Question

is carried out adiabatically in a CSTR. The reaction is second order in A. The feed, which is equal molar in water (which contains the catalyst) and A, enters the reactor at a temperature of 52 C and a total volumetric flow rate of 10 dm3/min. The concentration of A entering the reactor is 4 molar. a) What is the reactor volume to achieve 80% conversion b) What conversion can be achieved in a 1000 dm3 CSTR? What is the exit temperature? vo- 10 dn /min CAo -4 mol/dm3 To -325K Additional Information H 3000 cal/gmol Cpy 18 cal/gmol /pc Cp 15 cal/gmol /DC k-0.0005 at 25DC CP, 15 cal/emol /DC E-15,000 cal/mol

Reaction is A-->B

Answer 1

A rightarrow B t = 0 C_Ao o t = t C_Ao(1 - x_A) C_Aox_A volumetric flow rate(v_0) = 10 dm^3/min C_Ao = 4 mol/dm^3 T_0 = 325l x_A = 0.80 energy balance (C_Ao v_0 C_PA + C_w C_Pw v_0)T_R - (C_Ao(1 - x_A)C_PA v_0 + C_Ao x_A C_PB + C_w C_Pw v_0)T - (Delta H_R)C_Ao v_0 x_A = 0 (4 times 15 + 4 times 18)times 10 times 325 - (4(1 - 0.80) times 15 + 4 times 0.80 times 15 + 4 times 18) times 10 times T + 3000 times 4 times 10 times 0.80 = 0 \r\n 429000 - (1320T) + 96000 = 0 T = 397.72 k ln(k_2/k_1) = E/rho(1/T_1 - 1/T_2) k_1 = 0.0005 (dm^3/mol. min) at = 29 degree c = 298k k_2 = ? at T = 397.72k E = 15000 cal/mol = 62760 J/mole ln(k_3/0.0005) = 1/8.314(1/298 - 1/397.72) ln(k_3/0.0005) = 62760/8.314(1/298 - 1/397.72) k_2 = 0.286(dm^3/mol.min) for CSTR v/v_0 = C_Ao x_A/k_2[C_Ao(1 - x_A)]^2 v/10 = 4 times 0.80/0.286 times 4^2(1