Answer to Question #140075 in General Chemistry for Khushal

Question

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.

Answer 1

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