A chemical manufacturer produces ethylene oxide by burning ethylene gas with air

Question

A chemical manufacturer produces ethylene oxide by burning ethylene gas with air in the presence of a catalyst. If the conditions are carefully controlled a substantial fraction of the ethylene is converted to ethylene oxide, while some ethylene remains unconverted and some is completely oxidized to form carbon dioxide and water. Formation of CO is negligible. After the gases leave reactor, they are passed through the adsorber in which the ethylene oxide is removed. A typical Orsat (dry basis) analysis of the gases leaving the absorber is as follows : 9.6% CO2 , 3.0 % O2 and 6.4 % ethylene. Of the ethylene entering the reactor, what percent is converted to ethylene oxide?

Explanation / Answer

C2H4+0.5O2---> C2H4O and C2H4+3O2 ------>2CO2 + 2H2O

Gases leaving the absorber : basis : 100 moles

moles of CO2= 9.6 moles. Moles of C2H4 unreacted to form CO2= 9.6/3= 3.2 moles

since the analysis of gases is on dry basis

out of 100 moles, 9.6 moles is CO2, 3 moles is O2, 6.4 moles is ethylene and rest has to be N2=100-9.6-3-6.4=81 moles

air contains 79%N2 and 21% O2,

the source of N2 is air. Hence moles of air supplied = 81/0.79 =102.53

total oxygen supplied = 102.53*0.21=21.53 moles

Oxygen remaining = 21.53-3= 18.53 oxygen utlized for formation of CO2= (3/2)* 9.6 = 14.4

oxygen reated to form C2H4O must be = 18.53- 14.4 = 4.13 moles

moles of C2H4 charged = 4.13*2= 8.26

moles of ethylene remaining after absorption = 6.4

moles of C2H4 reacted = 8.26- 6.4= 1.86 moles, % conversion = 100*(1.86/8.26) =22.52%

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