ChE 101 – Chemical Reaction Engineering Winter 2014 Homework #6 Due Tuesday, February 25, 2014 in class Read Schmidt, Chapter 7, Read Wolfenden and Snider, and Sheldon on Handouts page of the website. NOTE: Please list all of your assumptionsand cite any external sources used when solving each problem. 1)Computational Problem of the Week:In this problem, you are evaluating a system in which nitrous oxide reacts with carbon monoxide in the presence of a ceria-promoted rhodium catalyst to form dinitrogen and carbon dioxide. A proposed sequence for this reaction is shown as follows: Note that (g) represents a gas phase species, while (s) indicates that a species is bound in a catalytic surface site. (a)Assuming that the surface coverage of oxygen is very small, derive the following rate expression: 𝑟𝑟=𝐾𝐾1𝜃𝜃0𝑃𝑃𝑁𝑁2𝑂𝑂1 +𝐾𝐾2𝑃𝑃𝑁𝑁2𝑂𝑂+𝐾𝐾3𝑃𝑃𝐶𝐶𝑂𝑂What are the parameters K1, K2, and K3in your result?
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(b)Linearize the rate expression derived in part (a) with respect to the reactants. Then use linear regression on the following data to determine the kinetic parameters found in part (a). PCO(torr) PN2O(torr) Turnover rate (s-1) 45.6 11.4 0.00566 45.6 22.8 0.00953 45.6 45.6 0.0194 45.6 68.4 0.0242 45.6 91.2 0.0367 11.4 45.6 0.0386 22.8 45.6 0.0255 68.4 45.6 0.0157 91.2 45.6 0.00927 (c)Use a nonlinear regression to fit the data above. How do the two methods compare? Under what conditions might each fit be better? It may help to use a graphical argument in your analysis. None of the reactions are stated as rate limiting, thus we can only use the pseudo steady state approximation to derive the rate law. Starting with N2O, 0 =𝑟𝑟𝑁𝑁2𝑂𝑂(𝑠𝑠)=𝑘𝑘1𝑓𝑓𝑃𝑃𝑁𝑁2𝑂𝑂𝜃𝜃𝑠𝑠−(𝑘𝑘1𝑏𝑏+𝑘𝑘2)𝜃𝜃𝑁𝑁2𝑂𝑂Grouping together the constants into a single term, We have Looking at oxygen, we have Finally looking at carbon monoxide, The rate of product formation is given by And using a mass balance on the free sites, we have However, we are told in the problem statement that the surface concentration is negligible, thus the surface balance becomes: Making the proper substitutions, we get a rate constant of the form