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Develop a stoichiometric table for the reaction, showing the inlet molar flow rates as well as the outlet molar flow rates and the conversion-based reactant and product concentrations in the CSTR.  Calculate the reactor volume necessary to achieve 24 % conversion of A.

Develop a stoichiometric table for the reaction, showing the inlet molar flow rates as well as the outlet molar flow rates and the conversion-based reactant and product concentrations in the CSTR.  Calculate the reactor volume necessary to achieve 24 % conversion of A.

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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The nitration of aromatic compounds is an exothermic reaction. With N2O5 (reactant A) employed as the nitrating agent the chemical reaction is given as A + 2 B → 2 C + D. The reaction of nitration of benzene is carried out in the liquid phase of an adiabatic CSTR. The reaction is first order in A and second order in B. The molar flow rates of A and B entering the reactor at 330 K are FA0= 25 mol min–1and FB0= 32 mol min–1, respectively. The volumetric flow rate is v= 1200 L min–1. 

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Develop a stoichiometric table for the reaction, showing the inlet molar flow rates as well as the outlet molar flow rates and the conversion-based reactant and product concentrations in the CSTR. 
Calculate the reactor volume necessary to achieve 24 % conversion of A.
Other data: AHrx = -360.1 kJ mol-1 CPB 137 J mol-1 K-1 CP,D=80 J mol-1 K-1 CPA = 84.5 J mol-¹ K-1 Cp,c 140 J mol-1 K-1 8.314 J mol-¹ K-1 Rgas = 40 kJ mol-1 1 k = 0.11 L² mol-²min-¹ exp -1 Rgas 330 K - )|
Transcribed Image Text:Other data: AHrx = -360.1 kJ mol-1 CPB 137 J mol-1 K-1 CP,D=80 J mol-1 K-1 CPA = 84.5 J mol-¹ K-1 Cp,c 140 J mol-1 K-1 8.314 J mol-¹ K-1 Rgas = 40 kJ mol-1 1 k = 0.11 L² mol-²min-¹ exp -1 Rgas 330 K - )|
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