Acetic acid is hydrolysed in three stirred tank reactors operated in series. The feed flows to the first reactor (V = 1 lit) at a rate of 400 cm3/min. The second and third reactors have volumes of 2 and 1.5 litres respectively. The first order irreversible rate constant is 0.158 min-1. Calculate the fraction hydrolysed in the effluent from the third reactor.
Calculations:
The design equation for series, steady flow mixed reactor is
Vi =FAo (XA,i - XA,i-1) / (-rA)i
Where Vi = volume of reactor i
FAo = molal flow rate of A into the first reactor
XA,i = fractional conversion of A in the reactor i
XA,i-1 = fractional conversion of A in the reactor i-1
For first order reaction, -rA,i = kCA,i = kCAo(1 - XA,i)
v = volumetric flow rate of A = 400 cm3/min = 0.4 lit/min
For the first reactor: (V = 1 lit)
(-rA)1 = (kCA)1 = k CA,1 = k CAo ( 1- XA,1)
CAo= FAo / v
i.e., FAo = v CAo
XA,i-1 = XA,0 = 0
Therefore,
Vi =FAo (XA,i - XA,i-1) / (-rA)i
1 = 0.4 (XA,1 - 0) / (0.158 x ( 1 - XA,1 ) )
XA,1 = 0.283
For the second reactor: (V = 2 lit)
(-rA)2 = (kCA)2 = k CA,2 = k CAo ( 1- XA,2)
Therefore,
(-rA)2 = k CAo ( 1- XA,2)
XA,1 = 0.283
FAo = v CAo
Vi =FAo (XA,i - XA,i-1) / (-rA)i
2 = 0.4 (XA,2 - 0.283) / ( k ( 1- XA,2) )
5 k = (XA,2 - 0.283) / ( 1- XA,2)
0.79 - 0.79 XA,2 = XA,2 - 0.283
1.073 = 1.79 XA,2
XA,2 = 0.60
For the third reactor: (V = 1.5 lit)
(-rA)3= (kCA)3 = k CA,3 = k CAo ( 1- XA,3)
XA,2 = 0.6
FAo = v CAo
Vi =FAo (XA,i - XA,i-1) / (-rA)i
1.5 = 0.4 (XA,3 - 0.60) / ( k ( 1- XA,3) )
0.5925 = (XA,3 - 0.60) / ( 1- XA,3)
0.5925 - 0.5925 XA,3 = XA,3 - 0.60
1.1925 = 1.5925 XA,3
XA,3 = 0.749
The fraction hydrolyzed in the effluent from the third reactor = 0.749
Last Modified on: 04-Feb-2022
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