CH5010 CRT. Total Marks: 25 Quiz I 01.Sep.2014 1. (3 marks) The following data are available for a reaction. Determine the activation energy and the pre exponent for the rate constant, assuming Arrhenius temperature dependency. Temp (K) Rate Const (s-1) 500 8 x 10-3 400 5 x 10-5 2. (5 marks) A liquid phase first order reaction A B is carried out in a batch reactor of 200 lit size, at 75˚C. The rate constant at the operating conditions is 0.003 min-1. Initially, the reactor is charged with 100 moles of pure A. Determine the time necessary to obtain a conversion of 80%. 3. (5 marks) A first order liquid phase reaction A B is to be conducted in an ideal CSTR under isothermal conditions. At the operating temperature, the reaction rate constant is 0.1 min-1. If pure A is fed to the reactor at a concentration of 1 mol/lit, and at a flow rate of 10 lit/min, and if the conversion desired is 80%, determine the volume of reactor needed. 4. (5 marks) A gas phase reaction A B + C is carried out isothermally in a PFR, with negligible pressure drop. The following data are available. Incoming total molar flow rate is 100 mol/min, and the volumetric flow rate = 25 lit/min. The reaction is first order with respect to A. The rate constant is k = 0.05 min-1. Mole percentage of A in the feed is 50% and the rest are inert gases. The desired conversion is 50%. Under steady state conditions, determine (i) the parameter epsilon (ε) and (ii) the volumetric flow rate at the outlet. (You don’t need to calculate the volume of the reactor). 5. (7 marks) Consider the liquid phase reaction A B occurring at constant temperature in a recycle reactor. The fresh feed contains pure A at a concentration of 5 mol/lit, and is coming at a flow rate of 10 lit/min. The reaction is first order with a rate constant of 0.05 min-1. The conversion required is 50%. Determine the volume of the reactor needed, when the recycle ratio R= 1. 0 1 2 R 3 Answers: −E 1. Using Arrhenius expression k = k0 e RT , with R = 8.314 J mole-1 K-1, we get E = 84,390 J mole-1 and k0 = 5.24 × 106 s-1. 2. V = 200 lit, k = 0.003 min-1, NA0 = 100 mol, x = 0.8 and T = 75 C. For a batch reactor, with first order kinetics, N A = N A0 (1 − x ) = N A0e − kt . Therefore, 1 − x = e− kt or t= − ln (1 − x ) − ln (1 − 0.8 ) = ≈536.5 min ≈ 9 h. k 0.003 3. k = 0.1 min-1, CA0 = 1 mol/lit, Q = 10 lit/min, FA0 = 10 mol/min, x = 0.8, V = ?. The design equation gives FA 0 + V ( rA out ) = FA out = FA0 (1 − x ) . i.e. V= FA0 x F x FA0 x 10 × 0.8 = A0 = = = 400 lit − rA out k C A out kC A0 (1 − x ) 0.1×1× 0.2 4. FT-in = 100 mol/min, Qin = 25 lit/min, k = 0.05 min-1, x = 0.5 Stoichiometric table Species A (FA) B (FB) C (FC) Inerts (FI) Total (FT) Inlet (x=0) 0.5 0 0 0.5 1 Outlet (x=1) 0 0.5 0.5 0.5 1.5 1.5 − 1 = 0.5 1 Volumetric flow rate at the outlet = Qin (1+ε x) = 25 × (1+0.5 ×0.5) =31.25 lit/min Parameter ε = 5. CA,0 = 5 mol/lit, Q = 10 lit/min, k = 0.05 min-1, Xo = 0.5, R = 1. R 1 X o = × 0.5 = 0.25 and FA,0 = 5 × 10 = 50 mol/min R +1 2 X0 V dX The design equation is = ∫ − rA ( R + 1) FA0 R Therefore, R +1 0.5 0.5 X0 V dX 1 dX −1 1 − 0.5 1 0.75 = ∫ = = ln ln = ∫ 2 × 50 0.25 kC A0 (1 − x ) 0.05 × 5 0.25 (1 − x ) 0.25 1 − 0.25 0.25 0.5 0.75 V = 400 × ln ≈ 162 lit 0.5
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