Material Balance (Corrected & Completed)

Production of 2-ethylhexanol Material Balance The following are the calculations done so as to produce one ton of 2-ethylhexanol in one day.

Oxo-Reactor We will start by determining the amount of fed required to achieve this Let, R= (synthesis gas + unreacted propylene)/kmol of propylene feed x= mole fraction of propylene in stripped gas. Let us take 24 kmol of fresh propylene

(Determined by trial & error)

Total propylene feed = (24+24Rx) kmol Unreacted propylene = 24(1+Rx)0.02

(assuming reactor has 98% conversion)

= 0.48(1+Rx) kmol Synthesis gas = 24R(1-x) kmol Propylene: synthesis gas = 1:2 [24(1+Rx)] / [24R(1-x)] = ½ Therefore x =(R-2)/3R

…(1)

Also x = (unreacted propylene)/( unreacted propylene + synthesis gas) x= [0.48(1+Rx)]/[0.48(1+Rx)+24R(1-x)]

…(2)

Solving for x we get x = 0.01 & R = 2.06 Therefore total propylene feed = 24+24Rx = 24.49 kmol Unreacted propylene = 0.48(1+Rx) = 0.49 kmol Synthesis gas = 24R(1-x) = 48.95 kmol Unreacted synthesis gas = 48.95-24.49 = 24.46 kmol Now 98% proplene is converted. Also butyraldehyde formed is a mixture of n-butyraldehyde and iso-butyraldehyde in the ratio 4:1 So n-butyraldehyde = 4/5 * 0.98 * 24.49 = 19.20 kmol iso-butyraldehyde = 1/5 *0.98 *24.49 = 4.80 kmol

6% of n-butyraldehyde is converted to n-butanol and 4% of iso-butyraldehyde is converted to iso-butanol. Therefore n-butanol = 0.06*19.20 = 1.15 kmol iso-butanol = 0.04*4.80 = 0.192 kmol Effectively n-butyraldehyde = 19.2-1.15 = 18.05 kmol iso-butyraldehyde = 4.8-0.19 = 4.61 kmol

Distillation column 1 5%w/w of butyraldehyde will be in alcohol stream in stripping section. Feed = 18.05 kmol n-butyraldehyde + 4.61 kmol of iso-butyraldehyde + 0.19 kmol isobutanol +1.15 kmol n-butanol = 1631.52 kg butyraldehyde + 99.16 kg butanol Residue = 5.22 kg butyraldehyde + 99.16 kg butanol Distillate = 1626.3 kg butyraldehyde = 17.992kgmol n-butyraldehyde + 4.56kgmol isobutyraldehyde

Distillation column 2 Feed = 17.992kmol n-butyraldehyde + 4.596 kmol of iso-butyraldehyde xf= 0.2 xd= 0.99 xw= 0.01 F= D+W 22.588= D + W Fxf = Dxd + Wxw 22.588*0.2= (22.588-W)0.99 + W*0.01 W= 18.163 kmol D= 4.425kmol n-butyraldehyde in residue = 18.163*(1-0.01) =17.945kmol iso-butyraldehyde in residue = 0.218 kmol n-butyraldehyde in distillate=0.404kmol iso-butyraldehyde in distillate= 4.3395kmol

Aldol Condensation (90% conversion) 2-ethylhex-2-enal = (17.945/2)*0.9 =8.075 kmol Water= 8.075 kmol

Unreacted n-butyraldehyde = 0.1*17.945 = 1.7945 kmol iso butyraldehyde = 0.218 kmol

Hydrogenation (99% conversion) Hydrogen fed (Considering 100% excess) = 2*2*8.075 =32.3 kmol 2-ethylhexanol = 0.99*8.075 = 7.994 kmol =1039.220 kg

Thermodynamic Feasibility 1. Reaction of Propylene with synthesis gas :

∆Greaction =∆Gproducts - ∆Greactants ∆G= ∆G(n-butyraldehyde+i-butyraldehyde+n-butanol+i-butanol)- ∆G(propylene+syn gas) = (-149.02-151.46-178.96-158.57)-(2*150.81-2*148.661) =-642.308kJ/mol

•

The Aldolization and subsequent Conversion to 2-ethylhexenal :

∆Greaction =∆Gproducts - ∆Greactants  ∆G=∆G(intermediate)-∆G(n-butyraldehyde) ∆G= -89.36-(2*-178.961) = 268.562kJ/mol  ∆G=∆G(2-ethylhex-2-enal+water)-∆G(intermediate) ∆G= (-70.19-237.14) – (-89.36) =-217.970kJ/mol

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The Hydrogenation to 2-ethylhexanol :

∆Greaction =∆Gproducts - ∆Greactants ∆G= ∆G(2-ethylhexanol)- ∆G (2-ethylhex-2-enal+Hydogen) ∆G=-140.15-(-70.19+0) =-69.96kJ/mol