A second order irreversible aqueous phase reaction, A rightarrow R, is to be car

Question

A second order irreversible aqueous phase reaction, A rightarrow R, is to be carried out in a CSTR, either operated isothermally or adiabatically. The rate expression for the reaction is -r_A = kC_A^2, fmole/L min where k = 10^13 e^-10.000/T, L/gmol min and T is in K. If the reactor is fed 100 L/min of an aqueous solution of 2 gmole/L a at 50 degree C and the conversion of A is 90%, calculate the size of the reactor based on: a. isothermal operation b. adiabatic operation The exothermic heat of reaction is 10,000 cal/gmole of A. Assume the physical properties of water for the solution in the reactor.

Explanation / Answer

V=FA0X/-rA

RATE LAW,

-rA =kCA2

LIQUID PHASE V=V0

FEED RATE V0=100L/MIN

a) Isothermal operation

FA0= V0CA0

FA= VCA= V0CA

CA= FA/V= FA/V0= FA0(1-X)/ V0

CA= CA0(1-X)

COMBINE,

V= V0CA0X/k CA02(1-X)2

= V0X/k CA0(1-X)2

=100X 0.90/(1013e-10000/323)(2)(1-0.9)2

=90/.00716=12556.31L

b)adiabatic operation

cp of water75.30 J/(mol K) at 50 °C

HRxn at temperature T = HRxn at 323K + Cp (T-323)

0=10000+75.30(T-323)

-10000=75.30T-24321.9

T=390.19K

V /V0=X/K(1-X)

V=100x .90/(1013e-10000/390)x(1-.9)

=12L

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