1. 16-1 KINETICS: RATES AND MECHANISMS OF CHEMICAL REACTIONS

2. 16-2 Chemical Kinetics – the study of the rate of reactions, the factors that affect reaction rates, and the mechanisms by which the reactions occur. Rates of reaction describes how fast reactants are consumed and products are formed.

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4. 16-4 Under a specific set of conditions, every reaction has its own characteristic rate, which depends upon the chemical nature of the reactants. Four factors can be controlled during the reaction: 1.Concentration - molecules must collide to react 2.Physical state - molecules must mix to collide 3.Temperature - molecules must collide with enough energy to react 4.The use of a catalyst Factors That Influence Reaction Rate

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8. 16-8 Collision energy and reaction rate.

9. 16-9 Expressing the Reaction Rate Reaction rate - changes in the concentrations of reactants or products per unit of time. Reactant concentrations decrease while product concentrations increase. Rate of reaction = - General reaction, A B change in concentration of A change in time = - conc A 2 - conc A 1 t 2 - t 1  (conc A) - tt

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11. 16-11 The concentration of O 3 vs. time during its reaction with C 2 H 4. -  [C 2 H 4 ] tt rate = -  [O 3 ] tt =

12. 16-12 Plots of [C 2 H 4 ] and [O 2 ] vs. time.

13. 16-13 In general, for the reaction, aA + bB cC + dD Rate = 1 a -= -  [A] tt 1 b  [B] tt 1 c  [C] tt = + 1 d  [D] tt = + The numerical value of the rate depends upon the substance that serves as the reference. The changes in concentration of the other reaction components are relative to their coefficients in the balanced chemical equation.

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17. 16-17 Each reaction has its own equation that gives its rate as a function of reactant concentrations.  this is called its Rate Law To determine the rate law we measure the rate at different starting concentrations. CONCENTRATION AND THE RATE LAW EXPRESSION

18. 16-18 The rate law expression for a particular reaction in which A, B are reactants has the general form: rate = k[A] x [B] y k = rate constant x and y = order of the reactants H 2 O 2 (aq) + 3I - (aq) + 2H + (aq)  2H 2 O (l) + I 3 - (aq) rate = k[H 2 O 2 ][I - ] CONCENTRATION AND THE RATE LAW EXPRESSION

19. 16-19 Determining Reaction Order from Rate Laws For each of the following reactions, use the given rate law to determine the reaction order with respect to each reactant and the overall order: (a) 2NO( g ) + O 2 ( g ) 2NO 2 ( g ); rate = k[NO] 2 [O 2 ] (b) CH 3 CHO( g ) CH 4 ( g ) + CO( g ); rate = k[CH 3 CHO] 3/2 (c) H 2 O 2 ( aq ) + 3I - ( aq ) + 2H + ( aq ) I 3 - ( aq ) + 2H 2 O( l ); rate = k[H 2 O 2 ][I - ]

20. 16-20 The Specific Rate Constant, k 1.Experimental data are used to determine the value of k 2.The value is for a specific reaction represented by a balanced equation 3.The units of k depend on the overall order of the reaction 4.The value does not change with concentrations of either reactants or products. It does not change with time. 5.The value was determined at a specific temperature and changes with temperature. 6.The value depends on whether a catalyst is present

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30. 16-30 Determining Reaction Orders Using initial rates - Run a series of experiments, each of which starts with a different set of reactant concentrations, and from each obtain an initial rate. See Table 16.2 for data on the reaction O 2 ( g ) + 2NO( g ) 2NO 2 ( g )rate = k [O 2 ] m [NO] n Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant. k [O 2 ] 2 m [NO] 2 n k [O 2 ] 1 m [NO] 1 n = rate 2 rate 1 = [O 2 ] 2 m [O 2 ] 1 m = 6.40 x 10 -3 mol/L s 3.21 x 10 -3 mol/L s [O 2 ] 2 [O 2 ] 1 m = 1.10 x 10 -2 mol/L 2.20 x 10 -2 mol/L m ;2 = 2 m m = 1 Do similar calculations for the other reactant(s).

31. 16-31 Determining Reaction Orders from Initial Rate Data PROBLEM:Many gaseous reactions occur in a car engine and exhaust systems. One of these is rate = k[NO 2 ] m [CO] n Use the following data to determine the individual and overall reaction orders: Experiment Initial Rate (mol/L s) Initial [NO 2 ] (mol/L)Initial [CO] (mol/L) 1 2 3 0.0050 0.080 0.0050 0.10 0.40 0.10 0.20 Solve for the order with respect to each reactant using the general rate law using the method described previously. rate = k [NO 2 ] m [CO] n First, choose two experiments in which [CO] remains constant and the [NO 2 ] varies.

32. 16-32 Determining Reaction Order from Initial Rate Data 0.080 0.0050 rate 2 rate 1 [NO 2 ] 2 [NO 2 ] 1 m = k [NO 2 ] m 2 [CO] n 2 k [NO 2 ] m 1 [CO] n 1 = 0.40 0.10 = m 16 = 4.0 m and m = 2.0 k [NO 2 ] 2 3 [CO] n 3 k [NO 2 ] 2 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = 0.0050 = 0.20 0.10 n 1 = 2.0 n and n = 0 The reaction is 2nd order in NO 2. The reaction is zero order in CO. rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2 The reaction is second order overall.

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39. 16-39 Integrated Rate Laws rate = -  [A] tt = k [A]rate = -  [A] tt = k [A] 0 rate = -  [A] tt = k [A] 2 First-order rate equation Second-order rate equation Zero-order rate equation ln [A] 0 [A] t = - ktln [A] 0 - ln [A] t = kt 1 [A] t 1 [A] 0 - = kt 1 [A] t 1 [A] 0 += kt [A] t - [A] 0 = - kt

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41. 16-41 PLAN: SOLUTION: Determining the Reactant Concentration at a Given Time PROBLEM:At 1000 o C, cyclobutane (C 4 H 8 ) decomposes in a first-order reaction, with the very high rate constant of 87 s -1, to two molecules of ethylene (C 2 H 4 ). (a) If the initial C 4 H 8 concentration is 2.00 M, what is the concentration after 0.010 s? (b) What fraction of C 4 H 8 has decomposed in this time? Find the [C 4 H 8 ] at time, t, using the integrated rate law for a 1 st order reaction. Once that value is found, divide the amount decomposed by the initial concentration. ; ln 2.00 [C 4 H 8 ] = (87 s -1 )(0.010 s) [C 4 H 8 ] = 0.83 mol/L ln [C 4 H 8 ] 0 [C 4 H 8 ] t = kt (a) (b)[C 4 H 8 ] 0 - [C 4 H 8 ] t [C 4 H 8 ] 0 = 2.00 M - 0.83 M 2.00 M = 0.58

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53. 16-53 Integrated rate laws and reaction orders. ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0

54. 16-54 Graphical determination of the reaction order for the decomposition of N 2 O 5.

55. 16-55 A plot of [N 2 O 5 ] vs. time for three half-lives. t 1/2 = for a first-order process ln 2 k 0.693 k =

56. 16-56 PLAN: SOLUTION: Determining the Half-Life of a First-Order Reaction Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 o bond angles reduce orbital overlap, its bonds are weak. As a result, it is thermally unstable and rearranges to propene at 1000 o C via the following first-order reaction: The rate constant is 9.2 s -1, (a) What is the half-life of the reaction? (b) How long does it take for the concentration of cyclopropane to reach one-quarter of the initial value? Use the half-life equation, t 1/2 = 0.693 k, to find the half-life. One-quarter of the initial value means two half-lives have passed. 2 t 1/2 = 2(0.075 s) = 0.15 s(b) t 1/2 = = 0.075 s(a) 0.693 9.2 s -1

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65. 16-65 Dependence of the rate constant on temperature.

66. 16-66 The Arrhenius Equation The Effect of Temperature on Reaction Rate ln k = ln A - E a /RT ln k2k2 k1k1 = EaEa R - 1 T2T2 1 T1T1 - where k is the kinetic rate constant at T E a is the activation energy R is the universal gas constant T is the Kelvin temperature A is the collision frequency factor

67. 16-67 Graphical determination of the activation energy. ln k = -E a /R (1/T) + ln A

68. 16-68 PLAN: SOLUTION: Determining the Energy of Activation PROBLEM:The decomposition of hydrogen iodide, 2HI( g ) H 2 ( g ) + I 2 ( g ) has rate constants of 9.51 x 10 -9 L/mol s at 500. K and 1.10 x 10 -5 L/mol s at 600. K. Find E a. Use the modification of the Arrhenius equation to find E a. ln k2k2 k1k1 = EaEa – R 1 T2T2 1 T1T1 – E a = – R (ln ) k2k2 k1k1 1 T2T2 1 T1T1 – 1 600 K 1 500 K – ln 1.10x10 -5 L/mol∙s 9.51x10 -9 L/mol∙s E a = – (8.314 J/mol∙K) E a = 1.76 x 10 5 J/mol = 176 kJ/mol

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77. 16-77 Information sequence to determine the kinetic parameters of a reaction.

78. 16-78 The dependence of number of possible collisions on the product of reactant concentrations.

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80. 16-80 The effect of temperature on the distribution of collision energies.

81. 16-81 Energy-level diagram for a reaction. The forward reaction is exothermic because the reactants have more energy than the products.

82. 16-82 The importance of molecular orientation to an effective collision. NO + NO 3 2 NO 2 A is the frequency factor A = pZ where Z is the collision frequency p is the orientation probability factor

83. 16-83 Nature of the transition state in the reaction between CH 3 Br and OH -. CH 3 Br + OH - CH 3 OH + Br - transition state or activated complex

84. 16-84 Figure 16.16 Reaction energy diagram for the reaction of CH 3 Br and OH -.

85. 16-85 Figure 16.17 Reaction energy diagrams and possible transition states for two reactions.

86. 16-86 SOLUTION: Drawing Reaction Energy Diagrams and Transition States PROBLEM:A key reaction in the upper atmosphere is O 3 ( g ) + O( g ) 2O 2 ( g ) The E a(fwd) is 19 kJ, and the  H rxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate E a(rev). PLAN:Consider the relationships among the reactants, products, and transition state. The reactants are at a higher energy level than the products and the transition state is slightly higher than the reactants.

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92. 16-92 REACTION MECHANISMS

93. 16-93 PLAN: SOLUTION: Determining Molecularity and Rate Laws for Elementary Steps PROBLEM:The following two reactions are proposed as elementary steps in the mechanism of an overall reaction: (1)NO 2 Cl( g )NO 2 ( g ) + Cl ( g ) (2)NO 2 Cl( g ) + Cl ( g )NO 2 ( g ) + Cl 2 ( g ) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (a) The overall equation is the sum of the steps. (b) The molecularity is the sum of the reactant particles in the step. 2NO 2 Cl( g )2NO 2 ( g ) + Cl 2 ( g ) (c) Write the rate law for each step. Rate 2 = k 2 [NO 2 Cl][Cl] (1)NO 2 Cl( g )NO 2 ( g ) + Cl ( g ) (2)NO 2 Cl( g ) + Cl ( g )NO 2 ( g ) + Cl 2 ( g ) (a) Step (1) is unimolecular. Step (2) is bimolecular. (b) Rate 1 = k 1 [NO 2 Cl](c)

94. 16-94 The Rate-Determining Step of a Reaction Mechanism The overall rate of a reaction is related to the rate of the slowest step, which is the rate-determining step. Correlating the Mechanism with the Rate Law The elementary steps must add up to the overall balanced equation. The elementary steps must be physically reasonable. The mechanism must correlate with the rate law.

95. 16-95 Reaction energy diagram for the two-step reaction of NO 2 and F 2.

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102. 16- 102 CATALYSTS  Each catalyst has its own specific way of functioning.  In general, a catalyst lowers the energy of activation.  Lowering the E a increases the rate constant, k, and thereby increases the rate of the reaction. A catalyst increases the rate of the forward and reverse reactions. A catalyzed reaction yields the products more quickly, but does not yield more product than the uncatalyzed reaction. A catalyst lowers E a by providing a different mechanism, for the reaction through a new, lower energy pathway.

103. 16- 103 Reaction energy diagram for a catalyzed and an uncatalyzed process.

104. 16- 104 Mechanism for the catalyzed hydrolysis of an organic ester.

105. 16- 105 The metal-catalyzed hydrogenation of ethylene.

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