In a process for the catalytic hydration of ethylene (C_2H_4 + H_2O) to ethyl al

Question

In a process for the catalytic hydration of ethylene (C_2H_4 + H_2O) to ethyl alcohol (C_2H_5OH) 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 m 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

Basis : 1 mole water and 0.55 moles of ethylene at the inlet to the reactor

The reaction is C2H4+H2O---àC2H5OH

Conversion of limiting reactant C2H4= 4.5%

Recycle consists of 6.5% water and 100-6.5= 93.5% C2H4

Conversion in the reactor = 4.5%

Ethylene is the limiting reactant entering the reactor

0.55*4.5/100 = moles of C2H5OH produced =0.02475 moles

Moles of C2H4 remaining = 0.55-0.02475= 0.52525

All the un reacted C2H4 is recycled and this correspond to 93.5% C2H4

Let R= Recylce

R*0.935= 0.52525

R = 0.5617 , moles of water recycled= 0.5617* 6.5/100 =0.036511 moles

Moles of C2H4 remaining = 0.52525

Moles of water remaining after reaction = 1-0.02475= 0.97525

Moles of water removed as liquid produced = 0.97525-0.036511 =0.938739 moles

Let F= Feed

Moles of water in feed + moles of water in recycle = 1

Moles of water in feed = 1-0.036511 =0.963489

Moles of C2H4 in feed + Moles of C2H4 recycled = 0.55

Moles of C2H4 in feed +.52525 =0.55

Moles of C2H4 in feed =0.55-0.52525= .02475

Composition : Feed : C2H4= 0.02475/(0.963489+0.02475) =0.025 , H2O= 1-0.025= 0.975

Conversion with respect to water = 100* (0.975-0.02475)/0.975=97.4%

With respect to C2H4= 100%

Water entering = 0.963489, water consumed =0.02475

Excess % water = 100*(0.963489-0.02475)/0.963489 =97.4%

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.