The cumene process is used to convert benzene and propene in the presence of oxy

Question

The cumene process is used to convert benzene and propene in the presence of oxygen to phenol and acetone via the following reaction. This reaction is carried out in with excess benzene and excess oxygen to ensure that no side reactions occur. A 1.19 ratio of benzene to propene is fed to a reactor in the presence of 35.0% excess air (N2 and 02) The first pass conversion of propene is 100%. The products leave the reactor and enter a condenser where the gaseous nitrogen and oxygen leave the process and the liquid organic products are fed to a distillation column. The distillation column separates the products from the reactants. Any unreacted reactants are fed into the fresh feed stream. The products are directed to a second distillation column that separates the acetone from the phenol. If 12.7 kg/hr of phenol is produced from this process, answer the following questions below, putting either a given numeric value or the words "unknown" or t, label the process flow diagram below, putting either a given numeric value or the words "unknown" or none" in each space. Do not leave any spaces blank. Variable names are shown for reference later

Explanation / Answer

Mass of phenol= 12.7 Kg/hr

Moles of phenol= mass/ molecular weight= 12.7/ 94 =0.135 Kgmoles/ hr

Moles of acetone formed = 0.135 kgmoles/ hr

Moles of mixture entering the distillation column =0.135+0.135= 0.27 kgmoles/hr

Moles of acetone/ mole of mixture =0.135/0.27 =0.5

Moles of phenol/moles of mixture= 0.135/0.27=0.5

Moles of porpene entering the reactor =0.135 kg moles/hr ( There is no contribution from recycle)

Mole ratio of propene to : benzene in the reactor 1:19

Moles of Benzene entering the reactor =19*0.135 =2.565 kgmoles/hr

Moles of Benzene reacted =0.135 moles

Moles of Benene returned from distillation column-1(n6) =18*0.135= 2.43 kg moles/hr

Moles of benzene/ mole of mixture (x12)=1

Mols of C3H6/ moles of mixture (x11) =0

Hence, feed enterign the distillation column-1(n5) =2.43 +0.27= 2.70 kg moles/hr

Condenser our let contains (n5) =2.7 kg moles/ hr of mixture and uncondensed nitrogen and oxygen

Moles of acetone/ mole of mixture (x13)= 1.35/2.7 =0.5 moles/mole

Moles of phenol/ moles of mixture(x14)= 0.5

Nitrogen entering the reactor = (0.135/0.21)*0.79*1.35 =0.685 kg moles/hr

this is not participating in the reactor hence remains in the uncondensed gas

Moles of oxygen reacted = 0.135 kg moles/hr

Excess oxygen =0.135*0.35= 0.04725 kg moles/hr

Total moles of oxygen and nitrogen (n4) =0.685+0.04725 =0.73225 kg moles/hr

n3=n4+n5= 0.73225+2.7=3.43225 kg moles/hr

Oxygen (x9)=0.04725/ 0.73225 =0.064 kg moles of oxygen/ kg mole

Nitrogen(x10)= 0.685/0.73225 =1-0.064 =0.96 kgmoles of nitrogen/ kg mole

Fresh feed

Benzene =0.135 kg moles/hr

Propene= 0.135 kg moles/hr

Nitrogen = (0.135/0.21)*1.35*0.79 ( Air is supplied 35% excess) = 0.685 kgmoles/hr

Oxygen = 0.135*1.35 kg moles/hr

n1= 0.135+0.135+0.685+0.18225=1.13725 kg moles/hr

x1= 0.135/ 1.13725 = 0.118 mol/mol

x2= 0.118 mol/mol ( same as x1)

x3= 0.18225/1.13725 = 0.160255 mol/mol

x4= 0.685/1.13725 = 0.60233 mol/mol

Moles of Propene required at the reactor inlet= 0.135 ( from stoichiometry) kgmoles/hr

Moles of mixture eneternig the distillation column-2 =0.135 +0.135 =0.27 kg moles/hr

Moles of Propene/ total moles =0.135/0.27=0.5

Moles entering the reactor (n2) = 0.18225 (Oxygen) + 0.685(Nitrogen) + 2.5625( Recycle + fresh)+0.135 propene= 3.56725 kg moles

x5= 2.5625/ 3.56725 =0.71834 mol/mol

x6= 0.135/ 3.56725 =0.037844 mol/mol

x7= 0.18225/ 3.56725= 0.05109 =0.192 mol/mol

x8= 0.686

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