5.111 Lecture 35 Notes

35.1 5.111 Lecture 35 Kinetics Topic: Catalysis Chapter 13 (Section 13.14-13.15) __________________________________________________________________________________ From Friday’s material __________________________________________________________________________________ Le Chatelier's Principle - when a stress is applied to a system in equilibrium, the equilibrium tends to adjust to minimize the effect of the stress. Increasing the temperature can cause reaction to shift in endothermic direction. P.E. Ea,r Ea,f Ea, f P.E. reactants Ea, r product exothermic endothermic reactants products reaction coordinate !E + (endo) = Ea,f - Ea,r = big number - small number increase temperature, easier to overcome Ea,f. Equilibrium shifts toward products for endothermic reaction. reaction coordinate !E - (exo) = Ea,f - Ea,r = small number - large number increase temperature, easier to overcome Ea,r. Equilibrium shifts toward reactants in the exothermic reaction. most molecules have enough energy to overcome small barriers increasing temperature allows more molecules to overcome larger barriers Recall, a large Ea means that the rate constant is very sensitive to changes in temperature. Big Ea - increasing the temperature makes a _______ difference Small Ea - increasing the temperature does not make much of a difference. End of Friday’s material 35.2 Kinetics of Catalysis A catalyst is a substance that takes part in a chemical reaction and speeds it up, but doesn't undergo any permanent change itself. Catalysts, therefore, don't appear in the overall balanced equation. energy barrier without catalysts energy barrier with catalysts Ea, f Ea, f P.E. reactants Ea, r Ea, r E Catalysts typically act by reducing the activation energy barrier, both for the forward and reverse reactions. Catalysts stabilize the activated complex or the transition state. products reaction coordinate Catalysts have no effect on the thermodynamics of the reaction. Free energy, ∆G, is a state function, independent of path, therefore the equilibrium constant is ______________________ by the presence of a catalyst. An inhibitor is the opposite of a catalyst. It slows the rate of a reaction, typically by increasing the activation energy. 35.3 Types of Catalysts Homogeneous catalysts: reactants and catalysts are in the same phase Example: chlorofluocarbons catalyze the depletion of O3 (all gas phase) Heterogeneous catalysts: different phase Example: catalytic converter Solid metals (platinum, palladium, and rhodium) catalyze oxidation of hydrocarbons and CO gases and the reduction of nitrogen oxide gases to reduce pollution. Catalysts of Life: enzymes An enzyme is a large protein molecule (typically 20,000 g/mol or more) that is capable of carrying out a specific reaction or series of reactions. Proteins are made up of amino acids O + H 3N CH C R= sidechain 20 different sidechains 20 different amino acids O- R Amino acids are connected by peptide bonds to form polypeptide chains (or proteins). O + H 3N CH R C N R O CH C H peptide bond amino acid #1 amino acid #2 O - 35.4 A long chain of amino acids folds up into a compact structure. Four polypeptide chains with 198 amino acids each fold to form this enzyme. Ribbons drawn through the alpha (α) carbons. Dimensions are ~90 Å x 70 Å x 50 Å. (1 Å equals 1 x 10-10 m) Enzyme catalysis Reactant molecules are called substrates. Substrates bind to an active site on the enzyme. enzyme (E) + substrate (S) ⇔ enzyme-substrate complex (ES) → enzyme (E) + product (P) S P + + E ES E Derive rate expression for E + S ⇔ ES → E + P: Step 1 Step 2 k1 E + S ⇔ ES k-1 rater = k2 ES → E + P Rate of product formation Solve for intermediate [ES] d[ES] = dt ratef = rate = = d[P] dt = k2[ES] 35.5 use steady-state approximation 0 = d[ES] = dt k1 [E][S] - k-1 [ES] - k2 [ES] Now a slight change. Instead of solving for [ES] in terms of [E], free enzyme, solve for [ES] in terms of [E]o, total enzyme. [E]0 = total enzyme [E] + free enzyme [ES] bound enzyme replace [E] with ([E]0 - [ES]) 0 = d[ES] = dt k1 [E]0 [S] - k1 [ES][S] - k-1 [ES] - k2 [ES] rearrange [ES] terms to one side of the equation k1 [ES][S] + k-1 [ES] + k2 [ES] = k1 [E]0 [S] [ES] (k1 [S] + k-1 + k2) = k1 [E]0 [S] [ES] ! = ! k1 [E]0 [S] ! k1 [S] + k-1 + k2 Introduce new term Km (michaelis-menten constant) Km ! =! k-1 + k2 k1 Substitute Km into [ES] expression [ES] ! = ! k1 [E]0 [S] ! k1 [S] + k-1 + k2 = k1 [E]0 [S] k1 ([S] + (k-1 + k2)) k1 [ES] ! = ! [E]0 [S] [S] + Km = k1 [E]0 [S] = k1 ([S] + (k-1 + k2)) k1 35.6 Substitute [ES] into the rate expression rate of product formation !=! k2[ES] !=! k2 [E]0 [S] Michaelis-Menten Equation [S] + Km At high [S], adding more substrate _______________ change the rate much. Rates level off. Enzyme is saturated. All active sites on enzyme are filled. dP dt At low [S], adding more substrate _____________the rate significantly. Lots of free enzyme is available. [S] When [S] >> Km, then rate of product formation !=! k2 [E]0 [S] = k2 [E]0 This is called Vmax [S] + Km small Maximal rate = Vmax = k2 [E] 0 When [S] = Km, then rate of product formation !=! k2 [E]0 [S] = 1 [S] + [S] 2 k2 [E]0 half maximal rate Definition of Km ≡ concentration of [S] for which the rate is half-maximal. Example: The conversion of CO2 in blood to HCO3- and H3O+ is catalyzed by the enzyme carbonic anhydrase. The Michaelis-Menten constants for this enzyme and substrate are Km = 8 x 10-5 M and k2 = 6 x 105 s-1. What is the maximum reaction rate if the enzyme concentration is 5 x 10-6 M? 35.7 Enzyme Inhibition If an inhibitor is bound in the active site, then substrate can't bind. + I S EI E Many pharmaceutical drugs work by blocking the action of enzymes. Enzymes catalyze reactions by stabilizing the "transition state" of the reaction. Compounds that resemble the _______________ should bind more tightly to the enzyme than reactants or products. These compounds can therefore serve as enzyme inhibitors (drugs). Example: Protease inhibitors are used in treatment of HIV. Enzymes that cleave peptide bonds are called proteases. Overall mechanism of peptide bond cleavage - a type of displacement reaction: Molecules with a stable tetrahedral atom at cleavage site will resemble the transition state and could bind more tightly to the enzyme than substrate does. O- O C C N H OH O C N H OH + NH2 tetrahedral intermediate OH- Examples of molecules with tetrahedral centers: Approved drug for HIV, indinavir sulfate (The New England J. Med. 338, 1285 (1998)). N OH Ph N N N O H N OH Ph=phenyl O Knowledge of reaction mechanism can lead to design of new therapeutic treatments. Interesting and important question - why doesn't this compound block all proteases? A second consideration is specificity. Tight binding is just one facet of drug design. Also must consider specificity and delivery.