Problem 2Ethylene oxide is produced by the catalytic oxidation of ethylene:2C₂H₂ +0₂ →2C₂H₂OAn undesired competing reaction is the combustion of ethyleneC₂H₁ +30₂ →2CO₂ +2H₂OThe feed to the reactor contains 3 moles of ethylene per mole of oxygen. The conversionof ethylene is 20%, and for every 100 moles of ethylene consumed in the reactor, 90moles of ethylene oxide emerges in the reactor products.(a)(b)Calculate the production rate of ethylene oxide for 100 moles of feed to thereactor.Calculate the molar flow rates of ethylene and oxygen in the feed needed toproduce 1 tonne per hour of ethylene oxide.

Answered: Calculate the production rate of…

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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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

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

conversion of ethylene= 20%

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