Question

A catalyst is a material that affect the rate of chemical reaction, yet emerges unchanged from the reaction. For example, consider the reaction: A + B C + D in which =-kCCB Where B is a catalyst. The reacrion is taking place in a semibatch reactor, in which 100 ft' of a solution containing 2 lb mol/ft of Ais initially present. No B is present initially. Starting at timet 0,5 ft/min of a solution containing 0.5 lb mol/ft of B is fed into the reactor. The reactor is isothermal, and k = 0.2 ft /Ib mol.min. How many moles of Care present in the reacter after half an hour?

Answer 1

Given that AtB CAB Here B is act as a codalyst Reactor is Semi batch Reactor? feed B Initial volume Given that VB 2 Ool dm3 1. min CBf = 0.2 mol dm3 The Initial volume of solution is vo= 100 dm3 The Initial concentraction of A in the Reactor dm3 * CAO = 0.5 md * Reactor is Esother mal Given that The rate of law = VA = RCA.CB T OA? RCA.CB No, we have to apply three balance on the reactor 1) Total mass balance 2) component 'B' balance 3) component 't balance

leto ensity of Input load stream and solution in the deactor is same. i Speed S solution = constant Total Mass balancet capus] - Coutput] - Accumulation Since in Semibatch reactor, no output is present. output zol let us take that dur. Vpcm Alterentegrating the Equation 0 Vz Vorvet) + Vo = volume Initially present : 100 dm3 Now component B balanced mole balance] (Ampay-fout pub] +( Generation of 7 - (Accumulation) Since 'B' is act as catalyst. So, it doy not consume in the geactor 51 (CBf.408): dode where, NB: Moley of B present in the Reactor

Now, we can write INB: CBXV Where CB= concentraction of 'B'in Reactor V: volume of solution in Reactor ... DNB do here (NB: (BND) die z1 dcCoru) dt 2) CB dy & V dCB OEDE al CBI UB from Equation ④ Wecan put the value dy and also we can put the put of CBf. and VB Cef.UB from Equation 086 Weget =) CRUB + (Not U8t) dce. CBp. UB -1 (0o+188) CK (Cefan CR ) UB. Applying Integration on both sides ce dick Sot ó voccep-CB) Ivo Hist) Now, Integrating the above terms at to, CB10 tit, CB2 CB moi a) de encontre) - Hors en vaste st) das

, Vo Vat Vot Vet CRF-CB No = CBf No a CBf - CB Votust con we can also write this above equation form 1 08Cef + CBf. VOⓇ Vot UBE above equations, and When we put values of CBS and UR in Weget Whese CBf = 0.2 Vo = 100 & Upt = Oolt. Substitute these value in equation ® CO2 %100) CB: 012 + (100+0.1t) He se, 'tis in minutes and CB in mol Toms CB: 0.2 + 20 Toutoulk I en . 40+0.02t 10otoilt À Now we can apply component balance of Since There is no input and output Value of A Pnput so output :

Mole of Balance À letely consider that (-y). V:- UNA and we can also write NAY NA - CAV, C-YA - KCACB) where KCACB V KCA CR-V = -d(Cav? s) - K CACB-V = CA day to da We can put du form in Equation . CR form in Equation and V formin equation ③ -) - KCA ( 40 +0:03) Clotott) - CAURT CVotust oleme -) -KCA (40+0.02%), CA VB +(Votuft) dem . =) CKCA (40+0-02 ) KAUB) = (Vest UBt) dea el - CA CUOXK+0.02*kt - VR] - (vot Ugt) dhe Now, we can put the value of 'k'in above equation where K = 0.8 dB3 molmin UB 2 0.1 dm3 Vo: loodm3 Substitute these value in above Equation

whese >) - CAC 40108 +0.02704E-0-] [lu 40114) de maig =) = A ( 31.9+0.016t] =(looto.it) dan 1319 +0.0166 \100 toilt Now, Integrate the above tering with tro, Ca - Cao - b oldn' 18 - dtt 0.016 soto cooto.lt t Gomin , CAL? - mere end 52349) at +0-0165 Coutor) out =lnCente, sua Can Ciotot) one shoulder Jat = Levering , 3196010582 +0.10 Jat touch tond) at 5-) - 10 ) 18.565 +0:16 (6) – OIGNOOC en cos tove)] 71 -en )28-65-160R 0.0582 schema18.853 or CA = 0's, -18.853 CA = 6.2451109 Mai

The volume of solution after one hour is V = Votube Where vo-100, Ust: 0.1860 :.= lout 001460 Tu: 106am a mole of A after one hour INA = CANI Wheao v=106 dm3 СА - 3 usus° NA = 3.2458109x106 TENA = 3.4397 Mo #moss Initial moles of A = No - CBo vo Where (Bozos Vo, 100 Substitute these two valgin above equation NHO=0.5x100 I NAO: 50moly & Conversion : NAO NA NAO 50-3.438 67 1. So , 99.994.

med When one hole of À consumy, one moleo i mole of c' in the Reactor after one hour is Moles of A consume - NAO NA 2 50-3.4597460 = 49.999 mole & moly of e'in Reactor after one hour 349,95 mol