Question

ethanol(the desired product) is produced commercially by the hydration of ethylene: C2H4 + H2O −→ C2H5OH

Some of the product is converted to the undesired product diethyl ether by the side reaction C2H5OH −→ (C2H5)2O + H2O

The feed to the reactor contains 53.7% mole C2H4 ,36.7% H2O,and the balance is non-reactive inert gases.these intert gases do not resct as they pass through the reactor. an ethylene conversion 5% and an efficiency as defined in your class notes of 90% are achieved . balance the chemical reactions and calculation the molar composition (mole%) of the reactor output stream ,and selectivity .

Answer 1

Selle CH4 + Hyo → & WOH SHOH → (645) 0 + H2O 100 moles feed. Basist - gasą R = 5.7 moly c H₂ + 36.7 moles H₂O + 9.6 moly ineet i oo Given: ethylene conversion = 5% mobs of CH4 converted - 537 x 5 = 2.685 moles moles of GHz OH produced before side reaction happended = 2.685 XI = 2.685 moly given': moles of G H oH produced after side reaction 0.9= mobs of a Hy consumed to GH OH @ reactor output .. mols of SH.OH . 100 - 0,9x2,685 = 29.4165 moly = = 1 moles of (cats)o a reactor output (2.685 - 2.4155) = 0,2685* Dul3 425 module 1

moles of S Hy @ reactor output = 58.7 x (1-0.05) - 53.7 x 0.95 = 51.0195 moly moly of Ho @ reactor output in ist = Hjo supplied - Ho consumed I reaktion + H₂O generated in and reaction = 36.7 - 2.685 X1 + 0.2685 36.7 - 2.685 + 0,13425 = 34.15 moles mols of 9.6 moly heat gases @ reactor output = + "Hot . :: Total moles = nC,H,OH & H4 + no neet = 2.4165 + 0.13425 + 34,15 +5015 + 9.6 = 97.315 moly Composition : 2.48% x 100 = У меня 47,315 0.13425 i 0,1387 4949)0 = 97.315 0.851% 34.15 9H - 9.6 X100 97.315 10 452.4% ; Sineet 51.105 Schy04.315 = 9.86%

Selectivity = moles of moles of Desired undesired product formed product formed QHpOH @ outlet "C& His )e outlet 2.4 165 0.13425