2 Reaction Mechanism and Rate Laws An elementary reaction is a simple reaction that proceeds in one step. Most reactions are complex and require more than one step. A reaction mechanism is the sequence of elementary steps that leads to product formation. An intermediate is a species that appears in the mechanism of the reaction, but not in the overall balanced reaction.

3 Most chemical reactions occur by mechanisms that involve more than one step. As a result, the rate law can not be directly deduced from the stoichiometry of a balanced chemical equation. Only elementary reactions (those which occur in one step) can give the rate law. The rate law of an elementary process does follow from the coefficients of its balanced equation.

4 Among the steps in a multi-step reaction, there is always one that acts like a bottleneck. There is usually one step that is slowest compared to the rest. This slowest step limits the overall reaction rate and is called the rate- determining step. The rate-determining step determines the rate law for the overall reaction.

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7 Here is another example:

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9 It is not always easy

10 The result is a sort of useless since it contains a term that does not correspond to either any of the reactants or the product. A rate law should only consist of concentrations of reactants and/or products, no intermediates. Since the intermediate, dinitrogen dioxide, reacts slowly with oxygen, the reverse reaction of Step 1 is possible.

11 To solve this problem we assume that the first production of N 2 O 2 from the first step is consumed and there is an equilibrium conditions where: 2NO  N 2 O 2 K equil = [N 2 O 2 ]/[NO] 2 Substitution in the rate equation gives Rate = K*K equil [NO] 2 [O 2 ] Rate = K obs [NO] 2 [O 2 ]

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13 By doing rate of reaction experiments, you find this rate equation: The reaction is first order with respect to the organic compound, and zero order with respect to the hydroxide ions. The concentration of the hydroxide ions isn't affecting the overall rate of the reaction.

14 Increasing the concentration of the hydroxide ions will speed up the fast step, but that won't have a noticeable effect on the overall rate of the reaction. That is governed by the speed of the slow step.

15 Rate law in presence of a catalyst Example: decomposition of H 2 O 2 in presence of I - Rate = k[H 2 O 2 ][I - ] Step1 (slow): H 2 O 2 + I -  H 2 O + IO - Step 2 (fast): H 2 O 2 + IO -  H 2 O + O 2 + I - Overall: 2H 2 O 2  2H 2 O + O 2 Therefore, rate depends on the concentration of the catalyst although catalyst does not show up in the overall equation.

16 Catalysts A catalyst is characterized by the ability to do each of the following: 1.Catalysts increase the rate of reaction. 2.Catalysts are not consumed by the reaction. 3.A small quantity of catalyst should be able to affect the rate of reaction for a large amount of reactant. 4.Catalysts do not change the equilibrium constant for the reaction.

17 Example: Decomposition of potassium chlorate 2KClO 3  2 KCl + 3O 2 This reaction is very slow except when a catalyst like MnO 2 where the reactions proceeds very fast. A catalyst speeds the reaction by providing a set of elementary steps with more favorable kinetics. In most cases E a is decreased

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19 Heterogeneous catalyst- the reactants and the catalyst are in different phases. catalyst = solidreactants = liquid/gas Example: Ostwald process for production of nitric acid. The role of the catalyst (Pt-Rh solid catalyst) is to speed up the first step of the reaction of ammonia with oxygen:

20 4NH 3(g) + 5O 2(g)  4NO (g) + 6H 2 O (g) (Catalysed step) 2NO (g) + O 2(g)  2NO 2(g) 2NO 2 (g) + H 2 O (l)  HNO 2(l) + HNO 3(l) Heating nitrous acid will produce nitric acid 3HNO 2(l)  HNO 3(l) + 2NO (g) + H 2 O (l)

21 Another example: Catalytic converters In the engine, nitrogen and oxygen from air will form NO N 2(g) + O 2(g)  2NO (g) After emission to atmosphere, NO 2, which is a known pollutant, is formed: 2NO (g) + O 2  2NO 2 Therefore, before emission to atmosphere, a catalytic converter containing Cr 2 O 3 or CuO is used to convert NO back to N 2 and O 2. 2NO  N 2(g) + O 2(g) (Catalytic step)

22 Catalytic Converters CO + Unburned Hydrocarbons + O 2 CO 2 + H 2 O catalytic converter

23 Homogeneous catalyst- catalyst and reactants are in the same phase, usually liquid. Example: Hydrolysis of an acetic acid ester is very slow in absence of a catalyst: Ester + water  acetic acid + alcohol Rate = K[ester] However, in presence of an acid catalyst, the reaction is fast and the rate law is: Rate = k obs [ester][H + ]

24 Enzyme Catalysts

25 Enzyme Catalysts

26 Enzyme Catalysis 13.6

27 Selected problems: 2, 5-7, 9, 10, 13, 15, 17-25, 27-29, 31, , 40, 44, 46, 47, 49,