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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: Excersises                                                        %%%
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\special{pdf: docinfo << /Author   (Oliver Schneider [cre AT assarbad DOT net])
                         /Title    (My CRE exam preparation [Part: Excersises] [v1.00])
                         /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 {\sffamily\textbf{bold sans-serif}} texts are the problems as given.
  \section{Stoichiometry}
%% ---------------------------------------------------------------------------
  \subsection*{Problem 1}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    Consider the following set of reactions
    \begin{eqnarray*}
      4CH_4 + O_2 & \stackrel{1}{\to} & 2C_2H_6 + 2H_2O \\
      2CH_4 + O_2 & \stackrel{2}{\to} & 2CH_3OH \\
      2CH_3OH + O_2 & \stackrel{3}{\to} & 2CH_2O + 2H_2O \\
      2C_2H_6 + 3O_2 & \stackrel{4}{\to} & 4CH_2O + 2H_2O
    \end{eqnarray*}
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Show, that there are 3 key components within the system!
    \\
  }}
  The number of key components is the number of species minus rank of the element species matrix.
  \begin{displaymath}
    N_{species} - Rg(\underline{B}) = N_{kc}
  \end{displaymath}
  First build the element species matrix:
  \begin{displaymath}
    \underline{B} =
    \left(\begin{array}{l|cccccc}
    {} & CH_4 & O_2 & C_2H_6 & H_2O & CH_3OH & CH_2O \\ \hline
    C & 1 & 0 & 2 & 0 & 1 & 1 \\
    O & 0 & 2 & 0 & 1 & 1 & 1 \\
    H & 4 & 0 & 6 & 2 & 4 & 2
    \end{array}\right)
  \end{displaymath}
  Calculate its rank using the Gauss method:
  \begin{displaymath}
    \underline{B} =
    \left(\begin{array}{rrrrrr}
    1 & 0 & 2 & 0 & 1 & 1 \\
    0 & 2 & 0 & 1 & 1 & 1 \\
    4 & 0 & 6 & 2 & 4 & 2
    \end{array}\right)%%\begin{array}{c}{}\\{}\\\mbox{- 4 times first row}\end{array}
    =
    \left(\begin{array}{rrrrrr}
    1 & 0 & 2 & 0 & 1 & 1 \\
    0 & 2 & 0 & 1 & 1 & 1 \\
    0 & 0 & -2 & 2 & 0 & -2
    \end{array}\right)
  \end{displaymath}
  This shows we have 3 linearly independent equations, that is $Rg(\underline{B}) = 3 = N_{nkc}$ (number of non-key compenents)
  \begin{displaymath}
    N_{species} - N_{nkc} = N_{kc} = 6 - 3 = 3
  \end{displaymath}
  The number of key components is 3.\\\hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    For the chosen key components $CH_4$, $O_2$ and $C_2H_6$ the following changes in amount of substance have been measured:
    \begin{displaymath}
      \Delta n_{CH_4} = -8 mol,\quad\Delta n_{O_2} = -12 mol,\quad \Delta n_{C_2H_6} = 3 mol
    \end{displaymath}
    Calculate the changes in amount of substance of the non-key components using the element species matrix!
    \\
  }}
  This can be solved by using the relation:
  \begin{eqnarray*}
    \underline{B} \cdot \Delta n & = & 0 \\
    & = &
    \left(\begin{array}{rrrrrr}
    1 & 0 & 2 & 0 & 1 & 1 \\
    0 & 2 & 0 & 1 & 1 & 1 \\
    4 & 0 & 6 & 2 & 4 & 2
    \end{array}\right)
    \cdot
    \left(\begin{array}{l}
    \Delta n_{CH_4} \\
    \Delta n_{O_2} \\
    \Delta n_{C_2H_6} \\
    \Delta n_{H_2O} \\
    \Delta n_{CH_3OH} \\
    \Delta n_{CH_2O}
    \end{array}\right) \\
     & = &
    \left(\begin{array}{rrrrrr}
    1 & 0 & 2 & 0 & 1 & 1 \\
    0 & 2 & 0 & 1 & 1 & 1 \\
    4 & 0 & 6 & 2 & 4 & 2
    \end{array}\right)
    \cdot
    \left(\begin{array}{l}
    -8 mol \\
    -12 mol \\
    3 mol \\
    \Delta n_{H_2O} \\
    \Delta n_{CH_3OH} \\
    \Delta n_{CH_2O}
    \end{array}\right)
  \end{eqnarray*}
  Now splitting this into equations of an equation system ...
  \begin{eqnarray*}
    -8mol + 2\cdot 3mol + \Delta n_{CH_3OH} + \Delta n_{CH_2O}& = & 0 \\
    2\cdot -12mol + \Delta n_{H_2O} + \Delta n_{CH_3OH} + \Delta n_{CH_2O} & = & 0 \\
    4\cdot -8mol + 6\cdot 3mol + 2\Delta n_{H_2O} + 4\Delta n_{CH_3OH} + 2\Delta n_{CH_2O} & = & 0
  \end{eqnarray*}
  ... one can solve the system. Rearranging gives:
  \begin{eqnarray*}
    \Delta n_{CH_3OH} + \Delta n_{CH_2O}& = & 2 \\
    \Delta n_{H_2O} + \Delta n_{CH_3OH} + \Delta n_{CH_2O} & = & 24 \\
    2\Delta n_{H_2O} + 4\Delta n_{CH_3OH} + 2\Delta n_{CH_2O} & = & 14
  \end{eqnarray*}
  This can be written as a matrix again which makes it easier to solve it by using the Gauss method:
  \begin{displaymath}
    \left(\begin{array}{rrrr}
    0 & 1 & 1 & 2 \\
    1 & 1 & 1 & 24 \\
    2 & 4 & 2 & 14
    \end{array}\right)
    \leadsto
    \left(\begin{array}{rrrr}
    1 & 1 & 1 & 24 \\
    0 & 1 & 1 & 2 \\
    0 & 2 & 0 & -34
    \end{array}\right)
  \end{displaymath}
  This gives:
  \begin{eqnarray*}
    2\Delta n_{CH_3OH} = -34mol & \leadsto & \Delta n_{CH_3OH} = -17mol \\
    -17mol + \Delta n_{CH_2O} = 2 & \leadsto & \Delta n_{CH_2O} = 19mol \\
    \Delta n_{H_2O} -17mol + 19 = 24 & \leadsto & \Delta n_{H_2O} = 22mol
  \end{eqnarray*}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    Calculate the number of key reactions and give a set of them!
    \\
  }}
  To do this, we need to write the given reactions in matrix form. Note: The
  stoichiometric coefficients on the left are to be negated!\\
  The Gauss method is used again to calculate the rank of the matrix.
  \begin{eqnarray*}
  \underline{N}^T & = &
    \left(\begin{array}{l|cccccc}
    {}  & CH_4 & O_2 & C_2H_6 & H_2O & CH_3OH & CH_2O \\ \hline
    I   &-4 & -1 &  2 &  2 &  0 &  0 \\
    II  &-2 & -1 &  0 &  0 &  2 &  0 \\
    III & 0 & -1 &  0 &  2 & -2 &  2 \\
    IV  & 0 & -3 & -2 &  2 &  0 &  4
    \end{array}\right) \\
  & = &
    \left(\begin{array}{rrrrrr}
    -4 & -1 &  2 &  2 &  0 &  0 \\
     0 &  1 &  2 &  2 & -4 &  0 \\
     0 & -1 &  0 &  2 & -2 &  2 \\
     0 & -3 & -2 &  2 &  0 &  4
    \end{array}\right)
    =
    \left(\begin{array}{rrrrrr}
    -4 & -1 &  2 &  2 &  0  &  0 \\
     0 &  1 &  2 &  2 & -4  &  0 \\
     0 &  0 &  2 &  4 & -6  &  2 \\
     0 &  0 &  4 &  8 & -12 &  4
    \end{array}\right) \\
  & = &
    \left(\begin{array}{rrrrrr}
    -4 & -1 &  2 &  2 &  0  &  0 \\
     0 &  1 &  2 &  2 & -4  &  0 \\
     0 &  0 &  2 &  4 & -6  &  2 \\
     0 &  0 &  0 &  0 & 0   &  0
    \end{array}\right)
  \end{eqnarray*}
  This shows us, that $Rg(\underline{N}^T) = 3 \leadsto $ hence we have 3 key reactions.
  We will build them from the last step of the Gauss elimination (negative $\nu$ on the left and positive on the right side).
  \begin{eqnarray*}
    4CH_4 + O_2 & \stackrel{1}{\to} & 2C_2H_6 + 2H_2O \\
    4CH_3OH & \stackrel{2}{\to} & O_2 + 2C_2H_6 + 2H_2O \\
    6CH_3OH & \stackrel{3}{\to} & 2C_2H_6 + 4H_2O + 2CH_2O
  \end{eqnarray*}
  \hrule
  \paragraph*{(d)}
  {\sffamily\textbf{
    Calculate the extents of reaction with the changes in amount of substance
    given in (1b) and use them to determine again the changes in number of
    moles of the non-key components!
    \\
  }}
  We know the relation $\underline{N} \cdot \underline{\xi_j} = \underline{\Delta n_i}$ and use the transposed
  matrix from (1c) skipping the 4th reaction which had been found to be linearly
  dependent with reaction 3. Since ${\underline{N}^T}^T = \underline{N}$ we can use it.
  \begin{eqnarray*}
    \underline{N} \cdot \underline{\xi_j} & = &
    \left(\begin{array}{rrr}
      -4 & -2 &  0 \\
      -1 & -1 & -1 \\
       2 &  0 &  0 \\
       2 &  0 &  2 \\
       0 &  2 & -2 \\
       0 &  0 &  2
    \end{array}\right)
    \cdot
    \left(\begin{array}{r}
      \xi_1 \\
      \xi_2 \\
      \xi_3
    \end{array}\right)
    =
    \left(\begin{array}{r}
      -8 mol \\
      -12 mol \\
       3 mol \\
       \Delta n_4 \\
       \Delta n_5 \\
       \Delta n_6
    \end{array}\right)
  \end{eqnarray*}
  Now writing this as an equation system in matrix form, we get:
  \begin{displaymath}
    \left(\begin{array}{rrrr}
      -4 & -2 &  0 & -8 \\
      -1 & -1 & -1 & -12 \\
       2 &  0 &  0 & 3 \\
       2 &  0 &  2 & \Delta n_4 \\
       0 &  2 & -2 & \Delta n_5 \\
       0 &  0 &  2 & \Delta n_6
    \end{array}\right)
    =
    \left(\begin{array}{r}
      {} \\
      \underline{N_1} \\
      {} \\
      {} \\
      \underline{N_2} \\
      {}
    \end{array}\right)
  \end{displaymath}
  Let's now split the matrix into the upper three ($N_1 \leadsto$ key components) and the lower
  three ($N_2 \leadsto$ non-key components) matrices. Of which the latter one needs to be calculated.
  Since it is possibly easier to read, let's write
  \begin{displaymath}
    \underline{N_1} =
    \left(\begin{array}{rrrr}
      -4 & -2 &  0 & -8 \\
      -1 & -1 & -1 & -12 \\
       2 &  0 &  0 & 3
    \end{array}\right)
    \leadsto
    \left\{
      \begin{array}{rcl}
        -4 \xi_1 -2 \xi_2 & = & -8 \\
        -1 \xi_1 -1 \xi_2 -1 \xi_3 & = & -12 \\
         2 \xi_1 & = & 3
      \end{array}
    \right.
  \end{displaymath}
  One can easily see the first solution and the other solutions just depend on
  that first one:
  \begin{eqnarray*}
     2 \xi_1 = 3 & \leadsto & \xi_1 = 1,5 \\
    -4 \cdot 1,5 -2 \xi_2 = -8 & \leadsto & \xi_2 = 1 \\
    -1 \cdot 1,5 -1 -1 \xi_3 = -12 & \leadsto & \xi_3 = 9,5
  \end{eqnarray*}
  Now it is easy to re-calculate the 3 non-key components again, using $\underline{N_2}$.
  \begin{eqnarray*}
    2 \cdot 1,5 + 2 \cdot 9,5 = \Delta n_4 & \leadsto & \Delta n_4 = 22 mol\\
    2 \cdot 1 -2 \cdot 9,5 = \Delta n_5 & \leadsto & \Delta n_5 = -17 mol\\
    2 \cdot 9,5 = \Delta n_6 & \leadsto & \Delta n_6 = 19 mol
  \end{eqnarray*}
  Since this is the same result as above, we are done ;-)
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 2}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    An unknown system with the species $CH_3OH$, $O_2$, $CH_2O$, $H_2O$ and $CO$ is investigated.
  }}
  \\\hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    For the so-called formula conversion $\Delta n_{ij} = \nu_{ij}$ the following formula is valid
    \begin{displaymath}
      \underline{B} \cdot \underline{N} = 0
    \end{displaymath}
    Assume the number of key reactions and key components being equal and give
    a set of key reactions!\\
    NOTE: Choose the stoichiometric coefficients of your chosen key components
    in such a way, that there is only one key component left in each equation!
    \\
  }}
  First let's set up the element species matrix and determine its rank:
  \begin{eqnarray*}
    \underline{B} & = &
    \left(\begin{array}{l|ccccc}
      {} & CH_3OH & O_2 & CH_2O & H_2O & CO \\ \hline
      C  &  1 &  0 &  1 &  0 &  1 \\
      O  &  1 &  2 &  1 &  1 &  1 \\
      H  &  4 &  0 &  2 &  2 &  0
    \end{array}\right) \\
    & = &
    \left(\begin{array}{rrrrr}
      1 &  0 &  1 &  0 &  1 \\
      0 &  2 &  0 &  1 &  0 \\
      0 &  0 &  -2 &  2 &  -4
    \end{array}\right)
  \end{eqnarray*}
  Hence $Rg(\underline{B}) = 3$ and therefore we have 2 key components ($N_{c} - Rg(\underline{B}) = N_{kc}$).
  Now we build up $\underline{N}$, which yields:
  \begin{eqnarray*}
    \underline{B} \cdot \underline{N} & = & 0 \\
    & = &
    \underline{B} \cdot
    \left(\begin{array}{rr}
      \nu_{1,1} & \nu_{1,2} \\
      \nu_{2,1} & \nu_{2,2} \\
      \nu_{3,1} & \nu_{3,2} \\
      \nu_{4,1} & \nu_{4,2} \\
      \nu_{5,1} & \nu_{5,2}
    \end{array}\right)
    =
    \left(\begin{array}{ccccc}
      1 &  0 &  1 &  0 &  1 \\
      1 &  2 &  1 &  1 &  1 \\
      4 &  0 &  2 &  2 &  0
    \end{array}\right) \cdot
    \left(\begin{array}{rr}
      \nu_{1,1} & \nu_{1,2} \\
      1 & 0 \\
      \nu_{3,1} & \nu_{3,2} \\
      \nu_{4,1} & \nu_{4,2} \\
      0 & 1
    \end{array}\right)
  \end{eqnarray*}
  The choice for $\nu_{2,1}$, $\nu_{2,2}$, $\nu_{5,1}$ and $\nu_{5,2}$ is done because of
  the condition that each key component shall only appear once per key reaction.
  \\\\
  We now split the problem into two parts (one per reaction) - and we also use the
  matrix ($\underline{B}$)we got by using the Gauss method (this allowed because the matrix is the
  same anyway, it just looks differently - sure, one could also take the original matrix, but this would require more work):
  \begin{eqnarray*}
    \underline{B} \cdot \underline{N_1} & = &
    \left(\begin{array}{rrrrr}
      1 &  0 &  1 &  0 &  1 \\
      0 &  2 &  0 &  1 &  0 \\
      0 &  0 &  -2 &  2 &  -4
    \end{array}\right) \cdot
    \left(\begin{array}{r}
      \nu_{1,1} \\
      1 \\
      \nu_{3,1} \\
      \nu_{4,1} \\
      0
    \end{array}\right) = 0 \\
    \underline{B} \cdot \underline{N_2} & = &
    \left(\begin{array}{rrrrr}
      1 &  0 &  1 &  0 &  1 \\
      0 &  2 &  0 &  1 &  0 \\
      0 &  0 &  -2 &  2 &  -4
    \end{array}\right) \cdot
    \left(\begin{array}{r}
      \nu_{1,2} \\
      0 \\
      \nu_{3,2} \\
      \nu_{4,2} \\
      1
    \end{array}\right) = 0
  \end{eqnarray*}
  Now writing the first one as an equation system:
  \begin{displaymath}
    \leadsto
    \left\{
      \begin{array}{rcl}
        \nu_{1,1} + \nu_{3,1} & = & 0 \\
        \nu_{4,1} & = & -2 \\
        - 2 \nu_{3,1} + 2 \nu_{4,1} & = & 0 \; \leadsto \; \nu_{4,1} = \nu_{3,1}
      \end{array}
    \right\}
    \leadsto
    \nu_{1,1} = 2
  \end{displaymath}
  This yields the first reaction equation now (check again if it's balanced $\to$ it is!):
  \begin{displaymath}
    2 CH_2O + 2 H_2O \to 2 CH_3OH + O_2
  \end{displaymath}
  Now do the same for the second reaction:
  \begin{displaymath}
    \leadsto
    \left\{
      \begin{array}{rcl}
        \nu_{1,2} + \nu_{3,2} & = & -1 \\
        \nu_{4,2} & = & 0 \\
        - 2 \nu_{3,2} + 2 \nu_{4,2} & = & 4 \; \leadsto \; \nu_{3,2} = -2
      \end{array}
    \right\}
    \leadsto
    \nu_{1,2} = 1
  \end{displaymath}
  The resulting reaction equation looks like this:
  \begin{displaymath}
    2 CH_2O \to CH_3OH + CO
  \end{displaymath}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Calculate the changes in amount of substance of $CH_3OH$, $CH_2O$ and $H_2O$
    with given $\Delta n_{O_2} = 10 mol$ and $\Delta n_{CO} = 2 mol$!
    \\
  }}
  Because the relation $\Delta n_{ij} = \nu_{ij}$ allows us to interchange the
  values for the stoichiometric coefficients $\nu$ and the change in amount of
  substance $\Delta n$, we can employ the already known relation:
  \begin{displaymath}
    \underline{B} \cdot \Delta n = 0
  \end{displaymath}
  \begin{displaymath}
    \left(\begin{array}{ccccc}
      1 &  0 &  1 &  0 &  1 \\
      1 &  2 &  1 &  1 &  1 \\
      4 &  0 &  2 &  2 &  0
    \end{array}\right)
    \cdot
    \left(\begin{array}{c}
      \Delta n_1 \\
      \Delta n_2 \\
      \Delta n_3 \\
      \Delta n_4 \\
      \Delta n_5
    \end{array}\right)
    =
    \left(\begin{array}{ccccc}
      1 &  0 &  1 &  0 &  1 \\
      1 &  2 &  1 &  1 &  1 \\
      4 &  0 &  2 &  2 &  0
    \end{array}\right)
    \cdot
    \left(\begin{array}{c}
      \Delta n_1 \\
      10 mol \\
      \Delta n_3 \\
      \Delta n_4 \\
      2 mol
    \end{array}\right)
    = 0
  \end{displaymath}
  Let's now develop the equation system from the matrix:
  \begin{displaymath}
    \left(\begin{array}{ccccc}
      1 &  0 &  1 &  0 &  1 \\
      1 &  2 &  1 &  1 &  1 \\
      4 &  0 &  2 &  2 &  0
    \end{array}\right)
    \leadsto
    \left\{
      \begin{array}{rcl}
        \Delta n_1 + \Delta n_3 & = & -2 mol \\
        \Delta n_1 + \Delta n_3 + \Delta n_4 & = & -22 mol \\
        4 \Delta n_1 + 2 \Delta n_3 + 2 \Delta n_4 & = & 0 mol
      \end{array}
    \right.
  \end{displaymath}
  Now writing this in matrix form while skipping column 2 and 5 (i.e. the given components $\Delta n_2$ and $\Delta n_5$)
  again will make things easier:
  \begin{eqnarray*}
    \underline{M} & = &
    \left(\begin{array}{r|ccccc}
    \mbox{Column} & \Delta n_1 & \Delta n_3 & \Delta n_4 & \mbox{Result} \\\hline
      & 1 &  1 &  0 & -2 \\
      & 1 &  1 &  1 & -22 \\
      & 4 &  2 &  2 & 0
    \end{array}\right)
    =
    \left(\begin{array}{ccccc}
      1 &  1 &  0 & -2 \\
      0 &  0 &  1 & -20 \\
      0 & -2 &  2 & 8
    \end{array}\right) \\
    & \leadsto &
    \left\{
      \begin{array}{rcr}
        \Delta n_4  & = & -20 mol \\
        \Delta n_3 & = & -24 mol \\
        \Delta n_1 & = & 22 mol
      \end{array}
    \right.
  \end{eqnarray*}
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 3}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    Consider the following set of reactions: \\
    \begin{center}\begin{tabular}{rl}
      RK 1: & $A_1 + A_2 \to A_3 + A_4$ \\
      RK 2: & $A_2 + A_3 \to A_4 + A_5$ \\
      RK 3: & $A_2 + A_5 \to A_4 + A_6$
    \end{tabular}\end{center}
    There are three key reactions within the system. Calculate the changes in
    number of moles $\Delta n_2$, $\Delta n_4$ and $\Delta n_5$, when the following
    changes have been measured:
    \begin{eqnarray*}
      \Delta n_1 & = & -1 mol \\
      \Delta n_3 & = & -2 mol \\
      \Delta n_6 & = & 2 mol
    \end{eqnarray*}
    HINT: Set up the matrix of stoichiometric coefficients and make use of the
    extents of reaction.
  }}
  We already know the relation:
  \begin{displaymath}
    \underline{N} \cdot \underline{\xi_j} = \underline{\Delta n_i}
  \end{displaymath}
  \begin{displaymath}
    \underline{N} =
    \left(\begin{array}{r|rrr}
      {}  & I & II & III \\ \hline
      A_1 & -1 &  0 &  0 \\
      A_2 & -1 & -1 & -1 \\
      A_3 &  1 & -1 &  0 \\
      A_4 &  1 &  1 &  1 \\
      A_5 &  0 &  1 & -1 \\
      A_6 &  0 &  0 &  1
    \end{array}\right);\;\;
    \underline{\xi_j} =
    \left(\begin{array}{r}
      \xi_1 \\
      \xi_2 \\
      \xi_3
    \end{array}\right);\;\;
    \underline{\Delta n_i} =
    \left(\begin{array}{r}
      \Delta n_1 \\
      \Delta n_2 \\
      \Delta n_3 \\
      \Delta n_4 \\
      \Delta n_5 \\
      \Delta n_6
    \end{array}\right)
  \end{displaymath}

  \begin{displaymath}
    \left(\begin{array}{rrr}
      -1 &  0 &  0 \\
      -1 & -1 & -1 \\
       1 & -1 &  0 \\
       1 &  1 &  1 \\
       0 &  1 & -1 \\
       0 &  0 &  1
    \end{array}\right)
    \cdot
    \left(\begin{array}{r}
      \xi_1 \\
      \xi_2 \\
      \xi_3
    \end{array}\right)
    =
    \left(\begin{array}{r}
      \Delta n_1 \\
      \Delta n_2 \\
      \Delta n_3 \\
      \Delta n_4 \\
      \Delta n_5 \\
      \Delta n_6
    \end{array}\right)
    =
    \left(\begin{array}{r}
      -1 mol \\
      \Delta n_2 \\
      -2 mol \\
      \Delta n_4 \\
      \Delta n_5 \\
      2 mol
    \end{array}\right)
  \end{displaymath}
  Writing this equation system as a matrix makes things easy again:
  \begin{displaymath}
    \left(\begin{array}{r|rrrr}
      A_1 & -1 &  0 &  0 & -1 \\
      A_3 &  1 & -1 &  0 & -2 \\
      A_6 &  0 &  0 &  1 &  2
    \end{array}\right)
    \leadsto
    \left\{\begin{array}{rcl}
      \xi_1 = 1 \\
      \xi_3 = 2 \\
      \xi_2 = 3
    \end{array}\right.
  \end{displaymath}
  Now let's calculate the missing (non-key) changes in amount of substance:
  \begin{displaymath}
    \left(\begin{array}{r|rrrr}
      A_2 & -1 & -1 & -1 & \Delta n_2 \\
      A_4 &  1 &  1 &  1 & \Delta n_4 \\
      A_5 &  0 &  1 & -1 & \Delta n_5
    \end{array}\right)
    \leadsto
    \left\{\begin{array}{rcr}
      - \xi_1 - \xi_2 - \xi_3 & = \Delta n_2 = & -6 mol \\
        \xi_1 + \xi_2 + \xi_3 & = \Delta n_4 = &  6 mol \\
                \xi_2 - \xi_3 & = \Delta n_5 = &  1 mol
    \end{array}\right.
  \end{displaymath}
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 4}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    The following species are involved in a set of reactions:
    \begin{center}\begin{tabular}{ll}
      $A_1 = C_6H_6$;     & $A_2 = Cl_2$ \\
      $A_3 = C_6H_5Cl$;   & $A_4 = HCl$ \\
      $A_5 = C_6H_4Cl_2$; & $A_6 = C_6H_3Cl_3$
    \end{tabular}\end{center}
    Calculate the changes in number of moles $\Delta n_1$, $\Delta n_2$ and $\Delta n_4$, when
    the following changes have been measured:
    \begin{center}\begin{tabular}{lr}
      $\Delta n_3 = $ & $-2 mol$ \\
      $\Delta n_5 = $ & $1 mol$ \\
      $\Delta n_6 = $ & $2 mol$
    \end{tabular}\end{center}
    HINT: Use the element-species matrix. The number of key components has \underline{not} to
    be calculated.
    \\
  }}
  Just set up the element-species matrix (\underline{B}):
  \begin{displaymath}
    \underline{B} =
    \left(\begin{array}{l|cccccc}
      {} & A_1 & A_2 & A_3 & A_4 & A_5 & A_6 \\ \hline
      C  & 6   & 0   & 6   & 0   & 6   & 6 \\
      H  & 6   & 0   & 5   & 1   & 4   & 3 \\
      Cl & 0   & 2   & 1   & 1   & 2   & 3
    \end{array}\right);\;\;
    \underline{\Delta n_i} =
    \left(\begin{array}{r}
      \Delta n_1 \\
      \Delta n_2 \\
      \Delta n_3 \\
      \Delta n_4 \\
      \Delta n_5 \\
      \Delta n_6
    \end{array}\right)
    =
    \left(\begin{array}{r}
      \Delta n_1 \\
      \Delta n_2 \\
      -2 mol \\
      \Delta n_4 \\
      1 mol \\
      2 mol
    \end{array}\right)
  \end{displaymath}
  The following formula is to be used:
  \begin{displaymath}
    \underline{B} \cdot \Delta n = 0
  \end{displaymath}
  Writing this as an equation system an rearranging gives:
  \begin{displaymath}
    \leadsto
    \left\{\begin{array}{rcrclr}
      6 \Delta n_1 & = & -6           & \leadsto & \Delta n_1 = & -1 mol \\
      6 \cdot -1 + \Delta n_4 & = & 0 & \leadsto & \Delta n_4 = & 6 mol \\
      2 \Delta n_2 & = & -12          & \leadsto & \Delta n_2 = & -6 mol
    \end{array}\right.
  \end{displaymath}
  \newpage
%% ---------------------------------------------------------------------------
  \section{Thermodynamics}
  \subsection*{Problem 1}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    During the production of synthesis gas, the conversion to water gas is
    a side reaction:
    \begin{center}\begin{tabular}{rcl}
      $CO (g) + H_2O (g)$ & $\rightleftharpoons$ & $CO_2 (g) + H_2 (g)$;
    \end{tabular}\end{center}
    the equilibrium is adjustable by the reaction conditions.\\\\
    The reaction is carried out at constant temperature in a closed vessel until
    equilibrium is reached. Following thermodynamic data is given:
    \begin{enumerate}
      \item Standard molar heats of formation:\\
        \begin{tabular}{llll}
          $CO(g)$:    & $\quad\Delta H_f^0$ & $=$ & $-111 \frac{kJ}{mol}$ \\
          $CO_2(g)$:  & $\quad\Delta H_f^0$ & $=$ & $-393 \frac{kJ}{mol}$ \\
          $H_2O(fl)$: & $\quad\Delta H_f^0$ & $=$ & $-283 \frac{kJ}{mol}$
        \end{tabular}
      \item Vaporization of water:\\
        \begin{tabular}{llll}
          $H_2O (fl) \to H_2O (g)$: & $\Delta H_{vap}^0$ & $=$ & $-+41 \frac{kJ}{mol}$
        \end{tabular}
      \item Constant pressure heat capacities:\\
        \begin{tabular}{llll}
          $\Delta c_p$ & $=$ & $A+BT+CT^2+DT^3$ & with $\left[T\right] = K$ and $\left[\Delta c_p\right] = \frac{cal}{mol \cdot K}$ \\
          $1 cal$      & $=$ & $4,184 J$ &
        \end{tabular}\\
        \begin{tabular}{|l|r|r|r|r|}\hline
          Species   & \multicolumn{4}{|c|}{Coefficients} \\
                    & A       & B                      & C                      & D \\\hline
          $CO_2$    & $4,728$ & $1,754  \cdot 10^{-2}$ & $-1,338 \cdot 10^{-5}$ & $4,097  \cdot 10^{-9}$ \\
          $CO$      & $7,373$ & $-0,307 \cdot 10^{-2}$ & $6,662  \cdot 10^{-6}$ & $-3,037 \cdot 10^{-9}$ \\
          $H_2O$    & $7,701$ & $4,595  \cdot 10^{-4}$ & $2,521  \cdot 10^{-6}$ & $-0,859 \cdot 10^{-9}$ \\
          $H_2$     & $6,483$ & $2,215  \cdot 10^{-3}$ & $-3,298 \cdot 10^{-6}$ & $1,826  \cdot 10^{-9}$
          \\\hline
        \end{tabular}
    \end{enumerate}
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Calculate the standard enthalpy change on reaction $\Delta H_R^0$ using the given
    thermodynamical data! (Consider the molecules' state of aggregation!)
    \\
  }}
  Using the following relation, we can easily calculate the reaction enthalpy:
  \begin{eqnarray*}
    \Delta H_R^0 & = & \sum_{i=1}^N \nu_i \Delta H_{f,i}\\
    & = & \underbrace{-(-111)}_{CO (g)} \underbrace{-(-283+41)}_{H_2O (g)} - \underbrace{393}_{CO_2 (g)} + \underbrace{0}_{H_2 (g)} \left[\frac{kJ}{mol}\right] \\
    & = & -40 \frac{kJ}{mol}
  \end{eqnarray*}
  Hence the reaction is exothermic (i.e. it produces heat).
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    In which direction is the equilibrium shifted in response to the following changes
    within the system?
    \\
    \begin{enumerate}
      \item Feed of further $CO$
      \item Removal of a fixed amount of $CO_2$ out of the vessel
      \item Feed of further $H_2$
      \item Addition of a catalyst
      \item Increase in temperature
      \item Increase in pressure
    \end{enumerate}
  }}
  The solution can be easily determined by using the different $K_x$ and some
  basic chemical knowledge:
  \begin{enumerate}
    \item $\rightarrow$ In favor for products, use $K_c$
    \item $\rightarrow$ In favor for products, use $K_c$
    \item $\leftarrow$  In favor for educts, use $K_c$
    \item $\rightleftharpoons$ No change. Catalysts only speed up or slow down reactions. The equilibrium is never affected!
    \item $\leftarrow$ In favor for educts, use $K_a$. The reaction is exothermic and therefore produces heat itself. Adding heat shifts the equilibrium!
    \item $\rightleftharpoons$ No change, use $K_p$. There is no change in volume during the reaction. Hence the partial pressures of all species should be the same and the quotient remains constant.
  \end{enumerate}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    Calculate the enthalpy change on reaction $\Delta H_{R, 1000}$ for a temperature of 1000 K
    using the given thermodynamical data!
    \\
  }}
  We will have to calculate the reaction enthalpy at a given temperature upon the
  basis of the already calculated standard reaction enthalpy:
  \begin{displaymath}
    \Delta H_R (T) = \Delta H_R^0 + \int_{T_0}^{T}{\Delta c_p(T) dT}
  \end{displaymath}
  Given the function for $\Delta c_p$ we have:
  \begin{eqnarray*}
    \Delta H_{R, 1000} & = & -40 \frac{kJ}{mol} + \int_{298}^{1000}{\left( A_R + B_R \cdot T + C_R \cdot T^2 + D_R \cdot T^3 \right) dT} \\
    & = & -40 \frac{kJ}{mol} + \left( A_R + B_R \cdot \left.T\right|_{298}^{1000} + C_R \cdot \left.T^2\right|_{298}^{1000} + D_R \cdot \left.T^3\right|_{298}^{1000} \right)
  \end{eqnarray*}
  For the constants $A$, $B$, $C$ and $D$, given for each species, we apply the following formula
  to get the overall constants $A_R$, $B_R$, $C_R$ and $D_R$:
  \begin{displaymath}
    Z_R = \sum_{i=1}^{N} \nu_i Z_i \qquad\qquad \mbox{where }Z=\left[A, B, C, D\right]
  \end{displaymath}
  Let's calculate:
  \begin{eqnarray*}
    A_R & = & - 7,373 - 7,701 + 4,728 + 6,483 \\
        & = & -3,863 \\
    B_R & = & -(-0,307 \cdot 10^{-2}) - 4,595 \cdot 10^{-4} + 1,754 \cdot 10^{-2} + 2,215 \cdot 10^{-3} \\
        & = & 0,0223655 \\
    C_R & = & -6,662 \cdot 10^{-6} - 2,521 \cdot 10^{-6} - 3,298 \cdot 10^{-6} + 1,338 \cdot 10^{-5} \\
        & = & -2,5861 10^{-5} \\
    D_R & = & 3,037 \cdot 10^{-9} + 0,859 \cdot 10^{-9} + 1,826 \cdot 10^{-9} + 4,097 \cdot 10^{-9} \\
        & = & 9,819 \cdot 10^{-9}
  \end{eqnarray*}
  Now it is possible to calculate the final result (partially without units):
  \begin{eqnarray*}
    &   & \int_{298}^{1000}{\left( A_R + B_R \cdot T + C_R \cdot T^2 + D_R \cdot T^3 \right) dT} \\
    & = & -3,863 \cdot \left.T\right|_{298}^{1000} + 0,0223655 \cdot \frac{1}{2}\left.T^2\right|_{298}^{1000} + -2,5861 \cdot 10^{-5} \cdot \frac{1}{3}\left.T^3\right|_{298}^{1000} \\
    &   & + 9,819 \cdot 10^{-9} \cdot \frac{1}{4}\left.T^4\right|_{298}^{1000} \\
    & = & -3,863 \cdot (1000 - 298) + \frac{1}{2}\left(0,0223655 \cdot (1000^2 - 298^2)\right) + \frac{1}{3}\left(-2,5861 \cdot 10^{-5} \cdot (1000^3-298^3)\right) \\
    &   & + \frac{1}{4}\left(9,819 \cdot 10^{-9} \cdot (1000^4-298^4)\right) \\
    & = & -2711,826 + 10189,67707 - 8392,208349 + 2435,391472 = 1521,034193 \frac{cal}{mol} \\
    & = & 1,521034193 \frac{kcal}{mol} = 6,364007065 \frac{kJ}{mol}
  \end{eqnarray*}
  Now plugging this into the above equation:
  \begin{eqnarray*}
    \Delta H_{R,1000} & = & \Delta H_R^0 + \int_{298}^{1000}{\Delta c_p(T) dT} \\
    & = & -40 \frac{kJ}{mol} + 6,364007065 \frac{kJ}{mol} \\
    & = & -33,635992935 \frac{kJ}{mol}
  \end{eqnarray*}

  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 2}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    The heat tonality of the experimentally not realizable formation of methanol
    from the elements (see the equation) cannot be measured directly:
    \begin{displaymath}
      C (s) + 2 H_2 (g) + \frac{1}{2} O_2 (g) \stackrel{\Delta H_R^0}{\to} CH_3 OH (fl),
    \end{displaymath}
    but the heats of combustion of carbon, hydrogen and methanol can be measured
    calorimetrically:\\
    \begin{eqnarray*}
      C (s) + O_2 (g) & \stackrel{\Delta H_{R,1}^0}{\to} & CO_2 (g) \\
      H_2 (g) + \frac{1}{2}O_2 (g) & \stackrel{\Delta H_{R,2}^0}{\to} & H_2 O (fl) \\
      CH_3 OH (fl) + \frac{3}{2}O_2 (g) & \stackrel{\Delta H_{R,3}^0}{\to} & 2 H_2 O (fl) + CO_2 (g)
    \end{eqnarray*}
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Calculate the enthalpy change on reaction $\Delta H_R^0$ of the formation of
    methanol from the elements!\\
    NOTE: Represent the methanol's formation reaction by a linear combination of
    the three combustion processes.
    \\
  }}
  We have know formula to calculate the enthalpy change on reaction from the formation
  enthalpies:
  \begin{displaymath}
    \Delta H_R^0 = \sum_{i=1}^{N} \nu_i \Delta H_{f, i}^0
  \end{displaymath}
  Furthermore we know, that $\Delta H_{f, i}^0$ for any element is zero. Let's use this, since
  it will cancel out several unknowns (following the sums in form $\nu_i \Delta H_{f, i}^0$):
  \begin{eqnarray*}
    \mbox{Form. from elem.:} & \Delta H_R^0     & = -1 \cdot 0 - 2 \cdot 0 - \frac{1}{2} \cdot 0 + 1 \cdot \Delta H_{f, CH_3OH}^0 \\
    \mbox{RK1:} & \Delta H_{R,1}^0 & = -1 \cdot 0 - 1 \cdot 0 + 1 \cdot \Delta H_{f, CO_2}^0 \leadsto \Delta H_{f, CO_2}^0 = -393 \frac{kJ}{mol} \\
    \mbox{RK2:} & \Delta H_{R,2}^0 & = -1 \cdot 0 - \frac{1}{2} \cdot 0 + 1 \cdot \Delta H_{f, H_2O}^0 \leadsto \Delta H_{f, H_2O}^0 = -285 \frac{kJ}{mol} \\
    \mbox{RK3:} & \Delta H_{R,3}^0 & = -1 \cdot \Delta H_{f, CH_3OH}^0 - \frac{3}{2} \cdot 0 + 2 \cdot \Delta H_{f, H_2O}^0 + 1 \cdot \Delta H_{f, CO_2}^0
  \end{eqnarray*}
  For the third equation we now know the two remaining unknowns $\Delta H_{f, H_2O}^0$ and $\Delta H_{f, CO_2}^0$, hence the equation look now like this:
  \begin{eqnarray*}
    \mbox{RK3:} & \Delta H_{R,3}^0 & = -1 \cdot \Delta H_{f, CH_3OH}^0 + 2 \cdot -285 \frac{kJ}{mol} + 1 \cdot -393 \frac{kJ}{mol} \\
     & & = -\Delta H_{f, CH_3OH}^0 - 570 \frac{kJ}{mol} -393 \frac{kJ}{mol} = -\Delta H_{f, CH_3OH}^0 - 963 \frac{kJ}{mol} \\
    \leadsto & \Delta H_{R,3}^0 + 963 \frac{kJ}{mol} & = -\Delta H_{f, CH_3OH}^0 = -724 \frac{kJ}{mol} + 963 \frac{kJ}{mol} \\
    \leadsto & \Delta H_{f, CH_3OH}^0 & = -239 \frac{kJ}{mol}
  \end{eqnarray*}
  Now we'll have to use the equation ("Form. from elem.") from above to get the result for $\Delta H_R^0$ of the formation of methanol from the elements:
  \begin{eqnarray*}
    \Delta H_R^0 & = & -1 \cdot 0 - 2 \cdot 0 - \frac{1}{2} \cdot 0 + 1 \cdot \Delta H_{f, CH_3OH}^0 \\
    \Delta H_R^0 & = & \Delta H_{f, CH_3OH}^0 = -239 \frac{kJ}{mol}
  \end{eqnarray*}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Is the reaction in problem 2a exothermic or endothermic?
    methanol from the elements!\\
    NOTE: Represent the methanol's formation reaction by a linear combination of
    the three combustion processes.
    \\
  }}
  The result ($\Delta H_R^0 = -239 \frac{kJ}{mol}$) clearly shows, that this reaction is
  exothermic.
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 3}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    The temperature dependence of the balanced reaction:
    \begin{displaymath}
      SO_2 (g) + \frac{1}{2} O_2 (g) \stackrel{\Delta H_R^0}{\rightleftharpoons} SO_3 (g), \quad \Delta H_R^0 = -99 \frac{kJ}{mol}
    \end{displaymath}
    is of decisive importance for the production of sulphuric acid.\\
    The standard molar Gibbs free energies of formation are known:
    \begin{eqnarray*}
      SO_2: & \Delta G_f^0 & = -300 \frac{kJ}{mol} \\
      SO_3: & \Delta G_f^0 & = -371 \frac{kJ}{mol}
    \end{eqnarray*}
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Calculate the standard Gibbs free energy change on reaction $\Delta G_R^0$!
    \\
  }}
  We use the following relation:
  \begin{eqnarray*}
    \Delta G_R^0 & = & \sum_{i=1}^N \nu_i \Delta G_{f,i} \\
    & = & -1 \cdot \Delta G_{f, SO_2} + \left(-\frac{1}{2}\right) \cdot \Delta G_{f, O_2} + 1 \cdot \Delta G_{f, SO_3} \\
    & = & -(-300 \frac{kJ}{mol}) - (\frac{1}{2} \cdot \underbrace{0 \frac{kJ}{mol}}_{\mbox{zero for elements!}}) + (-371 \frac{kJ}{mol}) \\
    & = & -71 \frac{kJ}{mol}
  \end{eqnarray*}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Calculate the standard activity equilibrium ratio $K_a^0$!
    \\
  }}
  \begin{eqnarray*}
    &     & G_R^0 = - RT \cdot \ln K_a  \\
    & \leadsto & -\frac{G_R^0}{RT} = \ln K_a \\
    & \leadsto & K_a = \exp\left(-\frac{G_R^0}{RT}\right) \\
    & \leadsto & K_a^0 = \exp\left(-\frac{G_R^0}{R \cdot 298 K}\right) = \exp\left(-\frac{-71 \frac{kJ}{mol}}{8,314 \frac{J}{mol \cdot K} \cdot 298 K}\right) = e^{28,6570884721009} = 2790071872271,74
  \end{eqnarray*}
  This means, that the standard activity equlibrium shifts the balance to the product's side:
  \begin{displaymath}
    K_a^0 = \frac{\left[SO_3\right]}{\left[SO_2\right]\left[O_2\right]} = 2,79007187227174 \cdot 10^{12}
  \end{displaymath}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    Calculate the activity equilibrium ratio $K_a$ for the following temperatures:
    \begin{center}
      T = 713K; 733K; 753K; 773K; 873K; 973K; 1073K; 1173K\\
    \end{center}
  }}
  Using the relation from (b), we could easily determine $K_a$ for changing $T$ if we knew $G_R (T)$:
  \begin{displaymath}
    K_a = \exp\left(-\frac{G_R^0}{RT}\right) = \exp\left(-\frac{G_R (T)}{RT}\right)
  \end{displaymath}
  This is not the case, hence we have to use another relation:
  \begin{displaymath}
    \frac{d \ln K_a}{dT} = \frac{\Delta H_R^0}{RT^2} = \quad \leadsto \quad \int_{K_a^0}^{K_a} d \ln K_a  = \int_{T_0}^{T}\frac{\Delta H_R^0}{RT^2} dT = \ln K_a - \ln K_a^0
  \end{displaymath}
  Rearranging it gives:
  \begin{displaymath}
    \ln K_a = \ln K_a^0 + \int_{T_0}^{T}\frac{\Delta H_R^0}{RT^2} dT
  \end{displaymath}
  Moving the constants out of the integral:
  \begin{displaymath}
    \ln K_a (T) = \ln K_a^0 + \frac{\Delta H_R^0}{R} \cdot \int_{T_0}^{T}\frac{dT}{T^2} = \ln K_a^0 + \frac{\Delta H_R^0}{R} \cdot -\left.\frac{1}{T}\right|_{T_0}^{T}
  \end{displaymath}
  Now plugging all known variables in and calculating for each given temperature yields:
  \begin{eqnarray*}
    \ln K_a (T) & = & 28,6570884721009 + \frac{-99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot -\left.\frac{1}{T}\right|_{298}^{T} \\
%% 28.6570884721009 + (99000/8.314) * (1/T-1/298)
    \ln K_a (713K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{713K}-\frac{1}{298K} = 5,39935029229595 \\
    \ln K_a (733K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{733K}-\frac{1}{298K} = 4,9436685107031 \\
    \ln K_a (753K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{753K}-\frac{1}{298K} = 4,51219293264772 \\
    \ln K_a (773K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{773K}-\frac{1}{298K} = 4,10304468075303 \\
    \ln K_a (873K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{873K}-\frac{1}{298K} = 2,33850496897352 \\
    \ln K_a (973K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{973K}-\frac{1}{298K} = 0,936666122924645 \\
    \ln K_a (1073K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{1073K}-\frac{1}{298K} = -0,203879368781479 \\
    \ln K_a (1173K) & = & 28,6570884721009 + \frac{99000 \frac{J}{mol}}{8,314 \frac{J}{mol \cdot K}} \cdot \frac{1}{1173K}-\frac{1}{298K} = -1,14995845098955 \\
%% pow(e, 28.6570884721009 + (99000/8.314) * (1/T-1/298))
    K_a (713K) & = &  \exp\left(5,39935029229595\right) = 221,262613469784 \\
    K_a (733K) & = &  \exp\left(4,9436685107031\right) = 140,283939879604 \\
    K_a (753K) & = &  \exp\left(4,51219293264772\right) = 91,1214227140278 \\
    K_a (773K) & = &  \exp\left(4,10304468075303\right) = 60,5242844733766 \\
    K_a (873K) & = &  \exp\left(2,33850496897352\right) = 10,365727887841 \\
    K_a (973K) & = &  \exp\left(0,936666122924645\right) = 2,55146096592772 \\
    K_a (1073K) & = & \exp\left(-0,203879368781479\right) = 0,815560747340286 \\
    K_a (1173K) & = & \exp\left(-1,14995845098955\right) = 0,316649925596806
  \end{eqnarray*}
  This shows us, that the equilibrium shifts to the left (educts) when adding heat to
  the exothermic reaction.
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 4}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    Consider the following gas-phase reaction in equilibrium
    \begin{displaymath}
      RK1:\quad\quad A_1 + 2A_2 \rightleftharpoons 3A_3
    \end{displaymath}
    with standard heat of reaction $\Delta H_{R,1}^0$, $A_1$ being $H_2$, and all
    species behaving like ideal gases.
    \\
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Calculate the standard heat of reaction $\Delta H_{R,1}^0$ when there is given
    a second reaction
    \begin{displaymath}
      RK2:\quad\quad A_1 + A_4 \rightleftharpoons A_2
    \end{displaymath}
    with the standard heat of reaction $\Delta H_{R,2}^0 = 100 \frac{kJ}{mol}$ and the
    following standard molar heats of formation for $A_3$ and $A_4$:
    \begin{displaymath}
      \Delta H_{f, 3}^0 = 20 \frac{kJ}{mol};\quad \Delta H_{f, 4}^0 = 50 \frac{kJ}{mol}
    \end{displaymath}
  }}
  We need to employ the already known relation:
  \begin{displaymath}
    \Delta H_R^0 = \sum_{i=1}^N \nu_i \cdot \Delta H_{f,i}^0
  \end{displaymath}
  This is a simple linear equation system. We know several of the variables, so let's
  start:
  \begin{displaymath}
    (RK2):\quad\Delta H_{R,2}^0 = 100 \frac{kJ}{mol} = \underbrace{-1 \cdot 0}_{A_1} + \underbrace{(-1) \cdot 50 \frac{kJ}{mol}}_{A_4} + 1 \cdot \Delta H_{f,2}^0 \quad\leadsto\quad \Delta H_{f,2}^0 = 150 \frac{kJ}{mol}
  \end{displaymath}
  So, why is $\Delta H_{f,1}^0 = 0$? Because $H_2$ is an element and this is one condition
  for elements!\\
  Now we can calculate $\Delta H_{R,1}^0$:
  \begin{displaymath}
    \Delta H_{R,1}^0 = \underbrace{-1 \cdot 0}_{A_1} + \underbrace{(-2) \cdot 150 \frac{kJ}{mol}}_{A_2} + \underbrace{3 \cdot 20 \frac{kJ}{mol}}_{A_3} = (-300 + 60) \frac{kJ}{mol} = -240 \frac{kJ}{mol}
  \end{displaymath}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Does it favor the formation of educts or products in the reaction (RK1), when the
    partial pressures $p_1$ and $p_2$ of the species $A_1$ and $A_2$ are both 25\%
    of the total pressure p?
    \\
  }}
  For the following relation
  \begin{displaymath}
    K_p = \prod_{i=1}^N p_i^{\nu_i},
  \end{displaymath}
  and under the condition, that the overall pressure must be 100\%
  ($p = p_1 + p_2 + p_3 = 1 = 0,25 + 0,25 + p_3 \leadsto p_3 = 0,5 = 50 \%$)
  we get
  \begin{displaymath}
    K_p = p_3^{\nu_3} \cdot p_1^{\nu_1} \cdot p_2^{\nu_2} = p_3^3 \cdot p_1^{-1} \cdot p_2^{-2} = \frac{p_3^3}{p_1^1 \cdot p_2^2} = \frac{0,5^3}{0,25^1 \cdot 0,25^2} = \frac{0,5^3}{0,25^3} = 8.
  \end{displaymath}
  Hence the equilibrium shifts in favor for the products (i.e. $\rightarrow$).\\
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    In which direction is the equilibrium shifted in the case the reaction is endothermic
    and a catalyst is added?
    \\
  }}
  First of all, the reaction is exothermic. So this question needs not to be answered.
  And anyway, the addition of a catalyst does not influence the equilibrium. It just
  speeds up or slows down the reaction.
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 5}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    The isomerisation reaction
    \begin{displaymath}
      A_1 \rightleftharpoons A_2 \quad\quad \Delta H_{f,1}^0 = 18,98 \frac{kJ}{mol} \quad \Delta H_{f,2}^0 = 17,93 \frac{kJ}{mol}
    \end{displaymath}
    is in equilibrium at standard conditions, the equilibrium conversion $X_1^*$ is 0,45.
    \\
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Calculate $K_c^0$ and $K_a^0$ using the stoichiometric table.
    \\
  }}
  The stoichiometric table can be found in the \textbf{Summary} or can be built.\\
  First of all we assume a batch system.
  \begin{eqnarray*}
    & \nu_1 A_1 + \nu_2 A_2 = 0 & \\
    & \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 \\\hline \rule{0mm}{6mm}
      \mbox{Sum}   & n_{\mbox{sum}0} & -                       & n_{\mbox{sum}} = n_{\mbox{sum}0} - (1 + \frac{\nu_2}{\nu_1})n_{10} X_1 \\ \hline
    \end{array} &
  \end{eqnarray*}
  We know, that $n_{20} = 0$ before the reaction starts. From the power law we know $K_a^0 = \prod_{i=1}^N c_i^{\nu_i}$ and $c_i = \frac{n_i}{V}$, hence:
  \begin{displaymath}
    K_a^0 = c_1^{-1} \cdot c_2^1 = \frac{c_2}{c_1} = \frac{n_2}{V} \cdot \frac{V}{n_1} = \frac{n_2}{n_1}
  \end{displaymath}
  Now from the stoichiometric table and the condition $c_{20} = n_{20} = 0$ we know:
  \begin{displaymath}
    K_a^0 = \frac{n_2}{n_1} = \frac{n_{20}-\frac{\nu_2}{\nu_1} \cdot n_{10} \cdot X_1}{n_{10} - n_{10} \cdot X_1} =
      \frac{-\frac{\nu_2}{\nu_1} \cdot n_{10} \cdot X_1}{(1 - X_1) \cdot n_{10}} =
      \frac{-\frac{\nu_2}{\nu_1} \cdot X_1}{1 - X_1} =
      \frac{1 \cdot X_1}{1 - X_1} =
      \frac{0,45}{1 - 0,45} = \frac{0,45}{0,55} = 0,81\overline{81}
  \end{displaymath}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Calculate the standard heat of reaction.
    \\
  }}
  We already know how to do that:
  \begin{displaymath}
     \Delta H_R^0 = \sum_{i=1}^N \nu_i \Delta H_{f,i} = -1 \cdot 18,98 \frac{kJ}{mol} + 1 \cdot 17,93 \frac{kJ}{mol} = -1,05 \frac{kJ}{mol}
  \end{displaymath}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    Calculate $K_a$ at a temperature of 400 K.
    \\
  }}
  \begin{eqnarray*}
    \ln K_a (T) & = & \ln K_a^0 + \int_{T_0}^T \frac{\Delta H_R^0}{RT^2} dT \\
    \ln K_a (400K) & = & \ln 0,81\overline{81} + \int_{298K}^{400K} \frac{-1,05 \frac{kJ}{mol}}{8,3145 \frac{J}{mol \cdot K} \cdot T^2} dT
  \end{eqnarray*}
  Moving the constants out of the integral we get:
  \begin{eqnarray*}
    \ln K_a (400K) & = & \ln 0,81\overline{81} + \frac{-1050 \frac{J}{mol}}{8,3145 \frac{J}{mol \cdot K}} \int_{298K}^{400K} \frac{1}{T^2} dT =
      \ln 0,81\overline{81} + \frac{-1050 \frac{J}{mol}}{8,3145 \frac{J}{mol \cdot K}} \left[ -\frac{1}{T} \right]_{298K}^{400K} \\
    & = & \ln 0,81\overline{81} + \frac{-1050 \frac{J}{mol}}{8,3145 \frac{J}{mol \cdot K}} \left( -\frac{1}{400K} - \left(-\frac{1}{298K}\right) \right) \\
    & = & \ln 0,81\overline{81} - 126,285405 \frac{1}{K} \left( \frac{1}{298K}-\frac{1}{400K} \right) = - 0,3087337099 \\
    K_a (400K) & = & 0,7343763011
  \end{eqnarray*}
  \newpage
%% ---------------------------------------------------------------------------
  \section{Microkinetics}
  \subsection*{Problem 1}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    In a laboratory experiment, $500 ml$ of a solution of $2 kmol/m^3$ ethylene oxide ($A_1$)
    in water was mixed with $500 ml$ of water ($A_2$) containing 0,9\%
    (w/w) sulfuric acid:
    \begin{displaymath}
      A_1 + A_2 \to A_3
    \end{displaymath}
    The concentration of ethylene glycol ($A_3$) was recorded as a function of time.
    From this data determine the reaction rate constant.\\
    \begin{center}\begin{tabular}{|c|c|}\hline\rule{0mm}{5mm}
      Time & Concentration of $A_3$ \\ \hline\rule{0mm}{5mm}
      0,0  & 0,000 \\
      0,5  & 0,145 \\
      1,0  & 0,270 \\
      1,5  & 0,376 \\
      2,0  & 0,467 \\
      3,0  & 0,610 \\
      4,0  & 0,715 \\
      6,0  & 0,848 \\
      10,0 & 0,957 \\ \hline
    \end{tabular}\end{center}
  }}
  From the definition of the rate of reaction we get:
  \begin{displaymath}
    r = \frac{1}{\nu_i} \cdot \frac{dc_i}{dt} = k \cdot c_1 c_2 = \frac{dc_3}{dt} = \frac{c_3(t_0 + \Delta t) - c_3(t_0)}{\Delta t}
  \end{displaymath}
  We can neglect $c_2$ because water is in excess.
  \begin{displaymath}
    r = k \cdot c_1
  \end{displaymath}
  From the stoichiometric table we know:
  \begin{displaymath}
    c_i = c_{i0} - \frac{\nu_i}{\nu_k} c_{k0} X_k \quad\leadsto\quad c_1 = c_{10} - c_{10} X_1
  \end{displaymath}
  From $c_{30} = 0$ we can derive:
  \begin{eqnarray*}
    c_3 = c_{30} - \frac{\nu_3}{\nu_1} c_{10} X_1 = c_{10} X_1 \quad\leadsto\quad c_1 = c_{10} - c_3 \\
    r = k \cdot (c_{10} - c_3)^{\kappa} \quad\leadsto\quad \underbrace{\log r}_{y} = \underbrace{\log k}_{a} + \underbrace{\kappa}_{b} \underbrace{\log (c_{10} - c_3)}_{x}
  \end{eqnarray*}
  The value for $r$ can be calculated at different points in time, since we
  have the data for $c_3$ and $c_{10}$ from the given table. Plotting $\log r$ against
  $\log (c_{10} - c_3)$ in an X/Y-diagram allows us to estimate $\log k$ (i.e. $a$) and
  $\kappa$ (i.e. $b$). Where $a$ is the distance from the X-axis to the curve and
  $b$ is the (mean-)inclination of the curve.\\
  \textsc{Note:} There is also the integral method available, but I believe that in
  this case it is better to use the differential method.
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 2}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    Consider the following reaction scheme, where $A^*$ is an activated molecule
    of the species $A$.
    \begin{eqnarray*}
      2A & \begin{array}{c}
        _1 \\
        \rightleftharpoons \\
        ^2
       \end{array}
       & A^* + A \\
      A^* & \stackrel{3}{\to} & \mbox{product}
    \end{eqnarray*}
    Derive the formula for the rate of the third reaction as a function of the
    concentration of $A$ and the rate constants by
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    formulating equations for the reaction rate of each partial reaction
    assuming elementary reaction,
    \\
  }}
  Reaction rates:
    \begin{eqnarray*}
      r_1 & = & k_1 c_A^2 \\
      r_2 & = & k_2 c_A c_{A^*} \\
      r_3 & = & k_3 c_{A^*}
    \end{eqnarray*}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    formulating an equation for the production rate $R_{A^*}$,
    \\
  }}
  Production rate for every species:
    \begin{eqnarray*}
      R_{A^*} & = & r_1 - r_2 - r_3= k_1 c_A^2 - k_2 c_A c_{A^*} - k_3 c_{A^*}
    \end{eqnarray*}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    formulating an expression for the concentration of $A^*$, assuming $R_{A^*}$ = 0 and
    \\
  }}
  Concentration of $A^*$:
    \begin{eqnarray*}
      0 & = & k_1 c_A^2 - k_2 c_A c_{A^*} - k_3 c_{A^*} \\
      k_1 c_A^2 & = & k_2 c_A c_{A^*} + k_3 c_{A^*} \\
      k_1 c_A^2 & = & c_{A^*} \cdot (k_2 c_A + k_3) \\
      c_{A^*} & = & \frac{k_1 c_A^2}{k_2 c_A + k_3}
    \end{eqnarray*}
  \hrule
  \paragraph*{(d)}
  {\sffamily\textbf{
    formulating the formula for $r_3$.
    \\
  }}
  Reaction rate for reaction 3:
    \begin{eqnarray*}
      r_3 & = & k_3 \cdot \frac{k_1 c_A^2}{k_2 c_A + k_3}
    \end{eqnarray*}

  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 3}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    Consider the following set of homogeneous reactions
    \begin{displaymath}
      A_1 + 2 A_2 \stackrel{1}{\to} A_3
      \begin{array}{ll}
        & A_4 \\
        \stackrel{2}{\nearrow} & \\
        \stackrel{\textstyle \searrow}{\scriptstyle 3} & \\
        & A_5
      \end{array}
    \end{displaymath}
    with the reaction rates
    \begin{displaymath}
      r_1 = 120 \frac{kmol}{m^3 \cdot s}; \quad r_2 = 5 r_3
    \end{displaymath}
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Formulate the equations for the reaction rate of each partial reaction assuming
    elementary reaction!
    \\
  }}
  Use the power law:
  \begin{eqnarray*}
    r_1 & = & k_1 c_1 c_2^2 = 120 \frac{kmol}{m^3 \cdot s} \\
    r_2 & = & k_2 c_3 = 5 r_3 \\
    r_3 & = & k_3 c_3 = \frac{1}{5} r_2
  \end{eqnarray*}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Calculate the net rates $R_i$ for all species assuming the concentration $c_3$ of
    species $A_3$ being constant!
    \\
  }}
  Calculate the net rate using $R_i = \sum_{j=1}^M \nu_{ij} \cdot r_j$
  \begin{eqnarray*}
    R_1 & = & - r_1 \\
    R_2 & = & -2r_1 \\
    R_3 & = & r_1 - r_2 - r_3 \\
    R_4 & = & r_2 \\
    R_5 & = & r_3
  \end{eqnarray*}
  \newpage
%% ---------------------------------------------------------------------------
  \subsection*{Problem 4}
%% ---------------------------------------------------------------------------
  \hrule\hrule\hrule\smallskip
%% ---------------------------------------------------------------------------
  {\sffamily\textbf{
    The description of a homogeneous catalyzed reaction
    \begin{displaymath}
      A_1 + A_2 \begin{array}{c}_1 \\ \rightleftharpoons \\ ^2\end{array} A_{12} \stackrel{3}{\to} A_3 + A_2
    \end{displaymath}
    follows enzyme kinetics, in which the homogeneous catalyst is represented by $A_{12}$.
  }}
  \hrule
  \paragraph*{(a)}
  {\sffamily\textbf{
    Form the equations for the reaction rate of every partial reaction assuming
    elementary reaction!
    \\
  }}
  \begin{eqnarray*}
    r_1 & = & k_1 c_1 c_2 \\
    r_2 & = & k_2 c_{12} \\
    r_3 & = & k_3 c_{12}
  \end{eqnarray*}
  \hrule
  \paragraph*{(b)}
  {\sffamily\textbf{
    Give the net rates of the changes in amount of substance of $A_1$, $A_2$, $A_3$ and $A_{12}$!
    \\
  }}
  \begin{eqnarray*}
    R_1    & = & -r_1 = -k_1 c_1 c_2 \\
    R_2    & = & -r_1 + r_2 + r_3 = k_2 c_{12} - k_1 c_1 c_2 + k_3 c_{12} \\
    R_3    & = & k_3 c_{12} \\
    R_{12} & = & r_1 - r_2 - r_3 = k_1 c_1 c_2 - k_2 c_{12} - k_3 c_{12}
  \end{eqnarray*}
  \hrule
  \paragraph*{(c)}
  {\sffamily\textbf{
    Determine the reaction rate of the formation of product assuming a (not?\footnote{"not" was in the original script, though it makes no sense. The text without "not" makes sense.})
    neglectable amount of $A_2$ being bound in the complex $A_{12}$ during the
    reaction and
    \begin{enumerate}
      \item step 3 to be rate determining!
      \item the steady state hypothesis to be valid for $A_{12}$!
    \end{enumerate}
  }}

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\end{document}

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