14.6 - Reaction Mechanisms

1. 14.6 - Reaction Mechanisms • A balanced chemical equation tells us what substances are present at the beginning of a reaction and what is formed as the reaction proceeds. • It does not tell us how a reaction occurs

2. 14.6 - Reaction Mechanisms • The process by which a reaction occurs is the reaction mechanism • A reaction mechanism can describe in detail the order in which bonds are broken and formed and the changes of relative positions of atoms in a reaction

3. 14.6 – Elementary Reactions • Reactions that occur in a single event or step are called elementary reactions • The number of molecules that participate as reactants determine the molecularity of a reaction

4. 14.6 – Elementary Reactions • A single molecule is a unimolecular reaction • The collision of two molecules is a bimolecular reaction • Three molecules is termolecular

5. bimolecular • dimolecular • termolecular • unimolecular

6. bimolecular • dimolecular • termolecular • unimolecular

7. Reaction Mechanisms The molecularity of a process tells how many molecules are involved in the process.

8. 14.6 – Multistep Mechanisms • Many balanced equations occur in multiple steps • For instance: NO2 + CO  NO +CO2 • Really occurs as two steps: NO2 + NO2  NO3 + NO NO3 + CO  NO2 + CO2

9. 14.6 – Multistep Mechanisms • Net equation: NO2 + CO  NO +CO2 • Equation showing intermediates: NO2 + NO2  NO3 + NO NO3 + CO  NO2 + CO2 • The chemical equations for a multistep mechanism must always add to give the chemical equation of the overall process • Notice that NO3 is neither a reactant nor a product in the overall reaction – it is called an intermediate

10. SAMPLE EXERCISE 14.12 Determining Molecularity and Identifying Intermediates It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism: (a) Describe the molecularity of each elementary reaction in this mechanism. (b) Write the equation for the overall reaction. (c) Identify the intermediate(s).

11. PRACTICE EXERCISE For the reaction the proposed mechanism is (a) Is the proposed mechanism consistent with the equation for the overall reaction? (b) What is the molecularity of each step of the mechanism? (c) Identify the intermediate(s). Answers:(a) Yes, the two equations add to yield the equation for the reaction. (b) The first elementary reaction is unimolecular, and the second one is bimolecular. (c) Mo(CO)5

12. 14.6 – Rate laws for Elementary Reactions • Rate laws are always determined experimentally • Elementary reactions are significant: if we know that a reaction is an elementary reaction, then we know its rate law

13. 14.6 – Rate laws for Elementary Reactions The rate law is based on its molecularity

14. SAMPLE EXERCISE 14.13 Predicting the Rate Law for an Elementary Reaction If the following reaction occurs in a single elementary reaction, predict the rate law: Rate = k[H2][Br2] Comment: Experimental studies of this reaction show that the reaction actually has a very different rate law: Rate = k[H2][Br2]1/2 Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism must involve two or more elementary steps.

15. PRACTICE EXERCISE Consider the following reaction: (a) Write the rate law for the reaction, assuming it involves a single elementary reaction. (b) Is a single-step mechanism likely for this reaction? Answers:(a) Rate = k[NO]2[Br2] (b) No, because termolecular reactions are very rare

16. 14.6 – Rate-determining Step for a Multistep Mechanism • What do you do when it is not a simple elementary reaction? • Most reactions are composed of two or more elementary steps • One step is usually slower, and it is called the rate-determining step

17. All reactions are not elementary. • Some information must be known about the rate constant to determine the rate law. • Concentrations of reactant must be known to determine the rate law. • The rate law depends not on the overall reaction, but on the slowest step in the mechanism.

18. All reactions are not elementary. • Some information must be known about the rate constant to determine the rate law. • Concentrations of reactant must be known to determine the rate law. • The rate law depends not on the overall reaction, but on the slowest step in the mechanism.

19. Slow Initial Step NO2(g) + CO (g) NO (g) + CO2(g) • The rate law for this reaction is found experimentally to be Rate = k [NO2]2 • CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. • This suggests the reaction occurs in two steps.

20. 14.6 – Mechanisms with a Slow Initial Step k1 Step 1: NO2 + NO2NO3 + NO (slow) Step 2:NO3 + CO NO2 + CO2(fast) • Step 1 must be rate determining and we write the rate of the overall reaction equaling the rate of Step 1 Rate = k1[NO2]2 k2

21. 14.6 – Mechanisms with a Slow Initial Step k1 Our two step mechanism: Step 1: NO2 + NO2NO3 + NO (slow) Step 2:NO3 + CO NO2 + CO2(fast) Overall: NO2 + CO NO +CO2 Step 2 is much faster than Step 1, k2 >> k1 The intermediate NO3is slowly produced in Step 1 & immediately consumed in Step 2 As CO is not involved in the slow, rate-determining step, it does not appear in the rate law. k2

22. SAMPLE EXERCISE 14.14 Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism: (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. (b) The rate law for the overall reaction is just the rate law for the slow, rate-determining elementary reaction. Because that slow step is a unimolecular elementary reaction, the rate law is first order: Rate = k[N2O]

23. PRACTICE EXERCISE Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: The reaction is believed to occur in two steps The experimental rate law is rate = k[O3][NO2]. What can you say about the relative rates of the two steps of the mechanism? Answer: Because the rate law conforms to the molecularity of the first step, that must be the rate-determining step. The second step must be much faster than the first one.

24. 14.6 – Mechanisms with a Fast Initial Step • It is difficult to derive the rate law for a mechanism in which an intermediate is a reactant in the rate-determining step. • The situation arises in multistep reactions when the first step is not rate determining.

25. Fast Initial Step 2 NO (g) + Br2(g) 2 NOBr (g) • The experimentally determined rate law for this reaction is second order in NO and first order in Br2: Rate = k [NO]2 [Br2]

26. 14.6 – Mechanisms with a Fast Initial Step • One possibility is the reaction occurs in a single termolecular step: NO + NO + Br2 2NOBr Rate = k[NO]2[Br2] • But this is unlikely because termolecular reactions are rare (why are they rare?)

27. Step 1: NO + Br2 NOBr2 (fast) Fast Initial Step • A proposed mechanism is kf kr k2 Step 2: NOBr2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions.

28. 14.6 – Mechanisms with a Fast Initial Step • Because step 2 is the slow, rate-determining step, the rate of the overall reaction is determined by that step: Rate = k[NOBr2][NO] • We have a problem though: NOBr2is a an unstable intermediate, which occurs in a low unknown concentration

29. 14.6 – Mechanisms with a Fast Initial Step • How do we deal with this? We make assumptions. • We assume the NOBr2is so intrinsically unstable that it does not accumulate to a significant extent

30. Fast Initial Step • NOBr2 can react two ways: • With NO to form NOBr • By decomposition to reform NO and Br2 • The reactants and products of the first step are in equilibrium with each other. • Therefore, Ratef = Rater

31. 14.6 – Mechanisms with a Fast Initial Step • Because step 2 is slow (forming NOBr), we assume that most of the NOBr2falls back to NO and Br2 • So both the forward and the reverse reactions of Step 1 are faster than Step 2 and the forward and reverse reactions form an equilibrium

32. k1 k−1 [NO] [Br2] = [NOBr2] Fast Initial Step • Because Ratef = Rater , k1 [NO] [Br2] = k−1 [NOBr2] • Solving for [NOBr2] gives us

33. k2k1 k−1 [NO] [Br2] [NO] Rate = Fast Initial Step Substituting this expression for [NOBr2] in the rate law for the rate-determining step gives = k [NO]2 [Br2] Where k is equal to k2(k1/k-1)

34. 14.6 – Mechanisms with a Fast Initial Step • This mechanism involves only unimolecular and bimolecular steps, which is far more likely than termolecular processes • In general, whenever a fast step proceeds a slow one, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.

35. Catalysis • Catalyst-substance that changes the rate of reactions, but is not permanently changed itself • Provides and easier way to react • Typically lowers activation energy. • Typically provides a completely different mechanism for a reaction.

36. Homogeneous Catalyst • A catalyst present in the same phase as the reacting molecules • Example: aqueous hydrogen bromide and hydrogen peroxide. 2H2O2 H2O+ O2

37. Heterogeneous Catalyst • A catalyst in a different phase from the reactant molecules. • Catalyst is usually a solid and the reactants gases or liquids.

38. Heterogeneous Catalysts Cont… • Terminology • Adsorption: binding of the reactant molecules to the surface of the metal. • Active Sites: places where reacting molecules may become adsorbed.

39. Catalysts • Catalysts increase the rate of a reaction by decreasing the activation energy of the reaction. • Catalysts change the mechanism by which the process occurs.

40. Catalysts One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

41. Enzymes • Enzymes are catalysts in biological systems. • The substrate fits into the active site of the enzyme much like a key fits into a lock.

42. Enzyme Inhibitor • Molecule that has a strong bond with the active site of an enzyme. • Blocks substrate entry and prevents reactions from occurring.