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: C2H4 (9) + H20 (v) = C2H5OH (v) Some of the product is converted to diethyl ether in the undesired side reaction 2C2H5OH (v) = (C2H5)20 (v) + H20 (v) The combined feed to the reactor contains 53.7 mole % C2H4, 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 5.00% is achieved, and the yield of ethanol (moles ethanol produced/mole ethylene consumed) is 0.800. Data for Diethyl Ether AO AĦ – 271.2 kJ/mol (for the liquid) Ah, = 26.05 kJ/mol (assume independent of T) Cp [kJ/mol. C)] = 0.08945 + 40.33 x 10-5T - 2.244 x 10-7T2 (T in °C) Calculate the reactor heating (positive Q) or cooling (negative Q) requirement in kJ/mol feed.

Answer 1

The two reactions occuring are

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

2C2H5OH(U) (C2H520 (0) + H2O(U)

Both reactions occur at T =310°C

From handbook at T = 25°C

∆H(H2O) (v) = -241. 83 KJ/mol

∆H(C2H5OH) (v) = -234 KJ/mol

∆H(C2H4) (g) = 52.47

For diethyl ether liquid

At T = 25°C given data

∆H°f(liquid) = -271. 2 KJ/mol

∆Hv = 26.05 KJ/mol

∆Hf(v) = ∆H°f(liquid) + ∆Hv = -271. 2+ 26.05 = -245. 15 KJ/mol

∆H_r = ∆H_products - ∆H_reactants

For reaction 1

∆H_r1 = (-234) -(52.47) -(-241.83)

∆H_r1 (25°C)= -44. 64 KJ/mol

Reaction occurs at 310°C

According to kirchoffs law

∆H(310°C) = ∆H(25°C) + ∆Cp(310-25)

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

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

Cp is taken from handbook at average temperature

Cp(H2O) (v) = 0.034622 KJ/mol°C

Cp(C2H4) (v) = 0.056889 KJ/mol°C

Cp(C2H5OH) (v) = 0.08713 KJ/mol°C

For diethyl ether

Cp equation from question is used

Cp(diethyl ether) (v) = 0.15070 KJ/mol°C

∆C_p1 = (0.08713) -(0.056889) -(0.034622) =

-0.004381 KJ/mol°C

∆H_r1(310°C) = -44. 64+ (-0.004381) (310-25) =

∆H_r1 = -45. 888 KJ/mol

For second reaction

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

∆H_r2 = -18. 98 KJ/mol

∆C_p2 = 0.15070 + (0.034622) -(2×0.08713)

∆C_p2 = 0.011062 KJ/mol°C

∆H_r2(310°C) = -18. 98 + (0.011062) (310-25)

∆H_r2 = -15. 8273 KJ/mol

Basis : 1 mole feed

Mole fraction of ethylene= 0.537

Ethylene conversion = 5 %

Ethylene reacted = (0.537) (0.05) = 0.02685 mol

Yield of ethanol =0.80 (ethanol produced/ethylene consumed  

Ethylene undergoing first reaction = (0.02685) (0.80) = 0.02148 mole

Ethylene undergoing second reaction = (0.02685) (0.20) = 0.00537 mole

∆H_r = 0.02148(∆H_r1) + (0.00537) (∆H_r2)

Q = ∆H_r = 0.02148(-45.888) +(0.00537) (-15.8273) = -1. 07066 KJ/mol feed

Q = -1. 07066 KJ/mol feed

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