A B Make a plot of equilibrium conversion as a function of temperature. i. ii. iii. Combine the rate law and stoichiometric table to write -гд as a function of k, CA, X and Xe. Determine the adiabatic equilibrium temperature and conversion when pure A is fed to the reactor at a temperature of 300 K. What is the CSTR volume to achieve 85% of the adiabatic equilibrium conversion for v₂ = dm³ ? 5 min Additional information: H°(298K)=-40 000 cal/mol H°(298K) = −60 000 cal/mol cal CpA = 50- mol. K CPB 50- cal mol. K 1 Kc 100 000 at 298 K, k = 10-3 exp - min with E = 10 000 cal/mol 298

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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Question
A B
Make a plot of equilibrium conversion as a function of temperature.
i.
ii.
iii.
Combine the rate law and stoichiometric table to write -гд as a function of k, CA, X and Xe.
Determine the adiabatic equilibrium temperature and conversion when pure A is fed to
the reactor at a temperature of 300 K.
What is the CSTR volume to achieve 85% of the adiabatic equilibrium conversion for v₂ =
dm³
?
5
min
Additional information:
H°(298K)=-40 000 cal/mol
H°(298K) = −60 000 cal/mol
cal
CpA = 50-
mol. K
CPB 50-
cal
mol. K
1
Kc 100 000 at 298 K,
k = 10-3 exp
-
min with E = 10 000 cal/mol
298
Transcribed Image Text:A B Make a plot of equilibrium conversion as a function of temperature. i. ii. iii. Combine the rate law and stoichiometric table to write -гд as a function of k, CA, X and Xe. Determine the adiabatic equilibrium temperature and conversion when pure A is fed to the reactor at a temperature of 300 K. What is the CSTR volume to achieve 85% of the adiabatic equilibrium conversion for v₂ = dm³ ? 5 min Additional information: H°(298K)=-40 000 cal/mol H°(298K) = −60 000 cal/mol cal CpA = 50- mol. K CPB 50- cal mol. K 1 Kc 100 000 at 298 K, k = 10-3 exp - min with E = 10 000 cal/mol 298
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