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
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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1
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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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