Problema 4.16 Smith-Ing.Reacciones Quimicas

The following conversion data were obtained in a tubular flow reactor for the gas-phase pyrolysis of acetone at 520oC and 1 atm.
CH3COCH3  CH3=C=O + CH4
Flow rate, g/hr 130 50 2110.8
Conversion of acetone 0.05 0.13 0.24 0.35
The reactor was 80 cm long with 3.3 cm inner diameter. What rate equation is suggested?
Calculations:
Volume of reactor = (/4)D2 L =684 cm3 = 0.684 lit
Molecular weight of acetone = 12 + 3 + 12 + 16 + 12 + 3 = 58
The above flow rate data in g/hr are converted into gmol/hr, by dividing them with Molecular weight.
Flow rate,g/hr 130 50 21 10.8
Molal flow rate, gmol/hr (FAo) 2.2414 0.8621 0.3621 0.1862
Conversion of acetone (XA) 0.05 0.13 0.24 0.35
For the (tubular flow) plug flow reactor the design equation is,

 1CAo = pAo/RT = 1.01325 x 10-5/(8314 x (273 + 520))
= 0.01537 kmol/m3 = 0.01537 gmol/lit
For the first order reaction, (variable density systems)
-rA = kCA = kCAo(1 - XA)/(1 + AXA)  2
substitutingfor -rA from equn.2 in equn.1,

for the given gas-phase reaction, A = (VXA = 1 - VXA = 0) / VXA = 0 = (2 - 1)/1 = 1
Therefore,

For the constant V and CAo,
FAo ( 2 ln(1 - XAf) + XAf) = -VCAok= constant.  3
For second order reaction, (by Analytical integration)
k CAo2 V/FAo = 2A(1 + A) ln(1- XAf) + A2XAf + (A + 1)2 XAf/(1 + XAf)
For the given problem, A = 1. Therefore,
(4 ln(1-XAf) + XAf + 4 XAf/(1 + XAf) ) FAo = k CAo2 V = constant.  4
We will check the above relations for various data available.
Molal flow rate, gmol/hr (FAo) | 2.2414 | 0.8621 | 0.3621 | 0.1862 |Conversion of acetone (XAf) | 0.05 | 0.13 | 0.24 | 0.35 |
For First order:FAo ( 2 ln(1 - XAf) + XAf) | -0.11787 | -0.12804 | -0.11184 | -0.09525 |
For Second order:(4 ln(1- XAf) + XAf + 4 XAf/(1 +XAf) ) FAo | 0.124069 | 0.14712 | 0.146799 | 0.145369 |
From the tabulated data, it could be seen that the second order mechanism is well fitting the data, than first order.
Therefore, the reaction...