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Polymerization kinetics • Stepwise polymerization: any two monomers present in the reaction mixture can link together at any time. The growth of the polymer is not confined to chains that are already formed. • Chain polymerization: an activated monomer attacks another monomer, links to it, then that unit attacks another monomer, and so on. 23.3 Stepwise polymerization • Commonly proceeds through a condensation reaction, in which a small molecule is eliminated in each step. • The formation of nylon-66 H2N(CH2)6NH2 + HOOC(CH2)4COOH → H2N(CH2)6NHOC(CH2)4COOH • HO-M-COOH + HO-M-COOH → HO-M-COO-M-COOH • Because the condensation reaction can occur between molecules containing any number of monomer units, chains of many different lengths can grow in the reaction mixture. Stepwise polymerization • The rate law can be expressed as d [ A] k[ A]2 dt • Assuming that the rate constant k is independent of the chain length, then k remains constant throughout the reaction. [ A]0 [ A] 1 kt[ A]0 p [ A]0 [ A] kt[ A]0 [ A]0 1 kt[ A]0 • The degree of polymerization: The average number of monomers per polymer molecule, <n> n [ A]0 1 [ A] 1 p 23.4 Chain polymerization • Occurs by addition of monomers to a growing polymer, often by a radical chain process. • Rapid growth of an individual polymer chain for each activated monomer. • The addition polymerizations of ethene, methyl methacrylate, and styrene. • The rate of polymerization is proportional to the square root of the initiator concentration. The three basic types of reaction step in a chain polymerization I → R. + R. vi = ki[I] M + R. → .M1 (fast) (b) Propagation: M + .M1→ .M2 M + .M2→ .M3 ░ vp = kp[M][.M] M + .Mn-1→ .Mn (c) Termination: Mutual termination: .Mn + .Mm→ Mn+m Disproportionation: .Mn + .Mm→ Mn + Mm Chain transfer: M + .Mn→ Mn + .M (a) Initiation: Influences of termination step on the polymerization • Mutual termination: two growing radical chains combine. vt = kt ([.M])2 • Disproportionation: Such as the transfer of a hydrogen atom from one chain to another, which corresponds to the oxidation of the donor and the reduction of acceptor. vt = kt ([.M])2 • Chain transfer: vt = ? • the net rate of change of radical concentration is calculated as d [.M ] 2 fk i [ I ] 2k t [.M ]2 dt production • Using steady-state approximation (the rate of production of radicals equals the termination rate) 1/ 2 fk [.M ] i kt [ I ]1 / 2 • The rate of polymerization 1/ 2 fk i . vp = kp[ M][M] = kp[M] k [I ]1 / 2 t • The above equation states that the rate of polymerization is proportional to the square root of the concentration of the initiator. • Kinetic chain length, v, v number of monomer units consumed k[ M ][ I ]1/ 2 number of activated centres produced where k • <n> = 2v 1 k p ( fki kt ) 1/ 2 2 (for mutual termination) • Example: For a free radical addition polymerization with ki = 5.0x10-5 s-1 , f = 0.5, kt = 2.0 x107 dm3 mol-1 s-1, and kp = 2640 dm3 mol-1 s-1 , and with initial concentrations of [M] = 2.0 M and [I] = 8x10-3 M. Assume the termination is by combination. (a) The steady-state concentration of free radicals. (b) The average kinetic chain length. (c) The production rate of polymer. Solution: (a) (b) 1/ 2 fk i . [ M ] kt v k[ M ][ I ] 1 / 2 [ I ]1 / 2 1 where k k p ( fki kt ) 1/ 2 2 (c) The production rate of polymer corresponds to the rate of polymerization is vp: vp = kp[.M][M] 23.5 Features of homogeneous catalysis • A Catalyst is a substance that accelerates a reaction but undergoes no net chemical change. • Enzymes are biological catalysts and are very specific. • Homogeneous catalyst: a catalyst in the same phase as the reaction mixture. • heterogeneous catalysts: a catalyst exists in a different phase from the reaction mixture. Example: Bromide-catalyzed decomposition of hydrogen peroxide: 2H2O2(aq) → 2H2O(l) + O2(g) is believed to proceed through the following pre-equilibrium: H3O+ + H2O2 ↔ H3O2+ + H2O [ H 3 O2 ] K [ H 2 O2 ][ H 3 O ] H3O2+ + Br- → HOBr + H2O v = k[H3O2+][Br-] HOBr + H2O2 → H3O+ + O2 + Br- (fast) The second step is the rate-determining step. Thus the production rate of O2 can be expressed by the rate of the second step. d [O2 ] k[ H 3O2 ][ Br ] dt The concentration of [H3O2+] can be solved [H3O2+] = K[H2O2][H3O+] Thus d [O2 ] Kk[ H 2O2 ][ H 3O ][ Br ] dt The rate depends on the concentration of Br- and on the pH of the solution (i.e. [H3O+]). • Exercise 23.4b: Consider the acid-catalysed reaction (1) HA + H+ ↔ HAH+ k1, k1’ , both fast (2) HAH+ + B → BH+ + AH k2, slow Deduce the rate law and show that it can be made independent of the specific term [H+] Solution: 23.6 Enzymes Three principal features of enzyme-catalyzed reactions: 1. For a given initial concentration of substrate, [S]0, the initial rate of product formation is proportional to the total concentration of enzyme, [E]0. 2. For a given [E]0 and low values of [S]0, the rate of product formation is proportional to [S]0. 3. For a given [E]0 and high values of [S]0, the rate of product formation becomes independent of [S]0, reaching a maximum value known as the maximum velocity, vmax. • Michaelis-Menten mechanism E + S → ES ES →E + S ES →P + E The rate of product formation: k1 k2 k3 d[ P ] k 3 [ ES ] dt To get a solution for the above equation, one needs to know the value of [ES] d [ ES ] k1[ E ][ S ] k 2 [ ES ] k 3 [ ES ] dt Applying steady-state approximation k1[ E ][ S ] k2[ ES ] k3[ ES ] 0 [ ES ] k1 [ E ][ S ] k2 k3 Because [E]0 = [E] + [ES], and [S] ≈ [S]0 [ ES ] [ E ]0 k k3 1 1 2 k1 [ S ]0 • Michaelis-Menten equation can be obtained by plug the value of [ES] into the rate law of P: dP dt k 3 [ E ]0 k k3 1 1 2 k1 [ S ]0 • Michaelis-Menten constant: KM k2 k3 k1 KM can also be expressed as [E][S]/[ES]. • Analysis: 1. When [S]0 << KM, the rate of product formation is proportional to [S]0: v k1k 3 [ S ]0 [ E ]0 k2 k3 2. When [S]0 >> KM, the rate of product formation reaches its maximum value, which is independent of [S]0: v = vmax = k3[E]0 With the definition of KM and vmax, we get v The above Equation can be rearranged into: v max K 1 M [ S ]0 K 1 1 M v v max v max 1 [ S ]0 Therefore, a straight line is expected with the slope of KM/vmax, and a yintercept at 1/vmax when plotting 1/v versus 1/[S]0. Such a plot is called Lineweaver-Burk plot, • The catalytic efficiency of enzymes Catalytic constant (or, turnover number) of an enzyme, kcat, is the number of catalytic cycles (turnovers) performed by the active site in a given interval divided by the duration of the interval. v max kcat k 3 • Catalytic efficiency, ε, of an enzyme is the ratio kcat/KM, k kk cat 1 3 k M k2 k3 [ E ]0 Example: The enzyme carbonic anhydrase catalyses the hydration of CO2 in red blood cells to give bicarbonate ion: CO2 + H2O → HCO3- + H+ The following data were obtained for the reaction at pH = 7.1, 273.5K, and an enzyme concentration of 2.3 nmol L-1. [CO2]/(mmol L-1) 1.25 2.5 5.0 20.0 rate/(mol L-1 s-1) 2.78x10-5 5.00x10-5 8.33x10-5 1.67x10-4 Determine the catalytic efficiency of carbonic anhydrase at 273.5K Answer: Make a Lineweaver-Burk plot and determine the values of KM and vmax from the graph. The slope is 40s and y-intercept is 4.0x103 L mol-1s 1 vmax = 4.0 103 Lmol 1 s = 2.5 x10-4 mol L-1s-1 KM = (2.5 x10-4 mol L-1s-1)(40s) = 1.0 x 10-2 mol L-1 4 1 1 s kcat = 2.5 10 molL = 1.1 x 105s-1 9 1 2.3 10 molL ε= k cat KM = 1.1 x 107 L mol-1 s-1 Mechanisms of enzyme inhibition • Competitive inhibition: the inhibitor (I) binds only to the active site. EI ↔ E + I • Non-competitive inhibition: binds to a site away from the active site. It can take place on E and ES EI ↔ E + I ESI ↔ ES + I • Uncompetitive inhibition: binds to a site of the enzyme that is removed from the active site, but only if the substrate us already present. ESI ↔ ES + I • The efficiency of the inhibitor (as well as the type of inhibition) can be determined with controlled experiments