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For the following reaction set, define or propose two types of reactor systems to maximize product D selectivity, sketch the reactor system and describe clearly the input reactant concentrations and separation requirements if necessary. The unit of reaction rate is in units (mol/dm^3.sec) and concentration in units (mol/dm^3). (question on picture)

For the following reaction set, define or propose two types of reactor systems to maximize product D selectivity, sketch the reactor system and describe clearly the input reactant concentrations and separation requirements if necessary. The unit of reaction rate is in units (mol/dm^3.sec) and concentration in units (mol/dm^3). (question on picture)

Introduction to Chemical Engineering Thermodynamics
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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For the following reaction set, define or propose two types of reactor systems to maximize product D selectivity, sketch the reactor system and describe clearly the input reactant concentrations and separation requirements if necessary. The unit of reaction rate is in units (mol/dm^3.sec) and concentration in units (mol/dm^3).

(question on picture)

a) (1) A+B → D -riA 3D 10 еxp (-8000 К/T) Сд Св n3/2 (2) A+B → U -Г2А 3D 100 ехp (-1000 К/T) Cс b) (1) A+B → D -riA 3D 100 еxp (-1000 К/T) Сд Св (2) A+B →U —Г2А 3D 10% ехp (—8000 К/T) СА Св с) (1) A+B → D -ria = 10 еxp (-1000 К/T) Сл Св (2) B+D → U -Г28 3 10° еxp (-10000 К/T) Св Ср d) (1) A+B → D -riA %3D 800 еxр (-8000 К/T) C Св (2) A+B → U1 r2B = 10 еxp (-300 К/т) СA Св (3) D+B > U2 ГзD %3D 106 еxp (-8000 К/T) Ср Св
Transcribed Image Text:a) (1) A+B → D -riA 3D 10 еxp (-8000 К/T) Сд Св n3/2 (2) A+B → U -Г2А 3D 100 ехp (-1000 К/T) Cс b) (1) A+B → D -riA 3D 100 еxp (-1000 К/T) Сд Св (2) A+B →U —Г2А 3D 10% ехp (—8000 К/T) СА Св с) (1) A+B → D -ria = 10 еxp (-1000 К/T) Сл Св (2) B+D → U -Г28 3 10° еxp (-10000 К/T) Св Ср d) (1) A+B → D -riA %3D 800 еxр (-8000 К/T) C Св (2) A+B → U1 r2B = 10 еxp (-300 К/т) СA Св (3) D+B > U2 ГзD %3D 106 еxp (-8000 К/T) Ср Св
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