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

t_{1} = V_{1}/v = (C_{0} - C_{1})/ (-r)_{1}

t_{2} = V_{2}/v = (C_{1} - C_{2})/ (-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 1^{st} order:

for first order reaction, -r_{A} = k C_{Ao}(1 - X_{A}) and

where, C_{Ao} = concentration of A in the feed to the reactor; and

X_{A} = fractional conversion of A in the reactor

for the second reactor (-r)_{2} = k 2 x 0.5 = k

since t_{2} = 65 sec, and C_{1 }- C_{2} = 2 - 1 = 1 gmol/lit

i.e., 65 = 1/k à 1

for the first reactor, V_{1}= V_{2}/2

and t_{1} = 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 à 2

from equn.1 and 2 it is seen that 1^{st} order mechanism is not suitable.

Test for 2^{nd} order:

for 2^{nd} order reaction, -r_{A} = k C_{Ao}^{2}(1 - X_{A})^{2} and

for the second reactor (-r)_{2} = k x 2^{2} x 0.5^{2} = k

t_{2} = V_{2}/v = (C_{1} - C_{2})/ (-r)_{2}

i.e.,

65 = (2 - 1)/k = 1/k

65 = 1/k à 3

for the first reactor (-r)_{1} = k x 4^{2} x 0.5^{2} = 4k

t_{1} = V_{1}/v = (C_{0} - C_{1})/ (-r)_{1}

32.5 = (4 - 2)/4k = 1/2k

i.e.,

65 = 1/k à 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, **-r _{A} = 0.0154 C_{A}^{2} gmol/(lit.sec)**