16-1 Chapter 16 Kinetics: Rates and Mechanisms of Chemical Reactions

16-2 Kinetics: Rates and Mechanisms of Chemical Reactions 16.1 Factors that influence reaction rates 16.2 Expressing the reaction rate 16.3 The rate law and its components 16.4 Integrated rate laws: Concentration changes over time 16.5 Reaction mechanisms: Steps in the overall reaction 16.6 Catalysis: Speeding up a chemical reaction

16-3 Chemical Kinetics The study of reaction rates, the changes in concentrations of reactants (or products) as a function of time Quantitative relationships through the derivation of a rate law Kinetics can reveal much about the mechanism of a reaction.

16-4 Figure 16.1 Reaction rate: the central focus of chemical kinetics Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-5 Figure 16.2 The wide range of reaction rates

16-6 Factors Influencing Reaction Rates Reactant Concentration: molecular collisions are required for reactions to occur reaction rate  collision frequency  concentration Physical State: molecules must mix to collide When reactants are in different phases, the more finely divided a solid or liquid reactant, the greater the surface area per unit volume, the more contact it makes with other reactants, and the faster the reaction. Temperature: molecules must collide with sufficient energy to react Higher T translates into more collisions per unit time and into higher-energy collisions reaction rate  collision energy  temperature

16-7 Figure 16.3 The effect of surface area on reaction rate Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. steel nail in O 2 steel wool in O 2

16-8 Figure 16.4 Collision energy and reaction rate Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Other factor: reaction trajectory (not all collisions with sufficient energy are productive)

16-9 Expressing Reaction Rate reaction rate = the changes in [reactants] or [products] per unit time; [reactant] decreases, [product] increases Average rate of reaction = - ([A 2 ] - [A 1 ]) / (t 2 - t 1 ) A B Measure [A 1 ] at t 1, then measure [A 2 ] at t 2 = -  [A] /  t rate units = mol/L/s Also, reaction rate = +  [B] /  t rate is always expressed as a positive value

16-10 C 2 H 4 ( g ) + O 3 ( g ) C 2 H 4 O( g ) + O 2 ( g ) time (s) concentration of O 3 (mol/L) x x x x x x x Table 16.1 Concentration of O 3 at various times in its reaction with C 2 H 4 at 303 K Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-11 Different Ways to Measure Reaction Rates Reaction rate varies with time as the reaction proceeds! C 2 H 4 (g) + O 3 (g) C 2 H 4 O(g) + O 2 (g) average rate = -  [O 3 ] /  t = - (1.10 x mol/L) - (3.20 x mol/L) 60.0 s s Similar calculations over earlier and later time intervals (e.g., s, s) reveals average rates of 7.80 x mol/L/s and 1.30 x mol/L/s, respectively. = 3.50 x mol/L/s Data show that the average rate decreases as the reaction proceeds!

16-12 Figure 16.5 Concentrations of O 3 vs time during its reaction with C 2 H 4 Curved line indicates change in reaction rate with time. The slope of a tangent at any point on the curve yields the instantaneous rate at that point. reaction rate = (instantaneous) reaction rate initial rates: measured to avoid complications due to back reactions

16-13 Figure 16.6 Plots of [C 2 H 4 ] and [O 2 ] vs reaction time Other Plots to Determine Reaction Rate C 2 H 4 (g) + O 3 (g) C 2 H 4 O(g) + O 2 (g)

16-14 Expressing rate in terms of changes in [reactant] and [product] H 2 (g) + I 2 (g) 2HI(g) rate = -  [H 2 ]/  t = -  [I 2 ]/  t = 1/2  [HI]/  t rate =  [HI]/  t = -2  [H 2 ]/  t = -2  [I 2 ]/  t or The mathematical expression for the rate, and the numerical value of the rate, depend on which substance is chosen as the reference. For the general reaction: aA + bB cC + dD rate = -1/a  [A]/  t = -1/b  [B]/  t = 1/c  [C]/  t = 1/d  [D]/  t

16-15 Sample Problem 16.1 PLAN: SOLUTION: Expressing rate in terms of changes in concentration with time PROBLEM:Because it generates a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by automobile engines in the near future. 2H 2 ( g ) + O 2 ( g ) 2H 2 O( g ) (a) Express the rate in terms of changes in [H 2 ], [O 2 ], and [H 2 O] with time. (b) If [O 2 ] decreases at 0.23 mol/L/s, at what rate is [H 2 O] increasing? Choose [O 2 ] as the reference since its coefficient is 1. For every molecule of O 2 that disappears, two molecules of H 2 disappear and two molecules of H 2 O appear, so [O 2 ] decreases at one-half the rates of change in [H 2 ] and [H 2 O]  [H 2 ] tt = -  [O 2 ] tt = +  [H 2 O] tt mol/L. s = +  [H 2 O] tt 1 2 ; = 0.46 mol/L. s  [H 2 O] tt rate = (a)  [O 2 ] tt -=(b)

16-16 The Rate Law The rate law expresses the reaction rate as a function of reactant concentrations, product concentrations, and temperature. It is experimentally determined. The derived rate law for a reaction must be consistent with the postulated chemical mechanism of the reaction! For a general reaction: aA + bB cC + dD rate law: rate = k[A] m [B] n k = rate constant m and n = reaction orders (not related to a, b,...) (for a unidirectional (one-way) reaction)!

16-17 How is the rate law determined experimentally? 1. Measure initial rates (from concentration measurements) 2. Use initial rates from several experiments to find the reaction orders 3. Calculate the rate constant The Procedure

16-18 Reaction Order Terminology rate = k[A] first order overall (rate  [A]) rate = k[A] 2 second order overall (rate  [A] 2 ) rate = k[A] 0 = k(1) = k zero order (rate independent of [A]) Examples NO(g) + O 3 (g) NO 2 (g) + O 2 (g) rate = k[NO][O 3 ] second order overall 2NO(g) + 2H 2 (g) N 2 (g) + 2H 2 O(g) rate = k[NO] 2 [H 2 ] 1 third order overall

16-19 Reaction orders are usually positive integers or zero. Reaction orders can be fractional or negative. Reaction orders cannot be deduced from the balanced chemical equation. Properties of Reaction Orders

16-20 Sample Problem 16.2 SOLUTION: Determining reaction order from rate laws PROBLEM:For each of the following reactions, determine the reaction order with respect to each reactant and the overall reaction order from the given rate law. (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 - ] PLAN:Inspect the rate law and not the coefficients in the balanced chemical reaction. (a) The reaction is 2nd order in NO, 1st order in O 2, and 3rd order overall. (b) The reaction is 3/2-order in CH 3 CHO and 3/2-order overall. (c) The reaction is 1st order in H 2 O 2, 1st order in I - and zero order in H +, and 2nd order overall.

16-21 Determining Reaction Orders O 2 (g) + 2NO(g) 2NO 2 rate = k[O 2 ] m [NO] n To find m and n: A series of experiments is conducted starting with different sets of reactant concentrations. The initial reaction rate is measured in each experiment.

16-22 experiment initial reactant concentrations (mol/L) initial rate (mol/L. s) O2O2 NO 1.10 x x x x x x x x x x x x x x x Initial rates in a series of experiments for the reaction between O 2 and NO Table 16.2 Initial rates in a series of experiments for the reaction between O 2 and NO Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-23 Manipulating the Data rate 2 k[O 2 ] 2 m [NO] 2 n rate 1 k[O 2 ] 1 m [NO] 1 n rate 2 [O 2 ] 2 m rate 1 [O 2 ] 1 m = == ([O 2 ] 2 / [O 1 ] 1 ) m 6.40 x mol/L/s 3.21 x mol/L/s (2.20 x mol/L/ 1.10 x mol/L) m = 1.99 = (2.00) m m = 1 Thus, the reaction is first order with respect to O 2. When [O 2 ] doubles, the rate doubles. ([NO] is held constant)

16-24 rate 3 k[O 2 ] 3 m [NO] 3 n rate 1 k[O 2 ] 1 m [NO] 1 n = ([O 2 ] is held constant) via same manipulations as before: 3.99 = (2.00) n n = 2 Thus, the reaction is second order with respect to NO. When [NO] doubles, the rate quadruples. The rate law is: rate = k[O 2 ][NO] 2 The reaction is third order overall.

16-25 Sample Problem 16.3 PLAN: SOLUTION: Determining reaction order from initial rate data PROBLEM:Many gaseous reactions occur in a car engine and exhaust system. One such reaction is as follows: 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) Solve for each reactant using the general rate law by applying the method described previously. rate = k [NO 2 ] m [CO] n First, choose two experiments in which [CO] remains constant and [NO 2 ] varies.

16-26 Sample Problem 16.3(continued) 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 = = m ;16 = 4 m and m = 2 k [NO 2 ] m 3 [CO] n 3 k [NO 2 ] m 1 [CO] n 1 [CO] 3 [CO] 1 n = rate 3 rate 1 = = n ;1 = 2 n and n = 0 The reaction is 2nd order in NO 2. The reaction is zero order in CO. Rate Law: rate = k [NO 2 ] 2 [CO] 0 = k [NO 2 ] 2 The reaction is second order overall.

16-27 Determining the Rate Constant The rate constant, k, is specific for a particular reaction at a particular temperature. k = rate 1 / ([O 2 ] 1 [NO] 1 2 ) 3.21 x mol/L/s 1.86 x mol 3 /L 3 =1.73 x 10 3 L 2 /mol 2 /s = k The units for k depend on the order of the reaction and the time unit.

16-28 overall reaction orderunits of k (t in seconds) 0 mol/L. s (or mol L -1 s -1 ) 11/s (or s -1 ) 2 L/mol. s (or L mol -1 s -1 ) 3 L 2 / mol 2. s (or L 2 mol -2 s -1 ) Table 16.3 Units of the rate constant, k, for several overall reaction orders Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-29 Integrated Rate Laws: Change in Concentration with Reaction Time Consider a general first order reaction: A B rate = -  [A]/  t = k[A] Upon integration over time, we obtain: ln ([A] 0 /[A] t ) = kt ln = natural logarithm [A] 0 = concentration of A at t = 0 [A] t = concentration of A at any time t ln [A] 0 - ln [A] t = kt

16-30 For a simple second order reaction (one reactant only): rate = -  [A]/  t = k[A] 2 Upon integration over time, we obtain: 1/[A] t - 1/[A] 0 = kt For a zero order reaction: rate = -  [A]/  t = k[A] 0 Upon integration over time, we obtain: [A] t - [A] 0 = -kt

16-31 Sample Problem 16.4 PLAN: SOLUTION: Determining reaction 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 yield two molecules of ethylene (C 2 H 4 ). (a) If the initial [C 4 H 8 ] is 2.00 M, what is the concentration after s of reaction? (b) What fraction of C 4 H 8 has decomposed in this time? Find [C 4 H 8 ] at time t using the integrated rate law for a 1st-order reaction. Once that value is found, divide the amount decomposed by the initial concentration. ; ln 2.00 M [C 4 H 8 ] t = (87 s -1 )(0.010 s)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 M 2.00 M = 0.58 = M / [C 4 H 8 ] t = e 0.87 = 2.4 [C 4 H 8 ] t = 0.83 M

16-32 Reaction Order from the Integrated Rate Law Graphical methods to determine reaction orders (when you don’t have initial rate data!) ln [A] t = -kt + ln [A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0 first order reaction simple second order reaction zero order reaction All take the form: y = mx + b Rearranged integrated rate laws:

16-33 Figure 16.7 Graphical method to determine reaction order ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t = -kt + [A] 0

16-34 Figure 16.8 Graphical determination of the reaction order for the decomposition of N 2 O 5 nonlinear linear! nonlinear

16-35 For the decomposition of N 2 O the reaction must be first order in N 2 O 5. Conclusion Graphical approach can likewise be applied to reactions involving multiple reactants.

16-36 Reaction Half-Life half-life (t 1/2 ) = the time required for the reactant concentration to reach one-half of its initial value For a first order reaction at fixed conditions: t 1/2 is a constant, independent of reactant concentration Just another way to express the speed of a reaction

16-37 Figure 16.9 A plot of [N 2 O 5 ] vs time for three half-lives A first order reaction

16-38 Why is t 1/2 independent of reactant concentration for a first order reaction? ln ([A] 0 /[A] t ) = kt After one half-life, t = t 1/2 and [A] t = 0.5[A] 0. Substituting..... ln ([A] 0 /0.5([A] 0 ) = kt 1/2 or ln 2 = kt 1/2 t 1/2 = (ln 2)/k = 0.693/k first order reaction: A B

16-39 Sample Problem 16.5 PLAN: SOLUTION: Determining the half-life of a first-order reaction PROBLEM:Cyclopropane is the smallest cyclic hydrocarbon. Because its 60 o bond angles allow poor 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: CH 2 H 2 CCH 2 ( g )  H 3 CCHCH 2 ( g ) The rate constant is 9.2 s - 1. (a) What is the half-life of the reaction? (b) How long does it take for [cyclopropane] to reach one-quarter of its initial value? Use the half-life equation, t 1/2 = k, to find the half-life. One-quarter of the initial value means two half-lives have passed. t 1/2 = / 9.2 s - 1 = s(a)2 t 1/2 = 2 (0.075 s) = 0.15 s(b)

16-40 t 1/2 = [A] 0 /2k t 1/2 = 1/k [A] 0 Half-Life for Zero and Second Order Reactions simple second order (rate = k[A] 2 ) zero order (rate = k) As a second order reaction proceeds, the half-life increases. For zero order reactions, higher initial reactant concentrations translate into longer half-lives.

16-41 zero orderfirst ordersecond order plot for straight line slope, y-intercept half-life rate law rate = k rate = k[A] rate = k[A] 2 units for k mol/L. s 1/s L/mol. s integrated rate law in straight-line form [A] t = kt + [A] 0 ln[A] t = -kt + ln[A] 0 1/[A] t = kt + 1/[A] 0 [A] t vs t ln[A] t vs t1/[A] t = t -k, [A] 0 -k, ln[A ] 0 k, 1/[A] 0 [A] 0 /2k(ln 2)/k 1/k[A] 0 Table 16.4 An overview of zero-order, first-order and simple second-order reactions Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-42 Temperature and Reaction Rate rate = k[A] Consider the rate law for a first order reaction: Where is the temperature dependence? Answer: It is embodied in the rate constant, k, that is, k depends on the temperature at which the reaction is conducted. What does the T-dependence of k look like? R-COOR’ + H 2 O R-COOH + R’OH esteracidalcohol organic ester hydrolysis test reaction

16-43 Figure Dependence of k on temperature for the hydrolysis of an organic ester k increases exponentially! Note that both reactant concentrations are held constant

16-44 The Arrhenius equation Quantitative treatment of the effect of temperature on k ln k = ln A - E a /R (1/T) ln k2k2 k1k1 = EaEa R - 1 T2T2 1 T1T1 - k = the kinetic rate constant at T E a = the activation energy R = the universal gas constant T = Kelvin temperature A = collision frequency factor k = Ae - E a /RT e = base of natural logarithms y = b + mx

16-45 Figure ln k = -E a /R (1/T) + ln A Graphical determination of the activation energy, E a Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. An Arrhenius plot

16-46 More about the Arrhenius Equation The activation energy = the minimum energy that the molecules must possess in order for reaction to occur The negative exponent suggests that, as T increases, the negative exponent becomes smaller (less negative), the value of k increases, and thus the reaction rate increases. higher T larger k increased rate

16-47 Sample Problem 16.6 PLAN: SOLUTION: Determining the energy of activation PROBLEM:The decomposition reaction of hydrogen iodide, 2HI( g ) H 2 ( g ) + I 2 ( g ) has rate constants of 9.51 x L/mol. s at 500. K and 1.10 x L/mol. s at 600. K. Find E a. Use a modification of the Arrhenius equation to find E a. ln k2k2 k1k1 = EaEa - R 1 T2T2 1 T1T1 - E a = - Rln k2k2 k1k1 1 T2T2 1 T1T1 - ln 1.10 x L/mol. s 9.51 x L/mol. s K K - E a = - (8.314 J/mol. K) E a = 1.76 x 10 5 J/mol = 176 kJ/mol

16-48 Figure series of plots of [ ] vs time initial rates reaction orders rate constant (k) and actual rate law integrated rate law (half-life, t 1/2 ) rate constant and reaction order activation energy, E a Find k at varied T Determine slope of tangent at t 0 of each plot Compare initial rates when [A] changes and [B] is held constant and vice versa Substitute initial rates, orders, and concentrations into general rate law: rate = k [A] m [B] n Rearrange to linear form and graph Find k at varied T Information Sequence to Determine the Kinetic Parameters of a Reaction A + B C + D

16-49 Using Collision Theory to Explain the Effects of [ ] and T on Reaction Rate Model: A + B products Why are [A] and [B] multiplied in the rate law? Consider several cases where there are a finite number of particles each of A and B and determine the number of possible A-B collisions.

16-50 Figure A A A A B B B B AA AA BB BB AA 4 collisions Add another molecule of A 6 collisions Add another molecule of B AA AA BB BB AABB The dependence of possible collisions on the product of reactant concentrations 9 collisions 2 x 2 = 4 3 x 2 = 6 3 x 3 = 9

16-51 Temperature and Reaction Rate increased T increased average speed of particles increased collision frequency increased reaction rate But, most collisions fail to yield products! Significance of activation energy: only those collisions with energy equal to, or greater than, E a can yield products. Increasing T enhances the fraction of productive collisions, f. f = e -E a /RT From this equation, we can see that both E a and T affect f, which in turn influences reaction rate.

16-52 E a (kJ/mol) f (at 298 K) x x x T f (E a = 50 kJ/mol) 25 o C (298 K) 1.70 x o C (308 K) 3.29 x o C (318 K) 6.12 x Effect of E a and T on the fraction (f) of collisions with sufficient energy to allow reaction Table 16.5 Effect of E a and T on the fraction (f) of collisions with sufficient energy to allow reaction Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.  T = 10 o ; f ~ doubles; reaction rate ~ doubles!

16-53 Figure The effect of temperature on the distribution of collision energies

16-54 Figure reactants products ACTIVATED STATE Collision Energy E a (forward) E a (reverse) The forward reaction is exothermic because the reactants have more energy than the products. Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. A + B C + D A + B C + D Energy-Level Diagram for an Equilibrium Reaction

16-55 Figure An energy-level diagram of the fraction of collisions exceeding E a A + B C + D

16-56 Effective Collisions Not all collisions that occur with energy equal to, or exceeding, the activation energy lead to products. Molecular orientation is critical! k = Ae - E a /RT the frequency factor = product of collision frequency Z and an orientation probability factor p (A = Zp)

16-57 Figure NO + NO 3 2 NO 2 Molecular orientation and effective collisions Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-58 Transition State Theory Addresses the limitations of collision theory in explaining chemical reactivity Key principle: During the transformation of reactant into product, one or more very short-lived chemical species form that resemble (but are different from) reactant or product; these transitional species contain partial bonds; they are called transition states (TS) or activated complexes. The activation energy is used to stretch/deform specific bonds in the reactant(s) in order to reach the transition state. Example reaction: CH 3 Br + OH - CH 3 OH + Br - What might the TS look like for this substitution reaction?

16-59 Figure The proposed transition state in the reaction between CH 3 Br and OH - The TS is trigonal bipyramidal; note the elongated C-Br and C-O bonds

16-60 Reaction Energy Diagrams Shows the potential energy of the system during a chemical reaction (a plot of potential energy vs reaction progress)

16-61 Figure Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Reaction energy diagram for the reaction between CH 3 Br and OH -

16-62 General Principles of TS Theory Every reaction (and each step in an overall reaction) goes through its own TS. All reactions are reversible. Transition states are identical for the individual forward and reverse reactions in an equilibrium reaction.

16-63 Sample Problem 16.7 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  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 relationships between reactants, products and transition state. The reactants are at a higher energy level than the products, and the transition state is slightly higher in energy than the reactants. transition state E a = 19 kJ  H rxn = -392 kJ E a(rev) = ( ) kJ = 411 kJ

16-64 Figure Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Reaction energy diagrams and possible transition states for three reactions endothermic exothermic

16-65 Reaction Mechanisms The Specific Steps in an Overall Reaction How a reaction works at the molecular level 2A + B E + F A + B C C + A D D E + F possible steps C and D are reaction intermediates. Mechanisms of reactions are proposed and then tested. All postulated steps must sum to yield the chemical equation for the overall reaction.

16-66 Elementary Reactions and Molecularity individual steps = elementary reactions (or steps) Elementary steps are characterized by their molecularity. molecularity = number of reactant particles involved in the step 2O 3 (g) 3O 2 (g) O 3 (g) O 2 (g) + O(g) O 3 (g) + O(g) 2O 2 (g) unimolecular bimolecular Termolecular elementary steps are rare! The rate law for an elementary reaction can be deduced from the reaction stoichiometry; equation coefficients are used as the reaction orders.

16-67 elementary stepmolecularityrate law A product 2A product A + B product 2A + B product unimolecular bimolecular termolecular rate = k [A] rate = k [A] 2 rate = k [A][B] rate = k [A] 2 [B] Table 16.6 Rate Laws for General Elementary Steps Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-68 Sample Problem 16.8 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)

16-69 The overall rate of a reaction is related to the rate of the rate-determining step. That is, the rate law for the rate-determining step represents the rate law for the overall reaction! The Rate-Determining Step of a Reaction Mechanism All of the elementary steps of a mechanism do not occur at the same rate. Usually one step is much slower than the others. This step is called the rate-determining (or rate-limiting) step.

16-70 An example: rate = k[NO 2 ] m [CO] n Rate law: rate = k[NO 2 ] 2 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) Step 1 is slow and rate-determining; Step 2 is fast. NO 3 is an intermediate! Rate 1 = k 1 [NO 2 ] 2 Rate 2 = k 2 [NO 3 ][CO] Thus, if k 1 = k, the rate law for Step 1 (rate-determining) is the same as the experimental rate law!

16-71 Correlating the Mechanism with the Rate Law 1. The elementary steps must add up to the overall equation. 2. The elementary steps must be physically reasonable. 3. The mechanism must correlate with the rate law. Chemical mechanisms can never be proven unequivocally! But potential chemical mechanisms can be eliminated based on experimental data. Three Key Criteria for Elementary Steps

16-72 Figure Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Reaction Energy Diagram for the two-step NO 2 -F 2 reaction 2NO 2 (g) + F 2 (g) 2NO 2 F(g) Two transition states and one intermediate are involved. The first step is rate-determining. The reaction is exothermic (thermodynamics). rate = k[NO 2 ][F 2 ]

16-73 What happens when a fast reversible step(s) occurs prior to the rate-determining step? 2NO(g) + O 2 (g) 2NO 2 (g) Rate law: rate = k[NO] 2 [O 2 ] Step 1: NO(g) + O 2 (g) NO 3 (g) fast and reversible Step 2: NO 3 (g) + NO(g) 2NO 2 (g) slow; rate-determining Proposed Mechanism Rate laws for the two elementary steps: Step 1: rate 1 (fwd) = k 1 [NO][O 2 ]; rate 1 (rev) = k - 1 [NO 3 ] Step 2: rate 2 = k 2 [NO 3 ][NO] Step 2 (rate-determining) contains [NO 3 ] which does not appear in the experimental rate law!

16-74 If Step 1 is at equilibrium, then.... rate 1 (fwd) = rate 1 (rev) k 1 [NO][O 2 ] = k - 1 [NO 3 ] or Solving for [NO 3 ]: [NO 3 ] = (k 1 /k - 1 )[NO][O 2 ] Substituting into the rate law for rate-limiting Step 2: rate 2 = k 2 [NO 3 ][NO] = k 2 (k 1 /k - 1 )[NO][O 2 ][NO] = [(k 1 k 2 )/ k - 1 ][NO] 2 [O 2 ] Conclusion: The proposed mechanism is consistent with the kinetic data.

16-75 Catalysis: Enhancing Reaction Rates Catalyst: increases reaction rate but is not consumed in the reaction A catalyst increases reaction rate (via increasing k) by lowering the activation energy (barrier) of the reaction. Both forward and reverse reactions are catalyzed; reaction thermodynamics is unaffected! The catalyzed reaction proceeds via a different mechanism than the uncatalyzed reaction.

16-76 Figure Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Energy diagram of an uncatalyzed and catalyzed reaction Note that both reactions exhibit the same thermodynamics! The catalyzed and uncatalyzed reactions occur via different pathways.

16-77 Uncatalyzed reaction: A + B product Catalyzed reaction: A + catalyst C C + B product + catalyst Types of Catalysts  Homogeneous: exists in solution with the reaction mixture  Heterogeneous: catalyst and reaction mixture are in different phases

16-78

16-79 R-COOR’ + H 2 O + H + R-COOH + R’OH + H + esteracidalcohol The Mechanism of Acid-catalyzed Organic Ester Hydrolysis The reaction rate is slow at neutral pH (pH 7.0), but increases significantly at acidic pH (pH 2). H + ion catalyzes the reaction at low pH; what is the molecular basis for the catalysis by H + ? Homogeneous Catalysis

16-80 Figure Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Proposed reaction mechanism for the H + -catalyzed hydrolysis of an organic ester +

16-81 Figure Heterogeneous Catalysis: Metal-catalyzed hydrogenation of ethylene H 2 CCH 2 ( g ) + H 2 ( g )H 3 CCH 3 ( g ) Ni, Pd or Pt

16-82 Figure B16.4 The widely separatedam ino acid groups that form the active site of an enzyme The enzyme, chymotrypsin A protease (hydrolyzes proteins)

16-83 Figure B16.5 Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Two Models of Enzyme Action lock and key induced fit

16-84 Tools of the Laboratory Figure B16.1 Spectrophotometric monitoring of a reaction Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

16-85 Figure B16.2 Figure B16.3 Conductometric monitoring of a reaction Manometric monitoring of a reaction Tools of the Laboratory Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.