
        \item Find the logarithms:
          \begin{displaymath}
            \log r = \log k + \sum_{i=1}^N (\kappa_i \log c_i)
          \end{displaymath}
      \end{enumerate}
      $\Rightarrow$ You get a linear set of equations, $\kappa_i$ can be estimated by solving it.\\
      $\Rightarrow$ Plot $\log r$ versus $\log c_i$ to determine $k$.
    \item Integral method\\
      \underline{Example:} $A_1 \longrightarrow A_2$
      \begin{enumerate}
        \item Use the power law and set it equal to the differential form:
          \begin{displaymath}
            r = k \cdot \prod_{i=1}^N c_i^{k_i} = k \cdot c_1^{\kappa_1} \quad\Longleftrightarrow\quad
              r = \frac{1}{\nu_i} \cdot \frac{dc_i}{dt} = \frac{1}{-1} \cdot \frac{dc_1}{dt} = - \frac{dc_1}{dt}
              \quad\leadsto\quad
              k \cdot c_1^{\kappa_1} = - \frac{dc_1}{dt}
          \end{displaymath}
        \item Solve the differential equation by separating the variables
          \begin{displaymath}
            \frac{dc_1}{c_1^{\kappa_1}} = - k  dt
              \quad\leadsto\quad
              c_1 = c_{10} \cdot e^{-kt}
          \end{displaymath}
        \item Find the logarithms
          \begin{displaymath}
            \ln c_1 = \ln c_{10} - kt
          \end{displaymath}
      \end{enumerate}
      $\Rightarrow$ Plot $\ln c_1$ versus $t$ to determine $k$.\\
      \textsc{In general for} $\quad\kappa_1 \not= 1\quad$ \textsc{it holds} $\quad\left( \frac{1}{c_1}\right)^{\kappa_1 - 1} - \left( \frac{1}{c_{10}}\right)^{\kappa_1 - 1} = (\kappa_1 - 1) k_{\kappa_1} \cdot t$
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Special reactions}
  General approach:
  \begin{enumerate}
    \item Get the number of species and of reactions.
    \item Set up the differential equations $R_i = \frac{dc_i}{dt} = \sum\limits_{j=1}^M \nu_{i,j} \cdot r_j$
    \item Substitute $r_j$ by the power law.
    \item Make any necessary assumptions.
    \item Solve the system of differential equations.
      \begin{enumerate}
        \item Reversible reactions
          \begin{displaymath}
            A_1 + A_2 \begin{array}{c}{r_+} \\ \rightleftharpoons \\ {r_-}\end{array} A_3 + A_4
          \end{displaymath}
          Considering equilibrium: $r_+ = r_-$
        \item Multiple concurrent reactions / parallel reactions (multiple reaction pathes)
          \begin{displaymath}
          A_1
            \begin{array}{ll}
              & A_2 \\
              \stackrel{1}{\nearrow} & \\
              \stackrel{\textstyle \searrow}{\scriptstyle 2} & \\
              & A_3
            \end{array}
          \end{displaymath}
        \item Consecutive reactions
          \begin{displaymath}
            A_1 \stackrel{1}{\longrightarrow} A_2 \stackrel{2}{\longrightarrow} A_3
          \end{displaymath}
        \item Homogeneous catalyzed reactions
          \begin{enumerate}
            \item General reaction equation:
              \begin{displaymath}
                A_1 + A_2 \begin{array}{c}_1 \\ \rightleftharpoons \\ ^2\end{array} A_{12} \stackrel{3}{\rightarrow} A_2 + A_3
              \end{displaymath}
              Where $A_2$ is the catalyst and $A_{12}$ is the activated complex.
            \item Set up the differential equations for $R_i$ using the power law for $r_j$.
            \item Assumptions:
              \begin{enumerate}
                \item The total amount of catalyst is constant: $c_2 + c_{12} = c_{20}$
                \item Reaction (3) is the rate determination step $\quad\leadsto\quad r_1 = r_2\quad$ or $\quad \frac{dc_{12}}{dt} = 0$
              \end{enumerate}
          \end{enumerate}
      \end{enumerate}
  \end{enumerate}
%% ---------------------------------------------------------------------------
  \newpage
  \section{Reactor design}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Stoichiometric table for batch systems}
  \begin{eqnarray*}
    & \nu_1 A_1 + \nu_2 A_2 + \nu_3 A_3 + \nu_4 A_4 = 0 & \mbox{with k=1} \\
    & \begin{array}{|c|c|c|c|}\hline\rule{0mm}{6mm}
      \mbox{Species} & n_{i0} & \Delta n_i = -\frac{\nu_i}{\nu_k}n_{k0} X_k & n_i = n_{i0} + \Delta n_i \\ \hline \rule{0mm}{6mm}
      A_1 & n_{10} & -n_{10} X_1                    & n_1 = n_{10} - n_{10} X_1 \\\rule{0mm}{6mm}
      A_2 & n_{20} & -\frac{\nu_2}{\nu_1}n_{10} X_1 & n_2 = n_{20} -\frac{\nu_2}{\nu_1}n_{10} X_1 \\\rule{0mm}{6mm}
      A_3 & n_{30} & -\frac{\nu_3}{\nu_1}n_{10} X_1 & n_3 = n_{30} -\frac{\nu_3}{\nu_1}n_{10} X_1 \\\rule{0mm}{6mm}
      A_4 & n_{40} & -\frac{\nu_4}{\nu_1}n_{10} X_1 & n_4 = n_{40} -\frac{\nu_4}{\nu_1}n_{10} X_1 \\\rule{0mm}{6mm}
      \mbox{Inert} & n_{I0} & -                                & n_I = n_{I0} \\\hline \rule{0mm}{6mm}
      \mbox{Sum}   & n_{\mbox{sum}0} & -                       & n_{\mbox{sum}} = n_{\mbox{sum}0} - (1 + \frac{\nu_2}{\nu_1} + \frac{\nu_3}{\nu_1} + \frac{\nu_4}{\nu_1})n_{10} X_1 \\ \hline
    \end{array} &
  \end{eqnarray*}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Stoichiometric table for flow systems}
  \begin{eqnarray*}
    & \nu_1 A_1 + \nu_2 A_2 + \nu_3 A_3 + \nu_4 A_4 = 0 & \mbox{with k=1}\\
    & \begin{array}{|c|c|c|c|}\hline\rule{0mm}{6mm}
      \mbox{Species} & F_{i0} & \Delta F_i = -\frac{\nu_i}{\nu_k}F_{k0} X_k & F_i = F_{i0} + \Delta F_i \\ \hline \rule{0mm}{6mm}
      A_1 & F_{10} & -F_{10} X_1                    & F_1 = F_{10} - F_{10} X_1 \\\rule{0mm}{6mm}
      A_2 & F_{20} & -\frac{\nu_2}{\nu_1}F_{10} X_1 & F_2 = F_{20} -\frac{\nu_2}{\nu_1}F_{10} X_1 \\\rule{0mm}{6mm}
      A_3 & F_{30} & -\frac{\nu_3}{\nu_1}F_{10} X_1 & F_3 = F_{30} -\frac{\nu_3}{\nu_1}F_{10} X_1 \\\rule{0mm}{6mm}
      A_4 & F_{40} & -\frac{\nu_4}{\nu_1}F_{10} X_1 & F_4 = F_{40} -\frac{\nu_4}{\nu_1}F_{10} X_1 \\\rule{0mm}{6mm}
      \mbox{Inert} & F_{I0} & -                                & F_I = F_{I0} \\ \hline\rule{0mm}{6mm}
      \mbox{Sum}   & F_{\mbox{sum}0} & -                       & F_{\mbox{sum}} = F_{\mbox{sum}0} - (1 + \frac{\nu_2}{\nu_1} + \frac{\nu_3}{\nu_1} + \frac{\nu_4}{\nu_1})F_{10} X_1 \\ \hline
    \end{array} &
  \end{eqnarray*}
%% ---------------------------------------------------------------------------
  \hrule\smallskip
%% ---------------------------------------------------------------------------
  \subsection{Design equations}
  \begin{enumerate}
    \item \textsc{Batch Reactor}:\\
      \begin{displaymath}
        \overbrace{\mbox{input}}^{=0} = \overbrace{\mbox{output}}^{=0} + \mbox{disappearance} + \mbox{accumulation}
      \end{displaymath}
    \item \textsc{Continuous Stirred Tank Reactor}:\\
      \begin{displaymath}
        \mbox{input} = \mbox{output} + \mbox{disappearance} + \overbrace{\mbox{accumulation}}^{=0}
      \end{displaymath}
    \item \textsc{Plug Flow Reactor}:\\
      \begin{displaymath}
        \mbox{input} = \mbox{output} + \mbox{disappearance} + \overbrace{\mbox{accumulation}}^{=0}
      \end{displaymath}
  \end{enumerate}
  \begin{displaymath}
    \begin{array}{|l|rcl|rcl|}\hline\rule{0mm}{8mm}
      \mbox{Batch Reactor (BR)}                     & n_{10} \cdot dX_1 & = & r \cdot V \cdot dt & t    & = & n_{10} \cdot \int\limits_0^{X_1} \frac{dX_1}{rV} \\\rule{0mm}{6mm}
      \mbox{Continuous Stirred Tank Reactor (CSTR)} & F_{10} \cdot X_1  & = & r \cdot V          & \tau & = & n_{10} \cdot \frac{dX_1}{r} \\\rule{0mm}{6mm}
      \mbox{Plug Flow Reactor (PFR)}                & F_{10} \cdot dX_1 & = &r \cdot dV          & \tau & = & c_{10} \cdot \int\limits_0^{X_1} \frac{dX_1}{r} \\ \hline
    \end{array}
  \end{displaymath}\\
  Where $t$ is the time that the mixture is in the batch reactor, $\tau$ is the space time.
  A space time of $5min$ denotes that every five minutes one reactor volume of feed is
  being treated by the reactor. $\tau = \frac{1}{s}$, where $s$ is the space velocity (in $min^{-1}$)
  and denotes how many reactor volumes of feed are being treated during a given time.\\\\
  \textsc{Note:} Volume of feed ($V$) is not necessarily the same as volume of reactor($V_R$). But we
  only consider volume of feed and how much time is needed to treat one reactor volume of
  feed. $\quad \leadsto V = \epsilon_R \cdot V_R \quad$ (with $\epsilon_R$ being the ratio of
  filling volume by reactor volume).

%% ---------------------------------------------------------------------------
  \newpage
  \begin{thebibliography}{99}
    \bibitem{1}\textsc{Baerns} Manfred, \textsc{Hofmann} Hanns, \textsc{Renken} Albert.\\
      \textsl{Chemische Reaktionstechnik}.\\
      Georg Thieme Verlag Stuttgart - New York.\\
      \texttt{German and available in the BTU library}\\
      \textsf{In my opinion the best book since it explains everything from the very scratch. BUT
      it also has a drawback: there is much information we simply don't need!}
    \bibitem{2}\textsc{Levenspiel} Octave.\\
      \textsl{Chemical Reaction Engineering}.\\
      John Wiley \& Sons, Inc..\\
      \texttt{English and available in the BTU library}\\
      \textsf{Very useful for Microkinetics and Reactor Design parts.}
    \bibitem{3}\textsc{Fogler} H. Scott.\\
      \textsl{Elements of Chemical Reaction Engineering}.\\
      Prentice Hall.\\
      \texttt{English and available in the BTU library}\\
      \textsf{This is partially useful for Reactor Design but not at all for the rest. Description
      of different reactors and the origin of their design equations good explained.}
    \bibitem{4}\textsc{Bechmann} Wolfgang, \textsc{Schmidt} Joachim.\\
      \textsl{Einstieg in die Physikalische Chemie f\"ur Nebenf\"achler }.\\
      B.G. Teubner Stuttgart - Leipzig.\\
      \texttt{German and available in the BTU library}\\
      \textsf{Microkinetics.}
    \bibitem{5}\textsc{M\"uller-Erlwein} Erwin.\\
      \textsl{Chemische Reaktionstechnik}.\\
      B.G. Teubner Stuttgart - Leipzig.\\
      \texttt{German}\\
      \textsf{Stoichiometry, Thermodynamics and partially Microkinetics and Reactor Design.}
    \bibitem{6}\textsc{Butt} John B..\\
      \textsl{Reaction Kinetics and Reactor Design}.\\
      Prentice Hall.\\
      \texttt{English and available in the BTU library}\\
      \textsf{Partially useful for Microkinetics and Reactor Design parts.}
    \bibitem{7}\textsc{Atkins} Peter William.\\
      \textsl{The Elements of Physical Chemistry}.\\
      Oxford University Press.\\
      \texttt{English and available in the BTU library}\\
      \textsf{Thermodynamics and partially Microkinetics.}
    \bibitem{8}\textsc{Kopka} Helmut.\\
      \textsl{{\LaTeX} Einf\"uhrung}. Band 1\\
      Addison-Wesley.\\
      \texttt{German}
    \bibitem{9}\textsc{Dalheimer} Mathias Kalle.\\
      \textsl{{\LaTeX} kurz \& gut}.\\
      O'Reilly.\\
      \texttt{German}
  \end{thebibliography}
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\end{document}

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  We need to employ the multiplication of two matrices. The only condition for this
  is, that the number of columns of the first matrix (A) be the same as the number of
  rows of the second matrix (B). Rule:
  \begin{displaymath}
    \begin{array}{c|c}
      A \cdot B & {
        \begin{array}{ccccc}
          b_{1,1} & \cdots & b_{1,k} & \cdots & b_{1,r} \\
          \vdots  &        & \vdots  &        & \vdots  \\
          b_{m,1} & \cdots & b_{m,k} & \cdots & b_{m,r}
        \end{array}
      } \\ \hline
      {
        \begin{array}{ccc}
          a_{1,1} & \cdots & a_{1,m} \\
          \vdots  &        & \vdots  \\
          a_{i,1} & \cdots & a_{i,m} \\
          \vdots  &        & \vdots  \\
          a_{n,1} & \cdots & a_{n,m} \\
        \end{array}
      } & {
        \begin{array}{ccc}
        & \downarrow & \\
        \to & c_{i,k} & \\
        & &
        \end{array}
      }
    \end{array}
  \end{displaymath}


  \section{CRE basics}
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  \hrule\smallskip
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  \subsection{Miscellaneous}
  \begin{enumerate}
    \item \textsc{Preliminary knowledge}\\
      The first basic relations is this $m_i = M_i \cdot n_i$, where $m$ is the mass
      of a species $[g]$, $M$ is the molar mass $[\frac{g}{mol}]$ and $n$ is the
      amount of substance $[mol]$ ($i$ is the index for a particalur species of the reaction
      system). The unit $mol$ is a basic unit and is defined as
      being $6,022136 \cdot 10^{23} mol^{-1}$ (i.e. $6,022136 \cdot 10^{23}$ atoms/molecules
      per mole of substance), the so called \textsc{Avogadro} constant. Anyway, this is not
      very important, since chemistry and
      hence CRE uses mostly mol as the basic unit (with no respect to any underlying
      definitions).
    \item \textsc{Principle of mass conservation}\\
      This principle states, that in a closed reaction system the overall mass will not
      change. This can be seen as a similarity to conservation of energy, which states
      that energy can be transformed (e.g. potential to thermal forms), but cannot be
      lost in a closed system.
    \item \textsc{Balanced reactions}\\
      In a certain reaction, the number of elements on each side has to be balanced.
      This is one of the basics of chemistry and is achieved by using methods of the
      so called stoichiometry.
    \item \Stoichio
  \end{enumerate}


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  \newpage
