1. CHE 107 Chemical Kinetics Sources: http://www.mikeblaber.org/oldwine/chm1046/chm1046.htm

2. Reaction Rates Kinetics is the study of both the rate of a chemical reaction as well as the reaction mechanism. 4* Factors That Affect Reaction Rate *Nature of the Reactants While different types of substances will behave differently in reaction systems - this is beyond human control. Instead, there are 4 other factors that we can manipulate to speed up/slow down chemical reactions.

3. Reaction Rates A.Physical State of Reactants Reactions depend on effective collisions and contact between the reacting particles. When reactions that involve liquids and gases are concerned – homogenous mixtures will proceed much faster. Reactions that involve substances of two different states – such as solid Zn reacting with HCl (aq)… these are heterogeneous and will proceed slower. We can increase this by increasing the surface area of the solid to increase the contact time.

4. Reaction Rates B.Reaction Concentrations Increasing the concentration increases the contact and collisions between reacting particles. Concentration can be increased for gases by increasing the pressure/decreasing the volume. For solids – increasing the surface area can also be considered increasing concentration because you are making more particles available for collisions.

5. Reaction Rates C. Reaction Temperature As temperature increases, rates increases because there is both a higher frequency of collisions and the collisions have more energy. D. Catalyst Catalysts change the mechanism by altering the collisions that occur to increase the rate of reaction. Not consumed.

6. Reaction Rates The rate of reactions are measured in the changes in concentration of reactants and products as a function of time. Unit is M/s.

7. Reaction Rates Average Rate of Appearance of B =  [B] /  T Average Rate of Disappearance of A = -  [A]/  T Because the change concentration of A will be a negative number, the overall equation is negated to give us a positive value for the rate.

8. Reaction Rates Notice, this reaction is 1:1 stoichiometry wise… as the amount of A drops from 1 to 0.75, the amount of B rises from 0 to 0.25. If the coefficients were not 1:1, how would this graph would look different?

9. Reaction Rates The rate is also NOT constant over the course of the reaction. The first 10 minutes consumes 0.25 moles of A; however from minute 50 to 60, much less is consumed. This tells us that the rate of a reaction is not linear and is not constant. As time passes, the rate often slows down.

10. Reaction Rates A sample set of data:

11. Reaction Rates We might want to calculate the rate at a particular moment, this is called the instantaneous rate. The instantaneous rate is useful because it tells us what is going on at a particular instant in time. Rather than having to two separate points. Calculating the average gives us a good ballpark, but we must acknowledge that the rate would even slow down a little bit over that “Delta T” time period.

12. Reaction Rates This is done by drawing a line through that point in time and constructing a right triangle to give us data for  B and  T. The height of the triangle represents the change in concentration and the width of the base represents the time elapsed. The height to width ratio is a good estimate of our rate at that particular interval.

13. Reaction Rates

16. For example: we can calculate the instantaneous rate at 350 seconds. Because in this case we are expressing the rate in terms of the disappearance of C 4 H 9 Cl, the rate should then be negated. Rates are always expressed as positive values. The negative sign is just helpful to give us a context, it tells us that this is something that is being consumed.

17. Reaction Rates and Stoichiometry When reactions are 1:1 in respect to reactant and products, the rate of appearance of B is equal to the rate is disappearance of A. When the stoichiometry is not 1:1, the rates are not equal: 2 HI (g)  H 2 (g) + I 2 (g) The rate of disappearance of HI will be twice as fast as the rate of appearance for H 2 for example. It takes 2 HI to make 1 H 2, so HI will be used up faster.

18. Reaction Rates In this case: In order to set the reaction rates equal, we correct the rates to match the stoichiometry: we divide the rate for each by the coefficient from the balanced equation.

19. Reaction Rates For any reaction:

20. Reaction Rates a)How is the rate at which ozone disappears related to the rate at which oxygen appears in the reaction: 2 O 3 (g)  3 O 2 (g)? b)If the rate at which O 2 appears is 6.0x10 -5 M/s at a particular instant, at what rate is O 3 disappearing at the same time? Answer:

21. Reaction Rates Example: If the rate of decomposition of N 2 O 5 in the reaction: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) at a particular instant is 4.2 x 10 -7 M/s, what is the rate of appearance of NO 2 and O 2 at that instant? Ans: 8.4 x 10 -7 M/s and 2.1 x 10 -7 M/s

22. Dependence of Rate on Concentration We have seen that when a reaction starts, it begins at a much faster rate and as time passes, the reaction proceeds much slower. The concentration of reactants is also decreasing over time as they are used up. If we wanted to speed the reaction back up, we know we can add more of the reactants. So how does the starting concentration of a reactant affect the initial reaction rate?

23. Dependence of Rate on Concentration For the following reaction: NH 4 + (aq) + NO 2 - (aq) -> N 2 (g) + 2H 2 O(l) For the reactants, NH 4 +, NO 2 - and N 2, the stoichiometry is all 1:1. Evaluating the rates by looking at the rate of consumption or production of these species, they would be equal. How will changing the starting concentrations of our reactants affect our initial rate?

24. For the reaction: NH 4 + (aq) + NO 2 - (aq) -> N 2 (g) + 2H 2 O(l) If we keep the nitrite ion constant, and change the concentration of ammonium (by adding a salt that also contains ammonium by does not contain nitrite… common ion effect) [NH 4 + ] (M)[NO 2 - ] (M)Initial reaction rate (M/s) 0.010.205.4 x 10 -7 0.020.2010.8 x 10 -7 0.040.2021.5 x 10 -7 0.060.2032.3 x 10 -7 As [NH 4 + ] doubles: the rate doubles. Rate is directly proportional. Dependence of Rate on Concentration

25. In a second situation, we can keep the ammonium constant and double the nitrite concentration and observe the initial rate. [NH 4 + ] (M)[NO 2 - ] (M)Initial reaction rate (M/s) 0.200.0210.8 x 10 -7 0.200.0421.5 x 10 -7 0.200.0632.3 x 10 -7 0.200.0843.3 x 10 -7 For the most part, as we double the nitrite ion, the rate also doubles. Directly proportional

26. Dependence of Rate on Concentration What would happen if we double both the ammonium and the nitrite? The rate would increase by a factor of 4. Because the rate changes proportional to the concentrations: We can write an expression that relates the rate to concentrations: Or – by using a proportionality constant, we can write the exact equation called a rate law with k being the rate constant.

27. Dependence of Rate on Concentration In order to calculate the rate from initial concentrations, we need the rate constant. This is determined experimentally by choosing a data point: For one of the data points:

28. Dependence of Rate on Concentration k is a constant – so it will be the same for any data point. Once we know k, we can finally calculate rate from any combination of initial concentrations without having to rely on conducting the full experiment.

29. Dependence of Rate on Concentration Example: What is the initial reaction rate for the previous reaction if we combine 0.5M NH 4 + with 1.0M NO 2 - ?

30. Rate Laws One useful application for the rate constant is to compare reactions to one another, especially in terms of their speed. A large rate constant (k ~ 10 9 or higher) indicates a relatively fast reaction while 10 or lower means a slow progressing reaction.

31. Rate Laws The general form for rate laws: Rate = k [reactant 1] m [reactant 2] n …. The exponents m and n are called reaction orders. They give us an indication of how the rate is affected by the change in the concentration. Overall reaction order is the sum of the reaction orders. M and N must be determine experimentally by inspecting how changes in concentration affect the rate. Not at all related to coefficients!

32. Rate Laws Reaction Order: determined experimentally, usually small whole numbers (0,1,2) but can be negative or a fraction. Reaction orders are assigned with respect to a specific reactant: Reaction Order of Zero: changing the concentration of that reactant does not change rate. Rate Order of One: Doubling the concentration, doubles the rate – directly proportional. Rate Order of Two: Exponential increases rate. Double concentration, quadruple the rate.

33. Rate Laws Units of Rate and Rate Constants: The reaction order in the rate law will determine the units of the rate constant: For a first order reaction: k = Reaction Rate  M s -1  s -1 [M a ] 1 [M b ] 0 M For a second order reaction: k = Reaction Rate  M s -1  M -1 s -1 [M a ] 1 [M b ] 1 M x M

34. Rate Laws ANS: CHOICE 3

35. Rate Laws Answer: Choice 4

36. Rate Laws Answer: Choice 1

37. Rate Laws The following data were measured for the reaction of nitric oxide with hydrogen: 2 NO(g) + 2 H 2 (g)  N 2 (g) + 2 H 2 O(g) EXPERIMENT [NO]M[H 2 ]M INITIAL RATE M/S 10.100.101.23 X 10-3 20.100.202.46 X 10 -3 30.200.104.92 X 10 -3 a)Determine the rate law b) determine the rate constant c) calculate the rate when [NO] = 0.050 M and [H 2 ] = 0.150 M ANSWER: rate = k [NO] 2 [H 2 ], k = 1.2 M -2 s -1, rate = 4.5 x 10 -4 M/s

38. Change in Concentration over Time The rate laws have showed that the reaction rate is dependent upon concentrations as well as the constant. We can manipulate rate laws to determine what the concentration of a given reactant might be at a particular time. How the concentrations are related to time depends on whether the reaction is 1 st, 2 nd or Zero order.

39. Concentration and Time First Order Reaction : A reaction whose rate depends on a single reactant: Ex: A  B Rate = -  [A] = k [A ]  t This is called a differential rate law and exclusively shows how the rate is related to concentration. Using calculus, that rate law is transformed into something called an integrated rate law that can relate concentration to time

40. Concentration and Time The resulting integrated rate law: This equation relates the concentration at the beginning of the reaction, to the concentration at any other time. 3 possible rearrangements:

41. Concentration and Time In this equation: [A] 0 = initial [ ] of A [A] t = [ ] of A at some time “t” during the reaction

42. Concentration and Time This form of the integrated rate law is written in y=mx+b form, which is helpful because we can study the reaction by graphing: For a first order reaction: plotting of ln[A] t vs. t, it is a straight line where: - the slope is –k - y intercept of the ln[A] 0

43. Concentration and Time

44. For this reaction, as they are gases, the data we collect for “concentration” is pressure. A graph of how the pressure of the reactant CH 3 NC changes is:

45. Concentration and Time If we graph the Ln(pressure) vs time: the resulting graph:

46. Concentration and Time The straight line when Ln(P) vs time confirms it is 1 st order. The graph can also be studied to determine the slope which gives us the rate constant.

47. Concentration and Time Summary: For a 1 st order reaction: You can use these to determine: 1.The concentration of a reactant remaining at any time if you know k, A 0 and t. 2.The time required for a given fraction of a sample to react. (If you know the ratio of A t /A 0 ) 3.The time required for a reactant concentration to reach a certain level.

48. Concentration and Time Example: The decomposition of a certain insecticide in water at 12 o C follows first order kinetics with a rate constant of 1.45 yr -1. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0 x 10 -7 g/cm 3. Assuming that the average temp is 12 o C: a)What is the concentration the following year? b)How long will it take for the insecticide concentration to decrease to 3.0 x 10 -7 g/cm 3 ? Ans: a) 1.2 x 10 -7 g/cm 3. b) 0.35 year

49. Concentration and Time Second Order Reactions: can be the result of: - A rxn involving a single reactant whose rate depends on the reactant concentration raised to the 2 nd power. Rate of reaction = k[A] 2 - A rxn involving two reactants, who rate depends on the 1 st power of each reactant. Rate of reaction = k[A][B] For a reaction that is 2 nd order with respect to one reactant:

50. Concentration and Time This equation is another linear equation in the form of y = mx +b Slope = rate constant: k X values = time Y values = 1/concentration of reactant (inverse) Y intercept = 1/ initial concentration (inverse). *To distinguished between 1 st and 2 nd order: Graph both ln[A] t and 1/[A] t.* If ln[A] t vs time gives straight line: 1 st order If 1/[A] t vs time gives straight line: 2 nd order

51. Concentration and Time

52. Zero Order reactions: The rate of disappearance of A is independent of how much A you start with… the rate is independent of [A]. [A]t = -kt + [A]0 This types of reactions are not encountered often and most often involves a gas decomposing, using the surface of a metal as a catalyst. The metal surface is constant and as long as there are gas molecules continually replenished the reaction will continue constantly at a constant rate.

53. Concentration and Time Half Life: The time (known as t 1/2 ) required for the concentration of a reactant to decrease to half of its original amount. Half life is the point in time where [A] t = ½ [A] 0 The first order equation can be manipulated to get us: *Half life is independent of concentration!*

54. Concentration and Time Half Life of Second Order Reactions: t 1/2 = 1 k [A] 0 For 2 nd order reaction: the half life does depend on the initial concentration. The higher the initial concentration, the shorter the half life.

59. Rate and Temperature As temperature increases, the rate of most reactions increases. However, temperature is not part of the rate expression. This can be explained by assuming the rate constant is temperature dependent. For most reactions: k increases with increasing temperature.

60. Rate and Temperature Collision Model: Overall reaction rates can be increased by: -increasing concentration - increasing temperature The Collision Model of Kinetics and the Kinetic Molecular Theory are used to explain these observations. Collision Model states that: -molecules must physically collide in order to react -the more collisions that occur: the faster the reaction proceeds.

61. Rate and Temperature Increasing the concentration increases the rate because you are increasing the number of molecules in a particular area. This increases the sheer number of collisions. The way that temperature affects reaction rates can be explained using the kinetic molecular theory (KMT).

62. Rate and Temperature KMT: Increasing the temperature leads to the speed of the molecules increasing. As molecules move faster, they have more collisions per unit time. In addition, not only does the number of collisions increase but also they have more energy which leads to more of them being effective.

63. Rate and Temperature Regardless of how many collisions that occur, not every collision results in a reaction.. This is due to several factors that will be discussed. 1. Activation Energy: Molecules must possess a minimum amount of energy in order to react. The kinetic energy of the collision is used to break bonds and begin to transform the reactants to become products. If molecules are moving too slowly, they will just bump into each other and not be able to break the bonds necessary for the reaction.

64. Rate and Temperature Activation Energy: E a Activation energy depends on the the reaction that you are studying. Even if a reaction is energetically favorable, (  H is neg), the rate of the reaction will still depend on the magnitude of the activation energy. The higher the activation energy, the slower the reaction.

65. Rate and Temperature Overall, the rxn is favorable because the product has less energy than the reactant. Molecule A will have to overcome the activation energy barrier in order for the reaction to happen. They will have to acquire enough energy through the collision to overcome E a.

66. Rate and Temperature For the reaction to rearrange: CH 3 NC  CH 3 CN It appears as though the reaction involves swapping the placement of the triple bond. However, there may be an intermediate step where the triple bond may be in the middle of the rotation.

67. Rate and Temperature

68. In the brackets, is one way of conceptualizing the rotation. In order for this C-N group to rotate into place, the CH 3 – N must be stretched and broken – which requires energy. This input of energy will be reflected in the activation energy barrier. The intermediate molecule has more energy than the reactant. The intermediate structure – in brackets above – is a high energy intermediate called the activated complex or transition state.

69. Rate and Temperature After the CH 3 -N bond is broken, the CH 3 -C bond forms. This releases energy, which is why the energy diagram decreases. In any energy diagram, the change in energy  E has no effect upon the rate of the reaction. The rate depends on the magnitude of the E a.

70. Rate and Temperature Will B convert back to A? The reaction of B  A is energetically unfavorable. It is endothermic. Also, the activation energy for the reverse reaction is greater. It is the E a +  E. This large energy barrier will lead to the rate being much slower.

71. Rate and Temperature What fraction of molecules has enough energy to overcome the activation energy barrier and how does temperature affect this?

72. Rate and Temperature The KE distribution of molecules at two different temps:

73. Rate and Temperature Increasing the temperature increases the percentage of molecules with higher speeds and more energy. If the activation energy is the minimum needed for the reaction to proceed, at higher temperatures more molecules will have that amount of energy. The rate is proportional to the number of molecules that have the minimum energy.

74. Rate and Temperature Although collisions with enough energy occur, why do only a small fraction result in a productive reaction. In addition to E a, collisions need a correct orientation for the reaction to occur. Atoms must have the correct geometry and position. f = e -Ea/RT

75. Rate and Temperature Example: Cl + NOCl -> NO + Cl 2 The O-Cl bond is broken and a Cl-Cl bond is formed. It is essential for the incoming Cl atom to hit the molecule at the right orientation with the Cl atom on the molecule.

76. Rate and Temperature The Arrhenius Equation gives us the relationship between reaction rate and temperature. A plot of reaction rate vs. temperature:

77. Rate and Temperature The increase in reaction rate is not linear with respect to temperature. The relationship was found to be: k is the rate constant, E a is activation energy and R is the gas constant (8.314 J/mol-K) and T is the Kelvin temperature. The variable “A” is the frequency factor. It is related to the frequency of collision and the likelihood that they are effective. It is reaction specific.

78. Rate and Temperature If we take the natural log on both sides of the Arrenhius equation: We will have a linear equation if we plot Lnk vs. 1/T. The slope is equal to –E a /R.

79. Rate and Temperature Manipulating the Arrenhius Equation If we know the reaction rate at two different temperatures, we can calculate the activation energy, even without knowing “A”. We can determine the rate constant at one temperature if know the activation energy and rate constant at some other temperature.

80. Reaction Mechanisms A balanced equation gives us no indication of how a reaction proceeds – it only gives us information about the net overall reaction. The process by which a reaction occurs is called the reaction mechanism. It gives us details about the order in which bonds are broken and reformed and the formation of intermediates.

81. Reaction Mechanisms Elementary Steps: Example: The reaction of NO and O 3 to form NO 2 and O 2 This occurs as a result of a single collision or correctly oriented molecules of NO and O 3 : Because it is a single collision event, there are no hidden reactions – this is called an elementary reaction step. The number of molecules that participate as reactants in an elementary reaction step defines the molecularity of the step. This reaction is bimolecular because there are two reactant molecules.

82. Reaction Mechanisms Unimolecular: single molecule involved. CH 3 NC  CH 3 CN Bimolecular: two reactant molecules Termolecular: three different reactant molecules in one elementary reaction step. This is very rare to occur in only one step.

83. Reaction Mechanisms Most reactions involve multi step processes that are a sequence of elementary reaction steps. This reaction proceeds in two elementary reaction steps: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) Step 1: 2 NO 2 molecules collide to produce NO 3 and NO. An oxygen atom changes position (is transferred to one of the NO 2 molecules). NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) Step 2: An NO 3 molecule collides with CO to produce CO 2 and NO 2. Oxygen is transferred. NO 3 (g) + CO (g)  CO 2 + NO 2

84. Reaction Mechanisms Each of these elementary reaction steps is bimolecular. When the reaction steps are added together – the intermediates cancel out and what you are left with should match the original balanced equation. NO 2 (g) + NO 2 (g) NO 3 (g) + CO (g)  NO 3 (g) + NO (g) + CO 2 + NO 2 By cancelling out species that appear on both sides, you are left with: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) In this case, NO 3 molecule is considered an intermediate. It was produced in the first step and then immediately consumed in the 2 nd step.

85. Reaction Mechanisms Relating Rate Laws to Elementary Reaction Steps: Each reaction is made of one of more elementary reaction steps. The rate laws and reaction rates of each step will lead to the overall rate law and reaction. The rate law of any elementary step is related to its molecularity.

86. Reaction Mechanisms Summary: Molecularity Elementary reaction stepRate law UnimolecularA  product(s)Rate = k[A] Bimolecular A + A  product(s)Rate = k[A] 2 Bimolecular A + B  product(s)Rate = k[A][B] Termolecular A + A + A  product(s)Rate = k[A] 3 Termolecular A + A + B  product(s)Rate = k[A] 2 [B] Termolecular A + B + C  product(s) Rate = k[A][B][C ]

87. Reaction Mechanisms If the following reaction occurs in a single elementary reaction, predict it’s rate law: H 2 (g) + Br 2 (g)  2 HBr (g) Because it is bimolecular and occurs in a single step, we can write the rate law as: Rate = k [H 2 ] [Br 2 ] Studies show that the experimental rate law is actually Rate = k [H 2 ] [Br 2 ] 1/2

88. Reaction Mechanisms Consider the following reaction: 2 NO (g) + Br 2 (g)  2 NOBr (g) a)Write the rate law assuming it involves a single elementary reaction. b)Is a single step mechanism likely for this reaction? ANS: a) Rate = k [NO] 2 [Br 2 ] b) no…. Single step reactions between three particles (termolecular) are very rare

89. Reaction Mechanisms You cannot tell just by looking at a chemical equation what the underlying elementary steps are. For reactions that take place with multistep mechanisms (most reactions) – the different elementary reaction steps will have different rates. The slowest of the elementary reaction steps will determine the overall rate of the reaction. This is called the rate determining step.

90. Reaction Mechanisms

91. Consider the reaction: NO 2 (g) + CO (g)  NO (g) + CO 2 (g) Based on experimental results: the reaction is 2 nd order with respect to NO 2 and zero order with respect to CO. Rate = k[NO 2 ] 2 The rate is entirely dependent on the NO 2 concentration. One proposed mechanism for this process: Step 1: NO 2 (g) + NO 2 (g)  NO 3 (g) + NO (g) Step 2: NO 3 (g) + CO (g)  NO 2 (g) + CO 2 (g) Based on what the experimental results tell us regarding reaction order, which step will most likely be the rate determining step?

92. Reaction Mechanisms The proposed underlying elementary steps includes an initial step that is bimolecular and the rate law would be Rate = k 1 [NO 2 ] 2. The second step is also bimolecular with the a rate law of: Rate = k 2 [NO 3 ][CO]. The rate constants for the two steps are quite different, k 1 is very slow, while k 2 is very fast. Therefore the 1 st step is the rate limiting step. The intermediate NO 3 is slowly produced in step 1, and used up quickly in step 2. NO 3 doesn’t have a chance to build up. CO is always present in vast excess. CO is never a limiting reagent, so the rate doesn’t depend on its concentration. Since step 1 is the rate limiting step, the overall reaction rate depends upon it.

93. Reaction Mechanisms Rates with Initial Fast Step: 2NO(g) + Br 2 (g)  2NOBr(g) The experimental rate law: Rate = k[NO] 2 [Br 2 ] Our goal is to find a reaction mechanism that is consistent with the reaction orders. A termoleuclar reaction step would match the experimental rate law, but those are quite rare.

94. Reaction Mechanisms One proposed mechanism: If the initial step if fast, and the 2 nd step is slow, the 2 nd step is the rate limiting step. The rate law would be: Rate = k 2 [NOBr 2 ][NO]

95. Reaction Mechanisms Mechanisms with a fast 1 st initial step: These are more difficult because an intermediate that is produced in the first “fast” reaction is now one of our reactants in the rate determining step. This makes it difficult because rate laws are often written in terms of concentration of reactants and products from the final reaction. It is incredibly difficult to get a reliable value for the concentration of the intermediate because as soon as it is produced, it is used up.

96. Reaction Mechanisms The problem with that rate law is the intermediate are unstable and have either very low or unstable concentrations. We need to find a way of expressing the rate law without using the concentration of the intermediate. In other words, expressing the [NOBr 2 ] in terms of NO and Br 2.

97. Reaction Mechanisms NOBr 2 is an intermediate and made quickly in the 1 st step. Because the 2 nd step is slow, the NOBr 2 will build up. We make some assumptions to come up with an equation to measure the concentration of this intermediate: As NOBr 2 builds up – it can do one of two things: a) collide with NO to produce NOBr (slow second step) b) it can decompose back to NO and Br 2.

98. Reaction Mechanisms Because choice b leads to the decomposition of NOBr 2 back to its components: we can rewrite step one as: Since Step 2 (the original rate limiting step) is slow, we can assume that enough NOBr 2 builds up to allow the rate of production and rate of decomposition to come to an equilibrium.

99. Reaction Mechanisms For this reaction: Rate of formation = Rate of decomposition k 1 [NO][Br2] = k -1 [NOBr 2 ] Using this we can solve for the concentration of NOBr 2 (so we can use it in our experimental rate law) [NOBr 2 ] = (k 1 /k -1 )[NO][Br 2 ]

100. Reaction Mechanisms Finally, we can substitute the concentration of NOBr 2 into our original rate law: Rate = k 2 [NOBr 2 ][NO] Rate = k 2 (k 1 /k -1 )[NO][Br 2 ][NO] Rate = k 2 (k 1 /k -1 )[NO] 2 [Br 2 ]

101. Reaction Mechanisms Rule: When we have a fast initial step, followed by a slow second step… we can determine the concentration of the intermediate by assuming that it achieves an equilibrium concentration.

102. Reaction Mechanisms Example: The decomposition of nitrous oxide, N2O, is believed to occur by a two step mechanism: N 2 O (g)  N 2 (g) + O (g) (SLOW) N 2 O (g) + O (g)  N 2 (g) + O 2 (g) (FAST) a)Write the equation for the overall reaction. b)Write the rate law for the overall reaction. a) To get the equation for the overall reaction, you need to list all the reactants on one side and the products on the other. Any species that appears twice, can be cancelled out. Sum of each reaction: N 2 O + N 2 O + O  N 2 + N 2 + O + O 2 Overall reaction: 2 N 2 O  2 N 2 + O 2 b) For the rate law: it is bimolecular with two nitrous oxide molecules colliding: Rate = k [N 2 O] 2

103. Reaction Mechanisms Ozone reacts with nitrogen dioxide to produce nitrogen pentoxide and oxygen: O 3 (g) + 2 NO 2 (g)  N 2 O 5 (g) + O 2 (g) Believed to occur in two steps: 1. O 3 (g) + NO 2 (g)  NO 3 (g) + O 2 (g) 2. NO 3 (g) + NO 2 (g)  N 2 O 5 (g) The experimental rate law is rate = k [O 3 ][NO 2 ]. What can be inferred about the relative rates of the two steps? ANS: The experimental rate law matches what would be written for the first step, therefore step 1 must be the rate determining step and the slower reaction.

104. Catalysis Catalyst: Substance the changes the rate of a chemical reaction without itself undergoing any chemical change in the process. Even reactions that are energetically favorable can still have slow rates if the energy barrier (E a ) is too high. Two Types: Homogenous and Heterogeneous Catalysts

105. Catalysis Homogenous Catalyst: catalyst that is present in the same phase as the reactants. The rate constant of a reaction is calculate using the Arrhenius equation: A catalyst can increase the rate by either increasing the frequency factor (A) or decrease the activation energy. Catalysts most often work by decreasing the activation energy. They facilitate the process of stretching and breaking bonds in way that requires less energy. They provide a completely different reaction mechanism.

106. Catalysis Heterogeneous Catalysts Exists in a different phase than the reactant: often a solid catalyst contacting gaseous reactants/products or the reactants in a liquid. (Catalytic Converter). The greater the surface area, the more reaction take place. Catalysts do their jobs in steps: 1.Adsorption: bind reactant molecules to catalyst surface. This occurs on the active sites of the catalyst surface. 2. Absorption: move molecules into the interior of the material.

107. Catalysis Example: C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g)  H = 137 kJ/mol Addition Reaction (Hydrogenation Reaction) – unsaturated  saturated hydrocarbon. Exothermic but very slow in the absence of a catalyst. When a metal catalyst is added – very quickly. 1.The reactants are adsorbed onto the surface of the metal catalyst. 2. Break the H 2 (g) bond and the H atoms move towards the C 2 H 4 (g). 3.When a H collides with the ethylene, it breaks the C-C  bond and forms a C-H  bond. 4.C 2 H 5 is bonded to the metal surface with a weak metal – carbon bond. Once the other H collides with the ethyl group, the C 2 H 6 molecule desorbs from the surface.

109. Catalysis Enzymes are biological catalysts – mostly large protein molecules. Incredibly efficient. They are compared by their turnover number – the number of catalytic events per unit time. The higher the turnover number, the lower the activation energy. The reaction takes place at the active site – a very specific location within the enzyme molecule. The substances that undergo these reactions are called substrates. One model to explain how enzymes catalyze reactions: lock and key model. The substrate binds with the active site and forms an enzyme-substrate complex. The shape of the active site only permits certain substrates to bind. Once the binding occurs – a reaction immediately follows. The products will get kicked out of the active site, freeing up the enzyme to catalyze another reaction.

110. Catalysis In some cases, molecules are able to bind to the active site in a way that substrate molecules that come along cannot displace them. These molecules are called inhibitors and block enzyme activity. This is how certain toxic substances and poison work. They either bind to the active site and block, or bind to another site on the enzyme and change the shape of the enzyme. This prevents future substrate molecules from being able to bind.