Unformatted text preview: CHE4473 Kinetics Spring 2015 Homework N°3: Design of Ideal Isothermal Reactors (Constant Volume) Due Tuesday JAN 27th 2015 _____________________________________________________________________________________________________ 1. We are operating a batch reactor to convert A into R. This is a liquid reaction, the stoichiometry is AèR, and the rate of reaction is given in the table below. How long must we react for the concentration to drop from CA0 = 1.3 to CAf = 0.3 mol/liter? CA mol/liter ‐ rA mol/liter min CA mol/liter ‐ rA mol/liter min 0.1 0.1 0.7 0.10 0.2 0.3 0.8 0.06 0.3 0.5 1.0 0.05 0.4 0.6 1.3 0.045 0.5 0.5 2.0 0.042 0.6 0.25 _____________________________________________________________________________________________________ 2. The aqueous decomposition of A is studied in an experimental CSTR. The results in the table below are obtained in steady‐state runs. To obtain 75% conversion of reactant in a feed, CA0 = 0.8 mol/liter, what space time is needed in a plug flow reactor? Concentration of A, mol/liter Space Time In Feed In Outlet (sec) 0.65 2.00 300 2.00 0.92 240 2.00 1.00 250 1.00 0.56 110 0.37 1.00 360 0.42 24 0.48 0.28 0.48 200 0.20 0.48 560 3. The reaction A + ½ B è 2 C has a rate law –rA = k * CA * CB and rate coefficient k = 0.05 liter/mol min. A. The reaction is conducted in a well‐mixed batch reactor (liquid phase) with initial concentration CA0 = 2 M; CB0 = 1 M. Obtain the time needed to reach 50 %, 95 % and 99 % conversion. Note that because the ratio of initial concentrations of A and B match their stoichiometric ratio, CA will always double CB., so this problem can be easily solved analytically. Why to reach 99 % conversion we need a time much longer than twice that needed to reach 50 % ? B. Use Polymath, EXCEL, or the software of your choice to develop a numerical solution for the problem of part A and compare with the analytical solutions for the three different conversions. C. Obtain the numerical solution to get 50 % conversion in the batch reactor, except that CA0 = 1.2 M; CB0 = 2 M. In this case, the analytical solution would have been much more complicated since CA0 ≠ 2 CB0 ...
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 Spring '14
 LanceL.Lobban
 1 m, 1.2 m, 2 M, 50 %, 0.05 liter

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