Question

Monochlorobenzene (C6H5Cl) is produced commercially by the direct catalytic chlorination of benzene (C6H6). In the process, dichlorobenzene(C6H4Cl2) is generated as a coproduct.

?6?6 + ??2 → ?6?5?? + ??? ?6?5?? + ??2 → ?6?4??2 + ???

Given that the molar ratio of benzene to chlorine in the feed is 5 to 1; and the feed contains only benzene and chlorine. Chlorine is the limiting reactant and completely consumed in the reaction. The fractional yield of C6H5Cl is 0.875

. a. Take a basis of 100 mol C6H5Cl produced, draw and label a flowchart. Do a degree of freedom analysis based on atomic species balances.

b. Calculate the amount of feed and product components of the reactor using the atomic species balances. Present the calculated values in a table to prove that the total mass to the reactor is equal to the total mass out from the reactor (mass conservation).

c. Calculate the percentage conversion of C6H6. The value that you have calculated should be low. Why do you think the reactor would be designed for low conversion? What additional processing steps are likely to take place downstream of the reactor in order to improve the conversion?

Answer 1

basis : 100 mols of C6H5Cl produced.

let the no of moles of feed = F

mol fraction of benzene in feed = 5/6 = 0.833

mol fraction of chlorine in feed = 1/6 = 0.167

the reaction are

the flowsheet is given by

the atomic species involved are C, H and Cl.

the products are C6H5Cl, C6H4Cl2, C6H6 and HCl.

the no of moles of C6H5Cl are denoted by n_C6H5Cl = 100 mols

no of moles of C6H4Cl2 are denoted by n_C6H4Cl2

no of moles of benzene in the outlet = n_C6H6.

no of moles of HCl in the outlet = n_HCl .

no of moles of benzene entering the reactor = 0.833 F

Carbon balance :

C entering the reactor = C leaving the reactor.

1 mol of C6H6 contains 6 mols of C = 72 g of C.

1 mol of C6H5Cl contains 6 mols of C = 72 g of C

1 mol of C6H4Cl2 contains 6 mols of C = 72 g of C

mass of Carbon entering the reactor = 6 x 12 x 0.833 x F

mass of carbon leaving the reactor = Carbon present in unreated benzene + C6H5Cl + C6H4Cl2

mass of carbon leaving the reactor = 6 x 12 x ( n_C6H5Cl + n_C6H4Cl2 + n_C6H6.)

6 x 12 x 0.833 x F = 6 x 12 x ( n_C6H5Cl + n_C6H4Cl2 + n_C6H6.)

0.833 F = 100 + n_C6H4Cl2 + n_C6H6 ----------------eq-1

Hydrogen balance :

1 mol of C6H6 contains 3 mols of H₂ i.e 6 H atoms = 6 g of H.

1 mol of C6H5Cl contains 2.5 mols of H₂ i.e 5 H atoms = 5 g of H.

1 mol of C6H4Cl2 contains 2 mols of H₂ i.e 4 H atoms = 4 g of H.

1 mol of HCl contains 0.5 mols of H2 i.e 1 H atom = 1 g of H

mass of Hydrogen entering the reactor = mass of Hydrogen leaving the reactor.

mass of Hydrogen entering the reactor = 6 x 0.833 F

mass of hydrogen leaving the reactor = Hydrogen present in ( unconverted C6H6 + C6H5Cl + C6H4Cl2 + HCl)

mass of hydrogen leaving the reactor = 6 n_C6H6 + 5 n_C6H5Cl + 4 n_C6H4Cl2 + 1 n_HCl

6 x 0.833 F = 6 n_C6H6 + 5 n_C6H5Cl + 4 n_C6H4Cl2 + 1 n_HCl

6 x 0.833 F = 6 n_C6H6 + 5 x 100 + 4 n_{C6H4Cl2 ​​​​​​​} + 1 n_HCl ------------------eq-2

Chlorine balance :

1 mol of C6H5Cl contains 1 atom of Cl = 35.453 g of Cl

1 mol of C6H4Cl2 contains 2 atoms of Cl = 70.906 g of Cl

1 mol of HCl contains 1 atom of Cl = 35.453 g of Cl

mass of Chlorine entering = mass of Chlorine leaving the reactor.

mass of Chlorine entering = 70.906 x 0.167 F

mass of Chlorine leaving = Chlorine present in ( C6H5Cl + C6H4Cl2 + HCl )

70.906 x 0.167 F = 35.453 n_C6H5Cl + 70.906 n_{C6H4Cl2 ​​​​​​​} + 35.453 n_HCl

70.906 x 0.167 F = 35.453 x 100 + 70.906 n_{C6H4Cl2 ​​​​​​​} + 35.453 n_HCl ----------------eq-3

now additional information :

fractional yield of C6H5Cl is 0.875

given total Cl2 is converted = 0.167 F

0.875 = no of mols of C6H5Cl formed / no of moles of Cl2 converted.

0.875 = (100 + n_{C6H4Cl2 ​​​​​​​}) / 0.167 F ------------------------eq-4

no of unknowns = n_C6H6 , n_{C6H4Cl2 ​​​​​​​} ,  n_HCl , F = 4

no of equations = 4

degrees of freedom = 0

b)​​​​​​​

from eq-4

0.1462125 F - n_{C6H4Cl2 ​​​​​​​} = 100 -------------eq-5

0.833 F = 100 + n_C6H4Cl2 + n_C6H6 ----------------eq-1

0.833 F - n_C6H4Cl2 - n_C6H6 = 100 ---------------eq-6

6 x 0.833 F = 6 n_C6H6 + 5 x 100 + 4 n_{C6H4Cl2 ​​​​​​​} + 1 n_HCl ------------------eq-2

4.998 F - 4 n_{C6H4Cl2 ​​​​​​​} + 1 n_HCl - 6 n_C6H6 = 500 ---------------eq-7

70.906 x 0.167 F = 35.453 x 100 + 70.906 n_{C6H4Cl2 ​​​​​​​} + 35.453 n_HCl ----------------eq-3

0.334 F - 2 n_{C6H4Cl2 ​​​​​​​}- n_HCl = 100 ----------------eq-8

on solving eq-5, eq-6, eq-7 and eq-8

we get

n_C6H4Cl2 = 16.7 mols

n_C6H6 = 548.4 mols

F = 798.4032 mols

n_HCl = 133.33 mols

molecular weight of benzene = 78.11 g/mol

molecular weight of Chlorine = 70.906 g/mol

molecular weight of = ?6?5?? = 112.56 g/mol

molecular weight of ?6?4??2 = 147.01 g/mol

mass of feed entering = 0.833 x 798.4032 x 78.11 + 0.167 x 798.4032 x 70.906 = 61402.74061 g = 61.4 kg

mass of product stream = 548.37 x 78.11 + 133.33 x 36.46 + 16.7 x 147.01 + 100 x 112.56 = 61.4 kg

mass balance is satisfied.

C6H6	Cl2	Feed	C6H5Cl	C6H4Cl2	HCl	C6H6	Product
665.07	133.33	798.4032	100	16.67	133.33	548.37	798.37
51.95 kg	9.453 kg	61.4 kg	11.256 kg	2.45 kg	4.874 kg	42.83	61.4 kg

c)

benzene entering = 665.07 mols

benzene leaving = 548.37 mols

conversion = (665.07 - 548.37) / 665.07 = 17.54 %

very less.

to increase the conversion, the unconverted benzene can be recycled along with makeup Cl2. this will increase the conversion.

separate formed products and benzene, recycle with make up Cl2 in to the reactor.