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 cm^{3}/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

V_{i} =F_{Ao} (X_{A,i} - X_{A,i-1}) / (-r_{A})_{i}

Where V_{i} = volume of reactor i

F_{Ao} = molal flow rate of A into the first reactor

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

X_{A,i-1} = fractional conversion of A in the reactor i-1

For first order reaction, -r_{A,i} = kC_{A,i} = kC_{Ao}(1 - X_{A,i})

v = volumetric flow rate of A = 400 cm^{3}/min = 0.4 lit/min

For the first reactor: (V = 1 lit)

(-r_{A})_{1} = (kC_{A})_{1} = k C_{A,1} = k C_{Ao} ( 1- X_{A,1})

C_{Ao}= F_{Ao} / v

i.e., F_{Ao} = v C_{Ao}

X_{A,i-1} = X_{A,0} = 0

Therefore,

V_{i} =F_{Ao} (X_{A,i} - X_{A,i-1}) / (-r_{A})_{i}

1 = 0.4 (X_{A,1} - 0) / (0.158 x ( 1 - X_{A,1 }) )

X_{A,1} = 0.283

For the second reactor: (V = 2 lit)

(-r_{A})_{2} = (kC_{A})_{2} = k C_{A,2} = k C_{Ao} ( 1- X_{A,2})

Therefore,

(-r_{A})_{2} = k C_{Ao} ( 1- X_{A,2})

X_{A,1} = 0.283

F_{Ao} = v C_{Ao}

V_{i} =F_{Ao} (X_{A,i} - X_{A,i-1}) / (-r_{A})_{i}

2 = 0.4 (X_{A,2} - 0.283) / ( k ( 1- X_{A,2}) )

5 k = (X_{A,2} - 0.283) / ( 1- X_{A,2})

0.79 - 0.79 X_{A,2} = X_{A,2} - 0.283

1.073 = 1.79 X_{A,2}

X_{A,2} = 0.60

For the third reactor: (V = 1.5 lit)

(-r_{A})_{3}= (kC_{A})_{3} = k C_{A,3} = k C_{Ao} ( 1- X_{A,3})

X_{A,2} = 0.6

F_{Ao} = v C_{Ao}

V_{i} =F_{Ao} (X_{A,i} - X_{A,i-1}) / (-r_{A})_{i}

1.5 = 0.4 (X_{A,3} - 0.60) / ( k ( 1- X_{A,3}) )

0.5925 = (X_{A,3} - 0.60) / ( 1- X_{A,3})

0.5925 - 0.5925 X_{A,3} = X_{A,3} - 0.60

1.1925 = 1.5925 X_{A,3}

X_{A,3 }= 0.749

The fraction hydrolyzed in the effluent from the third reactor = 0.749

Last Modified on: 30-Apr-2024

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