### Stirred Tank Reactors of Unequal Volumes in Series

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