4kd 742ab areactionvelocityv 00025ms

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Unformatted text preview: g/s =6π*0.010 g/cm s* = 1.52x10‐11s= 152 Svedbergs Then, s= = 7.3) = = 43375.1 g/mol = 43.4 kD 7.4) 2A B = ‐ A) reaction velocity v= ‐ = 0.0025 M/s = k[A]2 which is rearranged as B) For second order recation in A, v= ‐ Integrate both sides, get the 2nd order rate law in terms of [A]: = 2kdt. Then k= 7.5) = a) At steady state, the two intermediates are B and D b) Use the steady state assumption to obtain expressions for [B] and [D]: c) v= 7.6) Increasing velocity of a reaction a) First order in A: v=k[A] 10v=k*x[A] Increase [A] by a factor of 10. b) Second order in A: v=k[A]2 10v= k*(x[A])2 x2=10, x= Increase [A] by a factor of . 7.7) Uncatalyzed rxn: vUncat=kUncat[A] Catalyzed: vcat=kcat[A]=1,000,000* kU[A] So kcat/kUncat= 1,000,000 Equilibrium constant K of the transition state to the final product is related to rate constant k (refer to Eyring’s Theory on enzyme catalysis): The contribution to energy of the transition state, (...
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This note was uploaded on 01/22/2014 for the course CHEM 156 taught by Professor Zavodszky during the Fall '08 term at UCLA.

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