Chemical reaction engg problem- 6 the reaction 2a lt

Question

Chemical reaction engg Problem- 6 The reaction , 2A < =K1/K2(C+D is to be conducted in a CSTR DATA - Flow rate = 3 m3 /hr Initial concentration of A ,C and D = 22 Kg-mole /m³ , 0 and 0 respectively K1 = 0.7 m³/kg-mole -hr K2 = 16.0 m³/kg-mole-hr Conversion to be obtained = 80% The volume of the reactor =?

Solution- The reaction -- 2A <=K1/K2=(C +D The rate will be - R = r = K1.Ca^2 -K2*Cc*Cd = K1[Cao^2(1-xa)^2 -- K2*(Cao.xa/2)^2 Where, Xa = xa= conversion at any instant Let , xae = equilibrium conversion Ke = equilibrium constant At equilibrium rate = 0 K1 [Cao (1-xae)]^2 = K2 .

Cao^2.xae^2/4 Or, 0.7 *( 1- xae)^2 =16[xae^2/4] = 4xae^2 3.3*xae^2+1.4xae+0.7 =0 Solving we get- Xae =xae =0.295 Actual conversion = 0.8*0.295 = 0.236 =0.236*100 =23.6% Vo = feed rate , m3/hr V= volume of reactor , m³ V/Vo = Cao.xa/r = 22*0.236/[(0.7)[ 22(1-0.23)]^2 -16 [22*0.236/2]^2 =0.0577 Vo = 3 m³/hr V = 0.0577 *3 = 0.1732 m³ Volume =0.1732*1000=173.2 liters -16 [22*0.236/2]^2 =0.0577 Vo = 3 m³/hr V = 0.0577 *3 = 0.1732 m³

Volume =0.1732*1000=173.2 liters DATA - Flow rate = 3 m³ /hr Initial concentration of A ,C and D = 22 Kg-mole /m³ , 0 and 0 respectively K1 = 0.7 m3/kg-mole -hr K2 = 16.0 m3/kg-mole-hr Conversion to be obtained = 80% The volume of the reactor =?

Solution- The reaction -- 2A <=K1/K2=(C +D The rate will be - R = r = K1.Ca^2 -K2*Cc*Cd = K1[Cao^2(1-xa)^2 -- K2*(Cao.xa/2)^2 Where, Xa = xa= conversion at any instant Let , xae = equilibrium conversion Ke = equilibrium constant At equilibrium rate = 0 K1 [Cao (1-xae)]^2 = K2 .Cao^2.xae^2/4 Or, 0.7 *( 1- xae)^2 =16[xae^2/4] = 4xae^2 3.3*xae^2+1.4xae+0.7 ...