For T = 500 K: t (days) 40 80 20 60 120 X (%) 0.7 0.56 0.45 0.38 0.29 For T = 550 K: 1 (days) 10 15 20 30 40 X (%) 2 1.2 0.89 0.69 0.57 0.42 0.33
For T = 500 K: t (days) 40 80 20 60 120 X (%) 0.7 0.56 0.45 0.38 0.29 For T = 550 K: 1 (days) 10 15 20 30 40 X (%) 2 1.2 0.89 0.69 0.57 0.42 0.33
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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The decomposition of spartanol to wulfrene and CO2is often carried out at high temperatures[J. Theor. Exp., 15, 15 (2014)]. Consequently, the denominator of the catalytic rate law is easily approximated as unity, and the reaction is first order with an activation energy of 150 kJ/mol Fortunately, the reaction is irreversible. Unfortunately, the catalyst over which the reaction occurs decays with time on stream. The following conversion-time data were obtained in a differential reactor:
(a) If the initial temperature of the catalyst is 480 K, determine the temperature–time trajectory to maintain a constant conversion.
(b) What is the catalyst lifetime?
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Step 1: Steps to plot a graph
VIEWStep 2: Plot the graph for T =500K and calculate kd
VIEWStep 3:Plot the graph for T =550K and calculate kd
VIEWStep 4: Calculate activation energy
VIEWStep 5: Calculate kd for T= 500K
VIEWStep 6: a. Calculate temperature–time trajectory to maintain a constant conversion
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