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Industrial production of ethylene oxide

How is ethylene oxide produced on an industrialscale and how does the ethylene oxide process

qualitatively compare to the production of propyleneoxide?

Hans David Wendt, Lars Heuvels, Eline van Daatselaar and Tim van Schagen(Group B)

April 3, 2014


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Contents

1 Introduction. . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Ethylene oxide production . . . . . . . . . . . . . . . . . . . . 3

2.1 Reaction. . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Chlorohydrin process. . . . . . . . . . . . . . . . . . . 3

3 Distillation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53.1 Determination of the adiabatic temperature rise . . . 73.2 Temperature rise in the reactor . . . . . . . . . . . . . 93.3 Constant reactor temperature . . . . . . . . . . . . . . 9

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1 Introduction

In this project we researched the production process of ethylene oxide (ab-breviated EtO,C2H4O). Ethylene oxide is a frequently used intermediate inthe chemical industry. The reason for this is the easily broken ring (Figure1), because the breaking of these bonds releases a lot of energy. So ethyleneoxide is a very reactive and much used compound. Because of this, verymuch ethylene oxide is produced on a yearly basis. In earlier times, theproduction process involved a chlorohydrin process, which was less efficientthen the oxidation process that is used presently. We focussed on the directoxidation process, with pure oxygen and not with air. We made this choice,since the process with pure oxygen is more often used and has a higher pro-

duction rate. For this process we evaluated mass balances, where differentvariables are discussed, like the influence of the conversion, selectivity andfeed rate. We also had a look at the separator units, mainly the distillationcolumn. Those columns are important for these kind of processes, becausethese units improve the purity of the final product. At the end, the en-ergy balance is computed. To evaluate these three processes, we have usedcalculation programs like Matlab and Microsoft Excel.Our research questions are:

How is the mass balance of the production of EtO arranged?

How does the distillation column work for this process?

How are the energy flows in the system set up?

Besides the production process of ethylene oxide, the production of propy-lene oxide (abbreviated PO, C3H6O) also is evaluated. We qualitativelycompare this process to the production process of EtO. The differences andsimilarities will be discussed with the following question:

How much differs the production of PO from that of EtO?

O

Figure 1: The molecular structure of ethylene oxide

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2 Ethylene oxide production

We will start with discussing the production of ethylene oxide in a chemicalfactory.

2.1 Reaction

The reaction we are evaluating is the production of ethylene oxide fromethylene and oxygen. The reactions that occur in the reactor are depictedin Figure2, Figure3and Figure4. The reaction in Figure2is the desiredreaction to produce ethylene oxide. Unfortunately, there is a side reaction.Ethylene oxide reacts further to carbon dioxide and water (Figure 3). Thereaction in Figure4is the sum of2 and3and so the total reaction equationfor the byproducts.

2 H2C CH2 + O2Ag

2O

Figure 2: The formation of ethylene oxide out of ethylene and oxygen cat-alyzed by a silver catalyst

2O

+ 5 O2 4 CO2 + 4 H2O

Figure 3: Combustion reaction of ethylene oxide.

H2C CH2 + 3 O2 2 CO2 + 2 H2O

Figure 4: Combustion reaction of ethylene.

The process in which ethylene oxide is produced, is schematically representedin Figure5. In this figure, all mass flows and separators are pointed out andwill be computed in following paragraphs.[1]

2.2 Chlorohydrin processThe process to make propylene oxide by using chlorohydrins consists basi-cally of three steps: chlorohydration, epoxidation and purification. In thechlorohydration step, propylene gas and chlorine gas are mixed stoichiomet-ric, while reacting as following to a chloronium ion intermediate (Figure6):This intermediate can react in three ways to different products. The de-sired products are two isomers of propylene chlorohydrin, namely 1-chloro-2-propanol (Figure7) and 2-chloro-1-propanol (Figure8), which arise froma reaction with water:

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Figure 5: Schematic representatation of the ethylene oxide process.

CH3

CH2

+ ClCl CH3

Cl

+

Cl

Figure 6: The reaction of propylene with chorine to form the chloroniumion.

CH3

Cl

+ HO

H +

Cl

H3C

OH

Cl

+ HCl

Figure 7: Reaction of the chloronium ion with water to form 1-chloro-2-propanol.

CH3

Cl + HO

H +

Cl

H3C

Cl

OH

+ HCl

Figure 8: Reaction of the chloronium ion with water to form 2-chloro-1-propanol.

1-chloro-2-propanol emerges for about 85%, because this isomer is formed viaa secondary carbocation, while 2-chloro-1-propanol is formed via a primarycarbocation. Usually this desired main reaction represents 90% or moreof the yield. Besides, there are two side reactions, which together emergefor up to 10%. The intermediate can react with the chloride ion to form1,2-dichloropropane (Figure9). The other side reaction produces isomers of

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dichloro-dipropyl ether out of a reaction with propylene chlorohydrin (Figure

10).

CH3

Cl

+

Cl

H3C

Cl

Cl

Figure 9: Reaction of the chloronium ion with chloride to form 1,2-dichloropropanol.

CH3

Cl

+ H3C

O H

Cl

+

ClH3C

Cl

O

CH3Cl

+ HCl

Figure 10: The formation of dichloro-dipropyl ether.

The next step can be called epoxidation, dehydrochlorination or saponifica-

tion. The chloropropanol reacts with a base, which is added in excess. Thiscould be calcium hydroxide or sodium hydroxide, dissolved in water. Thereaction is as following (Figure11):

CH3

OH

Cl

+ HCl + 2OH CH3O

+ 2Cl + 2H2O

Figure 11: Reaction of chloropropanol with a base to form propylene oxide.

The issue with the saponification is that the propylene oxide decomposeswhen a base is present. Therefore...[3]

3 Distillation

Changing the Antoine parameters makes no sense in the final version, butwe will discuss that later. The parameters are only present to show us theAntoine parameters. The boiling temperatures are valid for 1 bar, but wesolved the assignment for 4.428 bar, since we could only find Vapor LiquidEquilibrium (VLE) data for water and ethylene oxide at this pressure. Wemade a plot and polynomial approximation of a VLE graph.[9]This graphcan be seen in Figure 12.

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Figure 12: VLE graph of EtO and water.

Next, we want to evaluate the temperature of the feed. This can be doneby solving equation1.[10]

Xeto p0

eto+ Xwater p0

H2O= 4.428 (1)

This can be done...Therefore, we made a polynomial approximation of the Antoine parametes ofranges of temperatures we could find. Then we extrapolated these. We putthis approximation in equation1to solve it. This makes a better polynomialapproximation than using only one set of parameters. The parameters used

are displayed in table1.[11]

Ethylene oxide 1 Ethylene oxide 2 Water

A 4.386 5.48696 4.6543B 1115.1 2022.83 1435.264C -29.015 62.656 -64.848

Temperature range 185-280 K 280-305 K 256-373 K

Table 1: Antoine parameters of water and ethylene oxide

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3.1 Determination of the adiabatic temperature rise

Because the oxidation of ethylene to ethylene oxide is an exothermal reac-tion, there is a lot of energy that is released in the reaction. We want toknow what the temperature rise of the reactor is when we dont let any heatflow in or out (adiabatic). Weve used an enthalpy balance to determine theadiabatic temperature rise. We use the steady-state open-system energybalance (equation2).[7]

H+ Ek+ Ep = Q Ws (2)

Ek (kinetic energy) and Ep (potential energy) can be neglected becausethese terms are very small in comparison to H (internal energy). In our

process the Ws (work) term is approximately zero and because we want theadiabatic temperature rise, the Q (heat flow) term is also zero. So we get

H= 0, which means:Hin = Hout (3)

We use the heat of formation method to calculate the enthalpy balance. Sowe can rewrite equation3 to:

in Hin =

out Hout (4)

In this equation we just take the sum of the enthalpy of each individualcomponent, which can be calculated by multiplying the molar flow with the

enthalpy per mol. In the mass balance section, we already calculated themolar flows of every single compound in our spreadsheet. These values aredisplayed in Table2.

Compound in (mol/s) out (mol/s)

Ethylene (g) 2114.286 1902.857Oxygen (g) 185.0 0

Methane (g) 6147.692 6147.692Carbon dioxide (g) 287.8681 351.2967

Water (g) 0 63.42857Ethylene oxide (g) 0 179.7143

Table 2: Molar flows over the reactor

To calculate the enthalpy per mole we need the heat of formation for thedifferent compounds. The heat of formation is given at 298 K and 1 atm.Weve used the data of the NIST Webbook[11] (Table3).Because we arent working at standard temperatures, we also have to knowthe heat capacities of the compounds, and because these arent constantat different temperatures we need the heat capacities as a function of thetemperature. Its true that we also didnt work at standard pressures, butbecause we work with gases and our working pressure isnt very extreme

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Compound Heat of formationfH (kJ/mol)

Ethylene (g) 52.47Oxygen (g) 0

Methane (g) -74.87Carbon dioxide (g) -393.52

Water (g) -241.83Ethylene oxide (g) -52.64

Table 3: Heat of formation at p0 and T0 of the compounds in the reactor.

Compound A B C D E

Ethylene (g) -6.387880 184.4019 -112.9718 28.49593 0.315540Oxygen (g) 31.32234 -20.23531 57.86644 -36.50624 -0.007374

Methane (g) -0.703029 108.4773 -42.52157 5.862788 0.678565Carbon dioxide (g) 24.99735 55.18696 -33.69137 7.948387 -0.136638

Water (g) 30.09200 6.832514 6.793435 -2.534480 0.082139Ethylene oxide (g) -23.25802 275.6997 -188.9729 51.03350 0.386930

Table 4: Parameters for the Shomate equation.

(compared to p0), we can neglect the VP term. We use the Shomateequation to express the heat capacity as a function of temperature (Equation5)

Cp =A + B t + C t2 + D t3 + E/t2 (5)

t is defined in Equation6, where T is the temperature in Kelvin.

t= T /1000 (6)

A, B, C, D and E are constants. These constants are compound specific.The constants for the compounds in our reactor are in Table 4.[11]

H= fH+

T2

T1

CpdT (7)

The adiabatic temperature rise depends also on the Tin , we set Tin = 523K[1] and want to calculate Tout. So Tout is our variable in equation 4. Weused our Matlab code Determination_of_adiabatic_Temperature_rise.mand the data above to calculate Tout. When we fill in the data with Tin= 523 K we get Tout = 628.385 K, so the adiabatic temperature rise =105.385 K.

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3.2 Temperature rise in the reactor

If the temperature in the reactor rises, the single pass conversion will rise,because the reaction will occur faster. When the conversion rises the selec-tivity will decrease. So a temperature rise will lead to a lower selectivity.This means that we produce more by-products. When our selectivity de-creases the production cost will increase so we dont want the selectivity todecrease. The temperature is allowed when the reactor is strong enough tohold these circumstances, but economically we dont want a temperaturerise in our reactor.

3.3 Constant reactor temperature

Because we want a constant temperature in our reactor and the reaction isexothermal we have to cool down the reactor. To calculate the amount ofheat we have to absorb we again use an enthalpy balance. This time the Q(heat flow) isnt zero so we get the equation:

in Hin

out Hout = Q (8)

In table 2 we can find the in and out again, because these dont changewhen we let the heat flow out. The Hin and Hout can be calculated becausetemperature in reactor is constant = 523 K. We just use equation 7 again(calculations done with Matlab file: Energybalance_ethyleenoxide.m). The

results are in table5.

Compound Hin (kJ/mol) Hout (kJ/mol)

Ethylene (g) 64.6064 64.6064Oxygen (g) 6.8060 6.8060

Methane (g) -65.5838 -65.5838Carbon dioxide (g) -384.4319 -384.4319

Water (g) -234.0780 -234.0780Ethylene oxide (g) -38.3405 -38.3405

Table 5: Hin and Hout of the compounds in the reactor.

in Hin = 4.8666 10

5 kJ/s (9)

out Hout = 5.3835 10

5 kJ/s (10)

This gives a heat flow Q= 5.1687 104kJ/sSo we have to absorb 5.1687 104

kJ per second. We use water at room temperature (298 K) to cool downthe reactor, because this is a cheap substance with very few risks. Anotheradvantage of cooling down the reactor with water is, that we can re-use the

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steam for heating up other streams of the process. To calculate the amount

of steam at 373 K that we can produce, we need the heat of vaporizationand the heat capacity of liquid water. To express the heat capacity as afunction of the temperature, we use the Shomate equation again (Equations5and 6). The constants for liquid water are depicted in table 6.[11]

A -203.6060B 1523.290C -3196.413D 2474.455E 3.855326

Table 6: Shomate parameters for liquid water.

The heat of vaporization of water Hvap = 40.68 kJ/mol. To calculate theamount of heat needed to vaporize 1 mol of water at room temperature weuse:

Hvap +

373

298

CpdT = 46.3401 kJ/mol (11)

When we divide Q by this amount we get the amount of water, we canvaporize by cooling down the reactor = 1.1154 103 mol/s.

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Bibliography

[1] Siegfried Rebsdat, Dieter Mayer, Ul lmans Encyclopedia of IndustrialChemistry - Article Ethylene Oxide. DOI: 10.1002/14356007.a10_117.

published: 15-03-2001, consulted: 21-01-2014.

[2] J. P. Dever, K. F. George, W. C. Hoffman, H. Soo, Kirk-Othmer:Encyclopedia of Chemical Technology - Artice Ethylene Oxide. DOI:10.1002/0471238961.0520082504052205.a01.pub2. published: 14-03-2004, consulted: 21-01-2014.

[3] David L. Trent, Kirk-Othmer: Encyclopedia of Chem-ical Technology - Artice Propylene Oxide . DOI:10.1002/0471238961.1618151620180514.a01.pub2. published: 4-01-2001, consulted: 21-01-2014.

[4] Dietmar Kahlich, Uwe Wiechern, Jrg Lindner, Ul lmans Ency-clopedia of Industrial Chemistry - Article Propylene Oxide. DOI:10.1002/14356007.a22_239.pub2. published: 15-10-2012, consulted:21-01-2014.

[5] Prof. Dr. Rainer Herges and Dr. Torsten Winkler, Epoxidationof Alkenes with Peracids. URL: http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_

1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/

oxidation_reduktion/oxidation/addition_sauerstoff/

epoxidierung_mit_persaeuren/mechanismus.vscml.html . pub-lished: 2000, consulted: 21-01-2014.

[6] Shell Oil Company, Patent Preparation Method SMPO Process. Publi-catio number: US6989469 B2. published: 24-01-2006, consulted: 21-01-2014.

[7] Richard M. Felder, Ronald W. Rousseau, Elementary Principles ofChemial Reactions, 2005 edition. ISBN: 987-0-471-68757-3. published:2005, consulted: 21-01-2014.

[8] University of Alicante, Conversion: Fundamentals and the differ-ence between single pass and overall conversion. URL: http://

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http://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.htmlhttp://www.chemgapedia.de/vsengine/vlu/vsc/en/ch/2/vlu/oxidation_reduktion/epxo_per_1.vlu/Page/vsc/en/ch/2/oc/reaktionen/formale_systematik/oxidation_reduktion/oxidation/addition_sauerstoff/epoxidierung_mit_persaeuren/mechanismus.vscml.html


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www.youtube.com/watch?v=Wgkv8uqQ3wk . published: 14-03-2004, con-

sulted: 21-01-2014.

[9] Y.-L. Huang , T. Merker , M. Heilig , H. Hasse , and J. Vrabec,Molec-ular Modeling and Simulation of Vapor-Liquid Equilibria of Ethylene

Oxide, Ethylene Glycol, and Water as Well as their Binary Mixtures.DOI: 10.1021/ie300248z. published: 30-04-2012, consulted: 22-01-2014.

[10] A.B. de Haan, H. Bosch,Fundamentals of Industrial Separations, Edi-tion 2. ISBN: 9789081097321. published: 2007, consulted: 21-01-2014.

[11] National Institute of Standards and Technology,NIST Chemistry Web-Book. URL: http://webbook.nist.gov/. published: different dates,

consulted: 21-01-2014.

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http://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://webbook.nist.gov/http://webbook.nist.gov/http://webbook.nist.gov/http://www.youtube.com/watch?v=Wgkv8uqQ3wkhttp://www.youtube.com/watch?v=Wgkv8uqQ3wk

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