Question

Exercise 4.4: Catalyst deactivation in a batch reactor Consider the irreversible, liquid-phase isomerization reaction carried out in a solvent containing dissolved catalyst at 25°C in a batch reactor Aka B The apparent first-order reaction-rate constant, ka, decreases with time because of cat- alyst deterioration. A chemist colleague of yours has studied the catalyst deactivation process and has proposed that it can be modeled by ka= Tt kat in which k is the fresh catalyst rate constant and ka is the deactivation-rate constant. (a) Write down the mole balance for this reactor. (b) Solve the mole balance for ca(t). Sketch your solution. (c) If it takes two hours to reach 50% conversion and the fresh catalyst has a rate 1 constant of 0.6 hr-1, what is ka? (d) How long does it take to reach 75% conversion?

Answer 1

Date : GIVEN: A - >B n= 1 kalk BATCH REACTOR SOLUTION : for more balance, 1 Moles of "A') - 1 Moles of 'A') + / Moles of A) =/ Modes of 'A' IN / OUT / generated (accuulation oro tr?a) V = d NA Cat Here, (ra) significs rate of consumption of 'A' N is the volume of reaction mixture NA are the moles of H at time t - ONA - ct O from mole balance, equation ( A) V = OINA - ct

Date : (RA) = d (NA/U). clt (CA) = dCA (since Ni-_ c) dit from rate law- we know that - Also (SA) = - kaca Theore, kereta Integrating, In CA = ok ln (ltkat) + In a where a is Some constat . At time to CA = CAO

Date : Lê Cao = a + a. = a - Co . KA Therefore, - ln CA = ek en (ltkat) + ln Cho On rearrangement, CA = Co. (1+ kat) * © CA = Coole ( since solo conveksion) 0.6 km -- E2hr - Therefore CA(+) from above equation of Cao e Coo [1+ ka (2) on edwingo The = o.pers ht"

Date : @ for conversion 75% CA = (CAO) Therefore, 0.6 0.8875 a | Cao : Có ( 1+ (0-8878) + ) Bí On solving t = "9.629 ho]