1. From the Arrhenius equation we have: 301

2. From the Arrhenius equation we have: 302

3. From the Arrhenius equation we have: Recall the equation of a straight line: where y = ln k, b= ln A, m = -E a /R, and x = T -1. 303

4. So if we have temperature dependent data for the rate constant, we can make the following plot to obtain E a : 304 T -1 ln k slope = -E a /R

5. 305

6. 306

7. 307

8. Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? 308

9. Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? Approach: Start with the Arrhenius equation applied to the two separate conditions, where T 2 is the higher temperature. 309

10. Sample problem: Some reactions double their reaction rates with every 10 o C rise in temperature. Assume a reaction to take place at 295 K and at 305 K. What must the activation energy be for the rate constant to exactly double? Approach: Start with the Arrhenius equation applied to the two separate conditions, where T 2 is the higher temperature. 310

11. Take the ratio 311

12. Take the ratio Now so that 312

13. Take the ratio Now so that 313

14. Now take the natural log of both sides. 314

15. Now take the natural log of both sides. 315

16. Now take the natural log of both sides. That is: 316

17. Now take the natural log of both sides. That is: (recall that ) 317

18. The preceding result simplifies to give: 318

19. The preceding result simplifies to give: The numerical value for E a is given as: = 52 kJ mol -1 319

20. The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form: 320

21. The Arrhenius equation is quite useful when studying reactions involving simple species (atoms or diatomic molecules). For more complex systems, the Arrhenius equation is modified to the form: where P, the probability factor, accounts for the fact that in order to react, molecules must be properly oriented with respect to each other during a collision. 321

22. This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. 322

23. This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. Example: NO 2 Cl + Cl NO 2 + Cl 2 323

24. This means that an effective collision not only has to satisfy an energy requirement, but an orientation requirement as well. Example: NO 2 Cl + Cl NO 2 + Cl 2 The two Cl atoms must come into “contact” for reaction to occur. 324

25. NO 2 Cl + Cl NO 2 + Cl 2 325

26. NO + NO 3 2NO 2 Red = oxygen blue = nitrogen 326

27. For reactions involving only atoms P = 1; for reactions involving simple small molecules, P varies between approximately 0.2 and 0.001. For reactions involving complex polyatomic molecules, P can be as small as 10 -6. 327

28. Reaction Mechanisms 328

29. Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. 329

30. Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written. 330

31. Reaction Mechanisms Reaction Mechanism: The sequence of elementary steps that leads to product formation. Elementary Step: (Single step reaction) A reaction that occurs on the molecular level exactly as written. An overall reaction may involve one or several elementary steps. 331

32. Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH 332

33. Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. 333

34. Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. O CH 3 C O CH 3 case 1 334

35. Example: Use of isotope labeling method. O O CH 3 C O CH 3 + H 2 O CH 3 C OH + CH 3 OH For this reaction there are two obvious ways to arrive at CH 3 OH. O O CH 3 C O CH 3 or CH 3 C O CH 3 case 1 case 2 335

36. The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. 336

37. The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. case 1 O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH 337

38. The activation energies required for these two possibilities will be different. Consequently, the corresponding rates for these two processes must also be different. case 1 O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH case 2 O O CH 3 C O CH 3 + H 2 18 O CH 3 C OH + CH 3 18 OH 338

39. Question: What might case 3 be? 339

40. No CH 3 18 OH is found in the experiment, that means case 1 is the correct bond breaking step. O O CH 3 C O CH 3 + H 2 18 O CH 3 C 18 OH + CH 3 OH 340

41. Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: 341

42. Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate 342

43. Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) 343

44. Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) N 2 O + O N 2 + O 2 rate 344

45. Most reactions involve more than one step in the formation of products. As an example, the gas phase decomposition of dinitrogen monoxide is believed to occur in two elementary steps: N 2 O N 2 + O rate (the O atom is a reaction intermediate) N 2 O + O N 2 + O 2 rate Overall reaction: 2 N 2 O 2 N 2 + O 2 345

46. and are the rate constants for the two individual steps. 346

47. and are the rate constants for the two individual steps. Key Point : The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process. 347

48. and are the rate constants for the two individual steps. Key Point : The exponents in the rate law for an elementary process are equal to the coefficients obtained from the chemical equation for that elementary process. Important reminder : You cannot get the rate law exponents for the overall reaction by looking at the balanced equation. 348

49. For the reaction: 2 N 2 O 2 N 2 + O 2 349

50. For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate 350

51. For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate Notice that this is exactly the same as the rate law for the first elementary step. 351

52. For the reaction: 2 N 2 O 2 N 2 + O 2 the experimental rate law is: overall rate Notice that this is exactly the same as the rate law for the first elementary step. The observed rate can be explained by assuming that the second step is faster than the first step, i.e. 352

53. Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step. 353

54. Thus, the overall rate of decomposition is then completely controlled by the rate of the first step, which is called the rate-determining step. Rate-determining step : The slowest step in the sequence of steps leading to the formation of products. 354

55. Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: 355

56. Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step 356

57. Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step NO 2 + F NO 2 F fast step 357

58. Second example: 2 NO 2 + F 2 2 NO 2 F The accepted mechanism for the reaction is: NO 2 + F 2 NO 2 F + F slow step NO 2 + F NO 2 F fast step Note that the two elementary steps add to the overall chemical equation. 358

59. 359 2 NO 2 + F 2 2 NO 2 F

60. Summary comments on mechanism For a reaction mechanism to be viable, two main conditions apply. 360