Question

The catalytic reaction A+C D , can be adequately described by the following mechanism: A B (with ki = forward rate constant, k-1 = backward rate constant) B+C D (with rate constant k) Use the quazi (pseudo) steady state approach to find the rate of formation of D. Consider the following liquid phase catalytic reaction: A+B → C. The reaction has been found to follow the mechanism: A+D AD (very fast) AD +BBAD BAD→D+C (very fast) The rate of reaction is given by: -1A = KCACAD (mol/m'.s). CAO = 60 mol/mº, CDo = 30 mol/m”, k= 2.5*10 m /mol.s Calculate the rate for this reaction at 60% conversion of A.

Answer 1

Solution:- reaction below : Let, us consider following catalyst Let, At catal Now, the the above above reaction is mechanism described by following Now, Kis zB Now, B is the intermediate formed. write sate can of 'B' as follows: Now, 8B = K, CA-K., G3 - Kq CB Cc Now, from Pseudo steady state approximation, 8B=0 Now , k; Co-, C8 - Kg CRCE=0 K, CA = Cg (kytka cc) - CB= KICA [x, + kg Ce) i concentration of intermediate is CB = KCA Now, write the rate of formation of our (katkace) Let, %= kable 1 Kg K, CA B (kuptka (c) = K, ka CACC (K-1 + xx cc) .. Hence rate of formation of 'i' is к, К. Се со (K, + kace) b) now, we consider following liquid phase catalytic reaction. ; Ate - D o catalyst

has follows below: (very fast reaction) Now, the above reaction Let, A+D Ky AD AD+B Kay BAD BADK3y Now, the rate of reaction reaction). (very fast is given -Во - ксе Сер Now , Rate of formation for AD as follows: From is Жер - ксесо - ка Cop cs steady state approximation, rate of intermedinte eva) 40 геко. , Көр го K, Caco - ка Сер се = 0 Ka Sop C = к, со со сер - к, сосы к, со Now, K =k неnсе, Сөр - Ксе со rate ean as follows; Now, Hence, sub CAD in - Кө = ксе Сөр - ксе се ср of CA, CB & G - к сфер Now, calculate the concentrations as follows; Сө = no С- Хо) = о том, - о. 6) . - 84 . вам по mod ms

- Co as B Now, we calculate of CA, CB & CA = CAO (1-XA) = 60 mod mi (1-0.6) = 24 mod NOW CAICOI ate the the concentration concentration of B of 'B' as follows: C₂ = CBO (1-xo) here, a, d are stiochiometric co-eff. NOW, CB = CBO -(+) CAO XA = Bo-CAOXA = (200 - 60x0.6.) mol |: 164 mode now, we calculate Conc. ot as follows; met, conc. of D' remains the same at indet & outlet, since D aets as a intermediate in reaction ... = CDO =30 modo m3 Hence, sub the respective valves in rate as follows; i. -85 kc ma co 245x10 mois x (24 mol yoxso poleg 164 mad m3 = 2.634 x 103 mail 1. m3.5 the rate of reaction .. Hence is follows 2.634x11 mol m. " Thank you,