1. Chapter 13 Chemical Kinetics 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2. Tro, Chemistry: A Molecular Approach2 kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds. experimentally it is shown that there are 4 factors that influence the speed of a reaction: nature of the reactants, temperature, catalysts, concentration Kinetics

3. Tro, Chemistry: A Molecular Approach3 Defining Rate rate is how much a quantity changes in a given period of time the speed you drive your car is a rate – the distance your car travels (miles) in a given period of time (1 hour) so the rate of your car has units of mi/hr

4. Tro, Chemistry: A Molecular Approach4 Defining Reaction Rate the rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time or product concentration increases for reactants, a negative sign is placed in front of the definition

5. Tro, Chemistry: A Molecular Approach5 Reaction Rate Changes Over Time as time goes on, the rate of a reaction generally slows down because the concentration of the reactants decreases. at some time the reaction stops, either because the reactants run out or because the system has reached equilibrium.

6. Tro, Chemistry: A Molecular Approach6 at t = 0 [A] = 8 [B] = 8 [C] = 0 at t = 0 [X] = 8 [Y] = 8 [Z] = 0 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1

7. Tro, Chemistry: A Molecular Approach7 at t = 16 [A] = 4 [B] = 4 [C] = 4 at t = 16 [X] = 7 [Y] = 7 [Z] = 1 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2

8. Tro, Chemistry: A Molecular Approach8 at t = 32 [A] = 2 [B] = 2 [C] = 6 at t = 32 [X] = 6 [Y] = 6 [Z] = 2 at t = 48 [A] = 0 [B] = 0 [C] = 8 at t = 48 [X] = 5 [Y] = 5 [Z] = 3

9. 9 Hypothetical Reaction Red  Blue Time (sec) Number Red Number Blue 01000 58416 107129 155941 2050 254258 303565 353070 402575 452179 501882 in this reaction, one molecule of Red turns into one molecule of Blue the number of molecules will always total 100 the rate of the reaction can be measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time

10. Tro, Chemistry: A Molecular Approach10 Hypothetical Reaction Red  Blue

11. Tro, Chemistry: A Molecular Approach11 Hypothetical Reaction Red  Blue

12. Tro, Chemistry: A Molecular Approach12 Reaction Rate and Stoichiometry in most reactions, the coefficients of the balanced equation are not all the same H 2 (g) + I 2 (g)  2 HI (g) for these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another for the above reaction, for every 1 mole of H 2 used, 1 mole of I 2 will also be used and 2 moles of HI made therefore the rate of change will be different in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient

13. Tro, Chemistry: A Molecular Approach13 Average Rate the average rate is the change in measured concentrations in any particular time period linear approximation of a curve the larger the time interval, the more the average rate deviates from the instantaneous rate

14. 14 Hypothetical Reaction Red  Blue Avg. Rate Time (sec) Number Red Number Blue (5 sec intervals) (10 sec intervals) (25 sec intervals) 01000 584163.2 1071292.62.9 1559412.4 2050 1.82.1 2542581.6 2.3 3035651.41.5 3530701 40257511 4521790.8 5018820.60.71

15. 15 H2H2 I2I2 HI Avg. Rate, M/s Time (s)[H 2 ], M[HI], M -  [H 2 ]/  t1/2  [HI]/  t 0.0001.000 10.0000.819 20.0000.670 30.0000.549 40.0000.449 50.0000.368 60.0000.301 70.0000.247 80.0000.202 90.0000.165 100.0000.135 Avg. Rate, M/s Time (s)[H 2 ], M[HI], M -  [H 2 ]/  t1/2  [HI]/  t 0.0001.0000.000 10.0000.8190.362 20.0000.6700.660 30.0000.5490.902 40.0000.4491.102 50.0000.3681.264 60.0000.3011.398 70.0000.2471.506 80.0000.2021.596 90.0000.1651.670 100.0000.1351.730 Stoichiometry tells us that for every 1 mole/L of H 2 used, 2 moles/L of HI are made. Assuming a 1 L container, at 10 s, we used 0.181 moles of H 2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles At 60 s, we used 0.699 moles of H 2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles Avg. Rate, M/s Time (s)[H 2 ], M[HI], M -  [H 2 ]/  t 0.0001.0000.000 10.0000.8190.3620.0181 20.0000.6700.6600.0149 30.0000.5490.9020.0121 40.0000.4491.1020.0100 50.0000.3681.2640.0081 60.0000.3011.3980.0067 70.0000.2471.5060.0054 80.0000.2021.5960.0045 90.0000.1651.6700.0037 100.0000.1351.7300.0030 The average rate is the change in the concentration in a given time period. In the first 10 s, the  [H 2 ] is -0.181 M, so the rate is Avg. Rate, M/s Time (s)[H 2 ], M[HI], M -  [H 2 ]/  t1/2  [HI]/  t 0.0001.0000.000 10.0000.8190.3620.0181 20.0000.6700.6600.0149 30.0000.5490.9020.0121 40.0000.4491.1020.0100 50.0000.3681.2640.0081 60.0000.3011.3980.0067 70.0000.2471.5060.0054 80.0000.2021.5960.0045 90.0000.1651.6700.0037 100.0000.1351.7300.0030

16. Tro, Chemistry: A Molecular Approach16 average rate in a given time period =  slope of the line connecting the [H 2 ] points; and ½ +slope of the line for [HI] the average rate for the first 10 s is 0.0181 M/s the average rate for the first 40 s is 0.0150 M/s the average rate for the first 80 s is 0.0108 M/s

17. Tro, Chemistry: A Molecular Approach17 Instantaneous Rate the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function  for you calculus fans

18. 18 H 2 (g) + I 2 (g)  2 HI (g) Using [H 2 ], the instantaneous rate at 50 s is: Using [HI], the instantaneous rate at 50 s is:

19. Ex 13.1 - For the reaction given, the [I  ] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the  [H + ]. H 2 O 2 (aq) + 3 I  (aq) + 2 H + (aq)  I 3  (aq) + 2 H 2 O (l) Solve the equation for the Rate (in terms of the change in concentration of the Given quantity) Solve the equation of the Rate (in terms of the change in the concentration for the quantity to Find) for the unknown value

20. Tro, Chemistry: A Molecular Approach20 Measuring Reaction Rate in order to measure the reaction rate you need to be able to measure the concentration of at least one component in the mixture at many points in time there are two ways of approaching this problem (1) for reactions that are complete in less than 1 hour, it is best to use continuous monitoring of the concentration, or (2) for reactions that happen over a very long time, sampling of the mixture at various times can be used when sampling is used, often the reaction in the sample is stopped by a quenching technique

21. Tro, Chemistry: A Molecular Approach21 Continuous Monitoring polarimetry – measuring the change in the degree of rotation of plane-polarized light caused by one of the components over time spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time the component absorbs its complimentary color total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction

22. Tro, Chemistry: A Molecular Approach22 Sampling gas chromatography can measure the concentrations of various components in a mixture for samples that have volatile components separates mixture by adherence to a surface drawing off periodic aliquots from the mixture and doing quantitative analysis titration for one of the components gravimetric analysis

23. Tro, Chemistry: A Molecular Approach23 Factors Affecting Reaction Rate Nature of the Reactants nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules; gases tend to react faster than liquids which react faster than solids; powdered solids are more reactive than “blocks”  more surface area for contact with other reactants certain types of chemicals are more reactive than others  e.g., the activity series of metals ions react faster than molecules  no bonds need to be broken

24. Tro, Chemistry: A Molecular Approach24 increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles  for many reactions there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later Factors Affecting Reaction Rate Temperature

25. Tro, Chemistry: A Molecular Approach25 catalysts are substances which affect the speed of a reaction without being consumed. most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts. homogeneous = present in same phase heterogeneous = present in different phase how catalysts work will be examined later Factors Affecting Reaction Rate Catalysts

26. Tro, Chemistry: A Molecular Approach26 generally, the larger the concentration of reactant molecules, the faster the reaction i ncreases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the gas  higher pressure = higher concentration concentration of solutions depends on the solute to solution ratio (molarity) Factors Affecting Reaction Rate Reactant Concentration

27. Tro, Chemistry: A Molecular Approach27 The Rate Law the Rate Law of a reaction is the mathematical relationship between the rate of the reaction and the concentrations of the reactants and homogeneous catalysts as well the rate of a reaction is directly proportional to the concentration of each reactant raised to a power for the reaction aA + bB  products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant

28. Tro, Chemistry: A Molecular Approach28 Reaction Order the exponent on each reactant in the rate law is called the order with respect to that reactant the sum of the exponents on the reactants is called the order of the reaction The rate law for the reaction: 2 NO(g) + O 2 (g)  2 NO 2 (g) is Rate = k[NO] 2 [O 2 ] The reaction is second order with respect to [NO], first order with respect to [O 2 ], and third order overall

29. Tro, Chemistry: A Molecular Approach29 Sample Rate Laws The reaction is autocatalytic, because a product affects the rate. Hg 2+ is a negative catalyst, increasing its concentration slows the reaction.

30. Tro, Chemistry: A Molecular Approach30 Reactant Concentration vs. Time A  Products

31. Tro, Chemistry: A Molecular Approach31 Half-Life the half-life, t 1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value the half-life of the reaction depends on the order of the reaction

32. Tro, Chemistry: A Molecular Approach32 Zero Order Reactions Rate = k[A] 0 = k constant rate reactions [A] = -kt + [A] 0 graph of [A] vs. time is straight line with slope = -k and y-intercept = [A] 0 t ½ = [A 0 ]/2k when Rate = M/sec, k = M/sec [A] 0 [A] time slope = - k

33. Tro, Chemistry: A Molecular Approach33 First Order Reactions Rate = k[A] ln[A] = -kt + ln[A] 0 graph ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A] 0 used to determine the rate constant t ½ = 0.693/k the half-life of a first order reaction is constant the when Rate = M/sec, k = sec -1

34. Tro, Chemistry: A Molecular Approach34 ln[A] 0 ln[A] time slope = −k

35. Tro, Chemistry: A Molecular Approach35 Half-Life of a First-Order Reaction Is Constant

36. Tro, Chemistry: A Molecular Approach36 Rate Data for C 4 H 9 Cl + H 2 O  C 4 H 9 OH + HCl Time (sec)[C 4 H 9 Cl], M 0.00.1000 50.00.0905 100.00.0820 150.00.0741 200.00.0671 300.00.0549 400.00.0448 500.00.0368 800.00.0200 10000.00.0000

37. Tro, Chemistry: A Molecular Approach37 C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl

38. Tro, Chemistry: A Molecular Approach38 C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl

39. Tro, Chemistry: A Molecular Approach39 C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl slope = -2.01 x 10 -3 k = 2.01 x 10 -3 s -1

40. Tro, Chemistry: A Molecular Approach40 Second Order Reactions Rate = k[A] 2 1/[A] = kt + 1/[A] 0 graph 1/[A] vs. time gives straight line with slope = k and y-intercept = 1/[A] 0 used to determine the rate constant t ½ = 1/(k[A 0 ]) when Rate = M/sec, k = M -1 ∙ sec -1

41. Tro, Chemistry: A Molecular Approach41 l/[A] 0 1/[A] time slope = k

42. 42 Rate Data For 2 NO 2  2 NO + O 2 Time (hrs.) Partial Pressure NO 2, mmHgln(P NO2 )1/(P NO2 ) 0100.04.6050.01000 3062.54.1350.01600 6045.53.8170.02200 9035.73.5760.02800 12029.43.3810.03400 15025.03.2190.04000 18021.73.0790.04600 21019.22.9570.05200 24017.22.8470.05800

43. Tro, Chemistry: A Molecular Approach43 Rate Data Graphs For 2 NO 2  2 NO + O 2

44. Tro, Chemistry: A Molecular Approach44 Rate Data Graphs For 2 NO 2  2 NO + O 2

45. Tro, Chemistry: A Molecular Approach45 Rate Data Graphs For 2 NO 2  2 NO + O 2

46. Tro, Chemistry: A Molecular Approach46 Determining the Rate Law can only be determined experimentally graphically rate = slope of curve [A] vs. time if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope initial rates by comparing effect on the rate of changing the initial concentration of reactants one at a time

47. Tro, Chemistry: A Molecular Approach47

48. Tro, Chemistry: A Molecular Approach48 Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod

49. Tro, Chemistry: A Molecular Approach49 Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod

50. Tro, Chemistry: A Molecular Approach50

51. Tro, Chemistry: A Molecular Approach51

52. Tro, Chemistry: A Molecular Approach52

53. Tro, Chemistry: A Molecular Approach53 Practice - Complete the Table and Determine the Rate Equation for the Reaction A  2 Prod the reaction is second order, Rate = -  [A]  t = 0.1 [A] 2

54. Ex. 13.4 – The reaction SO 2 Cl 2(g)  SO 2(g) + Cl 2(g) is first order with a rate constant of 2.90 x 10 -4 s -1 at a given set of conditions. Find the [SO 2 Cl 2 ] at 865 s when [SO 2 Cl 2 ] 0 = 0.0225 M the new concentration is less than the original, as expected [SO 2 Cl 2 ] 0 = 0.0225 M, t = 865, k = 2.90 x 10 -4 s -1 [SO 2 Cl 2 ] Check: Solution: Concept Plan: Relationships: Given: Find: [SO 2 Cl 2 ][SO 2 Cl 2 ] 0, t, k

55. Tro, Chemistry: A Molecular Approach55 Initial Rate Method another method for determining the order of a reactant is to see the effect on the initial rate of the reaction when the initial concentration of that reactant is changed for multiple reactants, keep initial concentration of all reactants constant except one zero order = changing the concentration has no effect on the rate first order = the rate changes by the same factor as the concentration  doubling the initial concentration will double the rate second order = the rate changes by the square of the factor the concentration changes  doubling the initial concentration will quadruple the rate

56. 56 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033 Write a general rate law including all reactants Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033 Comparing Expt #1 and Expt #2, the [NO 2 ] changes but the [CO] does not

57. Tro, Chemistry: A Molecular Approach57 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Determine by what factor the concentrations and rates change in these two experiments. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033

58. Tro, Chemistry: A Molecular Approach58 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Determine to what power the concentration factor must be raised to equal the rate factor. Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033

59. Tro, Chemistry: A Molecular Approach59 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Repeat for the other reactants Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033 Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033

60. Tro, Chemistry: A Molecular Approach60 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Substitute the exponents into the general rate law to get the rate law for the reaction Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033 n = 2, m = 0

61. Tro, Chemistry: A Molecular Approach61 Ex 13.2 – Determine the rate law and rate constant for the reaction NO 2(g) + CO (g)  NO (g) + CO 2(g) given the data below. Substitute the concentrations and rate for any experiment into the rate law and solve for k Expt. Number Initial [NO 2 ], (M) Initial [CO], (M) Initial Rate (M/s) 1.0.10 0.0021 2.0.200.100.0082 3.0.20 0.0083 4.0.400.100.033

62. Tro, Chemistry: A Molecular Approach62 Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 10.02000.20010.8 20.06000.20032.3 30.2000.020210.8 40.2000.040421.6

63. 63 Practice - Determine the rate law and rate constant for the reaction NH 4 +1 + NO 2 -1      given the data below. Expt. No. Initial [NH 4 + ], M Initial [NO 2 - ], M Initial Rate, (x 10 -7 ), M/s 10.02000.20010.8 20.06000.20032.3 30.2000.020210.8 40.2000.040421.6 Rate = k[NH 4 + ] n [NO 2  ] m

64. Tro, Chemistry: A Molecular Approach64 The Effect of Temperature on Rate changing the temperature changes the rate constant of the rate law Svante Arrhenius investigated this relationship and showed that: R is the gas constant in energy units, 8.314 J/(mol∙K) where T is the temperature in kelvins A is a factor called the frequency factor E a is the activation energy, the extra energy needed to start the molecules reacting

65. Tro, Chemistry: A Molecular Approach65

66. Tro, Chemistry: A Molecular Approach66 Activation Energy and the Activated Complex energy barrier to the reaction amount of energy needed to convert reactants into the activated complex aka transition state the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy because partial bonds

67. Tro, Chemistry: A Molecular Approach67 Isomerization of Methyl Isonitrile methyl isonitrile rearranges to acetonitrile in order for the reaction to occur, the H 3 C-N bond must break; and a new H 3 C-C bond form

68. 68 Energy Profile for the Isomerization of Methyl Isonitrile As the reaction begins, the C-N bond weakens enough for the C  N group to start to rotate the collision frequency is the number of molecules that approach the peak in a given period of time the activation energy is the difference in energy between the reactants and the activated complex the activated complex is a chemical species with partial bonds

69. Tro, Chemistry: A Molecular Approach69 The Arrhenius Equation: The Exponential Factor the exponential factor in the Arrhenius equation is a number between 0 and 1 it represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier the higher the energy barrier (larger activation energy), the fewer molecules that have sufficient energy to overcome it that extra energy comes from converting the kinetic energy of motion to potential energy in the molecule when the molecules collide increasing the temperature increases the average kinetic energy of the molecules therefore, increasing the temperature will increase the number of molecules with sufficient energy to overcome the energy barrier therefore increasing the temperature will increase the reaction rate

70. Tro, Chemistry: A Molecular Approach70

71. Tro, Chemistry: A Molecular Approach71 Arrhenius Plots the Arrhenius Equation can be algebraically solved to give the following form: this equation is in the form y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line (-8.314 J/mol∙K)(slope of the line) = E a, (in Joules) e y-intercept = A, (unit is the same as k)

72. Tro, Chemistry: A Molecular Approach72 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: Temp, Kk, M -1 ∙s -1 Temp, Kk, M -1 ∙s -1 6003.37 x 10 3 13007.83 x 10 7 7004.83 x 10 4 14001.45 x 10 8 8003.58 x 10 5 15002.46 x 10 8 9001.70 x 10 6 16003.93 x 10 8 10005.90 x 10 6 17005.93 x 10 8 11001.63 x 10 7 18008.55 x 10 8 12003.81 x 10 7 19001.19 x 10 9

73. Tro, Chemistry: A Molecular Approach73 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: use a spreadsheet to graph ln(k) vs. (1/T)

74. Tro, Chemistry: A Molecular Approach74 Ex. 13.7 Determine the activation energy and frequency factor for the reaction O 3(g)  O 2(g) + O (g) given the following data: E a = m∙(-R) solve for E a A = e y-intercept solve for A

75. Tro, Chemistry: A Molecular Approach75 Arrhenius Equation: Two-Point Form if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:

76. Ex. 13.8 – The reaction NO 2(g) + CO (g)  CO 2(g) + NO (g) has a rate constant of 2.57 M -1 ∙ s -1 at 701 K and 567 M -1 ∙ s -1 at 895 K. Find the activation energy in kJ/mol most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable T 1 = 701 K, k 1 = 2.57 M -1 ∙s -1, T 2 = 895 K, k 2 = 567 M -1 ∙s -1 E a, kJ/mol Check: Solution: Concept Plan: Relationships: Given: Find: EaEa T 1, k 1, T 2, k 2

77. Tro, Chemistry: A Molecular Approach77 Collision Theory of Kinetics for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other. once molecules collide they may react together or they may not, depending on two factors - 1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; 2. whether the reacting molecules collide in the proper orientation for new bonds to form.

78. Tro, Chemistry: A Molecular Approach78 Effective Collisions collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions the higher the frequency of effective collisions, the faster the reaction rate when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state

79. Tro, Chemistry: A Molecular Approach79 Effective Collisions Kinetic Energy Factor for a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide it can form the activated complex

80. Tro, Chemistry: A Molecular Approach80 Effective Collisions Orientation Effect

81. Tro, Chemistry: A Molecular Approach81 Collision Theory and the Arrhenius Equation A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)

82. Tro, Chemistry: A Molecular Approach82 Orientation Factor the proper orientation results when the atoms are aligned in such a way that the old bonds can break and the new bonds can form the more complex the reactant molecules, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p slightly less than 1 for most reactions, the orientation factor is less than 1 for many, p << 1 there are some reactions that have p > 1 in which an electron is transferred without direct collision

83. Tro, Chemistry: A Molecular Approach83 Reaction Mechanisms we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism

84. Tro, Chemistry: A Molecular Approach84 An Example of a Reaction Mechanism Overall reaction: H 2(g) + 2 ICl (g)  2 HCl (g) + I 2(g) Mechanism: 1)H 2(g) + ICl (g)  HCl (g) + HI (g) 2)HI (g) + ICl (g)  HCl (g) + I 2(g) the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps

85. Tro, Chemistry: A Molecular Approach85 H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) 1) H 2 (g) + ICl(g)  HCl(g) + HI(g) 2) HI(g) + ICl(g)  HCl(g) + I 2 (g) Elements of a Mechanism Intermediates notice that the HI is a product in Step 1, but then a reactant in Step 2 since HI is made but then consumed, HI does not show up in the overall reaction materials that are products in an early step, but then a reactant in a later step are called intermediates

86. Tro, Chemistry: A Molecular Approach86 Molecularity the number of reactant particles in an elementary step is called its molecularity a unimolecular step involves 1 reactant particle a bimolecular step involves 2 reactant particles though they may be the same kind of particle a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps

87. Tro, Chemistry: A Molecular Approach87 Rate Laws for Elementary Steps each step in the mechanism is like its own little reaction – with its own activation energy and own rate law the rate law for an overall reaction must be determined experimentally but the rate law of an elementary step can be deduced from the equation of the step H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) 1) H 2 (g) + ICl(g)  HCl(g) + HI(g) Rate = k 1 [H 2 ][ICl] 2) HI(g) + ICl(g)  HCl(g) + I 2 (g) Rate = k 2 [HI][ICl]

88. Tro, Chemistry: A Molecular Approach88 Rate Laws of Elementary Steps

89. Tro, Chemistry: A Molecular Approach89 Rate Determining Step in most mechanisms, one step occurs slower than the other steps the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy the rate law of the rate determining step determines the rate law of the overall reaction

90. Tro, Chemistry: A Molecular Approach90 Another Reaction Mechanism NO 2(g) + CO (g)  NO (g) + CO 2(g) Rate obs = k[NO 2 ] 2 1) NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Rate = k 1 [NO 2 ] 2 slow 2) NO 3(g) + CO (g)  NO 2(g) + CO 2(g) Rate = k 2 [NO 3 ][CO] fast The first step in this mechanism is the rate determining step. The first step is slower than the second step because its activation energy is larger. The rate law of the first step is the same as the rate law of the overall reaction.

91. Tro, Chemistry: A Molecular Approach91 Validating a Mechanism in order to validate (not prove) a mechanism, two conditions must be met: 1. the elementary steps must sum to the overall reaction 2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law

92. Tro, Chemistry: A Molecular Approach92 Mechanisms with a Fast Initial Step when a mechanism contains a fast initial step, the rate limiting step may contain intermediates when a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related and the product is an intermediate substituting into the rate law of the RDS will produce a rate law in terms of just reactants

93. Tro, Chemistry: A Molecular Approach93 An Example 2 NO (g)  N 2 O 2(g) Fast H 2(g) + N 2 O 2(g)  H 2 O (g) + N 2 O (g) SlowRate = k 2 [H 2 ][N 2 O 2 ] H 2(g) + N 2 O (g)  H 2 O (g) + N 2(g) Fast k1k1 k-1k-1 2 H 2(g) + 2 NO (g)  2 H 2 O (g) + N 2(g) Rate obs = k [H 2 ][NO] 2 for Step 1 Rate forward = Rate reverse

94. Tro, Chemistry: A Molecular Approach94 Ex 13.9 Show that the proposed mechanism for the reaction 2 O 3(g)  3 O 2(g) matches the observed rate law Rate = k[O 3 ] 2 [O 2 ] -1 O 3(g)  O 2(g) + O (g) Fast O 3(g) + O (g)  2 O 2(g) Slow Rate = k 2 [O 3 ][O] k1k1 k-1k-1 for Step 1 Rate forward = Rate reverse

95. Tro, Chemistry: A Molecular Approach95 Catalysts catalysts are substances that affect the rate of a reaction without being consumed catalysts work by providing an alternative mechanism for the reaction with a lower activation energy catalysts are consumed in an early mechanism step, then made in a later step mechanism without catalyst O 3(g) + O (g)  2 O 2(g) V. Slow mechanism with catalyst Cl (g) + O 3(g)  O 2(g) + ClO (g) Fast ClO (g) + O (g)  O 2(g) + Cl (g) Slow

96. Tro, Chemistry: A Molecular Approach96 Ozone Depletion over the Antarctic

97. Tro, Chemistry: A Molecular Approach97 Energy Profile of Catalyzed Reaction polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

98. Tro, Chemistry: A Molecular Approach98 Catalysts homogeneous catalysts are in the same phase as the reactant particles Cl (g) in the destruction of O 3(g) heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system

99. Tro, Chemistry: A Molecular Approach99 Types of Catalysts

100. Tro, Chemistry: A Molecular Approach100 Catalytic Hydrogenation H 2 C=CH 2 + H 2 → CH 3 CH 3

101. Tro, Chemistry: A Molecular Approach101 Enzymes because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate protein molecules that catalyze biological reactions are called enzymes enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction

102. Tro, Chemistry: A Molecular Approach102 Enzyme-Substrate Binding Lock and Key Mechanism

103. Tro, Chemistry: A Molecular Approach103 Enzymatic Hydrolysis of Sucrose