1000 mol h^-1 of ethylene oxide is produced by the catalytic oxidation of ethyle

Question

1000 mol h^-1 of ethylene oxide is produced by the catalytic oxidation of ethylene C_2H_4 + 1/2 O_2 C_2H_4O An undesired competing reaction is the combustion of ethylene to carbon dioxide and water. C_2H_4 + 3O_2 rightarrow 2 CO_2 + 2H_2O The feed enters to the rector contains ethylene and oxygen at the feed ratio of 3 2 mol ethylene per mol oxygen The conversion of ethylene is 25% The yield from the reactor is 0 80 mol C_2H_4O produced/mol C_2H_4 consumed Perform the degree-of-freedom analysis and determine the product composition from the reactor.

Explanation / Answer

The reactions are C2H4+O2----> C2H4O and C2H4+3O2---> 2CO2+2H2O

moles of C2H4O formed/ moles of C2H4 consumed = 0.8

moles of C2H4 consumed = 1000/0.8= 1250 moles

Moles of C2H4 reacted is only 25%. moles of C2H4 added = 1250/0.25=5000 moles

Moles of C2H4 added = 5000 moles

Moles of C2H4 converted = 1250 moles

Moles of C2H4O formed = 1000 moles, moles of C2H4O reacted for forminng CO2 and H2O= 250 mole

Moles of oxygne fed =3.2*5000 =16000 moles

Products : C2H4O= 1000 moles, CO2= 2*250 =500moles. H2O= 2*250 =500moles. O2 consumed = 1000/2( for C2H4O formation)+3*250( for formation of CO2 and H2O)= 1250 moles

Oxygen remaining = 16000-1250= 14750 moles

Products :

C2H4O= 1000 moles, CO2= 500moles, H2O= 500 moles, O2= 14750 moles, C2H4= 5000-1250 = 3750

total moles of products = 16750

composition =moles of products/ total moles

composition : C2H4O= 1000/16750=0.0597, CO2= 500/16750 =0.0298, H2O= 0.0298, O2= 14750/16750= 0.8806

C2H4= 3750/16750= 0.2238

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