QUESTION  -  1 . The rate  of a liquid phase  reaction  of the type ,A+B --àproducts  is  found to be  independent  of concentration of A and B  and  equal  to 1 KMOL/m3  ( min)   at 300  ok  .Find  the conversion  in a  mixed flow reactor  having equal   to 2 m3 with feed   concentration of A and B   equal  to 5 kmol/m3 , feed  flow rate   equal  to  1 m3/min and reactor    temperature  equal to 300 ok.

If the   activation energy of the reaction   is  given   as 63.1 kg /mol, what  is the volume  of the isothermal   plug flow reactor  for the  same conversion   and feed  conditions as in  case  of above   mentioned  reactor  but  with reactor temperature kept  at 325 ok.

Solution-  Vo = 1  m3/min

Cao = Cno  =5 kmol/m3

-r =  1  kmol/ m3-min

V= 2 m3

T=  300 ok

Vo*Cao  = Vo*Ca+ (-ra*V)

Vo ( Cao - Ca) = -ra*V

Vo*CaO*xa  = - ra*V

XA  = Xa  =  - ra*V/Vo*Cao =1*2/1*5 = 0.4

K1  =Ko e^-E/R.T1

K2  =Ko^e^-E/R*T2

K2/K1  = e^-E/R*(1/T2-1/T1)

K2  =K1*e^-63.1*10^3/6.314*(1/320-1/300)

K2 =1*e^-63.1*10^3/6.314*( 1/320-1/300)  =8.023 kmol/m3 -min

For the  plug flowreactor-

V/Vo = Cao ∫( limit xa  to  0 )  dxa/k =Cao *xa/k

V = Vo.Cao *xa/k = 0.249 m3