In a process for the catalytic hydration of ethylene (C_2 H_4 + H_2 O) to ethyl

Question

In a process for the catalytic hydration of ethylene (C_2 H_4 + H_2 O) to ethyl alcohol (C_2 H_5 OH) the once-through conversion is only 4.5%, so that a recycle is necessary. The separator after the reactor removes all the alcohol produced, and most of the water as a liquid product. The gas from the separator, containing 6.5 % water is recycled. The molar ratio of water to ethylene in the reactor feed is 0.55: 1.0. Calculate, the amount and composition for each stream in the system, the overall conversion with respect to ethylene, the overall conversion with respect to water. The percentage of excess reactant entering the overall system. The yield of alcohol based on the amount of ethylene entering the system.

Explanation / Answer

The reaction is C2H4+H2O----->C2H5OH

Basis : 0.55 moles of water and 1 mole of C2H4. So as per the reaction. lmiting reactant is H2O.

Feed to reactor composition   mole fraction : C2H4 = 0.55/1.55=0.35 H2O= 1-0.35=0.65

4.5% of conversion has taken place. So moles of water converted = 0.55*4.5/100=0.02475 moles. Moles of C2H5OH formed = 0.02475

moles of C2H4 converted = 0.02475 moles. Moles of C2H4 remaining after reaction =1-0.02475 =0.97525 moles

All the moles of C2H4 will have to be there in the recycle,. Let R= Recycle % of C2H4 in the recycle = R*(100-6.5)/100=R*0.935

R*0.935= 0.97525

R = 1.043 moles. Moles of water in recycle = 1.043*6.5/100 =0.068, moles of C2H4 in recycle = 1.043-0.068=0.975 moles

Fresh feed of C2H4 ( in moles)= moles to the reactor -moles of recycle = 1-0.975= 0.025 moles

Moles of water in the Fresh feed = 0.55-0.068=0.482

Composition of fresh feed : C2H4= 100*0.025/(0.482+0.025)=3.4% and H2O =100-3.4 = 96.6%

moles of water condensed = moles of water remaining-moles ofwater in recycle = 0.55*(1-4.5/100)-0.068= 0.45725

excess moles of water = 0.484-0.02475 =0.45925

% exces water in fresh feed = 100*0.45925/0.484)=94.88%

Yiled of alcohol = 0.02475 moles

moles of fresh water entering = 0.482 , moles of water converted = 0.02475

% conversion =100*0.02475/0.484 =5.13%

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