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