Question

An aqueous solution containing reactant A is fed to a batch reactor, operated as part of a continuous process involving other equipment. The feed rate to the reactor is 500 L/min. The reaction, A + 2B rightarrow C is second order irreversible, with the rate expression given as rr_A = 0.04 C_AC_B, gmole/L middot min If C_A0 = 1.7 gmole/L and C_B0 = 3.0 gmole/L, what is the size of the reactor required to obtain an 80% conversion of A? The maintenance time between batches is 5 min.

Answer 1

Initial Conc of A = 1.7 gmol/L

Initial Conc of B = 3 gmol/L

Rate = 0.04C_AC_B = 0.04 C_A0 (1-x) (C_B0 - C_A0x) = 0.04 C_A0 (1-x) C_A0(M- x) = 0.1156 (1-x)(1.765-x)

Here M = C_{B0 /} C_A0 3/1.7 =1.765

For batch reactor

V₀C_A0 (dX/dt) = Rate x volume

V₀C_A0 integral ( 1/ rate ) dx = Volume x dt

(500 L/min x 1.7 gmol/L )* integral(1/ 0.1156 (1-x)(1.765-x) ) = Volume x integral(1 dt)

At t=0, x=0 and t=5 , x =0.8

Volume ,V = 437 L