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: 2H 2 O 2 (aq) →2H 2 O(l) + O 2 (g) is believed to proceed through the following pre-equilibrium: H 3 O + + H 2 O 2 ↔H 3 O H 2 O H 3 O Br - →HOBr + H 2 O v = k[H 3 O 2 + ][Br - ] HOBr + H 2 O 2 → H 3 O + + O 2 + Br - (fast) The second step is the rate-determining step. Thus the production rate of O 2 can be expressed by the rate of the second step. The concentration of [H 3 O 2 + ] can be solved [H 3 O 2 + ] = K[H 2 O 2 ][H 3 O + ] Thus The rate depends on the concentration of Br - and on the pH of the solution (i.e. [H 3 O + ]).

Exercise 23.4b: Consider the acid-catalysed reaction (1) HA + H + ↔ HAH + k 1, k 1 ’, both fast (2) HAH + + B → BH + + AH k 2, 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] For a given [E] 0 and low values of [S] 0, the rate of product formation is proportional to [S] 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, v max.

Michaelis-Menten mechanism E + S → ES k 1 ES→ E + Sk 2 ES→ P + E k 3 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

Michaelis-Menten equation can be obtained by plug the value of [ES] into the rate law of P: Michaelis-Menten constant: K M can also be expressed as [E][S]/[ES]. Analysis: 1. When [S] 0 << K M, the rate of product formation is proportional to [S] 0 : 2. When [S] 0 >> K M, the rate of product formation reaches its maximum value, which is independent of [S] 0 : v = v max = k 3 [E] 0

With the definition of K M and v max, we get The above Equation can be rearranged into: Therefore, a straight line is expected with the slope of K M /v max, and a y- intercept at 1/v max 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, k cat, 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 k cat /K M,

Example: The enzyme carbonic anhydrase catalyses the hydration of CO 2 in red blood cells to give bicarbonate ion: CO 2 + H 2 O →HCO 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. [CO 2 ]/(mmol L -1 ) rate/(mol L -1 s -1 )2.78x x x x10 -4 Determine the catalytic efficiency of carbonic anhydrase at 273.5K Answer: Make a Lineweaver-Burk plot and determine the values of K M and v max from the graph. The slope is 40s and y-intercept is 4.0x10 3 L mol -1 s v max = = 2.5 x10 -4 mol L -1 s -1 K M = (2.5 x10 -4 mol L -1 s -1 )(40s) = 1.0 x mol L -1 k cat = = 1.1 x 10 5 s -1 ε = = 1.1 x 10 7 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

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