Conversion Problem

So a reaction between benzene and chlorine occurs forming dichlorbenzene and monochlorobenzene. The question states:

Assume that the molar ratio of the feed of chlorine to that of benzene is 0.9:1 at the inlet to the reactor. For a feed of 100 kmol/h of benzene to the inlet of the reactor, calculate the molar flow of all the species leaving the reactor.

The overall conversion of benzene is 55.3%, of which 73.6% reacts to form monochlorobenzene and the remainder form dichlorobenzene.

So first I got the reaction equations:

C₆H₆ + Cl₂ --> C₆H₆Cl + HCl
C₆H₆ + 2Cl₂ --> C₆H₄Cl₂ + 2HCl

Then using the mole ratios I got:

•

Moles of benzene in: 47.4 kmol/h

•

Moles of chlorine in: 52.6 kmol/h

Then, if I remember correctly, conversion is:

X = consumed/fed

so the amount of benzene consumed is: 25.122 kmol/h and hence 47.4-25.122=22.3 kmol/h leaves the reactor.

Here's where I get a bit stuck. If 55.3% of benzene is consumed, how much Cl₂ is consumed? Do I have to take into consideration the the amount of moles (i.e. the coefficient of the Cl₂ is the reaction equations)?

And also, would the amount of monochlorobenzene out simply be 73.6% of the benzene that is reacted and the rest be dichlorobenzene?

Thanks!

(edited 8 years ago)

Reply 1

Infraspecies

1 mol of Cl₂ is consumed per every monochlorobenzene.
1 further mol of Cl₂ is consumed per every disubstitution.

I would;

1. Work out the total number of moles of benzene consumed
2. =n₁(Cl₂)
3. n₂(Cl₂)= moles of dichlorobenzene
4. n_total(Cl₂)=n₁(Cl₂)+n₂(Cl₂)