The kinetics of an aqueous phase decomposition of A is investigated in two CSTR's in series, the first having half the volume of the second. At steady state with a feed concentration of 4 gmol/lit., and mean residence time of 65 sec in the second reactor, the concentration of feed to second from the first is found to be 2 gmol/lit., while that of the stream leaving the second is 1 gmol/lit. Find the suitable kinetic rate expression.
Calculations:
Data:
Residence time of second reactor = 65 sec
τ1 = V1/v = (C0 - C1)/ (-r)1
τ2 = V2/v = (C1 - C2)/ (-r)2
Conversions based on the feed to the reactors = 50% = 0.5
(4 to 2 gmol/lit in first reactor, and 2 to 1 gmol/lit in second reactor)
Test for 1st order:
for first order reaction, -rA = k CAo(1 - XA) and
where, CAo = concentration of A in the feed to the reactor; and
XA = fractional conversion of A in the reactor
for the second reactor (-r)2 = k 2 x 0.5 = k
since τ2 = 65 sec, and C1 - C2 = 2 - 1 = 1 gmol/lit
i.e., 65 = 1/k → 1
for the first reactor, V1= V2/2
and τ1 = 65/2 = 32.5 sec.
for the first reactor (-r)1 = k 4 x 0.5 = 2k
i.e., 32.5 = (4 - 2)/2k = 1/k → 2
from equn.1 and 2 it is seen that 1st order mechanism is not suitable.
Test for 2nd order:
for 2nd order reaction, -rA = k CAo2(1 - XA)2 and
for the second reactor (-r)2 = k x 22 x 0.52 = k
τ2 = V2/v = (C1 - C2)/ (-r)2
i.e.,
65 = (2 - 1)/k = 1/k
65 = 1/k → 3
for the first reactor (-r)1 = k x 42 x 0.52 = 4k
τ1 = V1/v = (C0 - C1)/ (-r)1
32.5 = (4 - 2)/4k = 1/2k
i.e.,
65 = 1/k → 4
On comparing equn.3 and 4 it could be seen that the given data is for a second order reaction.
k = 1/65 lit/(gmol.sec) = 0.0154 lit/(gmol.sec)
and, -rA = 0.0154 CA2 gmol/(lit.sec)
Last Modified on: 04-Feb-2022
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