Question

Synthetically produced ethanol is an important industrial commodity used for various purposes including: as a solvent (especially for substances intended for human contact or consumption); in coatings, inks, and personal-care products; for sterilization; and as a fuel. Although a great deal of attention has been given to the production of ethanol by fermentation of sugars, industrial ethanol is a petrochemical synthesized by the hydrolysis of ethylene: CzH4 (9) + H20 (v) = C2H5OH (V) Some of the product is converted to diethyl ether in the undesired side reaction 2C2H5OH (v) (C2Ms)20 (v) + H20 (v) The combined feed to the reactor contains 53.7 mole % CzHa, 36.7% H20 and the balance nitrogen which enters the reactor at 310.0°C. The reactor operates isothermally at 310.0°C. An ethylene conversion of 35.0% is achieved, and the yield of ethanol (moles ethanol produced/mole ethylene consumed) is 0.800. Data for Diethyl Ether All, = - 271.2 kJ/mol (for the liquid) AĦ, = 26.05 kJ/mol (assume independent of T) C [kJ/mol.°c)] = 0.08945 + 40.33 x 10-ST - 2.244 x 10-7 (T in °C) Calculate the reactor heating (positive Q) or cooling (negative () requirement in kJ/mol feed. Q= kJ/mol feed.

Answer 1

The reactions occuring are

$C_{2}H_{4}(g) + H_{2}O(v) \rightarrow C_{2}H_{5}OH(v)$

$2C_{2}H_{5}OH(v)\rightarrow (C_{2}H_{5})_{2}O(v) + H_{2}O(v)$

Given

∆Hf(diethyl ether) (L) = -271. 2 KJ/mol

∆Hv = 26.05 KJ/mol

∆H°f(diethyl ether ) (v) (25°C)= ∆Hf(C2H5OH) (L) + ∆Hv

∆Hf°(diethyl ether)(v) = -271. 2 + 26.05

= -245. 15 KJ/mol

Standard heat of formation are at 25°C

From handbook at T = 25°C

∆H(H2O) (v) = -241. 83 KJ/mol

∆H(C2H5OH) (v) = -234 KJ/mol

∆H(C2H4) (v) = 52.47 KJ/mol

∆H_r = ∆H_products - ∆H_reactants

For Reaction 1

∆H_r1 (25°C) =

-234-(52.47-241.83) = -44. 64 KJ/mol

Both reactions occur at 310°C

Average temperature = (25+310) /2 = 167.5°C

Cp is taken from handbook at average temperature

For diethyl ether the eqaution in the question is used to calculate Cp at average temperature

Cp(diethyl ether) (v) = 0.163298 KJ/mol°C

Cp(C2H4) = 0.056889 KJ/mol°C

Cp(C2H5OH) (v) = 0.087138 KJ/mol° C

Cp(H2O) (v) = 0.034622 KJ/mol°C

According to kirchoffs law

∆H(310°C) = ∆H(25°C) + ∆Cp(310-25)

∆Cp = (nCp) _products -(nCp) _reactants

∆Cp₁ = (0.087138) -(0.056889) -(0.034622) =

-0. 004373 KJ/mol°C

∆H_r1(310°C) = -44. 64+ (-0.004373) (310-25)

∆H_r1(310°C) = -45. 8863 KJ/mol

For second reaction

∆H_r2(25°C) = (-245.15) +(-241.83) -(2×-234)

∆H_r2(25°C) = -18. 98 KJ/mol°C

∆C_p2 = (0.163298) +(0.034622) -(2×0.087138) =

= 0.023644 KJ/mol°C

∆H_r2(310°C) = -18. 98+ (0.023644) (310-25) = -12. 2414 KJ/mol

Basis :

1 mole of feed

Mole fraction of ethylene in feed = 0.537

Conversion of ethylene = 0.35

Ethylene reacted = (0.35) (0.537) = 0.18795

0.18795 moles reacts in first reaction

Yield of ethanol

(moles of ethanol/ethylene consumed) = 0.8

Moles of ethanol = (0.18795) (0.8) = 0.15036 moles

Moles of ethanol undergoing second reaction = (0.18795-0.15036) = 0.03759 moles

The reactor is isothermal and hence only heat of reaction needs to be removed

Q = ∆H_r1(0.18795) + ∆H_r2(0.03759)

Q = -45. 8863(0.18795) + (-12.2414) (0.03759) =

Q = -9. 08448 KJ/mol feed

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