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An elementary step is a reaction that proceeds as written This important because we know the rate law for an elementary process. RECALL: REACTION MECHANISM: The process or series of elementary steps by which a reaction occurs. MJ Bojan Chem 112 1 If a step is elementary, then the rate of that reaction is proportional to the concentration of the reactants. Examples: Rate Law Unimolecular Bimolecular termolecular PROBLEM: I can’t tell if a reaction is an elementary process unless I do an experiment. Experiments can be used to support a proposed reaction mechanism or be proof that a proposed mechanism is incorrect. MJ Bojan Chem 112 2 Goal is to use kinetics to find the Mechanism of a Reaction What is the Mechanism for the following reaction? NO2 +CO è NO + CO2 What would the rate law be IF it was an elementary process? "# [NO 2 ] rate = = #t What is the Experimental Result? ! MJ Bojan Chem 112 3 Example: NO2 +CO è NO + CO2 Experimental shows: Propose a mechanism: NO2 + NO2 è NO3 + NO 1 NO3 + CO è NO2 + CO2 2 NO2 + CO è NO + CO2 Find the Rate Law for each step in a multi-step mechanism !" [ NO2 ] 2 rate!1 = = k [ NO2 ] "t !" [CO ] rate!!2!= = k [ NO3 ] [CO ] "t Then compare these with the experimental result. 4 MJ Bojan Chem 112 How do you know which step is the Rate Determining Step MJ Bojan Chem 112 5 Some Key points related to Mechanisms • Rate determining step = slow step transition states Energy • Elementary steps in a mechanism must add up to give the balanced equation for the overall process. • NO3 is produced in step 1 and consumed in step 2 Intermediate: a stable molecule Note: intermediate reactant product Reaction coordinate • Intermediates do not (should not) appear in the rate law. (Concentration dependence of intermediates cannot be measured.) 6 MJ Bojan Chem 112 To find mechanisms 1. Find the experimental rate law 2. Postulate elementary steps 3. Find the rate law predicted by the mechanism and compare to experiment. No rate can be written in terms of intermediates 7 MJ Bojan Chem 112 Mechanism Example Problem Cl2 + CHCl3 è HCl + CCl4 Rate = kobs [Cl2]1/2 [CHCl3] Postulate the following mechanism: is it consistent with the experimental rate law? Cl2 2Cl fast Cl + CHCl3 è HCl + CCl3 Cl + CCl3è CCl4 8 slow fast MJ Bojan Chem 112 CATALYSIS Catalyst is a substance that speeds ups a reaction without undergoing permanent change How: Thermodynamic state functions (ΔE, Δ H, Δ G, Δ S…) are unaffected by catalysis (Changes are path-independent) 9 MJ Bojan Chem 112 Catalysis Homogeneous catalysis: when catalyst is in the same phase as the reactant. Heterogeneous catalysis: when catalyst is in a different phase from the reactants • usually a solid catalyst and gas or solution reactants • reaction happens on the surface of the catalyst 10 MJ Bojan Chem 112 Demonstrate catalysis by studying the decomposition of hydrogen peroxide. Example: 2H2O2 (aq) è 2H2O () + O2(g) Use 3 different catalysts: Br− (I−) MnO2 catalase Homogeneous catalysis: (I−, catalase) Heterogeneous catalysis: MnO2 Enzyme catalysis (catalase) 11 MJ Bojan Chem 112 Catalyze the decomposition reaction via HOMOGENOUS CATALYSIS Energy profiles for catalyzed and uncatalyzed H2O2 decomposition 12 MJ Bojan Chem 112 Catalyze the decomposition reaction via HETEROGENOUS CATALYSIS EXAMPLE: H2 + 1/2 O2 è H2O reaction requires breaking strong H−H and O=O bonds é é 435 kJ 498 kJ negligible rate without catalyst solid H-H O=O Pt surface H diffusion on surface H H H O = O = H-H = adsorbed atoms H H O O gas molecules 13 MJ Bojan Chem 112 CATALYTIC CONVERTER Example of heterogeneous catalysis O2 1) CO è CO2(g) + H2O Hydrocarbons 2) NO, NO2 è N2(g) Catalysts: CuO, Cr2O3, Pt, Rh Usually, the stronger the bonds in reactants, the more we need a catalyst. e.g. 3H2 + N2 è 2NH3 ΔG° = −33 kJ/mol N−N triple bond (D = 946 kJ) 14 MJ Bojan at 298 K Chem 112 HOW DOES LOWERING Ea AFFECT RATES? EXAMPLE. H2O2 è H2O + 1/2 O2 hydrogen peroxide is toxic Uncatalyzed reaction has Ea = 72 kJ Catalase (enzyme in liver) lowers Ea to 28 kJ What is the ratio of kcat/kuncat at 37°C? (body temperature) Ea,cat ! RT kcat Acat e !=! Ea,uncat ! kuncat Auncat e RT Assume Acat = Auncat (Q: is this a good assumption?) 15 MJ Bojan Chem 112 CATALYZED VS UNCATALYZED kcat !=!e kuncat ! (Ea,cat !Ea,uncat ) RT kcat !(28 ! 72)!kJ / mol ln !=! !=!17.1 kuncat (0.0083!kJ / mol ! K )(310K ) kcat "!! !=!3#!10 7 kuncat speeds up by a factor of 30 million! Peptidase enzymes – break up proteins into amino acids (in your stomach) have a similar effect on Ea Without these it would take ~ 300 years to digest a steak 16 MJ Bojan Chem 112