Can someone help me with a chemical engineering question? Benzene C_6H_6 is conv

Question

Can someone help me with a chemical engineering question? Benzene C_6H_6 is converted to cyclohexane C_6H_12 by direct reaction with H_2. The fresh feed to the process is 260 L/min of C_6H_6 plus 950 L/min of H_2 at 100 degree C and 150 kPa. The single pass conversion of H_2 in the reactor is 48% while the overall conversion of H_2 in the process is 75%. The recycle stream contains 90% H_2 and the remainder benzene (no cyclohexane). a) Write a balanced equation for the above reaction. b) Determine the partial pressure of H_2 and C_6H_6 in the fresh feed. c) Determine the molar flow rates of H_2, C_6H_6 and C_6H_12 in the exiting product. d) Determine the volumetric flow rate of the components in the product streams if it exists at 100 kPa and 200 degree C. e) Determine the molar flow rate of the recycle stream. f) Determine the volumetric ratio of the recycle stream at 100 degree C and 100 kPa to the fresh stream at STP.

Explanation / Answer

1.The balanced reaction for production of cyclohexane (C6H12) from Benzene (C6H6) is

C6H6+ 3H2--------àC6H12

2. From gas law equation at constant temperature and pressure, PV=nRT becomes

V is proportional to no of moles.

Hence volume fraction = mole fraction

Mole fraction of Benzene = 260/(260+950)= 0.215, mole fraction of H2= 950/(260+950)= 0.785

Partial pressure= mole fraction* total pressure

Partial pressures : Benzene = 0.215*150 Kpa =32.25 and that of H2= 150-32.25= 117.75 Kpa ( total pressure = sum of partial pressures of individual gases)

3. Moles of Benzene in the fresh feed, n= PV/RT P= 150/101.3 atm=1.48 atm V= 260 L/min, R=0.0821 L.atm/mole.K, T= 150+273=423 K, hence, n= 1.48*260/(0.0821*423) =11.1 moles/min

Moles of hydrogen in the fresh feed = 1.48*950/(0.0821*423)= 40.5 moles/min

Overall conversion of hydrogen in only 75%. Moles of hydrogen converted= 40.5*0.75= 30.4 moles/min

Moles of cyclohexane formed per moles of hydrogen= 1/3

Moles of cyclohexane formed= 30.4/3= 10.33 moles/min

Moles of benzene consumed = 10.33 moles/min

Product contains : 10.33 moles/min Cyclohexane, 11.1- 10.33 moles/min = 0.77 moles/min benzene and hydrogen= 40.5*0.25= 10.125 moles/min. Total moles of product= 10.33+0.77+10.125 =21.225 moles/min

Flow rate of each product, V= nRT/P, for cyclohexane, V= 10.33*0.0821*(200+273)/100/101.3 =406 L/min, benzene= 406*0.77/10.33= 30.3 L/min and hydrogen = 406*10.125/10.33= 398 L/min

Let R= Recycle flow rate. Hydrogen entering the reactor = R*0.9+40.5

Hydrogen converted per pass = (R*0.9+40.5)*0.48

Moles of cyclohexane formed = 10.33

Moles of cyclohexane formed per moles of cyclohexane= (R*0.9+40.5)*0.48/3= (R*0.9+40.5)*0.16

But moles of cyclohexane formed= 10.33

(R*0.9+40.5)*0.16= 10.33, R*0.9+40.5= 64.56

R*0.9 = 64.56-40.50= 24.06, R= 24.06/0.9= 26.7 moles/min

Total volume of fresh stream= 260+950=1210 L/min at 100+273= 373K and 150/101.3 atm =1.48 atm

Converting this to standard conditions of 273 K and 1 atm pressure

And noting that P1V1/T1= P2V2/T2, V1= 1.48*1210*273/373*1 = 1311 L/min

Volumetric flow rate of recycle= 26.7*0.0821*373/(100/101.3) L/min=828 L/min

Rati of volumetric flow rate of recycle/ fresh feed at STP= 1311/828 =1.58

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