Unformatted text preview: CHE4473 Kinetics Spring 2015 Homework N°7: Multiple Reactions Due Tuesday MAR 3rd 2015 IMPORTANT: You must include copies of the graphs of all the numerical solutions (not the whole spreadsheet, please !). 1.
An organic molecule reacts in a batch reactor that operates isothermally in the liquid phase at constant volume. Two reversible reactions occur, one desirable the other undesirable A ó D -‐ r1A = k1 ( CA – CD/K1A) A ó U -‐ r2A = k2 ( CA – CU/K2A) • Plot the evolution of CA , CD and CU as a function of reaction time. • At what time would you stop the reaction to maximize the overall selectivity (CD/CU)? • Is this the same time that maximizes the yield of D? • Find the equilibrium concentrations of A, D, and U. • If the reaction is carried out in a CSTR operating at a space time of 1.0 min, find the concentrations of A, D, and U. Calculate conversion, yield, and product selectivity (Sp = CD/(CD+CU) ). Kinetic parameters: k1 = 1 min-‐1; K1A = 10 k2 = 100 min-‐1; K2A = 2.5 CA0 = 1 mol / Liter 2.
Two reactants (A and B) are cofed into 50 L tubular reactor at a rate of 10 L/min. The initial concentrations or A and B are 1.5 and 2 mol/L, respectively. They undergo a complex set of reactions, which limit the production of D (desirable product). •
A + 2 B è C + D r1 = k1 CA CB2 k1 = 0.25 •
2D + 3 A è C + E r2 = k2 CA CD k2 = 0.10 • (a)
(c) B + 2 C è D + F r3 = k3 CB CC2 k3 = 5.00 Plot the evolution of CA , CB , CC , CD , CE and CF along the reactor volume. Determine the reactor volume that maximizes CD and calculate the maximum product selectivity (Sp). Compare with the product selectivity obtained with the 50 L tubular reactor. What would be the product selectivity and conversion from a 50 L CSTR? 3.
A stream of 2000 L/h containing 1.8 mol/L of reactant A is processed in a tandem of 2 equal-‐volume CSTRs. Two products are obtained D (desirable) and U (undesirable), neither of them is present in the feed. The reaction system describing what happens inside the reactor is the following A + A è D r1 = k1 CA2 k1 = 1.5x1016 e-‐100/RT A + D è U r2 = k2 CA CD k2 = 1.0x1012 e-‐75/RT Activation energies are in kJ/mol, T in K The operating temperature is such that k1 = k2. Determine the volume of each reactor (they are equal), which maximizes the concentration of D. (a) Determine the product selectivity (moles of A ending up as D/total moles of A converted) (b) Determine the volume that maximizes the yield of the desirable product (moles of A ending up as D/total moles of A fed) 4.
An aromatic compound (B) is alkylated with an alcohol (A), producing the desirable alkylate (C) plus water (W). An undesirable side reaction is the over-‐
alkylation by which the desirable product (C) is further alkylated by A to yield a lower value di-‐alkylate (D) and a second water molecule. The first (desirable) step is first order in each, the aromatic and alcohol. The rate of the second (undesirable) step is proportional to both concentration of the alcohol and the monoalkylate. • Write down the reaction scheme and the kinetic rate laws, including the units for the reaction constants (k1 and k2) • When the reaction was conducted in a batch reactor in liquid phase, the following data resulted. Time (sec) CA (mol/L) CB (mol/L) CC (mol/L) (-‐rA) (mol/L sec) rC (mol/L sec) 0 3.000 1.000 0.0 3.00 X 10-‐2 ? -‐3 100 1.359 ? ? 4.85 X 10 0 • Calculate the values of k1 and k2 ...
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