%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%% %%% %%% Copyright (c) 2003 by Oliver Schneider %%% %%% Portions Copyright (c) Lehrstuhl Chemische Reaktionstechnik BTU Cottbus %%% %%% %%% %%% Topic: %%% %%% My exam preparations for CRE. Most parts taken from the original %%% %%% script, but slightly changed, corrected and commented. %%% %%% %%% %%% Part: Excersises %%% %%% %%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[a4paper]{article} \usepackage{latexsym} \pagestyle{plain} \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) >>} \setlength{\paperwidth}{210mm} \setlength{\paperheight}{297mm} \setlength{\oddsidemargin}{0mm} \setlength{\evensidemargin}{0mm} \setlength{\textwidth}{170mm} \setlength{\textheight}{240mm} \setlength{\headheight}{0mm} \setlength{\headsep}{0mm} \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} }} %% --------------------------------------------------------------------------- \end{document} %% --------------------------------------------------------------------------- %% ---------------------------------------------------------------------------