Answer to Question #140075 in General Chemistry for Khushal

Question #140075
mixture of 5% ethylene and 95% air is passed through a suitable catalyst in a reactor.
Some of the ethylene does not react, some form oxide, some turn to CO2 and water.
The entire gas mixture enters an absorption tower where water is sprayed. The oxide is
converted to glycol. The gas leaving the absorber analyzes C2H4 : 1.085%, CO2 :
4.345%, O2 : 13.055% and N2 : 81.515% on dry basis. The partial pressure (pp) of H2O
in this gas is 15.4 mm Hg while total pressure is 745 mm Hg. If one mole of water is
sprayed per 100 mole of gas mixture, calculate the composition of ethylene glycol–
water product formed.
1
Expert's answer
2020-11-19T08:50:00-0500


1. 2C2H4+ O2 "\\rightarrow" 2C2H4O

2. C2H4 + 3O2 "\\rightarrow" 2CO2 + 2H2O

3. C2H4O + H2O "\\rightarrow" (CH2OH)2


Basis: 100 kmoles of feed gas.

 C2H4 : 5 kmoles; O2 : (95 ×0.21) = 20 kmoles; N2: 75 kmoles


N2 balance: Moles leaving the absorber = (75/0.81515) = 92 kmoles


C2H4 : 92 ×0.01085 = 1 kmole

CO2 : 92 ×0.04345 = 4 kmoles

O2: 92 ×0.13055 = 12 kmoles


H2O = [92 ×15.4 /(745 -15.4)]


 = 1.94 kmoles


C2H4reacted : 5 – 1 = 4 kmoles

C2H4converted to CO2 : 4-2= 2 kmoles


C2H4 converted to C2H4O : 4 – 2 = 2 kmoles

Water formed by reaction 2 = 4 mole"\\equiv" CO2 kmole


[oxygen sent] – [used up oxygen] = 20 – [1 + 6] = 13 kmoles


Moles of gases entering absorber: C2H4 : 1; CO2: 4; C2H4O : 2; H2O : 4; O2 : 12; N2 : 75

Total number of moles = 98


Water sprayed at the top of absorber = 98 ×(1/100)= 0.98

Total moles of water entering absorber = (4 + 0.98) = 4.98 kmoles


Water in exit gas = 1.94 kmoles

Water reacted in absorption column = 2.00 kmoles


Ethylene glycol formed = 2 kmoles

Water leaving along with ethylene glycol = (4.98– 2 – 1.94) = 1.04Kmoles


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