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23.5 Features of homogeneous catalysis

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.

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23.5 Features of homogeneous catalysis

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  1. 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.

  2. 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 (i.e. [H3O+]).

  3. 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:

  4. 23.6 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.

  5. 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

  6. Michaelis-Menten equation can be obtained by plug the value of [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

  7. 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,

  8. 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 20.0rate/(mol L-1 s-1) 2.78x10-5 5.00x10-5 8.33x10-5 1.67x10-4Determine 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 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

  9. 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

  10. 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

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