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homework solutions6

# homework solutions6 - ChE 101 Chemical Reaction Engineering...

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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 assumptions and 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 = 𝐾𝐾 1 𝜃𝜃 0 𝑃𝑃 𝑁𝑁 2 𝑂𝑂 1 + 𝐾𝐾 2 𝑃𝑃 𝑁𝑁 2 𝑂𝑂 + 𝐾𝐾 3 𝑃𝑃 𝐶𝐶𝑂𝑂 What are the parameters K 1 , K 2 , and K 3 in 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). P CO (torr) P N2O (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 N 2 O, 0 = 𝑟𝑟 𝑁𝑁 2 𝑂𝑂 ( 𝑠𝑠 ) = 𝑘𝑘 1𝑓𝑓 𝑃𝑃 𝑁𝑁2𝑂𝑂 𝜃𝜃 𝑠𝑠 ( 𝑘𝑘 1𝑏𝑏 + 𝑘𝑘 2 ) 𝜃𝜃 𝑁𝑁2𝑂𝑂 Grouping 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