Consider the pair of reactions in which ethylene (C 2 H 4 ) is oxidized either t

Question

Consider the pair of reactions in which ethylene (C2H4) is oxidized either to ethylene oxide (C2H4O, desired) or to carbon dioxide (undesired).

(hint: assume 100 moles feed entering in reactor)

                                    C2H4 + ½ O2   C2H4O

                                    C2H4 + 3 O2 2 CO2 + 2 H2O

The feed to the reactor contains 25% mole ethylene and 75% mole oxygen. The conversion of ethylene is 75% and the selectivity of ethylene oxide over carbon dioxide is 12.

a) determine 1 and 2 (18 mole, 0.75mole)

b) Calculate the molar composition (mole fractions) of the outlet stream. (0.069, 0.7, 0.198, 0.016, 0.016)

b) Calculate the fractional yield of ethylene oxide.(0.72)

Explanation / Answer

(a)

Feed:

F=100 moles

C2H4=25 moles

O2=75 moles

Now fractional conversion f=0.75 =moles reacted/moles fed

1 = extent of reaction in reaction producing C2H4O

2 = extent of reaction in reaction producing CO2

So

(1+2)/25=0.75------(1) (moles of reacted/moles of ethylene fed)

selectivity=12

selectivity=moles of desired product/moles of undesired product=12

1/22=12 ( 1mole of C2H4O/2 moles of CO2)

1=242------(2)

So solving these two equations:

2=0.75 moles

1=18 moles

(b)

Now doing species balance:

Product stream:

C2H4=n1 mole

C2H4O=n2 mole

O2=n3 mole

CO2=n4 mole

H2O=n5 mole

25 – n2 – 1 – 2 = 0

C2H4=6.25 mol

C2H4O :– n1 + 1 = 0

C2H4O=18 mole

O2 :75 – n3 – 1/2 – 32 = 0

O2 :n3=63.75 mole

CO2: – n4 + 22 = 0

n4=1.5 mole

H2O:– n5 + 22 = 0

n5=1.5 mole

Total mole=91 mole

Now composition

C2H4=6.25/91=0.069

C2H4O=18/91=0.198

O2=63.75/91=0.7

H2O=1.5/91=0.016

CO2=1.5/91=0.016

(c)

Yield=moles of desired product/Moles of desired product made if the limiting reacting reacts completely

Yield= 1/25=18/25=0.72

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