Hydrodealkylation is a reaction that can be used to convert toluene (C7H8) into

Question

Hydrodealkylation is a reaction that can be used to convert toluene (C7H8) into benzene (C6H6), which is historicaly more valuable than toluene. The reaction is C7H8 + H2 -> C6H6 + CH4

Zimmerman and York have studied this reaction between 700 and 950 C in the absence of any catalyst. They found that the rate of toluene disappearance was well correlated by

-rT = kT [H2]1/2 [C7H8]
kT = 3.5*1010 exp(-E/RT) (L/mol)1/2 /s
E=50,900 cal/mol

The reaction is essentially irreversible at the conditions of the study, and the ideal gas laws are valid.
Consider a feed stream that consists of 1 mol of H2 per mole of toluene. A reactor is to be designed that operates at atmospheric pressure and a temperature of 850 C. What volume (in liters) of reactor is required to achieve a fractional conversion of toluene of 0.50 with a toluene feed rate of 1000 mol/h?
First assume that the reactor is an ideal, plug-flow reactor. Then repeat the calculation for an ideal CSTR

Explanation / Answer

kT = 3.5*1010exp(-E/RT)

kT = 3.5*1010 exp(-50900/ 1.986 x 1123.15 K) = 4.303

The mole ratio of H2 and toluene = 1:1

So we can replace H2 with toluene in rate law

rate = 4.303 x (C7H8)1.5

Conversion is 50%

dF/dV = 4.303 x (C7H8)1.5

Q dC/dV = 4.303 x (C7H8)1.5

dV = (Q/4.303) x (dC/(C7H8)1.5 )

V = (Q/4.303) x- ( 1/1.5xC in2.5  - 1 / (1.5xCout2.5 ))

=(Q/4.303) x- ( 1/1.5xC in2.5  - 1 / (1.5xCout2.5 ))

In question Q(volumtric flow rate is not given) so keep suitable number and calculate Volume.

CSTR:

Q(C in - C out ) = V x rate = 4.303 x (C7H8)1.5

If u get Q value , first calculte C in = 1000 mol/h / Q

Then C out = C7H8 = 0.5 x C in

Then easily u can calculte Volume of PFR and CSTR

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