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. C. H C, H This reaction is carried out with excess benzene and excess oxygen to ensure that no side reactions occur A 1.18 ratio of benzene to propene is fed to a reactor in the presence of 26.0% excess air (N2 and O2). 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 17.5 kg/hr of phenol is produced from this process, answer the following questions First, 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 produced. m9= 17.5 kg/hr

Molar mass of Phenol =94, moles of phenol= 17.5*1000/94 gmoles/hr=186 gmoles/hr

same no of C3H6O moles are produced. Moles of C3H6O produced, n8= 186 gmoles/hr

molar mass of C3H6O= 3*12+6+16= 58, mass of C3H6O produced= 186*58=10788 kg/hr, Mass of C3H6O produced,

Balance across the second distillation column gives n7= 186+ 186= 372 moles/hr

x13= 186/372=0.5= x14

fresh feed as per the reaction contains 186 moles/hr of propene ( liminting reactant), ratio of benzene/ propene =1.18

moles of benzene = 1.18*186= 219.48 moles/hr, moles of air to be supplied= 186 moles/hr, Since air contains 21% O2 and 79% N2, moles of air to be supplied= 186/0.21 =886 moles/hr, Air supplied is 26% excess. Moles of air supplied= 1.26*886= 1116 moles/hr, N2= 1116*0.79= 882 and O2= 1116-882= 234

total flow rate of feed= 186( propene)+219.48( benzene)+ 882 (N2)+234(O2)= 1520.48

composition of feed : propene = 186/1520.48= 0.122, Benzene= 219.48/152048=0.1443, O2= 234/1520.48=0.1538 and N2= 882/1520.48=0.58

The excess oxygen along with unreacted N2 leaves from the bottom of condenser.

flow rate of bottoms ( moles/hr):O2=234-186=48 , N2= 882

flow rate of bottoms= 48+882= 930moles/hr

Composition of bottoms : O2= 48/930= 0.052 and N2= 882/930=0.948

let R= recycle flow rate. Moles of propene entering the reactor = R*x11+157.63

all the proene got converted. Moles of phenol that can be produced = R*x11+157.63

hence n6*x11+157.63= 0, x11= 0

since all the propene got converted, x11=0, x12=1-x11=1

moles of benzene entering the reactor= n6+219.48

moles of propene got converted = n6+180= moles of benzene got converted

moles of benzene remaining = n6+219.48-n6-180= 39.48 moles/hr. This has to be recycled. n6*x12=39.48 moles/hr

n6= 39.48 moles/hr

The reactor exit contsins : 186 moles/hr phenol, 186 moles/hr of C3H6O, 39.48 moles/hr benzene,48 moles/hr of gaseous O2 and 882 moles/hr of unreacted N2.

the gaseous stream of O2 and N2 are removed at the condesner and the outlet from condenser liquid stream contains 186 moles/hr of C3H6O and 186 moles/hr of phenol and 39.48 moles/hr of Benzene.

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