Question

"7 ITS O a ana m. rConsider an enzyme suríace catalyzed bimolecular reaction betweeri molecules A and B, which has where A is the fraction of surface sites occupied by reactant A a rate law of the form v k200 and e is the fraction of surface sites occupied by reactant B. The mechanism proposed as follows: A(a)+surface sites AS [fast oquilibrium, k,(A) forward rate and k (A) reverse rate}- -(1) B(g)+surface sites= B3 [fas' equilibrium, k, (B) forward rate and ka (B) reverse rate]-- 2 AS+BS-Product(s) (k3, the rate constant for this step)--- -(3) if KA and Kg are the equilibrium constants for reactions 1 and 2 respectively, Show that the rate law can be written as Cs Scanned with CamScanner A k3KAKp[A][B] (1+KA[A]K[B])2 72

Answer 1

Answer Tt is abimdleulan &urlace Yeaclino A+S AS product As +8s .3 st 4he Yate law iA given teat , = fraction of tee Suface Sitescoveved by A yadion of tee Suvface STI es Covered by B -3) Yate of forword veaclion in O Va = kal n£] = kalAJi-(ontn fraction of va cont Sites availabe to coccupy tuhere fs Cs total antentralion of Ses va cant Ste fraclionj Ya - ka(A)Csa Yate of back oovd veaction in Yd 5pCsJP

rate of dis sociation depends tva ctim of Occupied on of e uibrium , Yo = Yd Ca CAJav kd Rete of for word readion n Ya va lant Stes availoble for ocuplion av fraction tf vacovt Siten v C-GA-O8) ca (8] C3 Qu Ya Tate of dissocialion. în Yd A equilibyium Ya Yd Ύa - ka (8Jv 5 QA ACAJv

Sukaig (ACA)( &v 1-v v t v +ka (A)+kBlB) (a] a - KA(A)+KB(B) tfe rate of Re veation is Y B CB) 'kA (A) 1+KA(A)t KB(B) 3 kn kg (AB Vate, C1R (AŁB(BJ2