For the liquid-phase reaction, the entering concentration of ethylene oxide and water, after mixing the inlet streams, are 16.13 mol/dm3, respectively. The specific reaction rate is k = 0.1 dm/mol*s at 300 K with E = 12,500 cal/mol. 3. CH2--OH CH2- CH2 + H20 → CH2--OH A + B > C CAO = 16.13mol/dm3 CBo = 55.5 mol/dm³ The stoichiometric table and rate law are as follows; Remaining CA= CAO(1-X) = (1-X) mol/dm3 Ca = CAo( 03-X) =(3.441-X) mol/dm² Cc= CAOX mol/dm Symbol Initial Change Species Ethylene CAO=16.13 mol/dm - CAOX A oxide CBO= 55.5 mol/dm, O3 =3.441 Water -CAOX %3D Glycol C CAOX Rate law: -TA = KCĄCB Therefore, -TA = k Co(1-X) (0g-X) = k (16.13)*(1-X) (3.441-X) Calculate the CSTR space-time, T, for 80% conversion at 350 K.

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
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For the liquid-phase reaction, the entering concentration of ethylene oxide and water,
after mixing the inlet streams, are 16.13 mol/dm³, respectively. The specific reaction rate is k
= 0.1 dm/mol*s at 300 K with E = 12,500 cal/mol.
3.
%3D
CH2--OH
CH2- CH2 + H20 → CH2--OH
+ B > C
A
3
CAO = 16.13mol/dm3
CBo = 55.5 mol/dm³
%3D
%3D
The stoichiometric table and rate law are as follows;
Change
- CAOX
Remaining
CA= CAO(1-X)
Symbol
Initial
Species
Ethylene
A
CAO=16.13 mol/dm
= (1-X) mol/dm
CB = CAo( O3-X)
oxide
%3D
CBO= 55.5 mol/dm',
O3 =3.441
Water
-CAOX
(3.441-X) mol/dm3
3.
Cc= CAOX mol/dm
Glycol
C
0.
CAOX
Rate law:
-TA = KCACB
%3D
Therefore,
-TA = k Co(1-x) (0g-X) = k (16.13)*(1-X) (3.441-X)
Calculate the CSTR space-time, T, for 80% conversion at 350 K.
Transcribed Image Text:For the liquid-phase reaction, the entering concentration of ethylene oxide and water, after mixing the inlet streams, are 16.13 mol/dm³, respectively. The specific reaction rate is k = 0.1 dm/mol*s at 300 K with E = 12,500 cal/mol. 3. %3D CH2--OH CH2- CH2 + H20 → CH2--OH + B > C A 3 CAO = 16.13mol/dm3 CBo = 55.5 mol/dm³ %3D %3D The stoichiometric table and rate law are as follows; Change - CAOX Remaining CA= CAO(1-X) Symbol Initial Species Ethylene A CAO=16.13 mol/dm = (1-X) mol/dm CB = CAo( O3-X) oxide %3D CBO= 55.5 mol/dm', O3 =3.441 Water -CAOX (3.441-X) mol/dm3 3. Cc= CAOX mol/dm Glycol C 0. CAOX Rate law: -TA = KCACB %3D Therefore, -TA = k Co(1-x) (0g-X) = k (16.13)*(1-X) (3.441-X) Calculate the CSTR space-time, T, for 80% conversion at 350 K.
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