1 c34 For the liquid-phase reaction:\[ \text{CH}_2\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_2\text{OHCH}_2\text{OH} \]The initial concentrations of ethylene oxide and water are 1 lb-mol/ft³ and 3.47 lb-mol/ft³ (62.41 lb/ft³ ÷ 18), respectively. If $ k = 0.1 \, \text{dm}^3/\text{mol} \cdot \text{s} $ at 300 K with $ E = 12,500 \, \text{cal/mol} $, calculate the space-time volume for 90% conversion at 300 K and at 350 K.

Answered: (a) For the liquid-phase reaction H₂SO4…

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 c34

For the liquid-phase reaction:

\[ \text{CH}_2\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_2\text{OHCH}_2\text{OH} \]

The initial concentrations of ethylene oxide and water are 1 lb-mol/ft³ and 3.47 lb-mol/ft³ (62.41 lb/ft³ ÷ 18), respectively. If $ k = 0.1 \, \text{dm}^3/\text{mol} \cdot \text{s} $ at 300 K with $ E = 12,500 \, \text{cal/mol} $, calculate the space-time volume for 90% conversion at 300 K and at 350 K.

Expert Solution

Step 1: Space time required for 90 % conversion in CSTR at 300 K

Rate law is given by

$negative r subscript A equals k C subscript A C subscript B$

A - Ethylene oxide and B - water

Where, C_A- Concentration of ethylene oxide

C_B- Concentration of water

$C subscript A 0 end subscript equals 1 space fraction numerator l b m o l over denominator f t cubed end fraction space C subscript B 0 end subscript equals 3.47 space fraction numerator l b m o l over denominator f t cubed end fraction space space k subscript 1 left parenthesis 300 K right parenthesis equals 0.1 space fraction numerator d m cubed over denominator m o l space s end fraction space space E equals 12500 space fraction numerator c a l over denominator m o l end fraction space X subscript A equals 0.9$

As we know 1 dm³ = 0.0353 ft³

Thus, rate constant at 300 K is

$k space equals space 0.1 space cross times 0.0353 space equals space 0.00353 space fraction numerator f t cubed over denominator l b m o l space s end fraction$