Reaction Mechanisms Glenn V. Lo, Ph.D. Department of Physical Sciences Nicholls State University Revised May 27, 2013

Lesson Objectives Students will be able to Define reaction mechanism, elementary and complex reactions, and molecularity Identify reaction intermediates and catalysts in a mechanism Relate rate law to reaction mechanism

Reaction Mechanisms Reaction Mechanism: a proposed series of molecular events to explain the conversion of reactants to products Must be consistent with experimentally observed rate laws We can experimentally obtain evidence to support (not prove) or to disprove Each step is called an “elementary reaction” Most reactions are multi-step or complex reactions. They involve more than one elementary step.

Potential energy for elementary step Typically, each step in a mechanism involves a “transition state” (T.S.) where reactant bonds are partially broken and product bonds are partially formed. Ea = Energy of activation; typically less than energy needed to break bonds Ea Ea’ H

Test Yourself Which of these is true about reaction mechanisms and experimental data? A. experimental data can disprove a proposed mechanism B. experimental data can prove a proposed mechanism C. neither

Test Yourself Consider the reaction H2(g) + I2(g)  2 HI(g) True or False. Without any experimental data, we can say for sure that the reaction is elementary.

Test Yourself How many elementary steps are involved in the mechanism associated with the potential energy diagram shown below? A. 1 B. 2 C. 3 R P

Molecularity Molecularity = number of molecules involved in an elementary reaction Unimolecular: fragmentation or rearrangement of one molecule Bimolecular: collision of two molecules Termolecular or trimolecular: collision of three molecules Molecularity > 2; very unlikely

Test Yourself Consider the hypothetical elementary reaction: A + A  B This reaction is A. unimolecular B. bimolecular C. termolecular

Molecularity and Reaction Order The reaction order of an elementary reaction is equal to the molecularity. A  B, rate = k[A] A + A  B, rate = k[A]2 A + B  C, rate = k[A][B] BUT… reaction Order is experimentally determined, ALWAYS! Example: If rate = k[A]2, it doesn’t mean that the reaction occurs in one bimolecular step (with two A molecules colliding)

Test Yourself Consider the hypothetical elementary reaction: X + X  Y + Z What is the rate law for this reaction? A. rate = k[X]2 B. rate = k[X] C. rate = k[Y][Z] D. insufficient information

Test Yourself Consider the reaction: 2 I(g)  I2(g) What is the rate law for this reaction? A. rate = k[I]2 B. rate = k[I2] C. insufficient information

Test Yourself Consider the reaction: 2 NO2 + F2  2 FNO2 This reaction is most probably A. an elementary reaction B. a multi-step reaction

Rate law for multi-step reactions Reaction Order is experimentally determined, ALWAYS! A proposed mechanism must predict a rate law consistent with the experimentally determined rate law. There may be more than one mechanism consistent with a rate law. Multi-step reactions may not always have simple rate laws.

Rate law for multi-step reactions Case I. First step is slowest (rate-deterimining); overall rate law is rate law of first step. Example: 2 NO2 + F2  2 FNO2, Experimental rate law: rate = k[NO2][F2] Proposed mechanism: Step 1. NO2 + F2  FNO2 + F, slow Step 2. F + NO2  FNO2, fast Step 1 is called the “rate determining step” Rate = rate of step 1 = k[NO2][F2] F here is an example of a “reaction intermediate”

Test Yourself Consider the reaction: CO + NO2  CO2 + NO and the following proposed mechanism: Step 1 (slow): NO2 + NO2  NO3 + NO Step 2 (fast): NO3 + CO  NO2 + CO2 Which of these is the reaction intermediate? A. CO, B. NO2, C. NO3, D. NO

Test Yourself Consider the reaction: CO + NO2  CO2 + NO and the following proposed mechanism: Step 1 (slow): NO2 + NO2  NO3 + NO Step 2 (fast): NO3 + CO  NO2 + CO2 Which experimental rate law would be consistent with this mechanism? A. rate = k[NO2]2 B. rate = k[CO][NO2] C. rate = k[NO3][CO]

Rate law for multi-step reactions Case II. There is one slow step, preceded by fast reversible steps. Example 2 NO + Br2  2 NOBr, Experimental rate law: rate = k[NO]2[Br2] A one-step termolecular reaction is not likely. Proposed Mechanism Step 1. NO + Br2  NOBr2, fast reversible Step 2. NOBr2 + NO  2 NOBr, slow Predict: Rate = k2[NOBr2][NO] Assuming: k1[NO][Br2] = k-1[NOBr2] Rate = k2(k1[NO][Br2]/k-1) [NO]

Test Yourself Consider the reaction: CO + NO2  CO2 + NO and the following proposed mechanism: Step 1 (fast, rev.): NO2 + NO2  NO3 + NO Step 2 (slow): NO3 + CO  NO2 + CO2 Which experimental rate law would be consistent with this mechanism? A. rate = k[NO2]2 B. rate = k[CO][NO2] C. rate = k[NO2]2[CO]/[NO]

Catalysts A catalyst speeds up a reaction by providing a different mechanism with a lower energy of activation. A catalyst is not consumed; it is regenerated at the end of the reaction. Typical mechanism for an enzyme-catalyzed reaction. Step 1. E + S  ES Step 2. ES  E + P The enzyme (E) is the catalyst. S is the reactant (“substrate”), P is the product. ES is a reaction intermediate (“enzyme-substrate” complex). Overall reaction: S  P

Test Yourself Consider the mechanism for the destruction of ozone (O3): M + O3  MO + O2 (slow) MO + O  M + O2 (fast) What are the reactants? What is the catalyst? What is the product? What is the reaction intermediate? What is the rate law?