VI. How Reactions Occur Most chemical reactions occur through several small steps, not one big step. A chemical equation typically shows the overall reaction, not the intermediate steps. e.g. H 2(g) + 2ICl (g)  2HCl (g) + I 2(g) only shows what’s at the beginning and what you end up with.

VI. Reaction Mechanisms A reaction mechanism is a series of individual chemical steps through which an overall chemical reaction occurs. A proposed mechanism for the reaction H 2(g) + 2ICl (g)  2HCl (g) + I 2(g) is: Step 1H 2(g) + ICl (g)  HI (g) + HCl (g) Step 2HI (g) + ICl (g)  HCl (g) + I 2(g)

VI. Elementary Steps The reactions in a mechanism are called elementary steps; what’s implied in these steps is exactly what happens. Proposed reaction mechanisms must add up to the overall reaction! Does the previous mechanism add up? Species that are formed in one step and then consumed in another are known as intermediates. What is/are the intermediate(s) in the previous mechanism?

VI. Elementary Step Rate Laws Elementary steps are characterized by their molecularity, i.e. the # of reactant particles involved in the step. Rate laws for elementary steps can be written directly from their stoichiometry! e.g. If A + B  C + D is an elementary step, then the rate law for this step is: Rate = k[A][B].

VI. Energy Diagram, 2-Step Mechanism

VI. The Rate-Determining Step The slow step in the mechanism will determine the overall rate of reaction. This step is known as the rate- determining step. It’s the bottleneck of the reaction.

VI. Valid Mechanisms Valid mechanisms satisfy 2 criteria:  Elementary steps add up to overall reaction.  Rate law predicted by mechanism must be consistent with experimental rate law. Note that a valid mechanism is not a proven mechanism.

VI. Example Consider the reaction: NO 2(g) + CO (g)  NO (g) + CO 2(g). Experimentally, Rate = k[NO 2 ] 2. This implies it’s not a single-step reaction. Why? Is the mechanism below valid? NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Slow NO 3(g) + CO (g)  NO 2(g) + CO 2(g) Fast

VI. Rate Laws w/ Intermediates Rate laws must always be written from the rate-determining step. However, rate laws cannot contain intermediates. Rate laws from other steps can be used to substitute for intermediates. We look at fast first steps.

VI. Fast 1 st Steps When the 1 st step is fast, its products will build up and reverse reaction starts. Eventually, an equilibrium is set up. Thus, for A + B  C + D (Fast), we can write A + B  C + D.  Rate = k[A][B] and Rate = k -1 [C][D].  At equilibrium, k[A][B] = k -1 [C][D].  This can be used to rewrite rate laws.

VI. Sample Problem What is the overall reaction and rate law for the mechanism below? Identify the intermediates as well. Cl 2(g)  2Cl (g) Fast Cl (g) + CHCl 3(g)  HCl (g) + CCl 3(g) Slow CCl 3(g) + Cl (g)  CCl 4(g) Fast

VII. Catalysts We know we can change reaction rates by changing the temperature or changing reactant concentrations. However, there are limits to these tactics. If available, can use catalysts, substances that increase reaction rate, but aren’t used up in the reaction.

VII. Catalytic Destruction of O 3 Uncatalyzed: O 3(g) + O (g)  2O 2(g) Catalyzed: Cl (g) + O 3(g)  ClO (g) + O 2(g) ClO (g) + O (g)  Cl (g) + O 2(g) Atomic chlorine from photodissociated CFC’s is the catalyst. O 3(g) + O (g)  2O 2(g)

VII. How Do Catalysts Work? Catalysts provide a lower-energy mechanism for the reaction.

VII. Types of Catalysts There are homogeneous and heterogeneous catalysts.

VII. Biological Catalysts A biological catalyst is called and enzyme. An enzymes has an active site into which a specific substrate fits – like a lock and key.