This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: CHE 349, Lecture 26 4/23/2003 CHE 349: Kinetics and Reactor Design Lecture 26 Announcement
TBA Comprehensive Project:
Jing Wu 350 Tiernan Department of Chemical Engineering [email protected]
4/23/2003 CHE 349 - Lecture 26 by Jing Wu 2 Due on the last day of class. No late project will be accepted. Today …
Review: P10-3, 10-4, 10-11 Discuss 10-10 How to “guess” a mechanism from its rate law. Problem …
For catalytic reaction E + H Given a rate law of the form:
− r 'C = k1 ( PE PH − PA / K P ) (1 + k E PE + k H PH ) 2 A What is the surface catalytic mechanism? 4/23/2003 CHE 349 - Lecture 26 by Jing Wu 3 4/23/2003 CHE 349 - Lecture 26 by Jing Wu 4 By Jing Wu 1 CHE 349, Lecture 26 4/23/2003 Guideline Only (Not 100% Foolproof)
− r 'C = k1 ( PE PH − PA / K P ) (1 + k E PE + k H PH )2 Important …
Summary on Table 10-3 and Table 10-4 How to “guess” a mechanism from its rate law. Note: Decomposition probably follows a dual site mechanism. For instance, as in P 10-10, the reaction A B+C may follow a dual-site mechanism. After obtaining a rate law, you should be able to deduce the its lumped constants given experimental data. The power of the denominator is 2, suggesting that there are two sites (either occupied or unoccupied) are involved in the rate limiting step. The denominator contains terms such as k E PE + kH PH This suggests that surface complex ES and HS are involved in the mechanism, not necessarily in the rate-limiting step. The numerator contains terms such as PE PH − PA / K P The minus sign suggests that the R.L.S is reversible. The reactants of the R.L.S are E or ES and H or HS, the product is A or AS.
4/23/2003 CHE 349 - Lecture 26 by Jing Wu 5 4/23/2003 CHE 349 - Lecture 26 by Jing Wu 6 By Jing Wu 2 ...
View Full Document