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Puzzle Problem What four things are wrong with this solution? The gas-phase reaction 3A+2B-3C+5D follows an elementary rate law as written and is carried out in a flow reactor operated isothermally at 427°C and 28.7 atmospheres. Pressure drop can be neglected. Express the rate law and the con- centration of each species solely as a function of conversion. The specific reaction rate is 200 dm¹2/mols and the feed is equal molar in A and B. Solution 3A+2B 3C+5D Because A is the limiting reactant, we divide through by its stoichiometry coefficient B⇒C+D So the elementary rate law is A + ²B- 3 Сло = уло -T₁=kC₁ C²³ Equal molar yo = 1: £ = YA08=3+5-2=3=3 C₁= Equal molar feed in A and B, therefore Co(1-X) = Cao 1+ EX _ C₁=CA = (1+3X) (1-X) AO 1+3X 28.7 (1.987)(427) RT = = Po - (0.5) -T₁ = k₁C₂ C²/³= k₁C5/3 (1-x)(1 − x) (1+3X)573 mol dm = 0.17- 5/3 (1-X)(1-X) (1+3X)³³ = (200)(0.17) 5/3

Puzzle Problem What four things are wrong with this solution? The gas-phase reaction 3A+2B-3C+5D follows an elementary rate law as written and is carried out in a flow reactor operated isothermally at 427°C and 28.7 atmospheres. Pressure drop can be neglected. Express the rate law and the con- centration of each species solely as a function of conversion. The specific reaction rate is 200 dm¹2/mols and the feed is equal molar in A and B. Solution 3A+2B 3C+5D Because A is the limiting reactant, we divide through by its stoichiometry coefficient B⇒C+D So the elementary rate law is A + ²B- 3 Сло = уло -T₁=kC₁ C²³ Equal molar yo = 1: £ = YA08=3+5-2=3=3 C₁= Equal molar feed in A and B, therefore Co(1-X) = Cao 1+ EX _ C₁=CA = (1+3X) (1-X) AO 1+3X 28.7 (1.987)(427) RT = = Po - (0.5) -T₁ = k₁C₂ C²/³= k₁C5/3 (1-x)(1 − x) (1+3X)573 mol dm = 0.17- 5/3 (1-X)(1-X) (1+3X)³³ = (200)(0.17) 5/3

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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em.pdf 1 100% Solution B P4-18 What four things are wrong with this solution? The gas-phase reaction 3A+2B-A3C+5D follows an elementary rate law as written and is carried out in a flow reactor operated isothermally at 427°C and 28.7 atmospheres. Pressure drop can be neglected. Express the rate law and the con- centration of each species solely as a function of conversion. The specific reaction rate is 200 dm ¹2/mols and the feed is equal molar in A and B. Puzzle Problem 3A+2B-3C+5D Because A is the limiting reactant, we divide through by its stoichiometry coefficient 3—>C+ ²D So the elementary rate law is A+=B_ 3 Equal molar yÃo = 1: ε = y^o8=3+5−2−3=3 CA Equal molar feed in A and B, therefore Сло CAO = YAO # KC CH -T₁=kC/C²¹³ CAD(1-X) = CAO 1+3X 1+ex C₁=C₁ = RTO = (0.5) Col-X) (1+3X) 28.7 (1.987)(427) 2/3 -T₁ = k₁C₁ C²/¹³ = k₁C5/³3 (1-X)(1-X) (1+3X) 5/³ -0.17 mol dm 5/3 (1-X)(1-X) (1+3X) =(200)(0.17) ។
Transcribed Image Text:em.pdf 1 100% Solution B P4-18 What four things are wrong with this solution? The gas-phase reaction 3A+2B-A3C+5D follows an elementary rate law as written and is carried out in a flow reactor operated isothermally at 427°C and 28.7 atmospheres. Pressure drop can be neglected. Express the rate law and the con- centration of each species solely as a function of conversion. The specific reaction rate is 200 dm ¹2/mols and the feed is equal molar in A and B. Puzzle Problem 3A+2B-3C+5D Because A is the limiting reactant, we divide through by its stoichiometry coefficient 3—>C+ ²D So the elementary rate law is A+=B_ 3 Equal molar yÃo = 1: ε = y^o8=3+5−2−3=3 CA Equal molar feed in A and B, therefore Сло CAO = YAO # KC CH -T₁=kC/C²¹³ CAD(1-X) = CAO 1+3X 1+ex C₁=C₁ = RTO = (0.5) Col-X) (1+3X) 28.7 (1.987)(427) 2/3 -T₁ = k₁C₁ C²/¹³ = k₁C5/³3 (1-X)(1-X) (1+3X) 5/³ -0.17 mol dm 5/3 (1-X)(1-X) (1+3X) =(200)(0.17) ។
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