Note: Chemical Engineering major course called Chemical Reaction Engineering. Th

Question

Note: Chemical Engineering major course called Chemical Reaction Engineering.

The following is the reaction for isothermal hydrogenation of o-cresol to 2-methylcyclohexane in a packed bed reactor. The reaction is zero order in A and first order in B (k = 1.74 mole/kg cat. min. atm.): A + 2B rightarrow C, where A = o-cresol. B = hydrogen, and C = 2-methylcyclohexane. The feed, which is at 5 atm., is stoichiometric in A and B with the total flow rate of 40 mole/min. a) Calculate the weight of catalyst needed to achieve 70% conversion of A. You may assume a to be 0.34 kg^-1. b) For the same weight of catalyst, what will be conversion if you assume that elementof X

Explanation / Answer

Solution:

Given Data:

k=1.74 mol/Kg.cat.min.atm

FA0= 40 mol/min

P= 5atm

Reaction = A+2B--> C

A is first order and B is zero order so rate equation wll be -rA= kCA

a) XA=0.7 alpha= 0.34 kg-1

Taking integration XA=0 to XA=0.7

dXA/dW= kCAo(1-XA)x (1-alphaW)^(1/2)/FAO

Integrating

ln(1/1-XA)= kcAo*2(1-(1-apha.W^3/2)/FAO*3 alpha

ln(1/1-0.7)=1.74(mol/Kg.cat.min.atm)*3*(1-(1-0.34*W^3/2)/40*3*0.34

1.204=3.48*(1-(1-0.34*W^3/2)/40.8

49.123=3.48*(1-(1-0.34*W^3/2)

14.1159=(1-(1-0.34*W^3/2)

41.517=W^3/2

W= 11.99 Kg

weight of catalyst at XA=0.7 is 11.99Kg

b) W= 11.99 Kg to calculate XA

dW= FAO*dXA/ -rA

Integrating with 0 to XA

W= FAO*(XA-0)/-rA

11.99= 40.XA/1.74

XA=0.522

For the same weight of catalyst conversion will be 52.2%

dXA/dW = -rA/FA0 Rate Law    -rA = kCA Stoich. CA = CA0(1-X)*P/P0 P/P0 = (1-W)1/2 CA = CA0(1-X)(1-W)1/2

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