a) Adiabatic Operation. Plot X, Xe, p, T, and the rate of disappearance as a function of V up to V = 40 dm³. Explain why the curves look the way they do.
a) Adiabatic Operation. Plot X, Xe, p, T, and the rate of disappearance as a function of V up to V = 40 dm³. Explain why the curves look the way they do.
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
Related questions
Question

Transcribed Image Text:The gas-phase reversible reaction
AZB
is carried out under high pressure in a packed-bed reactor with pressure drop. The feed consists of both
inerts I and Species A with the ratio of inerts to the species A being 2 to 1. The entering molar flow

Transcribed Image Text:rate of A is 5 mol/min at a temperature of 300 K and a concentration of 2 mol/dm³. Work this problem
in terms of volume. Hint: V = W/PB, ₁ = PBA ·
Additional information:
FA0 = 5.0 mol/min
ΑΟ
Cao =2mol/dm3
C=2 CAO
C₁ =18 cal/mol/K
CPA
= 160 cal/mol/K
E = 10,000 cal/mol
AHRX = -20,000 cal/mol
Rx
Kc = 1,000 at 300 K
CPB = 160 cal/mol/K
To = 300 K
T₁ = 300 K
k₁ = 0.1 min¯¹ at 300 K
Ua = 150 cal/dm³/min/K
Tao = 300 K
V = 40 dm³
-3
apb=0.02 dm¯
Coolant
mc = 50 mol/min
CP Cool
= 20 cal/mol/K
PB = 1.2 kg/dm³
(a) Adiabatic Operation. Plot X, Xê, p, T, and the rate of disappearance as a function of V up to
V = 40 dm³. Explain why the curves look the way they do.
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