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Lecture 16

Lecture 16. Chain polymerization. the net rate of change of radical concentration is calculated as Using steady-state approximation The rate of polymerization v p = k p [ . M][M] = k p [M]

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Lecture 16

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  1. Lecture 16

  2. Chain polymerization • the net rate of change of radical concentration is calculated as • Using steady-state approximation • The rate of polymerization vp = kp[.M][M] = kp[M] • 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, • <n> = 2v

  3. 26.5 Features of homogeneous catalysis • 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.

  4. 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 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. The concentration of [H3O2+] can be solved [H3O2+] = K[H2O2][H3O+] Thus The rate depends on the concentration of Br- and on the pH of the solution (e.g [H3O+]).

  5. 26.6 Enzymes Three principal features of enzyme-catalyzed reactions: • 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. • For a given [E]0 and low values of [S]0, the rate of product formation is proportional to [S]0. • 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.

  6. Michaelis-Menten mechanism E + S → ES k1 ES → E + S k2 ES → P + E k3 The rate of product formation: To get a solution for the above equation, one needs to know the value of [ES] Applying steady-state approximation Because [E]0 = [E] + [ES], and [S] ≈ [S]0

  7. Michaelis-Menten equation can be obtained by plug the value od [ES] into the rate law of P: • Michaelis-Menten constant: 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: 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

  8. With the definition of KM and vmax, we get The above Equation can be rearranged into: Therefore, a straight line is expected with the slope of KM/vmax, and a y-intercept 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. • Catalytic efficiency, ε, of an enzyme is the ratio kcat/KM,

  9. Example 26.6a 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 20.0rate/(mol L-1 s-1) 2.78x10-5 5.00x10-5 8.33x10-5 1.67x10-4 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 vmax = = 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 kcat = = 1.1 x 105s-1 ε = = 1.1 x 107 L mol-1 s-1

  10. 26.7 Autocatalysis • Autocatalysis: the catalysis of a reaction by its products A + P → 2P The rate law is = k[A][P] To find the integrated solution for the above differential equation, it is convenient to use the following notations [A] = [A]0 - x; [P] = [P]0 + x One gets = k([A]0 - x)( [P]0 + x) integrating the above ODE by using the following relation gives or rearrange into with a=([A]0 + [P]0)k and b = [P]0/[A]0

  11. 26.8 Oscillating reactions • The lotka-Volterra mechanism (a) A + X → X + X (b) X + Y → Y + Y (c) Y → B

  12. The brusselator (a) A → X (b) B + X → Y + C (c) 2X + Y → 3X (d) X → D

  13. The Oregonator A simple three variable model developed based on the Belousov-Zhabotinsky reaction system. (a) A + Y → X + C (b) X + Y → 2C (c) A + X → 2X + 2Z (d) X + X → A + C (e) B + Z → Y + P A: NaBrO3; Y: Br-; X: HBrO2; C: HOBr Z: Ce4+; B: Malonic acid

  14. 26.9 Bistability Consider the reaction: A + 2B → 3B, Assuming the above reaction is run in an open system with the inflow rate k0 Apply steady-state approximation to the above eqn. Because [B] = [A]0 –[A] -k[A]( [A]0 –[A])2 + k0([A]0 - [A]) = 0 ([A]0 - [A])(ko – k[A] ([A]0 - [A])) = 0 solutions of the above equation are: [A]0 - [A] = 0 (the initial state) and ko – k[A] ([A]0 - [A]) = 0 (this equation can be reorganized in the familiar format: k[A]2 - k[A]0[A] + k0 = 0, which has the following two solutions: In order to have any physical meaning, k[A]02 > 4k0

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