1. A Radioactivity Example Carbon-14 is unstable. It decays very slowly by a process called beta-decay. A beta particle is an electron. It is ejected from the nucleus. In the process a neutron becomes a proton.

2. The rate that Carbon-14 decays is an inherent property of the nucleus. Thus the rate of decay is proportional to the number of 14 C atoms present. It is a perfect example of a first order reaction.

8. Decomposition of N 2 O 5 in CCl 4 solution Rate determining step is: First order reaction

13. Second order reaction First order reaction

14. What is the order of the reaction?

15. 2nd order 1st order

16. Reaction is Second Order

17. Zero order

18. The rate does not depend on any reactant! For this to be true something else must be determining the rate. A catalyst for example

19. The rate depends only on the surface area of the Pt not on the concentration of N 2 O

20. Second order reaction First order reaction Zero order reaction

21. Things are usually not so simple

22. Consider the famous reaction. Assume the reaction in both directions is first order in each reactant Rate forward = k f [A][B] Rate reverse = k r [C][D] The two reactions will compete.

23. Rate forward = k f [A][B] Rate reverse = k r [C][D] High [A] or [B] favor fast forward reaction High [C] or [D] favor fast reverse reaction At some set of concentrations the rates will be equal. When the forward and reverse reactions occur at the same rate there will be no change in concentrations.

24. k r [A][B] Rate forward = k f [A][B] = Rate reverse = k r [C][D] This is equilibrium. The forward and reverse rates are the same at equilibrium. k f [A][B] = k r [C][D] k f [C][D] == K equil

25. k r [A][B] k f [A][B] = k r [C][D] k f [C][D] == K equil The equilibrium constant is determined by the relative forward and reverse rates.

26. The rate law does not necessarily reflect the stoichiometry of the reaction. Rate = k[NO 2 ] 2 What mechansim can explain this?

27. Over all reaction Elementary steps slow fast Rate = k[NO 2 ] 2

28. Reactions that start with a fast equilibrium How can you explain this? Why is [O 2 ] in the rate?

29. Overall Reaction Elementary Steps Fast Slow

30. Forward = Reverse Rate= k 1 [O 3 ] = k -1 [O 2 ][O] Rate of slow step = k 2 [O][O 3 ] Rate of slow step = =

31. Rate = k[Cl 2 ] 1/2 [CHCl 3 ] This is really getting weird. fast slow fast

32. Rate of slow step = Rate of slow step = k [Cl 2 ] 1/2 [CHCl 3 ]

33. Transition State

34. Energy distribution of molecules versus Temperature

35. Fraction of Collisions with Energy > E a -E a /RT e As -E a /RT approaches zero the Fraction of Collisions with Energy > E a approaches unity. This would happen at high T or low E a

36. Fraction of Collisions with Energy > E a -E a /RT k = Ae -E a /RT e A Pre-exponential factor. fraction of collisions with proper steric orientation Arrhenius Equation

37. -E a /RT k = Ae ln(k) = -E a /RT + ln(A) Arrhenius Equation

38. ln(k) = -E a /RT + ln(A)

39. Slope = -E a /R E a = -R (-1.2 x 10 4 )K E a = 100 kJ/mol E a = - 8.31J/K (-1.2 x 10 4 )K

40. ln(k 1 ) = -E a /RT 1 + ln(A) ln(k 2 ) = -E a /RT 2 + ln(A) ln(k 2 / k 1 ) = E a /R(1/T 1 - 1/T 2 ) This equation allows you to compare rates at different T

41. ln(k 2 / k 1 ) = E a /R(1/T 1 - 1/T 2 ) Assume E a = 100 kJ/mol and T = 300 K What temperature would be required to double the rate? T 2 = ? K

42. ln(k 2 / k 1 ) = E a /R (1/T 1 - 1/T 2 ) Assume E a = 100 kJ/mol and T = 300 K ln(2) = 10 5 / R (1/300 - 1/T 2 ) 5.8 x 10 -5 = 1/300 - 1/T 2 T 2 = 305

43. Catalysis A catalyst changes the mechansim of a reaction to one with a lower energy of activation.

44. ? E a too high thus no reaction Try to find a new mechanism with a lower E a How do you do that? Use a catalyst.

45. Amide hydrolysis

48. Uncatalyzed reaction very slow. Acid or base catalyzed reaction faster, but still requires high temperature and long time and it is unselective. Enzymes can do it very fast at body temperature.

49. Carboxypeptidase-A Hydrolyzes the last amide bond on a peptide

50. Carboxypeptidase-A Enzyme Zn

51. Carboxypeptidase-A Enzyme

52. - d [CH 3 Br] d t = k [CH 3 Br] [ - OH]Rate = S N 2 reaction This is a bimolecular reaction, first order in each reactant. Second order over all. Can you integrate such an equation? Sure, but it is too complicated for us. So how can you study it?

53. - d [CH 3 Br] d t = k [CH 3 Br] [ - OH]Rate = To study the effect of [CH 3 Br] use a large excess of - OH. Thus [ - OH] will be effectively constant. If [ - OH] >> [CH 3 Br] then k [ - OH] = k’ - d [CH 3 Br] d t = k’ [CH 3 Br] Pseudo first order reaction in [CH 3 Br]

54. - d [CH 3 Br] d t = k [CH 3 Br] [ - OH]Rate = The to study the effect of [ - OH] use a large excess of CH 3 Br. Now [CH 3 Br] will be effectively constant. If [CH 3 Br] >> [ - OH] then k [CH 3 Br] = k’’ - d [ - OH] d t = k’’ [ - OH] Pseudo first order reaction in [ - OH]