11/2/ and the not so familiar. Chemical kinetics is the study of how fast reactions take place.

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1 Familiar Kinetics...and the not so familiar Reaction Rates Chemical kinetics is the study of how fast reactions take place. Some happen almost instantaneously, while others can take millions of years. Increasing the rate of a reaction is important to many industrial processes. Chemical Kinetics Reaction Rates Collision Theory of Chemical Reactions Measuring Reaction Progress and Expressing Reaction Rate Average Reaction Rate Instantaneous Rate Stoichiometry and Reaction Rate Dependence of Reaction Rate on Reactant Concentration The Rate Law Experimental Determination of the Rate Law Dependence of Reactant Concentration on Time First-Order Reactions Second-Order Reactions Dependence of Reaction Rate on Temperature The Arrhenius Equation 1

2 Chemical Kinetics Reaction Mechanisms Elementary Reactions Rate-Determining Step Experimental Support for Reaction Mechanisms Catalysis Heterogeneous Catalysis Homogeneous Catalysis Enzymes: Biological Catalysts Collision Theory of Chemical Reactions Most reactions happen faster at higher temperature. Chemical reactions generally occur as a result of collisions between reacting molecules. According to collision theory of chemical kinetics, the reaction rate is directly proportional to the number of molecular collisions per second: rate number of collisions s Theoretical Models for Chemical Kinetics Kinetic-Molecular theory can be used to calculate the collision frequency. In gases collisions per second. If each collision produced a reaction, the rate would be about 10 6 Ms -1. Actual rates are on the order of 10 4 M s -1. Still a very rapid rate. Only a fraction of collisions yield a reaction. 2

3 Factors that Affect Collisions & Collision Effectiveness Concentration Temperature Orientation Surface Area Catalyst (Enzyme) Effect of Concentration Minimum energy (Ea) for reaction to occur Distribution of molecular kinetic energies 3

4 Activation Energy: The minimum energy above the average kinetic energy that molecules must bring to their collisions for a chemical reaction to occur. An analogy for a reaction profile and activation energy Svante Arrhenius k = Ae -Ea/RT ln k = -Ea 1 R T + lna Where else have you heard of Svante Arrhenius? -E k = Ae -Ea/RT a 1 ln k = + ln A R T -E ln k 2 lnk 1 = a 1 + ln A - -E a 1 - ln A R T 2 R T 1 k ln 1 -E = a k 2 R T 2 T 1 4

5 For a reaction to occur there must be a redistribution of energy sufficient to break certain bonds in the reacting molecule(s). E.g., CH 4 + 2O 2 CO 2 + 2H 2 O What does this really mean? break 4 C-H bonds break 2 O-O bonds make 2 C=O bonds make 4 O-H bonds Collision Theory of Chemical Reactions When molecules collide in an effective collision, they form an activated complex (also called the transition state). Transition State Theory A reaction profile for the reaction N 2 O(g) + NO(g) N 2 (g) + NO 2 (g) 5

6 Collision Theory of Chemical Reactions Collisions that result in a chemical reaction are called effective collisions. The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. Molecules must also be oriented in a way that favors reaction. Cl + NOCl Cl 2 + NO Correct orientation to facilitate reaction An effective collision results in reaction. Collision Theory of Chemical Reactions Collisions that result in a chemical reaction are called effective collisions. The activation energy (E a ) is the minimum amount of energy required to initiate a chemical reaction. Molecules must also be oriented in a way that favors reaction. Cl + NOCl Cl 2 + NO Incorrect orientation does not favor reaction An ineffective collision results in no reaction. COLLISION & ORIENTATION ENERGY PROFILE R H CH 3 C.. Cl:.. 6

7 ORIENTATION ENERGY PROFILE.. :Br:.. R H CH 3 C.. Cl:.. ORIENTATION ENERGY PROFILE.. :Br:.. R H CH 3 C.. Cl:.. ORIENTATION ENERGY PROFILE :Br:.... R H CH 3 C.. Cl:.. 7

8 ORIENTATION ENERGY PROFILE R.. :Br:.. H CH 3 C.. Cl:.. ORIENTATION ENERGY PROFILE R.... :Br.. C Cl: Transition State H CH3 Activated Complex ORIENTATION ENERGY PROFILE.. :Br.. C R CH 3 H :Cl:.... 8

9 ORIENTATION ENERGY PROFILE.. :Br.. C R CH 3 H.. :Cl:.. ORIENTATION ENERGY PROFILE.. :Br.. C R CH 3 H.. :Cl:.. ORIENTATION ENERGY PROFILE.. :Br.. C R CH 3 H 9

10 ORIENTATION.. :Br.. C R CH 3 H Importance of Molecular Orientation 0601_nuc_sub_sn2_04.swf sn2 04 swf 29 Importance of Molecular Orientation 0601_nuc_sub_sn2_05.swf 30 10

11 rate=11/2/2013 FIGURE 14-9 Molecular collisions and chemical reactions Copyright 2011 Pearson Canada Inc. General Chemistry: Chapter 14 Slide 31 of 56 Measuring Reaction Progress and Expressing Reaction Rate Chemical kinetics is the study of how fast reactions take place. A AB[][ rate= t t B][A] decreases Measuring Reaction Progress and Expressing Reaction Rate A B 11

12 r]braervagerate=tial2intial11/2/2013 Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g)2[b] [Bt tfinaltinifinal[ ]red-brown Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time t 1 < t 2 < t nm Detector light [Br 2 ] Absorption 35 Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) 12

13 13 Measuring Reaction Progress and Expressing Reaction Measuring Reaction Progress and Expressing Reaction Rate Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) Measuring Reaction Progress and Expressing Reaction Measuring Reaction Progress and Expressing Reaction Rate Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) [B]averagerate= / M M s t s [B]M First 50 seconds: Fi t averagerate= / M M s t s First 100 seconds:

14 Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) Measuring Reaction Progress and Expressing Reaction Rate The instantaneous rate is the rate for a specific instant in time. Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) Time (s) [Br 2 ](M) Rate (M/s) x x x 10 5 Br5r2Bat50s r rateat250.0s s rarrate[br] 22eat0.50s3.5210te[B ]k 2 k is called the rate constant. 14

15 rate10srate=rate=11/2/2013 Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) Time (s) [Br 2 ](M) Rate (M/s) x x x 10 52[Br ]k s [Br]k M / 3.49k M at t = 50.0 s2rate3 1 Measuring Reaction Progress and Expressing Reaction Rate 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) 2[O]12P O t RT t Measuring Reaction Progress and Expressing Reaction Rate aa + bb cc + dd1[a]1[b]1[c]1[ a t b t c t d t D]15

16 rate=11/2/2013 Measuring Reaction Progress and Expressing Reaction Rate Write the rate expressions for the following reaction: CO 2 (g) + 2H 2 O(g) CH 4 (g) + 2O 2 (g) Solution: Use the equation below to write the rate expressions. p1[a]1[b]1[c]1[ a t b t c t d t D]rate= [CO]1[HO][CH]1[O] t t 2t t Worked Example 1 Write the rate expressions for each of the following reactions: (a) I - (aq) + OCl - (aq) Cl - (aq) + OI - (aq) (b) 2O 3 (g) 3O 2 (g) (c) 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Strategy For reactions containing gaseous species, progress is generally monitored by measuring pressure. Pressures are converted to molar concentrations using the ideal gas equation, and rate expressions are written in terms of molar concentrations. Solution (a) All the coefficients in this equation are 1. Therefore, Δ[I rate = - ] Δ[OCl = - ] Δ[Cl = - ] Δ[OI = - ] Δt Δt Δt Δt Worked Example 1 (cont.) Solution 1 Δ[O (b) rate = 3 ] = 2 Δt 1 3 Δ[O 2 ] Δt 1 Δ[NH (c) rate = 3 ] 1 Δ[O = 2 ] 1 Δ[NO] = = 4 Δt 5 Δt 4 Δt Δ[H 2 O] Δt Think About It Make sure that the change in concentration of each species is divided by the corresponding coefficient in the balanced equation. Also make sure that the rate expressions written in terms of reactant concentrations have a negative sign in order to make the resulting rate positive

17 Worked Example 2 Consider the reaction 4NO 2 (g) + O 2 (g) 2N 2 O 5 (g) At a particular time during the reaction, nitrogen dioxide is being consumed at the rate M/s. (a) At what rate is molecular oxygen being consumed? (b) At what rate is dinitrogen pentoxide being produced? Strategy Dt Determine the rate of reaction and, using the stoichiometry tihi t of fthe reaction, convert to rates of change for the specified individual species. Solution 1 Δ[NO rate = 2 ] Δ[O = 2 ] 1 Δ[N = 2 O 5 ] 4 Δt Δt 2 Δt We are given Δ[NO 2 ] = M/s Δt where the minus sign indicates that the concentration of NO 2 is decreasing the time. orked Example 2 (cont.) Solution The rate of reaction, therefore, is 1 Δ[NO rate = 2 ] 1 = ( M/s) 4 Δt 4 = M/s Δ[O (a) M/s = 2 ] Δt Δ[O[ 2 Think About 2] = It Remember -4 M/s that the negative sign in a rate Δt expression indicates that a species is being consumed rather than Molecular produced. oxygen Rates is being are always consumed expressed at a rate as of positive quantities. -4 M/s. Δ[N (b) M/s = 2 O 5 ] 2 Δt Δ[N 2( M/s) = 2 O 5 ] Δt Δ[N 2 O 5 ] = M/s Δt Dinitrogen pentoxide is being produced at a rate of M/s. Dependence of Reaction Rate on Reactant Concentration The rate law is an equation that relates the rate of reaction to the concentrations of reactants. aa + bb cc + dd rate = k[a] x [B] y k: the rate constant x: order with respect to A y: order with respect to B determined experimentally x + y: represents the overall reaction order. 17

18 Dependence of Reaction Rate on Reactant Concentration F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) The initial rate is the rate at the beginning of the reaction. Initial Rate Data for the Reaction between F 2 and ClO 2 Experiment [F 2 ](M) [ClO 2 ](M) Initial Rate (M/s) x x x 10 3 Dependence of Reaction Rate on Reactant Concentration F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k[f 2 ] x [ClO 2 ] y Initial Rate Data for the Reaction between F 2 and ClO 2 Experiment [F 2 ](M) [ClO 2 ](M) Initial Rate (M/s) x 10 3 [F 2 2 ] [ClO ] Rate x 10 doubles constant doubles x 10 3 F M rate M / s F 0.10 M rate M / s The reaction is first order in F 2 ; x = 1 rate = k[f 2 ][ClO 2 ] y 18

19 Dependence of Reaction Rate on Reactant Concentration F 2 (g) + 2ClO 2 (g) 2FClO 2 (g) rate = k[f 2 ][ClO 2 ] y Initial Rate Data for the Reaction between F 2 and ClO 2 Experiment [F 2 ](M) [ClO 2 ](M) Initial Rate (M/s) 1 [F 2 ] [ClO 2 ] Rate x constant 0.10 x x x x 10 3 ClO M 4 ClO M rate M / s 3 rate M / s 4 The reaction is first order in ClO 2 ; y = 1 rate = k[f 2 ][ClO 2 ] Dependence of Reaction Rate on Reactant Concentration aa + bb cc + dd rate = k[a] x [B] y Initial Rate Data for the Reaction between A and B Experiment [A](M) [B](M) Initial Rate (M/s) x x x 10 4 Dependence of Reaction Rate on Reactant Concentration aa + bb cc + dd rate = k[a] x [B] y Initial Rate Data for the Reaction between A and B Experiment [A](M) [B](M) Initial Rate (M/s) 1 [A] [B] Rate 21x x constant x x x 10 4 A M rate M / s A 0.20 M rate M / s The reaction is first order in A; x = 1 rate = k[a][b] y 19

20 Dependence of Reaction Rate on Reactant Concentration aa + bb cc + dd rate = k[a][b] y Initial Rate Data for the Reaction between A and B Experiment [A](M) [B](M) Initial Rate (M/s) x 10 4 [A] [B] Rate x 10 constant x 2 x x 10 4 B M rate M / s B M rate M / s The reaction is second order in B; y = 2 rate = k[a][b] 2 Dependence of Reaction Rate on Reactant Concentration Three important things to remember about the rate law: 1) The exponents in a rate law must be determined from a table of experimental data. 2) Comparing changes in individual reactant concentrations with changes in rate shows how the rate depends on each reactant concentration. 3) Reaction order is always defined in terms of reactant concentrations, never product concentrations. Dependence of Reaction Rate on Reactant Concentration The reaction of peroxydisulfate ion (S 2 O 2 8 ) with iodide ion (I ) is: S 2 O 2 8 (aq) + 3I (aq) 2SO 2 4 (aq) + I 3 (aq) Determine the rate law and calculate the rate constant, including its units. Initial Rate Data for the Reaction between S 2 2 O 8 and I Experiment [S 2 O 8 2 ](M) [I ](M) Initial Rate (M/s) x x x

21 Dependence of Reaction Rate on Reactant Concentration Solution: Step 1:In experiments 1 and 2, [S 2 O 8 2 ] is constant. The [I ] is doubled, and rate doubles. Initial Rate Data for the Reaction between S 2 O 8 2 and I Experiment [S 2 O 2 8 ](M) [I ](M) Initial lrate (M/s) x x x I 2 I M M 4 rate M / s 4 rate M / s 2 The reaction is first order in I Dependence of Reaction Rate on Reactant Concentration Solution: Step 2:In experiments 2 and 3, [S 2 O 8 2 ] is doubled, [I ] is constant, and rate doubles. Initial Rate Data for the Reaction between S 2 O 8 2 and I Experiment [S 2 O 2 8 ](M) [I ](M) Initial lrate (M/s) x x x SO M 2 2 SO M 2 4 rate M / s 2 4 rate M / s 2 The reaction is first order in S 2 O 8 2 Dependence of Reaction Rate on Reactant Concentration Solution: Step 2:In experiments 2 and 3, [S 2 O 8 2 ] is doubled, [I ] is constant, and rate doubles. Initial Rate Data for the Reaction between S 2 O 8 2 and I Experiment [S 2 O 2 8 ](M) [I ](M) Initial lrate (M/s) x x x 10 4 The rate law is: rate = k [S 2 O 8 2 ] [I ] 21

22 Dependence of Reaction Rate on Reactant Concentration Solution: Step 3:Use the data from any experiment to calculate k. Initial Rate Data for the Reaction between S 2 O 2 8 and I Experiment [S 2 O 2 8 ](M) [I ](M) Initial lrate (M/s) x x x rate M / s k M s SO (0.16 M)(0.017 M) I Worked Example 3 The gas-phase reaction of nitric oxide with hydrogen at 1280 C is 2NO(g) + 2H 2 (g) N 2 (g) +2H 2 O(g) From the following data collected at 1280 C, determine (a) the rate law, (b) the rate constant, including units, and (c) the rate of the reaction when [NO] = M and [H 2 ] = M. Experiment [NO] (M) [H 2 ](M) Initial rate (M/s) Strategy Compare two experiments at a time to determine how the rate depends on the concentration of each reactant. The rate law is rate = k[no] x [H 2 ] y Worked Example 3 (cont.) Solution The rate of reaction, therefore, is rate = -5 2 M/s k( = -2 M) x ( M) y rate M/s k( M) x ( M) y Canceling identical terms in the numerator and denominator gives ( M) x ( M) x = 2 x = 4 Therefore, x = 2. The reaction is second order in NO. Dividing the rate from experiment 3 by the rate from experiment 2, we get rate = -4 3 M/s k( = 2 = -2 M) x ( M) y rate M/s k( M) x ( M) y Canceling identical terms in the numerator and denominator gives ( M) y ( M) y = 2 y = 2 Therefore, y = 1. The reaction is first order in H 2. The overall rate law is rate = k[no] 2 [H 2 ] 22

23 Worked Example 3 (cont.) Solution We can use data from any of the experiments to calculate the value and units of k. Using the data from experiment 1 gives rate k = = -5 M/s [NO] 2 ( M) 2 ( = [H 2 ] M) 2 M-2 s-1 (c) Using the rate constant determined in part (b) and the concentrations of NO and H 2 given in the problem statement, we can determine the reaction rate as follows: rate = ( M -2 s-1 )( M) 2 ( M) = M s -1 Think About It The exponent for the concentration of H 2 in the rate law is 1, whereas the coefficient for H 2 in the balanced equation is 2. It is a common error to try to write a rate law using the stoichiometric coefficients as the exponents. Remember that, in general, the exponents in the rate law are not related to the coefficients in the balanced equation. Rate laws must be determined by examining a table of experimental data. Dependence of Reactant Concentration on Time The rate law can be used to determine the rate of a reaction using the rate constant and the reactant concentrations: rate law rate = k[a] x [B] y rate rate constant A rate law can also be used to determine the concentration of a reactant at a specific time during a reaction. Dependence of Reactant Concentration on Time A first-order reaction is a reaction whose rate depends on the concentration of one of the reactants raised to the first power. C 2 H 6 2 CH 3 rate = k[c 2 H 6 ] 2N 2 O 5 (g) 2NO 2 (g) + O 2 (g) rate = k[n 2 O 5 ] 23

24 Dependence of Reactant Concentration on Time In a first-order reaction of the type A products The rate can be expressed as the rate of change in reactant concentration, A rate = t as well as in the form of the rate law: rate = k[a] Setting the two expressions equal to each other yields: A t k A Dependence of Reactant Concentration on Time Using calculus, it is possible to show that: A 0 t ln kt A ln is the natural logarithm [A] 0 and [A] t refer to the concentration of A at times 0 and t The equation above is sometimes called the integrated rate law for a first order reaction. Dependence of Reactant Concentration on Time The rate constant for the reaction 2A B is 7.5 x 10 3 s 1 at 110 C. The reaction is first order in A. How long (in seconds) will it take for [A] to decrease from 1.25 M to 0.71 M? Solution: Step 1:Use the equation below to calculate time in seconds. t 0.71 ln s 1.25 t = 75 seconds 0 A 0 t ln kt A 3 1 t 24

25 Worked Example 4 The decomposition of hydrogen peroxide is first order in H 2 O 2. 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) The rate constant for this reaction at 20 C is s -1. If the start concentration of H 2 O 2 is 0.75 M, determine (a) the concentration of H 2 O 2 remaining after 3 h and (b) how long it will take for the H 2 O 2 concentration to drop to 0.10 M. Strategy Use ln ([A] t /[A] 0 ) = kt to find [H 2 O 2 ] t where t = 3 h, and then solve for t to determine how much time must pass for [H 2 O 2 ] t to equal 0.10 M. [H 2 O 2 ] 0 = 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 10,800 s. Solution [H (a) ln 2 O 2 ] t = kt [H 2 O 2 ] 0 (b) ln [H 2 O 2 ] t 0.75 M = ( s -1 )(10,800 s) = Worked Example 4 (cont.) Solution Take the inverse natural logarithm of both sides of the equation to get [H 2 O 2 ] t 0.75 M = e = [H 2 O 2 ] t = (0.823)(0.75 M) = 0.62 M The concentration of H 2 O 2 after 3 h is 0.62 M M (b) ln M = = ( s )t s -1 = t = s The time required for the peroxide concentration to drop to 0.10 M is s or about 31 h. Think About It Don t forget the minus sign. If you calculate a concentration at time t that is greater than the concentration at time 0 (or if you get a negative time required for the concentration to drop to a specified level), check your solution for this common error. Dependence of Reactant Concentration on Time The rate law can be used to determine the rate of a reaction using the rate constant and the reactant concentrations: rate law rate = k[a] x [B] y rate rate constant A rate law can also be used to determine the concentration of a reactant at a specific time during a reaction. 25

26 Dependence of Reactant Concentration on Time Rearrangement of the first-order integrated rate law gives: At ln kt A ln[a] t = kt + ln[a] 0 0 Rearrangement in this way has the form of the linear equation y = mx + b. ln[a] t = kt + ln[a] 0 Slope = k Intercept = ln[a] 0 Dependence of Reactant Concentration on Time The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time. CH 3 N=N CH 3 (g) N 2 (g) + C 2 H 6 (g) The data obtained at 300 C are listed in the following table: Time (s) P azomethane (mmhg) Dependence of Reactant Concentration on Time Plotting the data gives a straight line, indicating the reaction is first order. ln P Time (s) ln[a] t = kt + ln[a] 0 Slope = 2.55 x 10 3 s 1 Intercept =

27 Dependence of Reactant Concentration on Time Ethyl iodide (C 2 H 5 I) decomposes at a certain temperature in the gas phase as follows: C 2 H 5 I(g) C 2 H 4 (g) + HI(g) Determine the rate of the reaction, after verifying that the reaction is first order. Time (s) [C 2 H 5 I] (M) Dependence of Reactant Concentration on Time Solution: Plot ln[c 2 H 5 I] vs time. If a straight line results, the reaction is first order. The slope is equal to k. Time (s) [C 2 H 5 I] (M) ln[c 2 H 5 I] ln [C 2 H 5 I] Time (s) Slope = 1.3 x 10 2 s 1 ; k = 1.3 x 10 2 s 1 Worked Example 5 The rate of decomposition of azomethane is studied by monitoring the partial pressure of the reactant as a function of time: CH 3 N=N CH 3 (g) N 2 (g) + C 2 H 6 (g) The data obtained at 20 C are listed in the following table: Time (s) P azomethane (mmhg) Strategy We can use ln ([A] t /[A] 0 ) = kt only for first-order reactions, so we must first determine if the decomposition of azomethane is first order. We do this by plotting ln P against time. If the reaction is first order, we can use ln ([A] t /[A] 0 ) = kt and the data at any two of the times in the table to determine the rate constant. 27

28 Worked Example 5 (cont.) Solution The table expressed in ln P is Time (s) ln P Plotting these data gives a straight line, indicating that the reaction is indeed first order. Thus, we can use ln ([A] t /[A] 0 ) = kt in terms of pressure. ln P t P 0 = kt P t and P 0 can be pressures at any two times during the experiment. P 0 need not be the pressure at 0 s it need only be at the earlier of the two times. Worked Example 5 (cont.) Solution Using data from times 100 s and 250 s of the original table (P azomethane versus t), we get 150 mmhg ln 220 mmhg = kt ln = k(150 s) k = s -1 Think About It We could equally well have determined the rate constant by calculating the slope of the plot of ln P versus t. Using the two points labeled on the plot, we get slope = = s -1 Remember that slope = k, so k = s -1. Dependence of Reactant Concentration on Time The half-life (t 1/2 ) is the time required for the reactant concentration to drop to half its original value t1/2 k 28

29 Worked Example 6 The decomposition of ethane (C 2 H 6 ) to methyl radicals (CH 3 ) is a first-order reaction with a rate constant of s -1 at 700 C. C 2 H 6 CH 3 Calculate the half-life of the reaction in minutes. Strategy Use t ½ = 0.693/k to calculate t ½ in seconds, and then convert to minutes Solution t ½ = = = 1293 s k s -1 1 min 1293 s 60 s = 21.5 min The half-life of ethane decomposition at 700 C is 21.5 min. Think About It Half-lives and rate constants can be expressed using any units of time and reciprocal time, respectively. Track units carefully when you convert from one unit of time to another. Dependence of Reactant Concentration on Time The half-life (t 1/2 ) is the time required for the reactant concentration to drop to half its original value. A 0 ln A t kt rearranges A 0 1 t ln k A t t = t 1/2 when [A] t = ½[A] 0. t 1/2 1 A ln k 1 A 2 0 1/2 0 simplifies t k Dependence of Reactant Concentration on Time Calculate the half-life of the decomposition of azomethane, k = s -1. Solution: Step 1:Use the equation below to calculate half-life: t1/2 k x 10 s t1/ s 29

30 Dependence of Reactant Concentration on Time rate law rate = k[a] x [B] y rate rate constant Dependence of Reactant Concentration on Time Plotting the data gives a straight line, indicating the reaction is first order. CH 3 N=N CH 3 (g) N 2 (g) + C 2 H 6 (g) ln P Time (s) ln[a] t = kt + ln[a] 0 Slope = 2.55 x 10 3 s 1 Intercept = 5.65 Dependence of Reactant Concentration on Time The half-life (t 1/2 ) is the time required for the reactant concentration to drop to half its original value. A 0 ln A t kt rearranges A 0 1 t ln k A t t = t 1/2 when [A] t = ½[A] 0. t 1/2 1 A ln k 1 A 2 0 1/2 0 simplifies t k 30

31 Dependence of Reactant Concentration on Time A second-order reaction is a reaction whose rate depends on the concentration of one reactant raised to the second power or on the product of the concentrations of two different reactants (first order in each). Second-order integrated rate law: 1 1 kt + A A t 0 Second-order half-life: t 1/2 1 k A 0 Worked Example 7 Iodine atoms combine to form molecular iodine in the gas phase: I(g) + I(g) I 2 (g) This reaction is second order and has a rate constant of M -1 s-1 at 23 C. (a) If the initial concentration of I is M, calculate the concentration after 2.0 min. (b) Calculate the half-life of the reaction when the initial concentration of I is 0.60 M and when the initial concentration of I is 0.42 M. Strategy Use 1/[A] t = kt + 1/[A] 0 to determine [I] t at t = 2.0 min; use t ½ = 1/k[A] 0 to determine t ½ when [I] 0 = 0.60 M and when [I] 0 = 0.42 M. Solution t = (2.0 min)(60 s/min) = 120 s 1 1 (a) = kt + [A] t [A] 0 = ( M -1 s 1-1 )(120 s) M Worked Example 7 (cont.) Solution = M -1 1 [A] t = = M M -1 The concentration Think About of atomic It (a) Iodine, iodine like after the 2 min other is halogens, exists M. as diatomic molecules at room temperature. It makes sense, therefore, (b) When that [I] atomic 0 =060M 0.60 iodine M, would react quickly, and essentially completely, to form 1 I 2 at room temperature. 1 The very low remaining concentration t ½ of = I k[a] after 2 = min = s 0 ( makes 9 M sense. -1 s-1 )(0.60 (b) As M) expected, the half-life of this second-order reaction is not constant. (A constant half-life is a When characteristic [I] 0 = 0.42 M, of first-order reactions.) 1 1 t ½ = = = s k[a] 0 ( M -1 s-1 )(0.42 M) 31

32 A second-order reaction is a reaction whose rate depends on the concentration Dependence of Reactant Concentration on Time The rate of a zero-order reaction is a constant. Third-order and higher are rare. Zero Orderst Order 1 st Od Orderst Od Order 2 nd Orderst Order 32

33 Dependence of Reaction Rate on Temperature The dependence of the rate constant on temperature can be expressed by the Arrhenius equation. k Ae Ea / RT A represents the collision frequency and is called the frequency factor. E a is the activation energy (in kj/mol). R is the gas constant (8.314 J/mol K). T is the absolute temperature. e is the base of the natural logarithm. Dependence of Reaction Rate on Temperature Taking the natural log of both sides, the Arrhenius equation may be written as: Ea lnk ln A RT Rearrangement gives the linear form of the Arrhenius equation: Ea 1 lnk ln A R T Dependence of Reactant Rate on Temperature Rate constants for the reaction CO(g) + NO 2 (g) CO 2 (g) + NO(g) were measured at four different temperatures. The data are shown in the table. Determine the activation energy for the reaction. k (M 1 s 1 ) T (K)

34 Dependence of Reactant Rate on Temperature Solution: Plot ln k versus 1/T and determine the slope of the line; slope = E a /R. k (M 1 s 1 ) T (K) ln k 1/T (K 1 ) ln k slope = 5.6 x 10 3 K = E a /R 1/T (K 1 ) E a = (5.6 x 10 3 K )(8.314 J/mol K) = 46 kj/mol Worked Example 8 Rate constants for the reaction CO(g) + NO 2 (g) CO 2 (g) + NO(g) were measured at four different temperatures. The data are shown in the table. Plot ln k versus 1/T, and determine the activation energy (in kj/mol) for the reaction. k (M -1 s-1 ) T(K) Strategy Plot ln k versus 1/T, and determine the slope of the resulting line. According to ln k = ( E a /R)(1/T) + ln A, slope = E a /R. R = J/mol K. Worked Example 8 (cont.) Solution Taking the natural log of each value of k and the inverse of each value of T gives ln k 1/T (K -1 ) A plot of these data yields the following graph: 34

35 Worked Example 8 (cont.) Solution The slope is determined using the x and y coordinates of any two points on the line. Using the points that are labeled on the graph gives 1.4 ( 2.5) slope = K K -1 = K The value of the slope is K. Because the slope = E a /R, E a = (slope)(r) = ( K)(8.314 J/K mol) = J/mol or 46 kj/mol Think About It Note that while k has units M -1 s -1, ln k has no units. Dependence of Reactant Rate on Temperature A two point form of the Arrhenius equation may be written: k1 E a 1 1 ln k2 R T2 T1 If the rate constants at two different temperatures are known, it is possible to calculate the activation energy. If the activation energy and the rate constant at one temperature are known, it is possible to determine the rate constant at any other temperature. Worked Example 9 The rate constant for a particular first-order reaction is given for three different temperatures: T (K) k (s -1 ) Using the data, calculate the activation energy of the reaction. Strategy Rearrange and solve for E a using the following E a = R k ln 1 k T 2 T 2 35

36 Worked Example 9 (cont.) Solution E a = J/K mol ln K K = 91,173 J/mol = 91 kj/mol The activation energy of the reaction is 91 kj/mol. Think About It A good way to check your work is to use the value of E a that you calculated (and Equation 14.11) to determine the rate constant at 500 K. Make sure it agrees with the value in the table. Worked Example 10 A certain first-order reaction has an activation energy of 83 kj/mol. If the rate constant for this reaction is s -1 at 150 C, what is the rate constant at 300 C? Strategy Rearrange and solve for k 2 using the following k 2 = Think About It Make sure that E a the 1 rate constant 1 you calculate at a R T 2 T higher temperature is in fact higher than the 2 original rate constant. e According to the Arrhenius equation, the rate constant always E a = increases 4 J/mol, with Tincreasing 1 = 423 K, temperature. T 2 = 573 K, If R you = get a J/K mol, smaller and k at a k 1 = higher -2 temperature, s -1. check your solution for mathematical errors. Solution k 2 = J/mol J/K mol s -1 The rate constant of 300 C is 10 s -1. e k K 423 K = s -1 Reaction Mechanisms A balanced chemical equation does not indicate how a reaction actually takes place. The sequence of steps that sum to give the overall reaction is called the reaction mechanism. Step 1: A + B C Step 2: C + B D Overall reaction: A + 2B D 36

37 Reaction Mechanism Chemical species that appear in the reaction mechanism, but not in the overall chemical equation are called intermediates. Step 1: NO + NO N 2 O 2 Step 2: N 2 O 2 + O 2 2NO 2 Overall reaction: 2NO + O 2 2NO 2 Reaction Mechanism Each step in a reaction mechanism represents an elementary reaction, one that occurs in a single collision of the reactant molecules. The molecularity of an elementary reaction is essentially the number of reactant molecules involved in the collision. unimolecular (one reactant molecule) A products rate = k[a] first order bimolecular (two reactant molecules) A + B products rate = k[a][b] second order A + A products rate = k[a] 2 second order termolecular (three reactant molecules) Reaction Mechanism In a reaction mechanism consisting of more than elementary step, the rate law for the overall process is given by the rate-determining step. The rate determining step is the slowest step in the sequence. A proposed mechanism must satisfy two requirements: The sum of the elementary reaction must be the overall balanced equation for the reaction. The rate determining step must have the same rate law as that determined from the experimental data. 37

38 Reaction Mechanism The decomposition of hydrogen peroxide can be facilitated by iodide ions: Step 1: (slow) H 2 O 2 + I H 2 O + IO Step 2: H 2 O 2 + IO H 2 O + O 2 + I Overall reaction: 2H 2 O 2 2H 2 O + O 2 Rate = k 1 [H 2 O 2 ][I ] Worked Example 11 The gas-phase decomposition of nitrous oxide (N 2 O) is believed to occur in two steps: k Step 1: N 2 O 1 N 2 + O k Step 2: N 2 O + O 2 N 2 + O 2 Experimentally the rate law is found to be rate = k[n 2 O]. (a) Write the equation for the overall reaction. (b) Identify the intermediate(s). (c) Identify the rate- determining step. Strategy Add the two equations, canceling identical terms on opposite sides of the arrow, to obtain the overall reaction. The canceled terms will be the intermediates if they were first generated and then consumed. Write rate laws for each elementary step; the one that matches the experimental rate law will be the rate-determining step. Worked Example 11 (cont.) k 1 Solution Step 1: N 2 O N 2 + O rate = k[n 2 O] Step 2: k N 2 O + O 2 N 2 + O 2 rate = k[n 2 O][O] (a) 2N 2 O 2N 2 + O 2 (b) O (atomic oxygen) is the intermediate. (c) Step 1 is the rate-determining step because its rate law is the same as the experimental law: rate = k[n 2 O]. Think About It A species that gets canceled when steps are added may be an intermediate or a catalyst. In this case, the canceled species is an intermediate because it was first generated and then consumed. A species that is first consumed and then generated, but doesn t appear in the overall equation, is a catalyst. 38

39 Worked Example 12 Consider the gas-phase reaction of nitric oxide and oxygen that was described at the beginning of Section NO(g) + O 2 (g) 2NO 2 (g) Show that the following mechanism is plausible. The experimentally determined rate law is rate = k[n 2 O] 2 [O 2 ]. k 1 Step 1: NO(g) + NO(g) N 2 O 2 (g) (fast) k Step 2: N 2 O 2 (g) + O 2 (g) 2 2NO 2 (g) (slow) k -1 Strategy To establish the plausibility of a mechanism, we must compare the rate law of the rate-determining step to the experimentally determined rate law. In this case, the rate-determining step has an intermediate (N 2 O 2 ) as one of its reactants, giving us a rate law of rate = k 2 [N 2 O 2 ][O 2 ]. Because we cannot compare this directly to the experimental rate law, we must solve for the intermediate concentration in terms of reactant concentrations. Worked Example 12 (cont.) Solution The first step is a rapidly established equilibrium. Both the forward and reverse of step 1 are elementary processes, which enables us to write their rate laws from the balanced equation. rate forward = k 1 [NO] 2 and rate reverse = k -1 [N 2 O 2 ] Because at equilibrium the forward and reverse processes are occurring at the same rate, Think we About can set It their Not rates all reactions equal to each have other a single and rate-determining solve for the intermediate concentration. step. Analyzing the kinetics of reactions with two or more comparably slow steps is k 1 beyond [NO] 2 = the k -1 scope [N 2 O 2 of ] this book. k [N 2 O 2 ] = 1 [NO] 2 k -1 Substituting the solution into the original rate law (rate = k[n 2 O 2 ][O 2 ]) gives k rate = k 1 [NO] 2 k 2 [O 2 ] = k[no] 2 [O 2 ] where k = 2 k 1 k -1 k -1 Catalysis A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. Step 1: (slow) Step 2: H 2 O 2 + I H 2 O 2 + IO H 2 O + IO H 2 O + O 2 + I Overall reaction: 2H 2 O 2 2H 2 O + O 2 A catalyst speeds up a reaction by providing a set of elementary steps with more favorable kinetics than those that exist in its absence. A catalyst usually speeds up a reaction by lowering the activation energy. 39

40 Catalysis In the presence of a catalyst the rate constant is k c, called the catalytic rate constant. uncatalyzed catalyzed rate catalyzed > rate uncatalyzed Catalysis In heterogeneous catalysis, the reactants and the catalysts are in different phases. Catalysis In homogeneous catalysis, the reactants and catalyst are dispersed in a single phase, usually liquid. Advantages: Reactions can be carried out under atmospheric conditions Can be designed to function selectively Are generally cheaper 40

41 Catalysis Enzymes are biological catalysts. Catalysis The mathematical treatment of enzyme kinetics is complex, but can be simplified: E + S k ES k 1 k2 ES E + P uncatalyzed catalyzed Catalysis Generally, the rate of an enzyme catalyzed reaction is given by the equation: P rate = t = k ES 41

42 Key Points Average Reaction Rates Instantaneous Rate Stoichiometry and Reaction Rate The Rate Law Experimental Determination of the Rate Law First-Order Reactions Second-Order Reactions Collision Theory Activation Energy and Arrhenius Theory Elementary Reactions Rate-Determining Step Experimental Support for Reaction Mechanisms Heterogeneous Catalysis Homogeneous Catalysis Enzymes: Biological Catalysts 42