1. Chemical Kinetics Chapter 14 Chemical Kinetics

2. Chemical Kinetics Studies the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the ___________________________ (exactly how the reaction occurs).

3. Chemical Kinetics Factors That Affect Reaction Rates ________________of the Reactants  In order to react, molecules must come in ________________ with each other.  The more ________________ the mixture of reactants, the faster the molecules can react.

4. Chemical Kinetics Factors That Affect Reaction Rates ________________ of Reactants  As the concentration of reactants ________________, so does the likelihood that reactant molecules will collide.

5. Chemical Kinetics Factors That Affect Reaction Rates Temperature  At ________________ temperatures, reactant molecules have more ________________ energy, move faster, and collide more often and with greater energy.

6. Chemical Kinetics Factors That Affect Reaction Rates Presence of a ________________  Catalysts ________________ up reactions by changing the mechanism of the reaction.  Catalysts are not ________________ during the course of the reaction.

7. Chemical Kinetics Reaction Rates Rates of reactions can be determined by monitoring the ________________ in concentration of either ________________ or ________________ as a function of time.

8. Chemical Kinetics Reaction Rates In this reaction, the concentration of butyl chloride, C 4 H 9 Cl, was measured at various times. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

9. Chemical Kinetics Reaction Rates The average rate of the reaction over each interval is the change in concentration divided by the change in time: Average rate =  [C 4 H 9 Cl]  t C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

10. Chemical Kinetics Reaction Rates Note that the average rate ________________ as the reaction proceeds. This is because as the reaction goes ________________, there are fewer collisions between reactant molecules. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

11. Chemical Kinetics Reaction Rates A plot of concentration vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the ___________________ _____________at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

12. Chemical Kinetics Reaction Rates All reactions ________________ down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the ________________. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq)

13. Chemical Kinetics Reaction Rates and Stoichiometry In this reaction, the ratio of C 4 H 9 Cl to C 4 H 9 OH is 1:1. Thus, the rate of disappearance of C 4 H 9 Cl is the same as the rate of appearance of C 4 H 9 OH. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Rate = -  [C 4 H 9 Cl]  t =  [C 4 H 9 OH]  t

14. Chemical Kinetics Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI (g)  H 2 (g) + I 2 (g) Therefore, Rate = − 1212  [HI]  t =  [I 2 ]  t

15. Chemical Kinetics Reaction Rates and Stoichiometry To generalize, then, for the reaction aA + bBcC + dD Rate = − 1a1a  [A]  t = − 1b1b  [B]  t = 1c1c  [C]  t 1d1d  [D]  t =

16. Chemical Kinetics Concentration and Rate One can gain information about the ________________ of a reaction by seeing how the rate changes with changes in ________________.

17. Chemical Kinetics Concentration and Rate Comparing Experiments 1 and 2, when [NH 4 + ] doubles, the initial rate ________________. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

18. Chemical Kinetics Concentration and Rate Likewise, comparing Experiments 5 and 6, when [NO 2 − ] doubles, the initial rate ________________. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

19. Chemical Kinetics Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2 − ] Rate  [NH + ] [NO 2 − ] or Rate = k [NH 4 + ] [NO 2 − ] This equation is called the ________________, and k is the ________________.

20. Chemical Kinetics Rate Laws A rate law shows the relationship between the ________________and the ________________ of reactants. The exponents tell the ________________ of the reaction with respect to each reactant. This reaction is First-order in [NH 4 + ] First-order in [NO 2 − ]

21. Chemical Kinetics Rate Laws The overall ________________can be found by adding the exponents on the reactants in the rate law. This reaction is ________________ overall.

22. Chemical Kinetics Integrated Rate Laws Using calculus to integrate the rate law for a first-order process gives us ln [A] t [A] 0 = −kt Where [A] 0 is the initial concentration of A. [A] t is the concentration of A at some time, t, during the course of the reaction.

23. Chemical Kinetics Integrated Rate Laws Manipulating this equation produces… ln [A] t [A] 0 = −kt ln [A] t − ln [A] 0 = − kt ln [A] t = − kt + ln [A] 0 …which is in the form y = mx + b

24. Chemical Kinetics First-Order Processes Therefore, if a reaction is ________________, a plot of ln [A] vs. t will yield a straight line, and the slope of the line will be -k. ln [A] t = -kt + ln [A] 0

25. Chemical Kinetics First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

26. Chemical Kinetics First-Order Processes This data was collected for this reaction at 198.9°C. CH 3 NCCH 3 CN

27. Chemical Kinetics First-Order Processes When ln P is plotted as a function of time, a straight line results. Therefore,  The process is ________________.  k is the negative slope: 5.1  10 -5 s −1.

28. Chemical Kinetics Second-Order Processes Similarly, integrating the ________________ for a process that is second-order in reactant A, we get 1 [A] t = −kt + 1 [A] 0 also in the form y = mx + b

29. Chemical Kinetics Second-Order Processes So if a process is ________________ in A, a plot of 1/[A] vs. t will yield a straight line, and the slope of that line is k. 1 [A] t = −kt + 1 [A] 0

30. Chemical Kinetics Second-Order Processes The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 1/2 O 2 (g) and yields data comparable to this: Time (s)[NO 2 ], M 0.00.01000 50.00.00787 100.00.00649 200.00.00481 300.00.00380

31. Chemical Kinetics Second-Order Processes Graphing ln [NO 2 ] vs. t yields: Time (s)[NO 2 ], Mln [NO 2 ] 0.00.01000−4.610 50.00.00787−4.845 100.00.00649−5.038 200.00.00481−5.337 300.00.00380−5.573 The plot is not a straight line, so the process is not first-order in [A].

32. Chemical Kinetics Second-Order Processes Graphing ln 1/[NO 2 ] vs. t, however, gives this plot. Time (s)[NO 2 ], M1/[NO 2 ] 0.00.01000100 50.00.00787127 100.00.00649154 200.00.00481208 300.00.00380263 Because this is a straight line, the process is ________________ in [A].

33. Chemical Kinetics Half-Life ________________ is defined as the time required for one-half of a ________________ to react. Because [A] at t 1/2 is one-half of the original [A], [A] t = 0.5 [A] 0.

34. Chemical Kinetics Half-Life For a first-order process, this becomes 0.5 [A] 0 [A] 0 ln = −kt 1/2 ln 0.5 = − kt 1/2 −0.693 = − kt 1/2 = t 1/2 0.693 k NOTE: For a first-order process, the half-life does not depend on [A] 0.

35. Chemical Kinetics Half-Life For a second-order process, 1 0.5 [A] 0 = kt 1/2 + 1 [A] 0 2 [A] 0 = kt 1/2 + 1 [A] 0 2 − 1 [A] 0 = kt 1/2 1 [A] 0 == t 1/2 1 k[A] 0

36. Chemical Kinetics Temperature and Rate Generally, as temperature ________________, so does the reaction rate. This is because k is temperature ________________.

37. Chemical Kinetics The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they ________________ with each other.

38. Chemical Kinetics The Collision Model Furthermore, molecules must collide with the correct ________________ and with enough ________________ to cause bond breakage and formation.

39. Chemical Kinetics Activation Energy In other words, there is a minimum amount of energy required for reaction: the ________________, E a. 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 ________________ to get over the activation energy barrier.

40. Chemical Kinetics Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a _______________ _______________ like this one for the rearrangement of methyl isonitrile.

41. Chemical Kinetics Reaction Coordinate Diagrams It shows the energy of the reactants and products (and, therefore,  E). The high point on the diagram is the ________________. The species present at the ________________ state is called the ________________. The energy gap between the reactants and the activated complex is the ________________________________.

42. Chemical Kinetics Maxwell–Boltzmann Distributions _______________ is defined as a measure of the average _______________ of the molecules in a sample. At any temperature there is a wide distribution of kinetic energies.

43. Chemical Kinetics Maxwell–Boltzmann Distributions As the temperature _______________, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has ________________ energy.

44. Chemical Kinetics Maxwell–Boltzmann Distributions If the dotted line represents the ________________, as the temperature increases, so does the fraction of molecules that can overcome the activation energy barrier. As a result, the reaction rate ________________.

45. Chemical Kinetics Maxwell–Boltzmann Distributions This fraction of molecules can be found through the expression where R is the gas constant and T is the Kelvin temperature. f = e −E a /RT

46. Chemical Kinetics Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e −E a /RT where A is the ________________, a number that represents the likelihood that collisions would occur with the proper orientation for reaction.

47. Chemical Kinetics Arrhenius Equation Taking the natural logarithm of both sides, the equation becomes ln k = -E a ( ) + ln A 1 RT y = mx + b Therefore, if k is determined experimentally at several ________________, E a can be calculated from the slope of a plot of ln k vs. 1/T.

48. Chemical Kinetics Reaction Mechanisms The sequence of events that describes the actual process by which reactants become products is called the _______________________________.

49. Chemical Kinetics Reaction Mechanisms Reactions may occur all at once or through several discrete ________________. Each of these processes is known as an _____________________________or _____________________________.

50. Chemical Kinetics Reaction Mechanisms The ________________ of a process tells how many molecules are involved in the process.

51. Chemical Kinetics Multistep Mechanisms In a multistep process, one of the steps will be ________________ than all others. The overall reaction cannot occur faster than this slowest, ________________.

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

53. Chemical Kinetics Slow Initial Step 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) The NO 3 ________________ is consumed in the second step. As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

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

55. Chemical Kinetics Fast Initial Step 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)

56. Chemical Kinetics Fast Initial Step The rate of the overall reaction depends upon the ________________ of the slow step. The rate law for that step would be Rate = k 2 [NOBr 2 ] [NO] But how can we find [NOBr 2 ]?

57. Chemical Kinetics Fast Initial Step NOBr 2 can react two ways:  With NO to form NOBr  By decomposition to reform NO and Br 2 The ________________ and ________________ of the first step are in equilibrium with each other. Therefore, Rate f = Rate r

58. Chemical Kinetics Fast Initial Step 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 ]

59. Chemical Kinetics Fast Initial Step 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 ]

60. Chemical Kinetics Catalysts ________________ increase the ________________ of a reaction by ________________ the activation energy of the reaction. Catalysts change the ________________ by which the process occurs.

61. Chemical Kinetics Catalysts One way a catalyst can speed up a reaction is by holding the ______________ together and helping bonds to break.

62. Chemical Kinetics Enzymes ________________ are catalysts in biological systems. The substrate fits into the active site of the enzyme much like a key fits into a lock.