Question 2.11: In the production of ethanol by the hydrolysis of ethylene, ...

Reactions: C_{2}H_{4}+ H_{2}O \longrightarrow C_{2}H_{5}OH (a)
2C_{2}H_{5}OH\longrightarrow \left(C_{2}H_{5} \right) _{2}O+ H_{2}O (b)

Basis: 100 mols feed (easier calculation than using the product stream)
Note: the flow of inerts will be constant as they do not react, and it can be used to
calculate the other flows from the compositions.

Feed stream ethylene 55 mol
inerts 5 mol
water 40 mol
Product stream
ethylene = \frac{52.26}{5.28}\times 5=49.49 mol
ethanol = \frac{5.49}{5.28}\times 5=5.20 mol
ether = \frac{0.16}{5.28}\times 5=0.15 mol
Amount of ethylene reacted = 55.0 – 49.49 = 5.51 mol
Yield of ethanol based on ethylene=\frac{5.2\times 1}{5.51}\times 100=\underline{\underline{94.4 per cent } }
As 1 mol of ethanol is produced per mol of ethylene the stoichiometric factor is 1.
Yield of ether based on ethylene =\frac{0.15\times 2}{5.51}\times 100=\underline{\underline{5.44 per cent} }
The stoichiometric factor is 2, as 2 mol of ethylene produce 1 mol of ether.
Note: the conversion of ethylene, to all products, is given by:
Conversion = \frac{mols fed – mols out}{mols fed}=\frac{55-49.49}{55} \times 100
=\underline{\underline{10 per cent} }
The yield based on water could also be calculated but is of no real interest as water
is relatively inexpensive compared with ethylene. Water is clearly fed to the reactor in
considerable excess.