1. Solutions © 2009, Prentice-Hall, Inc. Chapter 6 Chemical Kinetics Dr Nehdi Imededdine Arbi Associate Professor King Saud University Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten

2. Solutions © 2009, Prentice-Hall, Inc. Kinetics In kinetics we study the rate at which a chemical process occurs. Besides information about the speed at which reactions occur, kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs).

3. Solutions © 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates Physical State of the Reactants –In order to react, molecules must come in contact with each other. –The more homogeneous the mixture of reactants, the faster the molecules can react.

4. Solutions © 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates 1) Concentration of Reactants –As the concentration of reactants increases, so does the likelihood that reactant molecules will collide and the reaction proceed faster Fig: 6.1 Effect of concentration on rate. (a)When heated in air, steel wool glows red-hot but oxidizes slowly. (b) When the red-hot steel wool is placed in an atmosphere of pure oxygen, it burns vigorously, forming Fe 2 O 3 at a much faster rate. The difference in behavior is due to the different concentrations of O 2 in the two environments.

5. Solutions © 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates 2)Temperature –At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. Consequently, the rate of chemical reaction increases.

6. Solutions © 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates 3)Presence of a Catalyst –The rates of many reactions can be increased by adding a substance known as a catalyst. –Catalysts speed up reactions by changing the mechanism of the reaction –Catalysts are not consumed during the course of the reaction.

7. Solutions © 2009, Prentice-Hall, Inc. Factors That Affect Reaction Rates The surface area of solid or liquid reactants or catalysts –Reactions that involve solids often proceed faster as the surface area of the solid is increased. –For example, a medicine in the form of a tablet will dissolve in the stomach and enter the bloodstream more slowly than the same medicine in the form of a fine powder.

8. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates Rates of reactions can be determined by monitoring the change in concentration of either reactants or products as a function of time. Fig. 6.2 Progress of a hypothetical reaction A → B. Each red sphere represents 0.01 mol A, each blue sphere represents 0.01 mol B, and the vessel has a volume of 1.00 L. (a) At time zero the vessel contains 1.00 mol A (100 red spheres) and 0 mol B (no blue spheres). (b) After 20 s the vessel contains 0.54 mol A and 0.46 mol B. (c) After 40 s the vessel contains 0.30 mol A and 0.70 mol B.

9. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates The reaction rate is a measure of how quickly A is consumed or how quickly B is produced. Thus, for a given interval of time, we can express the average rate of reaction as the increase in the number of moles of B over that interval: Fig. 6.2 Progress of a hypothetical reaction A → B. Average rate = Change in the number of moles of B Change in time =  moles of   t

10. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates We can express also the average rate of reaction as the decrease In the number of moles of A over that interval: Fig. 6.2 Progress of a hypothetical reaction A → B. Average rate = - Change in the number of moles of A Change in time = -  moles of   t The negative sign in this equation is present because A is reactant that is consummed during the reaction.

11. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates Fig. 6.2 Progress of a hypothetical reaction A → B. For example, the average rate over the interval t = 0 s to t = 20 s is given by: Average rate = (moles of B at t = 20 s) - (moles of B at t = 0 s) 20 s – 0 s =  mol – 0 mol  s = 0.026 mol/s Average rate = - (moles of A at t = 20 s) - (moles of A at t = 0 s) 20 s – 0 s = -  mol – 1 mol  s = 0.026 mol/s

12. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates in Terms of concentration 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)

13. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates in Terms of concentration The average rate of the reaction over each interval (50 sec) is the change in concentration divided by the change in time: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Average rate = -  [C 4 H 9 Cl]  t ( unit: M/s) Concentration of C 4 H 9 Cl

14. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates in Terms of concentration Note that the average rate decreases as the reaction proceeds. This is because as the reaction goes forward, 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)

15. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates in Terms of concentration A plot of [C 4 H 9 Cl] vs. time for this reaction yields a curve like this. The slope of a line tangent to the curve at any point is the instantaneous rate at that time. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Instantaneous rate = -  [C 4 H 9 Cl]  t

16. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates in Terms of concentration For example, to determine the instantaneous rates at 600 s, we draw the tangent at this time, then construct horizontal and vertical lines to forme the right triangle shown. The slope is the ratio of the height of the vertical side to the horizontal side: C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Instantaneous rate = -  [C 4 H 9 Cl]  t = - (0.017 -0.042) M (800 -400)s = 6.2 x 10 -5 M/s

17. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates All reactions slow down over time. Therefore, the best indicator of the rate of a reaction is the instantaneous rate near the beginning of the reaction. C 4 H 9 Cl (aq) + H 2 O (l)  C 4 H 9 OH (aq) + HCl (aq) Exe 6.2.

18. Solutions © 2009, Prentice-Hall, Inc. 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

19. Solutions © 2009, Prentice-Hall, Inc. Reaction Rates and Stoichiometry What if the ratio is not 1:1? 2 HI (g)  H 2 (g) + I 2 (g) In such a case, Rate = − 1212  [HI]  t =  [I 2 ]  t

20. Solutions © 2009, Prentice-Hall, Inc. 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 = Exer 6.3.

21. Solutions © 2009, Prentice-Hall, Inc. Concentration and Rate One can gain information about the rate of a reaction by seeing how the rate changes with changes in concentration.

22. Solutions © 2009, Prentice-Hall, Inc. Concentration and Rate If we compare experiments 1 and 2, we see that when [NH 4 + ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l)

23. Solutions © 2009, Prentice-Hall, Inc. Concentration and Rate Likewise, when we compare experiments 5 and 6, we see that when [NO 2 − ] doubles, the initial rate doubles. NH 4 + (aq) + NO 2 − (aq) N 2 (g) + 2 H 2 O (l) Table 6.4.

24. Solutions © 2009, Prentice-Hall, Inc. Concentration and Rate This means Rate  [NH 4 + ] Rate  [NO 2 − ] Rate  [NH 4 + ] [NO 2 − ] which, when written as an equation, becomes Rate = k [NH 4 + ] [NO 2 − ] This equation is called the rate law, and k is the rate constant. Therefore,

25. Solutions © 2009, Prentice-Hall, Inc. Concentration and Rate If we known the rate law for a reaction and its rate for a set of reactant concentrations, we can calculate the value of the rate constant, k. For example, using the data in the table 6.4. and the results from experiment 1, we can subsitute into Rate = k [NH 4 + ] [NO 2 − ] 5.4 x 10 -7 M/s= k (0.01 M)(0.2 M) Solving for k, k = 2.7 x 10 -4 M -1 s -1

26. Solutions © 2009, Prentice-Hall, Inc. Units of Rate Constants The units of the rate constant depend on the overall reaction order of the rate law. For exapmle, in a reaction that is second order overall, the units of the rate constant must satisfy Units of rate = (units of rate constant) (units of concentration) 2 If we known the rate law for a reaction and its rate for a set of reactant concentrations, we can calculate the value of the rate constant, k. For example, using the data in the table 6.4. and the results from experiment 1, we can substitute into Rate = k [NH 4 + ] [NO 2 − ] 5.4 x 10 -7 M/s= k (0.01 M)(0.2 M) Solving for k, k = 2.7 x 10 -4 M -1 s -1 Exer 6.4.

27. Solutions © 2009, Prentice-Hall, Inc. Reaction Order A rate law shows the relationship between the reaction rate and the concentrations of reactants. The rate laws for most reactions have the general form Rate = k [reactant 1] m [reactant 2] n … The exponents tell the order of the reaction with respect to each reactant. For example, the rate law for the reaction of NH 4 + with NO 2 - Rate = k [NH 4 + ] 1 [NO 2 − ] 1 the reaction is First-order in [NH 4 + ] and First-order in [NO 2 − ].

28. Solutions © 2009, Prentice-Hall, Inc. Reaction Order The overall reaction order can be found by adding the exponents on the reactants in the rate law( m +n + ….). For the example Rate = k [NH 4 + ] 1 [NO 2 − ] 1 The overall reaction order is 1+1 = 2; we say the reaction is second-order overall. Rate = k [reactant 1] m [reactant 2] n …

29. Solutions © 2009, Prentice-Hall, Inc. Reaction Order The following are some further examples of rate laws: 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) Rate = k [N 2 O 5 ] CHCl 3 (g) + Cl 2 (g) CCl 4 (g) + HCl (g) Rate = k [CHCl 3 ] [Cl 2 ] 1/2 H 2 (g) + l 2 (g) 4 Hl(g) Rate = k [H 2 ] [l 2 ] Rate = k [reactant 1] m [reactant 2] n … Exer 6.5.

30. Solutions © 2009, Prentice-Hall, Inc. Using Initial Rates to Determine Rate Laws The rate law for any chemical reaction must be determined experimentally. By observing the effect of changing the initial concentrations of the reactants on the initial rate of the reaction (example table 6.4). If a reaction is zero order in a particular reactant, changing its concentration will have no effect on rate.

31. Solutions © 2009, Prentice-Hall, Inc. Using Initial Rates to Determine Rate Laws If a reaction is first order in a reactant, changes in the concentration of that substance will produce proportional changes in the rate. Thus, doubling the concentration will double the rate, and so forth. When the rate law is second order a reaction in a reactant, doubling its concentration increases the rate, by a factor of 2 2 = 4, tripling its concentration causes the rate increase by a factor of 3 2 = 9 It is important to realize that the rate of a reaction depends on concentration, but the rate constant does not The rate constant is affected by temperature and by the presence of a catalyst..

32. Solutions © 2009, Prentice-Hall, Inc. The change of Concentration with Time Integrated Rate Laws Using calculus to integrate the rate law for a first-order process: Rate = -  [A] = -k [A] gives us ln [A] t [A] 0 = −kt Where [A] 0 is the initial concentration of A, and [A] t is the concentration of A at some time, t, during the course of the reaction. tt

33. Solutions © 2009, Prentice-Hall, Inc. 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

34. Solutions © 2009, Prentice-Hall, Inc. First-Order Processes Therefore, if a reaction is first-order, 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

35. Solutions © 2009, Prentice-Hall, Inc. First-Order Processes Consider the process in which methyl isonitrile is converted to acetonitrile. CH 3 NCCH 3 CN

36. Solutions © 2009, Prentice-Hall, Inc. First-Order Processes This data was collected for this reaction at 198.9 C. CH 3 NC(g)CH 3 CN(g)

37. Solutions © 2009, Prentice-Hall, Inc. First-Order Processes When the natural logarithm (ln P) is plotted as a function of time, a straight line results. Therefore, –The process is first-order. –The slope of this line is -5.1 x10 -5 s -1. –k is the negative of the slope: 5.1  10 -5 s −1.

38. Solutions © 2009, Prentice-Hall, Inc. Half-Life (t 1/2 ) of a first order reaction Half-life is defined as the time required for one-half of a reactant to react. Because [A] at t 1/2 is one-half of the original [A], [A] t1/2 = 0.5 [A] 0.

39. Solutions © 2009, Prentice-Hall, Inc. Half-Life (t 1/2 ) of a first order reaction 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 NOTE: For a first-order process, then, the half-life does not depend on the initial concentration of reactant [A] 0. Thus, the half- life is the same at any time during the reaction. 0.693 k t 1/2 =

40. Solutions © 2009, Prentice-Hall, Inc. Half-Life (t 1/2 ) of a first order reaction The data for the first-order rearrangement of methyl isonitrile at 198.9  C show that the first half-life t 1/2 is at 13.300 s. At a time 13.300 s later, the isonitrile concentration has decreased to one half of one half, or one-fourth the original concentration. NOTE: In a first-order reaction, the concentration of the reactant decreases by 1/2 in each of a series of regularly spaced time intervals, namely, t 1/2.

41. Solutions © 2009, Prentice-Hall, Inc. Second-Order Processes Similarly, integrating the rate law for a process that is second-order in reactant A, Rate = -  [A] = -k [A] 2 we get 1 [A] t = kt + 1 [A] 0 also in the form y = mx + b tt

42. Solutions © 2009, Prentice-Hall, Inc. Second-Order Processes So if a process is second-order in A, a plot of vs. t will yield a straight line, and the slope of that line is k. 1 [A] t = kt + 1 [A] 0 1 [A] NOTE: In the ln [A] t plot is linear, the reaction is first-order; if the plot is linear, the reaction is second-order 1 [A] t

43. Solutions © 2009, Prentice-Hall, Inc. Second -Order Processes When is plotted as a function of time, a straight line results. Therefore, –The process is second-order. –The slope of this line is k. 1 [A] 1 [A]

44. Solutions © 2009, Prentice-Hall, Inc. Half-Life (t 1/2 ) of a second order reaction 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 = 1 k[A] 0 t 1/2 = NOTE: Unlike, the half-life a first-order process, the half-life of a second order reaction is depend on the initial concentration of reactant [A] 0.

45. Solutions © 2009, Prentice-Hall, Inc. Practice Exercise The decomposition of NO 2 at 300°C is described by the equation NO 2 (g) NO (g) + 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 1212 Indicate whether this reaction is first or second order ?

46. Solutions © 2009, Prentice-Hall, Inc. SOLUTION Plotting ln [NO 2 ] vs. t yields the graph at the right. 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].

47. Solutions © 2009, Prentice-Hall, Inc. Second-Order Processes Graphing 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 second- order in [A]. 1 [NO 2 ]

48. Solutions © 2009, Prentice-Hall, Inc. Temperature and Rate Generally, as temperature increases, so does the reaction rate. This is because k is temperature dependent. Fig. Temperature affects the rate of the chemiluminescence reaction in Cyalume ® light sticks.The light stick in hot water (left) glows more brightly than the one in cold water (right); at the higher temperature, the reaction is initially faster and produces a brighter light.

49. Solutions © 2009, Prentice-Hall, Inc. Temperature and Rate For example, this figure shows the rate constant as function of temperature for the first-order reaction CH 3 NC(g)CH 3 CN(g) The rate constant, and hence the rate of the reaction, increases rapidly with temperature, approximately doubling for each 10  C.

50. Solutions © 2009, Prentice-Hall, Inc. The Collision Model In a chemical reaction, bonds are broken and new bonds are formed. Molecules can only react if they collide with each other.

51. Solutions © 2009, Prentice-Hall, Inc. The Collision Model Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation. Cl + NOCl NO + Cl 2

52. Solutions © 2009, Prentice-Hall, Inc. Activation Energy In other words, there is a minimum amount of energy required for reaction: the activation energy, 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 energy to get over the activation energy barrier.

53. Solutions © 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams It is helpful to visualize energy changes throughout a process on a reaction coordinate diagram like this one for the rearrangement of methyl isonitrile.

54. Solutions © 2009, Prentice-Hall, Inc. Reaction Coordinate Diagrams The diagram shows the energy of the reactants and products (and, therefore,  E). The high point on the diagram is the transition state. The species present at the transition state is called the activated complex. The energy gap between the reactants and the activated complex is the activation energy barrier.

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

56. Solutions © 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions As the temperature increases, the curve flattens and broadens. Thus at higher temperatures, a larger population of molecules has higher energy.

57. Solutions © 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions If the dotted line represents the activation energy, then as the temperature increases, a much greater fraction of molecules has kinetic energy greater than E a, which leads to a much greater rate reaction. As a result, the reaction rate increases.

58. Solutions © 2009, Prentice-Hall, Inc. Maxwell–Boltzmann Distributions This fraction of molecules that has an energy equal to or greater than 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

59. Solutions © 2009, Prentice-Hall, Inc. Arrhenius Equation Svante Arrhenius developed a mathematical relationship between k and E a : k = A e 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

60. Solutions © 2009, Prentice-Hall, Inc. Evaluation of the Activation Energy (E a ) by graphical way Taking the natural logarithm of both sides, the equation becomes ln k = - ( ) + ln A 1T1T y = m x + b Therefore, if k is determined experimentally at several temperatures, E a can be calculated from the slope of a plot of ln k vs.. (E a in J/mol, R = 8.314 J/K.mol) EaREaR 1T1T

61. Solutions © 2009, Prentice-Hall, Inc. Using the equation ln k = - ( ) + ln A and suppose that at two different temperatures, T 1 and T 2, a reaction has rate constants k 1 and k 2. For each condition we have 1T1T EaREaR Evaluation of the Activation Energy (E a ) by nongraphical way ln k 1 = - ( ) + ln A 1T11T1 EaREaR, ln k 2 = - ( ) + ln A 1T21T2 EaREaR Subtracting ln k 2 – ln k 1 ln k 2 – ln k 1 = ( - ) and we determine E a 1T21T2 EaREaR 1T11T1 And from this equation we can calculate the rate constant, k 1, at some temperature, T 1, when we know E a, k 2 and T 2.

62. Solutions © 2009, Prentice-Hall, Inc. The Orientation Factor Not every collision in which the reactants have an energy E a or greater results in reaction. In addition they must be oriented in a certain way of the collision to lead reaction. The relative orientations of the molecules during their collisions determine whether the atoms are suitably positioned to form new bonds. For example, consider the reaction of Cl atoms with NOCl: Cl + NOCl NO + Cl 2

63. Solutions © 2009, Prentice-Hall, Inc. The Orientation Factor The reaction will take place if the collision brings Cl atoms together to form Cl 2,as shown in figure (a). In contrast, the collision shown in figure (b) will be ineffective and will not yield products. Figure : Molecular collisions and chemical reactions. Two possible ways that Cl atoms and NOCl molecules can collide are shown. (a) If molecules are oriented properly, a sufficiently energetic collision will lead to reaction. (b) If the orientation of the colliding molecules is wrong, no reaction occurs.