Question

An elementary reaction is operated in liquid phase in a perfect mixed flow (CSTR) reactor A 3B The feeding is composed of pure reactant A at an inlet molar flowrate of 1mol.h. The reaction advancement ? is 0.8 mol.h-1. a) Compute the partial molar flowrates of A and B at the reactor outlet b) ? (h) the space time, is defined as the ratio between the volume of the liquid mixture ver the volumetric flowrate of the feeding stream q(m.h), and k (h) is the kinetic constant of the elementary reaction What is the value of (k) in the conditions above? c) What would be the degree of advancement of the reaction if the inlet molar flowrate composed of pure A is reduced to 50%? Justify your response. i. 0.251 mol.h ii. 0.381 mol.h iii. 0.440 mol.h iv. 0.481 mol.h d) What would be the degree of advancement of the reaction i f instead of pure A, an equimolar mixture of reactant A (0.5 mol.h) and of an inert I (0.5 mol.h) is introduced into the reactor? Justify your response. i. 0.335 mol.h-1 ii. 0.400 mol.h" ii. 0.455 mol.h iv. 0.485 mol.h e) In which of these two last cases is B produced in the larger amount? f) By how much is then the conversion rate increased compared to the initial conditions? 8) Assuming a partial recirculation of B of 1 mol/h to the reactor entrance, added to the pure A feeding (1 mol.h), how will the advancement be affected? Justify.

Answer 1

As per Chegg policy and time limit, only first five parts a to e are answered.
A riight 3B Feed is pure 1 mol/h of A The reaction aadvancement epsilon is 0. 8 mol/h (a) epsilon = N_A0-N_A/-mu_A = N_BD - N_B/-mu_B N_A0 = 1 mol/h, N_B0 = 0, mu_A = -1, mu_B = 3, epsilon = 0. 8 N_A = 1-0. 8 = 0. 2 mol/h N_B = 3 times 0. 8 = 2. 4 mol/h partial molar flow racts os A = 0. 2 mol/h B = 0. 2 mol/h } Answer \r\n b) for a CSTR, tau/C_A0 = X_n/-r_A (Here r_A = -k C_A tau/C_A0 = X_A?K C_A {C_A = C_A0(1-X_A) } Ktau = C_A0 X_A/C_A = C_A0 X_A/C_A0(1-X_A) = X_A/1-X_A Here X_A = 0. 8 Ktua = 0. 8/1-0. 8 = 0. 8/0. 2 = 4 [Ktua = 4] Answer \r\n (c) V/N_A0 = X_A/K C_A = X_A/K C_A0(1-X_A) for case 1: N_A0 = 1 mol/h, X_A = 0. 8 case 2: N_A0 = 0. 5 mol/h, X_A(unknown) case 1 VK C_A0 = 4----(1) case 2 V/0. 5(X_A/1-X_A) X_A = 8/9 = 0. 88 N_A + N_A0(1-X_A) = 0. 06 mol/h epsion = N_A0-N_A/-mu_A = 0. 5-0. 06/-(-1) = 0. 44 mol/h [Answer(iii) 0. 440 mol h^-1] \r\n For case 1 VK C_Ao = 4 VK = 4---(1) { C_A0 = 1 pure v/0. 5 = X_A/K 0. 5(1-X_A) VK = X_A/1-X_A h = X_A/1-X_A X_A + 0. 8 N_A = N_A0(1-X_A) = 0. 1 mol;/h epsilon = N_A0-N_A/-mu = 0. 5-0. 1/(-1) = 0. 4 mol/h [Answer (ii) 0. 4 molh^-1] \r\n (e) For part (c) N_B = 0. 44 times 3 = 1. 32 mol/h For part (d) N_B = 0. 4 times 3 = 1. 2 mol/h For the case, where flow rate cy A is halved, the B will be produced more than with 50% inert in feed