HW 2 S - CHE4473 Kinetics Spring 2016 Homework N2...

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CHE4473 Kinetics Spring 2016 Homework N°2 : Conversion and Reactor Sizing - Levenspiel Graphs (Ch. 2) Due : Tuesday FEB 2 nd 2016 _____________________________________________________________________________________________________ Problem 1. A constant volume reaction A P is being carried out a. If a 1000 liter CSTR is used with C A0 = 2 M and inlet flow rate of 10 L/min, what reaction rate (-r A ) is required to achieve 70 % conversion of A? What is the concentration of A in the reactor? FAo.X=r.V 20*0.7=r*1000 r=FAoX/V=14/1000= 0.014 mol/L min CA= CAo(1-X) = 2 (1-0.7) = 0.6 mol/Liter b. What rate would be required to achieve 95 % conversion in the CSTR? r=FAo.X/V=20*0.95/1000=19/1000= 0.019 mol/L min c. For the case of 70 % conversion of A, if the reaction is first order in A, what is the reaction rate coefficient k (include units)? And if the reaction is 2 nd order in A? And if the reaction is ½ order in A? 1 st order k = r/CA = 0.014 mol/L min/0.6 mol/Liter = 0.023 min -1 2 nd order k = r/CA 2 = 0.014 mol/L min/(0.6 mol/Liter) 2 = 0.039 Liter mol -1 min -1 ½ order k = r/CA 0.5 = 0.014 mol/L min/(0.6 mol/Liter) 0.5 = 0.018 Liter -1/2 mol 1/2 min -1 d. Using the value of the first-order coefficient k found in (c), if a 1000 L PFR were used instead of the CSTR, what would be the conversion? In your own words, explain why this conversion is greater than 70 % FAo dX = r dV INT (1/ k CAo(1-X) ) dX = V/FAo (1/ k CAo) INT ( 1/(1-X) ) dX = V/FAo - (1/ k CAo) ln (1-X) = V/FAo - ln (1-X) = V k CAo /FAo = k (V/v) (1-X) = e -k TAU X = 1- e -k TAU = (1-exp(0.023*1000/10) = 1 – 0.1 X = 0.9 90 % > 70 %
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