1. Notes 14-4 Obj. 14.5, 14.6

2. The half-life of a first-order reaction is equal to _________, where k is the rate constant. a. 0.693/ k b. 0.693 k c. k /2 d. 2 k

3. 14.5 Temperature and Rate A.) Generally, as temperature increases, so does the reaction rate. B.) This is because k is temperature dependent. 1.) faster rate at a higher temperature is due to an increase in the rate constant

4. C.) The Collision Model 1.) In a chemical reaction, bonds are broken and new bonds are formed. 2.) Molecules can only react if they collide with each other. 3.) Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

5. D.) Activation Energy 1.) There is a minimum amount of energy required for reaction: the activation energy, E a. 2.) Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

6. E.) Reaction Coordinate Diagrams 1.) It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

7. 2.) The diagram shows the energy of the reactants and products (and, therefore,  E ). 3.) The high point on the diagram is the transition state. 4.) The species present at the transition state is called the activated complex. 5.) The energy gap between the reactants and the activated complex is the activation energy barrier. 6.) Generally the lower the E a, the faster the reaction.

8. © 2009, Prentice-Hall, Inc. F.) Maxwell-Boltzmann Distributions 1.) Temperature is defined as a measure of the average kinetic energy of the molecules in a sample. 2.) At any temperature there is a wide distribution of kinetic energies.

9. 3.) As the temperature increases, the curve flattens and broadens. 4.) Thus at higher temperatures, a larger population of molecules has higher energy.

10. 5.) If the dotted line represents the activation energy, then as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. 6.) As a result, the reaction rate increases.

11. 7.) This fraction of molecules that are greater than or equal to the E a can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e - E a RT

12. G.)Arrhenius Equation 1.) Svante Arrhenius developed a mathematical relationship between k and E a : k = A e 2.) where A is the frequency factor, a number that represents the likelihood that collisions would occur with the proper orientation for reaction. - E a RT

13. Sample Exercise 14.10 Relating Energy Profiles to Activation Energies and Speed of Reaction Imagine that these reactions are reversed. Rank these reverse reactions from slowest to fastest. Consider a series of reactions having the following energy profiles: Rank the reactions from slowest to fastest assuming that they have nearly the same frequency factors.

14. 1.) Taking the natural logarithm of both sides, the equation becomes ln k = - ( ) + ln A 1T1T y = m x + b 2.) Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs EaREaR 1T1T H.) Determining the Activation Energy

15. Sample Exercise 14.11 Determining the Energy of Activation The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperatures (these are the data in Figure 14.12): (a) From these data, calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K? c.) Using the data in Sample Exercise 14.11, calculate the rate constant for the rearrangement of methyl isonitrile at 280 °C.

16. 14.6 Reaction Mechanisms A.) The sequence of events that describes the actual process by which reactants become products is called the reaction mechanism. B.) Reactions may occur all at once or through several discrete steps. C.) Each of these processes is known as an elementary reaction or elementary process. D.) The molecularity of a process tells how many reactants are involved in the process.

17. D.) Multistep Mechanisms 1.) In a multistep process, one of the steps will be slower than all others. 2.) The overall reaction cannot occur faster than this slowest, rate-determining step.

18. E.) Slow Initial Step 1.) The rate law for this reaction is found experimentally to be Rate = k [NO 2 ] 2 2.) CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration. 3.) This suggests the reaction occurs in two steps. NO 2 (g) + CO (g)  NO (g) + CO 2 (g)

19. 4.) A proposed mechanism for this reaction is Step 1: NO 2 + NO 2  NO 3 + NO (slow) Step 2: NO 3 + CO  NO 2 + CO 2 (fast) 5.) The NO 3 intermediate is consumed in the second step. 6.) As CO is not involved in the slow, rate- determining step, it does not appear in the rate law.

20. Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates It has been proposed that the conversion of ozone into O 2 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).

21. Sample Exercise 14.12 Determining Molecularity and Identifying Intermediates 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). Practice Exercise

22. Sample Exercise 14.13 Predicting Rate Law for an Elementary Reaction Consider the following reaction: 2 NO(g) + Br 2 (g) → 2 NOBr(g). (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? If the following reaction occurs in a single elementary reaction, predict its rate law:

23. Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The decomposition of nitrous oxide, N 2 O, 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.

24. Sample Exercise 14.14 Determining the Rate Law for a Multistep Mechanism The reaction is believed to occur in two steps: The experimental rate law is rate = k[O 3 ][NO 2 ]. What can you say about the relative rates of the two steps of the mechanism? Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen :

25. F.) Fast Initial Step 1.) The rate law for this reaction is found to be Rate = k [NO] 2 [Br 2 ] 2.) Because termolecular processes are rare, this rate law suggests a two-step mechanism. 2 NO (g) + Br 2 (g)  2 NOBr (g)

26. A proposed mechanism is Step 2: NOBr 2 + NO  2 NOBr (slow) Step 1 includes the forward and reverse reactions. Step 1: NO + Br 2 NOBr 2 (fast)

27. © 2009, Prentic e-Hall, Inc. 3.) The rate of the overall reaction depends upon the rate of the slow step. 4.) The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

28. © 2009, Prentic e-Hall, Inc. 5.) NOBr 2 can react two ways: o With NO to form NOBr o By decomposition to reform NO and Br 2 6.) The reactants and products of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

29. © 2009, Prentic e-Hall, Inc. Because Rate f = Rate r, k 1 [NO] [Br 2 ] = k −1 [NOBr 2 ] Solving for [NOBr 2 ] gives us k1k−1k1k−1 [NO] [Br 2 ] = [NOBr 2 ]

30. © 2009, Prentic e-Hall, Inc. Substituting this expression for [NOBr 2 ] in the rate law for the rate-determining step gives k2k1k−1k2k1k−1 Rate =[NO] [Br 2 ] [NO] = k [NO] 2 [Br 2 ]

31. Sample Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the experimentally observed one:

32. The first step of a mechanism involving the reaction of bromine is What is the expression relating the concentration of Br(g) to that of Br 2 (g)?