1. 1 Chemical Kinetics Part 3: Reaction Mechanisms Chapter 13

2. 2 Reaction Mechanisms The balanced chemical equation provides information about the beginning and end of reaction. 2A + B → C This is an overall rxn; it doesn’t tell you how the rxn occurs. The reaction mechanism gives the path of the reaction. (or the how) Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. So it tells you the paths or steps in a rxn.

3. 3 Elementary Steps Elementary step: any process that occurs in a single step. Molecularity: the number of reactant molecules present in an elementary step. –Unimolecular: one molecule in the elementary step –bimolecular: two molecules in the elementary step –termolecular: three molecules in the elementary step It is not common to see termolecular processes (statistically improbable).

4. 4 Elementary Steps & Mechanisms A multistep mechanism consists of a sequence of elementary steps. The elementary steps must add to give the overall balanced chemical equation. Many mechanisms include intermediates. An intermediate is a species which appears in an elementary step but is not a reactant or product. So intermediates are produced in 1 elementary step and consumed in a later step. They do not appear in the overall equation.

5. 5 Elementary Steps & Mechanisms Step 1: A →B Step 2: B →C Overall: A →C B is an intermediate which is produced in Step 1 and consumed in Step 2. So it does not appear in the overall rxn.

6. 6 Rate Laws of Elementary Steps The rate laws of the elementary steps determine the overall rate law of the reaction. The rate law of an elementary step is determined by its molecularity. In other words, for an elementary step, the rate law may be written directly from the balanced equation for that step! Remember that the rate law for a rxn can’t be written from the overall balanced equation for the rxn. (Unless it is a 1-step rxn.)

7. 7 Rate Laws of Elementary Steps Unimolecular processes are first order A → Brate = k[A] Bimolecular processes are second order 2A → Brate = k[A] 2 OR A + B →Crate = k[A][B] Termolecular processes are third order (very rare) 3A →Brate = k[A] 3 OR 2A + B →Crate = k[A] 2 [B] OR A + B + C →Drate = k[A][B][C] Why are termolecular processes rare?

8. 8 Rate Laws of Multistep Mechanisms Most rxns occur by mechanisms with more than one elementary step. Often, 1 step is much slower than the other steps. This slow step limits or governs the overall rxn rate. The slow step is called the rate-determining step. Therefore, the rate-determining step governs the overall rate law for the reaction.

9. 9 Mechanisms with an Initial Slow Step Look at the following rxn: NO 2 (g) + CO(g) →NO(g) + CO 2 (g) The experimentally derived rate law is: Rate = k[NO 2 ] 2 Because chemists want to know how the rxn occurs, they then propose mechanisms which would yield the experimentally known rate law. They then test their proposed mechanism to see if it is correct.

10. 10 Mechanisms with an Initial Slow Step So for the previous rxn, the following mechanism is proposed: Note that NO 3 is an intermediate. If k 2 >> K 1, then Step 1 is the rate-determining step. So the overall rate law comes from Step 1.

11. 11 Mechanisms with an Initial Slow Step Does the mechanism support the known rate law of rate = k[NO 2 ] 2 ? As Step 1 is the rate-determining step, the mechanism yields the rate law.

12. 12 Mechanisms with an Initial Slow Step Do the mechanism steps sum up to give the known rxn equation of: NO 2 (g) + CO(g) → NO(g) + CO 2 (g) Yes!

13. 13 Mechanisms with an Initial Slow Step This means that the proposed mechanism is consistent with the experimentally known rate law. It does not prove that the mechanism is correct. Many, many experiments need to done to state with reasonable confidence that a proposed mechanism is probably correct.

14. 14 Mechanisms with an Initial Fast Step Not all mechanisms start with the slow step. And, it is possible for an intermediate to be in the slow step. Consider 2NO(g) + Br 2 (g) → 2NOBr(g)

15. 15 Mechanisms with an Initial Fast Step 2NO(g) + Br 2 (g) → 2NOBr(g) The experimentally determined rate law is Rate = k[NO] 2 [Br 2 ] Consider the following mechanism What are the step rate laws?

16. 16 Mechanisms with an Initial Fast Step The theoretical rate law is (based on Step 2): Rate = k 2 [NOBr 2 ][NO] But NOBr 2 is an intermediate! And we can hardly ever measure intermediates as they are unstable (don’t stick around long)! Lastly, we want to write the rate law in terms of the reactants in the balanced equation! How do we get rid of the intermediate NOBr 2 ?

17. 17 Mechanisms with an Initial Fast Step If NOBr 2 is unstable, then it doesn’t accumulate! This means that it breaks apart as soon as it forms. It can do this 2 ways: oIt can react with NO to produce the product NOBr; but this is a slow step so only a small fraction does this. oIt can fall apart and reform the reactants NO and Br 2 in the reverse of Step 1: this is a fast dynamic equilibrium step, so it happens very quickly.

18. 18 Mechanisms with an Initial Fast Step In a dynamic equilibrium, the forward rate, r f, equals the reverse rate, r r In our case, this means that r 1 = r -1 Therefore: k 1 [NO][Br 2 ] = k -1 [NOBr 2 ] Rearranging, we get:

19. 19 Mechanisms with an Initial Fast Step Now we can get rid of [NOBr 2 ]! We substitute this into our original rate law from Step 2: rate = k 2 [NO][NOBr 2 ] = k -1 [NOBr 2 ] Therefore, the overall rate law becomes Note the final rate law is consistent with the experimentally observed rate law.

20. 20 Catalysis A catalyst changes the rate of a chemical reaction. There are two types of catalyst: –homogeneous, and –heterogeneous. Catalysts are common in the body, in the environment, and in the chemistry lab! Homogeneous Catalysis The catalyst and reaction is in one phase. Hydrogen peroxide decomposes very slowly: 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g).

21. 21 Catalysis Homogeneous Catalysis 2H 2 O 2 (aq) → 2H 2 O(l) + O 2 (g). In the presence of the bromide ion, the decomposition occurs rapidly: 2Br - (aq) + H 2 O 2 (aq) + 2H + (aq) → Br 2 (aq) + 2H 2 O(l). Br 2 (aq) + H 2 O 2 (aq) → 2Br - (aq) + 2H + (aq) + O 2 (g). Note that the sum of these steps is the overall rxn. So Br - is a catalyst because it can be recovered at the end of the reaction. Generally, catalysts operate by lowering the activation energy for a reaction.

22. 22 Catalysis Homogeneous Catalysis

23. 23 Catalysis Homogeneous Catalysis Catalysts can operate by increasing the number of effective collisions. That is, from the Arrhenius equation: catalysts increase k be increasing A or decreasing E a. A catalyst usually changes the mechanism. Example: In the presence of Br -, Br 2 (aq) is generated as an intermediate in the decomposition of H 2 O 2. When a catalyst adds an intermediate, the activation energies for both steps must be lower than the activation energy for the uncatalyzed reaction.

24. 24 Catalysis Heterogeneous Catalysis The catalyst is in a different phase than the reactants and products. Typical example: solid catalyst, gaseous reactants and products (catalytic converters in cars). Most industrial catalysts are heterogeneous. First step is adsorption (the binding of reactant molecules to the catalyst surface). Adsorbed species (atoms or ions) are very reactive. Molecules are adsorbed onto active sites (red spheres) on the catalyst surface.

25. 25 Catalysis Heterogeneous Catalysis The number of active sites on a given amount of catalyst depends on: –Type of catalyst –How the catalyst was prepared –How the catalyst was treated prior to use

26. 26 Catalysis Heterogeneous Catalysis

27. 27 Catalysis Heterogeneous Catalysis Consider the hydrogenation of ethylene: C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g),  ΔH° = -136 kJ/mol. –The reaction is slow in the absence of a catalyst. –In the presence of a metal catalyst (Ni, Pt or Pd) the reaction occurs quickly at room temperature. –First the ethylene and hydrogen molecules are adsorbed onto active sites on the metal surface. –The H-H bond breaks and the H atoms migrate about the metal surface.

28. 28 Catalysis Heterogeneous Catalysis Consider the hydrogenation of ethylene: C 2 H 4 (g) + H 2 (g) → C 2 H 6 (g), ΔH° = -136 kJ/mol. –When an H atom collides with an ethylene molecule on the surface, the C-C π bond breaks and a C-H σ bond forms. –When C 2 H 6 forms it desorbs from the surface. –When ethylene and hydrogen are adsorbed onto a surface, less energy is required to break the bonds, the activation energy for the reaction is lowered, and the rate increases.

29. 29 Catalysis Enzymes Enzymes are biological catalysts. Most enzymes are protein molecules with large molecular masses (10,000 to 10 6 amu). Enzymes have very specific shapes. Most enzymes catalyze very specific reactions. Substrates undergo reaction at the active site of an enzyme. A substrate locks into an enzyme and a fast reaction occurs.

30. 30 Catalysis Enzymes

31. 31 Catalysis Enzymes The products then move away from the enzyme. Only substrates that fit into the enzyme lock can be involved in the reaction. If a molecule binds tightly to an enzyme so that another substrate cannot displace it, then the active site is blocked and the catalyst is inhibited (enzyme inhibitors). The number of events (turnover number) catalyzed is large for enzymes (10 3 - 10 7 per second).

32. 32 Catalysis Nitrogen Fixation and Nitrogenase Nitrogen gas cannot be used in the soil for plants or animals. Nitrogen compounds, NO 3, NO 2 -, and NO 3 - are used in the soil. The conversion between N 2 and NH 3 is a process with a high activation energy (the N-N triple bond needs to be broken). An enzyme, nitrogenase, in bacteria which live in root nodules of legumes, clover and alfalfa, catalyses the reduction of nitrogen to ammonia.

33. 33 Catalysis Nitrogen Fixation and Nitrogenase

34. 34 Catalysis Nitrogen Fixation and Nitrogenase The fixed nitrogen (NO 3, NO 2 -, and NO 3 - ) is consumed by plants and then eaten by animals. Animal waste and dead plants are attacked by bacteria that break down the fixed nitrogen and produce N 2 gas for the atmosphere.