1. Chapter 13 Chemical Kinetics Chemistry II

2. kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds Factors that influence the speed of a reaction: physical state of reactants, temperature, catalysts, concentration The Rate of a Chemical Reaction

3. The Rate of a Chemical Reaction Defining Rate rate is how much a quantity changes in a given period of time, e.g.

4. The Rate of a Chemical Reaction Chemical reaction - concentration change with time for reactants, a negative sign is used to show a decrease in concentration as time goes on, the rate of a reaction generally slows down and stops because the concentration of the reactants decreases

5. 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

6. 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

7. 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

8. 8

9. 9 rate of change reactants = rate of change products

10. 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

11. Hypothetical Reaction Red  Blue

13. The Rate of a Chemical Reaction Stoichiometry e.g. H 2 (g) + I 2 (g)  2 HI (g) 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

14. The Rate of a Chemical Reaction Stoichiometry e.g. H 2 (g) + I 2 (g)  2 HI (g)

15. The Rate of a Chemical Reaction 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

16. 16 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

17. H2H2 I2I2 HI 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 The average rate is the change in the concentration in a given time period. In the first 10 s, the Δ[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 0.0001.000 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 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 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 Ave. rate slows down as reaction proceeds Rate of loss reactant = Rate gain product

18. 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 Ave. rate slows down as reaction proceeds

19. The Rate of a Chemical Reaction Instantaneous Rate average rate becomes less accurate over longer time spans 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

20. 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: rate reactants = rate products

21. Generalized rate law: aA + bB → cC + dD

22. 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 Δ[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

23. Factors Affecting Reaction Rate - nature of 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. increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the speed of the reaction doubles 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. 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. generally, the larger the concentration of reactant molecules, the faster the reaction increases 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

28. The Rate Law: Effect of Concentration On Reaction Rate Mathematical relationship between the rate of the reaction and the concentrations of the reactants 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

29. The Rate Law: Effect of Concentration On Reaction Rate sum of the exponents is called the order of the reaction The rate law for the reaction: 2 NO(g) + O 2 (g) ⇌ 2 NO 2 (g) 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

30. The Rate Law: Effect of Concentration On Reaction Rate Sample Rate Laws

31. The Rate Law: Effect of Concentration On Reaction Rate Example: A → Products Rate = k[A] 1 k = 0.015 / 0.10 = 0.15 s -1 If concentration of A doubles, the new rate,Rate 2 = k[2A] 1 = 2 k[A] 1 = 2 x Rate [A] (M)Initial Rate (M/s) 0.100.015 0.200.030 0.300.060

32. The Rate Law: Effect of Concentration On Reaction Rate Example: Zero order:Second order: Concentration of A doubles, the rate is constantConc. A doubles, new rate Rate 1 = Rate 2 = kRate 2 = k[2A] 2 = 4 x k[A] 2 Rate 2 = 4 x Rate [A] (M)Initial Rate (M/s) 0.100.015 0.200.015 0.300.015 [A] (M)Initial Rate (M/s) 0.100.015 0.200.060 0.300.240

33. Reactant Concentration vs. Time A  Products Rate = k[A] 2 Rate = k[A] Rate = k 0: Concentration dec. linearly with time. Rate Is constant, reaction does not slow down as [A] dec. 1 and 2: Rate slows as reaction proceeds since [A] dec.

34. The Integrated 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

35. The Integrated Rate Law 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

36. Zero Order Reactions (n = 0) Rate = -d[A] = k[A] 0 = k dt constant rate reactions Solution: [A] = -kt + [A] 0 (= integrated rate law) y = mx + b graph of [A] vs. time is straight line with slope = -k and y-intercept = [A] 0 [A] = [A 0 ]/2, t ½ = [A 0 ]/2k Units: when Rate = M/sec, k = M/sec [A] 0 [A] time slope = - k ∫ -d[A]/dt = ∫ k ∫ d[A] = -∫ k dt [A] = -kt + C, where C = [A] 0

37. First Order Reactions (n = 1) Rate = -d[A] = k[A] dt Solution: ln[A] = -kt + ln[A] 0 graph of ln[A] vs. time gives straight line with slope = -k and y-intercept = ln[A] 0 used to determine the rate constant [A] = [A 0 ]/2, t ½ = ln 2[lna-lnb = ln(a/b)] k the half-life of a first order reaction is constant Units: when Rate = M/sec, k = sec -1 (dim. Analysis: M/s = k.M) Rate slows as reaction proceeds since [A] dec. ∫ -d[A]/dt = ∫ k[A] ∫ 1 d[A] = -∫ k dt [A] ln[A] = -kt + C, where C = ln[A] 0 ln[A] 0 ln[A] time slope = −k

38. Rate Data for hydrolysis of C 4 H 9 Cl 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 Show reaction is first-order and find k

39. C 4 H 9 Cl + H 2 O  C 4 H 9 OH + 2 HCl

41. slope = -2.01 x 10 -3 k = 2.01 x 10 -3 s -1

42. Second Order Reactions Rate = -d[A] = k[A] = k[A] 2 dt Solution: 1/[A] = kt + 1/[A] 0 y = mx + b 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 l/[A] 0 1/[A] time slope = k

43. Example Second-Order Reaction Time (hrs.) P NO2, (mmHg)ln(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 Show that the reaction: NO 2 → NO + O is incorrect according to the data below

44. Rate Data Graphs For NO 2 → NO + O Non-linear so not zero order

45. Rate Data Graphs For NO 2 ® NO + O Non-linear so not first order

46. Rate Data Graphs For NO 2 → NO + O We can deduce actual reaction should be: 2NO 2 → 2NO + O 2 k = 2 x 10 -4 M -1 s -1

47. Tro, Chemistry: A Molecular Approach47

48. 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

49. 49 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

50. 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

51. 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

52. 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

53. 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

54. 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

55. Practice - Determine the rate law and rate constant for the reaction NH 4 + + NO 2 -      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

56. Practice - Determine the rate law and rate constant for the reaction NH 4 + + NO 2 - → N 2 + 2 H 2 O 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

58. The Effect of Temperature on Reaction Rate Rate constant k is temperature dependent 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 kelvin A is a constant called the frequency factor E a is the activation energy, the extra energy needed to start the molecules reacting

59. As x (temperature) increases e -1/x will increase up to a maximum value of 1, k increases As E a increases k will decrease (follows e -x graph)

60. The Effect of Temperature on Reaction Rate Activation Energy and the Activated Complex E a is an 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

61. Tro, Chemistry: A Molecular Approach61 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

62. 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

63. The Effect of Temperature on Reaction Rate The Exponential Factor e -Ea/RT is a number between 0 and 1 it represents the fraction of reactant molecules with sufficient energy to make it over the energy barrier that extra energy comes from converting the KE of motion to PE in the molecule when the molecules collide e -Ea/RT decreases as E a increases Reaction rate inc.: -Increasing T increases the Ave. KE of the molecules -Increases no. of molecules with sufficient energy to overcome the energy barrier

64. The Effect of Temperature on Reaction Rate Arrhenius Plots the Arrhenius Equation can be algebraically solved: y = mx + b where y = ln(k) and x = (1/T) a graph of ln(k) vs. (1/T) is a straight line slope of the line = -E a /R so E a = -mR e y-intercept = A, (unit is the same as k …why?) k = Ae -Ea/RT lnk = ln(Ae -Ea/RT ) lnk = lnA + lne -Ea/RT lnk = lnA – E a /RT

65. 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

66. 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)

67. 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

68. The Effect of Temperature on Reaction Rate Arrhenius Plots: 2-Point Form if you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:

69. 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

70. The Effect of Temperature on Reaction Rate The Collision Model 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.

71. The Effect of Temperature on Reaction Rate The Collision Model Effective Collisions collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions the higher the A value (frequency of effective collisions), the higher k value and 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

72. The Effect of Temperature on Reaction Rate The Collision Model Orientation Effect

73. The Effect of Temperature on Reaction Rate The Collision Model 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)

74. The Effect of Temperature on Reaction Rate The Collision Model Orientation Factor proper orientation is when the atoms are aligned so that old bonds can break and the new bonds can form the more complex the reactants, 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 e.g. H(g) + I(g) → HI(g) H 2 (g) + I 2 (g) → 2HI(g) HCl(g) + HCl(g) → H 2 (g) + Cl 2 (g) Smallest p

76. 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

77. Reaction Mechanisms 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

78. 1) H 2 (g) + ICl(g)  HCl(g) + HI(g) 2) HI(g) + ICl(g)  HCl(g) + I 2 (g) H 2 (g) + 2 ICl(g)  2 HCl(g) + I 2 (g) Reaction Mechanisms 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

79. Reaction Mechanisms 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

80. Reaction Mechanisms 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]

81. Reaction Mechanisms Rate Laws for Elementary Steps

82. Reaction Mechanisms 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

83. Another Reaction Mechanism NO 2(g) + NO 2(g)  NO 3(g) + NO (g) Rate = k 1 [NO 2 ] 2 slow NO 3(g) + CO (g)  NO 2(g) + CO 2(g) Rate = k 2 [NO 3 ][CO] fast NO 2(g) + CO (g)  NO (g) + CO 2(g) Rate obs = k[NO 2 ] 2 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.

84. Reaction Mechanisms 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

85. Reaction Mechanisms Mechanisms with a Fast Initial Step when a mechanism contains a slow initial step, the rate law will not contain intermediates when a mechanism contains a fast initial step, the rate limiting step, and hence the rate law may contain intermediates We can express[intermediate] in terms of [reactant] If first step is fast, intermediate products build up (limited by slower step down the line) as they build up they react to re-form reactants reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related substituting into the rate law of the RDS will produce a rate law in terms of just reactants

86. 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) Slow(rate limiting) 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) Experimentally observed Rate obs = k [H 2 ][NO] 2 Proposed mechanism: 2 H 2(g) + 2 NO (g)  2 H 2 O (g) + N 2(g)

87. Reaction Mechanisms Mechanisms with a Fast Initial Step Is the mechanism valid? 1. steps must sum to the over all reaction 2. rate law predicted by mechanism must be consistent with exp. observation Since 2 nd step is rate limiting, Rate = k 2 [H 2 ][N 2 O 2 ] BUT! Contains intermediate [N 2 O 2 ], not consistent with observation Since 1 st step is in equilibrium we can express [intermediate] in terms of reactants

88. 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 Let k 2 k 1 /k -1 = k, we have obs rate law

89. 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 (Slow rate has intermediate)

90. 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 O 3(g) + O (g)  2 O 2(g)

91. Catalysts Demo http://academics.rmu.edu/~short/chem1215/1215- demos/oscillating-methanol-7mins.mov 2CH 3 OH(l) + 3O 2 (g) → 2CO 2 (g) + 4H 2 O(g)

92. Ozone Depletion over the Antarctic

93. Catalysts polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

94. Catalysts Homorgeneous and Heterogeneous 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

95. Catalysts 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

96. Enzymatic Hydrolysis of Sucrose