Question

QUESTION 2: 20 marks 20 points Kinetic studies on a solid-catalyzed reversible gas phase reaction, A B+C yielded the following rate law shown in Equation 1, which is a function of partial pressures of A, B and Ci.e. PA, PB and PC from the experimental observations, a) Propose a dual-site mechanism which is consistent with the rate law given above, using the standard guidelines (8 marks) b) Making use of the proposed mechanism, derive a rate law. Compare with the rate law in Equation 1 and comment. (12 marks) * Keq : -r' = (1 + KA PA +KpPg + KcPc)2 Equation 1: Experimentally determined Rate Law

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Answer 1

Solution Given Information A BTC reversible gus phase ron Adsorption of on the catalytic Surface „Kads Ats A.S ول 'Kads Surface reaction Rs A.S ts Bost cos 'k-s -122 Desorption B of from the surface ROL RDI B + S B.S - (3) Desorption of c from the surface kozo ets - 14 R-D2-1552 es

S = an actina site represents Adsorption rate given by , VAD RA PACO CAS -15) KA Whire, PA² perrtial pressure of A moler concentration og Vacant су sity conepstration adsorbed 4 CAS KA on the surface Kads adsorption equilibrions K-dds Const. Surface van vate can be given as os = Rs I CAS Cv CBS Cas ] - 167 Ks CBS Clis = concentration of product B formed on the surface of the centalyst concentration of product formed on the surface of the catalyst surface reaction equilibrium Constant ks RS Ks

Deserption B tak law can be given as, VDB Roil CBS PBCV e poi Kg -172 B in the PB = partial pressure of s bult gul KB' = ke is the reciprocal of adsorption equilibriom consternt for B =L KB Descrption of crate law con be gives as, 8pc = Rp2 cas Pc Co ket ] -18, balk Pc = partial pressure of a in the gay ke' = ý the reciprocal of adsorption equilibrium const for c. I Kc

be the rat - 197 Assuming Surface ran Conivuilly Step This means Rs, a very small and and ko2 are very large Roi RAI from equtis) get ТАР = PACO CA.S RA KA HCA'S = KAPAC - 19.00 Similarly from equin (7 and 18, we get VDB co - Roi CB.S PB Cr KB KB CB.S . KBPB.Co -1107 KB = Y FB Үрс RD2 Cc.s a Pc Co kol Ces = Kcpc co -(W

from equth (6) we get Ys - PACKA - KC KB Pp Pc Cv Blect Ks ks Cy² [KAPA - KC KB PB Pc ks [PACKA 2 TKAPA- [t ks ROKAG KAPA KckB Pola KSKA let kska = k and Keke Kg KA Keg Where Key auprell = kska Ke KB equilibrion const from here we get is k cz ( PA - PEPC) (12) Keq From Site balance herke a Cut Cast CB.Stce.s pa et = conetration of total site

using equth 1970) and (2) we get et co + KAPAC out KB PB Cot klaw ct ² (1H KAPA + KBPB + Kole ct? (1+ KAPA + KBPB + kopa) 2 substituting this eguth in eguth (12) get we rs R4² ( PA - Pape) (+ KAPA + K8PB + kepala det, k'cq² = k Then, (b) 8s = k ( Keg (1+KAPAT KBPBt kolc)² k ( PA - Pc PB w

so the derired rate law similar to equation 1. This means the surface ran of adsorbed A to product B and is rate- determining step