Calculate the production rate of ethylene oxide for 100 moles of feed to the reactor.

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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Problem 2
Ethylene oxide is produced by the catalytic oxidation of ethylene:
2C₂H₂ +0₂ →2C₂H₂O
An undesired competing reaction is the combustion of ethylene
C₂H₁ +30₂ →2CO₂ +2H₂O
The feed to the reactor contains 3 moles of ethylene per mole of oxygen. The conversion
of ethylene is 20%, and for every 100 moles of ethylene consumed in the reactor, 90
moles of ethylene oxide emerges in the reactor products.
(a)
(b)
Calculate the production rate of ethylene oxide for 100 moles of feed to the
reactor.
Calculate the molar flow rates of ethylene and oxygen in the feed needed to
produce 1 tonne per hour of ethylene oxide.
Transcribed Image Text:Problem 2 Ethylene oxide is produced by the catalytic oxidation of ethylene: 2C₂H₂ +0₂ →2C₂H₂O An undesired competing reaction is the combustion of ethylene C₂H₁ +30₂ →2CO₂ +2H₂O The feed to the reactor contains 3 moles of ethylene per mole of oxygen. The conversion of ethylene is 20%, and for every 100 moles of ethylene consumed in the reactor, 90 moles of ethylene oxide emerges in the reactor products. (a) (b) Calculate the production rate of ethylene oxide for 100 moles of feed to the reactor. Calculate the molar flow rates of ethylene and oxygen in the feed needed to produce 1 tonne per hour of ethylene oxide.
Expert Solution
Step 1: Provided information

fraction numerator m o l e s space o f space e t h y l e n e space i n space f e e d over denominator m o l e s space o f space o x y g e n space i n space f e e d end fraction equals 3 over 1
m o l e space f r a c t i o n space o f space e t h y l e n e space i n space f e e d comma space x subscript 1 space end subscript equals fraction numerator 3 over denominator 3 plus 1 end fraction equals 0.75
m o l e space f r a c t i o n space o f space o x y g e n space i n space f e e d comma space x subscript 2 space end subscript equals fraction numerator 1 over denominator 3 plus 1 end fraction equals 0.25

conversion of ethylene= 20%

For 100 moles of ethylene consumed 90 moles of ethylene oxide emerges.

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