2/23/2018. Familiar Kinetics. ...and the not so familiar. Chemical kinetics is the study of how fast reactions take place.

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1 CHEMICAL KINETICS & REACTION MECHANISMS Readings, Examples & Problems Petrucci, et al., th ed. Chapter 20 Petrucci, et al., 0 th ed. Chapter 4 Familiar Kinetics...and the not so familiar Reaction Rates Chemical kinetics is the study of how fast reactions take place. how fast rate at what speed and in what direction Instantaneous to millions of years. Increasing the rate of a reaction is important to many industrial processes.

2 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 First-Order Reactions Second-Order Reactions Dependence of Reaction Rate on Temperature The Arrhenius Equation 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. generally occur as a result of collisions between atoms and molecules. Collision theory: reaction rate is directly yproportional p to the number of molecular collisions per second: rate number of collisions s 2

3 Theoretical Models for Chemical Kinetics Kinetic Molecular theory can be used to calculate the collision frequency (i.e., rate). In gases 0 30 collisions per second. If each collision produced a reaction, rate would be about 0 6 M s. BUT actual rates on order of 0 4 M s. Still a very rapid rate. Only a fraction of collisions yield a reaction. Factors that Affect Collisions & Collision Effectiveness Concentration Temperature Orientation Surface Area Catalyst (Enzyme) Effect of Concentration 3

4 Minimum energy (E a ) for reaction to occur Distribution of molecular kinetic energies 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 Effect of Temperature on Reaction Rates Svante Arrhenius Equation k = Ae -Ea/RT ln k = -Ea R T + lna Where else have you heard of Svante Arrhenius? 4

5 -E k = Ae -Ea/RT a ln k = + ln A R T Two point form of the Arrhenius equation -E ln k 2 lnk = a + ln A - -E a - ln A R T 2 R T k ln -E = a - k 2 R T 2 T Recall the two point vapor pressure vs temperature equation from Chem A. Reaction requires 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). 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. Transition State Theory A reaction profile for the reaction N 2 O(g) + NO(g) N 2 (g) + NO 2 (g) 6

7 COLLISION & ORIENTATION ENERGY PROFILE 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.. :Br:.. R H CH 3 C.. Cl:.. ORIENTATION ENERGY PROFILE R.. :Br:.. H CH 3 C.. Cl:.. ORIENTATION ENERGY PROFILE R.... :Br.. C Cl: Transition State H CH3 Activated Complex 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 :Cl:.... 9

10 ORIENTATION ENERGY PROFILE.. :Br.. C R CH 3 H ORIENTATION.. :Br.. C R CH 3 H Importance of Molecular Orientation OH - + CH 3 Cl 060_nuc_sub_sn2_04.swf 30 0

11 rate=2/23/208 Importance of Molecular Orientation OH - + CH 3 ClH 060_nuc_sub_sn2_05.swf 3 Molecular collisions and chemical reactions Orientation Copyright 20 Pearson Canada Inc. Slide 32 of 56 Measuring Reaction Progress and Expressing Reaction Rate Chemical kinetics is the study of how fast (i.e., speed & direction) reactions take place. A AB[][ rate= t t B][A] decreases

12 Measuring Reaction Progress and Expressing Reaction Rate A B Measuring Reaction Progress and Expressing Reaction Rate Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) red-brown Br 2 (aq) + HCOOH (aq) 2Br - (aq) + 2H + (aq) + CO 2 (g) time t 3 < t 2 < t 393 nm Detector light [Br 2 ] Absorbance 36 2

13 3 Measuring Reaction Progress and Expressing Reaction 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 00 seconds: 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.

14 4 Measuring Reaction Progress and Expressing Reaction Rate Time (s) [Br 2 ](M) Rate (M/s) x x 0 5 Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) x x s2250sBr2Br 55rateat50.0s3.5202rateat250.0s.750 2rate[Br] 2rate[Br]k k is called the rate constant. Measuring Reaction Progress and Expressing Reaction Rate Time (s) [Br 2 ](M) Rate (M/s) x x 0 5 Br 2 (aq) + HCOOH(aq) 2Br (aq) + 2H + (aq) + CO 2 (g) x x 0 52rate[Br]k at t = 50.0 s2rate[br]k s3.490s0.00/ M k M Measuring Reaction Progress and Expressing Reaction Rate 2H 2 O 2 (aq) 2H 2 O(l) + O 2 (g) 2O2[O]rate=P t RT t

15 rate=rate=2/23/208 Measuring Reaction Progress and Expressing Reaction Rate aa + bb cc + dd[a][b][c][ a t b t c t d t D]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. p[a][b][c][ a t b t c t d t D]rate= [CO][HO][CH][O] t t 2t t Worked Example 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. Therefore, Δ[I rate = - ] Δ[OCl = - ] Δ[Cl = - ] Δ[OI = - ] Δt Δt Δt Δt 5

16 Worked Example (cont.) Solution Δ[O (b) rate = 3 ] = 2 Δt 3 Δ[O 2 ] Δt Δ[NH (c) rate = 3 ] Δ[O = 2 ] Δ[NO] = = 4 Δt 5 Δt 4 Δt 6 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. 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 Δ[NO rate = 2 ] Δ[O = 2 ] 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. Worked Example 2 (cont.) Solution The rate of reaction, therefore, is Δ[NO rate = 2 ] = ( M/s) 4 Δt 4 = M/s Δ[O (a) M/s = 2 ] Δ[O Δt [ 2] 2 = Δt M/s Molecular oxygen is being consumed at a rate of M/s. (b) Δ[N 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. 6

17 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. 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 0 3 7

18 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 [F 2 ] [ClO ] Rate doubles constant doubles 4.8 x x 0 3 F M 2 F 0.0 M 2 3 rate M / s 3 rate.20 M / s 2 The reaction is first order in F 2 ; x = rate = k[f 2 ][ClO 2 ] y 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) [F 2 ] [ClO 2 ] Rate 2.2 x constant 0.0 x x x x 0 3 ClO M 4 ClO 0.00 M 2 3 rate M / s 3 rate.20 M / s 4 The reaction is first order in ClO 2 ; y = rate = k[f 2 ][ClO 2 ] Dependence of Reaction Rate on Reactant Concentration The systematic way to approach and solve rate law relationships 8

19 In summary rate ratio = (concentration ratio) n 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 0 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) [A] [B] Rate 2x x constant 0.05 x x x 0 4 A 0.0 M 0.5 A 0.20 M 2 4 rate 2.0 M / s 4 rate M / s 0.5 The reaction is first order in A; x = rate = k[a][b] y 9

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 [A] [B] Rate constant x 2 x x x B M 2 B 0.05 M 4 rate M / s 4 rate 2.0 M / s 4 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: ) 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 2 8 ](M) [I ](M) Initial Rate (M/s) x x x

21 Dependence of Reaction Rate on Reactant Concentration Solution: Step :In experiments 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 lrt Rate (M/s) x x x 0 4 I 2 I M M 4 rate 2.20 M / s 4 rate2.0 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 lrt Rate (M/s) x x x SO M 2 2 SO M 2 4 rate M / s 2 4 rate.0 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 lrt Rate (M/s) x x x 0 4 The rate law is: rate = k [S 2 O 8 2 ] [I ] 2

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 lrt Rate (M/s) x x x 0 4 k 4 rate 2.20 M / s 0.08 M s 3 2 SO (0.6 )(0.07 ) 2 8 I M M 3 3 Worked Example 3 The gas-phase reaction of nitric oxide with hydrogen at 280 C is 2NO(g) + 2H 2 (g) N 2 (g) +2H 2 O(g) From the following data collected at 280 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.0 0 = -4 3 M/s k(.0 0 = 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 =. 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 gives rate.3 0 k = = -5 M/s [NO] 2 ( M) 2 ( = [H 2 ] M) 2 M-2 s- (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- )( M) 2 ( M) = M s - Think About It The exponent for the concentration of H 2 in the rate law is, 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. 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. 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 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 Using calculus, it is possible to show that: A 0 ln A t kt 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. The rate constant for the reaction 2A B is 7.5 x 0 3 s at 0 C. The reaction is first order in A. How long (in seconds) will it take for [A] to decrease from.25 M to 0.7 M? Solution: Step :Use the equation below to calculate time in seconds. t 0.7 ln s.25 t = 75 seconds 0 A 0 t ln kt A 3 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 Cis s -. 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 0.0 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 0.0 M. [H 2 O 2 ] 0 = 0.75 M; time t for part (a) is (3 h)(60 min/h)(60 s/min) = 0,800 s. Solution [H (a) ln 2 O 2 ] t = kt [H 2 O 2 ] 0 (b) [H ln 2 O 2 ] t 0.75 M = ( s - )(0,800 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. 0.0 M 205= (8 0-5 (b) ln M = 2.05 (.8 0 s )t s - = t =.2 05 s The time required for the peroxide concentration to drop to 0.0 M is. 0 5 s or about 3 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. 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 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 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) 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 0 3 s Intercept =

27 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) STOPPED 3/25 Begin with board example where only of 2 Experiments is held constant t 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 =.3 x 0 2 s ; k =.3 x 0 2 s 27

28 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. 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 00 s and 250 s of the original table (P azomethane versus t), we get 50 mmhg ln 220 mmhg = kt ln = k(50 s) k = s - 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 - Remember that slope = k, so k = s -. 28

29 The half-life (t /2 ) is the time required for the reactant concentration to drop to half its original value. t / k 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 - 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 = = = 293 s k s - min 293 s = 2.5 min 60 s The half-life of ethane decomposition at 700 C is 2.5 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. The half-life (t /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 t ln k A t t = t /2 when [A] t = ½[A] 0. t /2 A ln k A 2 0 /2 0 simplifies t k 29

30 Calculate the half-life of the decomposition of azomethane, k = s -. Solution: Step :Use the equation below to calculate half-life: t / k x 0 s t/ s rate law rate = k[a] x [B] y rate rate constant 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 0 3 s Intercept =

31 The half-life (t /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 t ln k A t t = t /2 when [A] t = ½[A] 0. t /2 A ln k A 2 0 /2 0 simplifies t k 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: kt + A A t 0 Second-order half-life: t /2 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 - s- 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 /[A] t = kt + /[A] 0 to determine [I] t at t = 2.0 min; use t ½ = /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) = 20 s (a) = kt + [A] t [A] 0 = ( M - s - )(20 s) M 3

32 Worked Example 7 (cont.) Solution = M - [A] t = = M M - 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 I 2 at room temperature. The very low remaining concentration t ½ of = I k[a] after 2 = min = s 0 (7.0 0 makes 9 M sense. - s- )(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.) t ½ = k[a] = = s 0 ( M - s- )(0.42 M) A second-order reaction is a reaction whose rate depends on the concentration The rate of a zero-order reaction is a constant. Third-order and higher are rare. 32

33 Zero Orderst Order st Od Orderst Od Order 2 nd Orderst Order 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 (8.34 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 lnk ln A R T 33

34 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 s ) T (K) Dependence of Reactant Rate on Temperature Solution: Plot ln k versus /T and determine the slope of the line; slope = E a /R. k (M s ) T (K) ln k /T (K ) ln k slope = 5.6 x 0 3 K = E a /R /T (K ) E a = (5.6 x 0 3 K )(8.34 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 /T, and determine the activation energy (in kj/mol) for the reaction. k (M - s- ) T(K) Strategy Plot ln k versus /T, and determine the slope of the resulting line. According to ln k = ( E a /R)(/T) + ln A, slope = E a /R. R = 8.34 J/mol K. 34

35 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 /T (K - ) A plot of these data yields the following graph: 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.4 ( 2.5) slope = K K - = K The value of the slope is K. Because the slope = E a /R, E a = (slope)(r) = ( K)(8.34 J/K mol) = J/mol or 46 kj/mol Think About It Note that while k has units M - s -, ln k has no units. Dependence of Reactant Rate on Temperature A two point form of the Arrhenius equation may be written: k E a ln k2 R T2 T 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. 35

36 Worked Example 9 The rate constant for a particular first-order reaction is given for three different temperatures: T (K) k (s - ) 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 k 2 T 2 T 2 Worked Example 9 (cont.) Solution E a = 8.34 J/K mol ln K K = 9,73 J/mol = 9 kj/mol The activation energy of the reaction is 9 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 4.) to determine the rate constant at 500 K. Make sure it agrees with the value in the table. Worked Example 0 A certain first-order reaction has an activation energy of 83 kj/mol. If the rate constant for this reaction is s - at 50 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 rate constant 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 = 423 K, temperature. T 2 = 573 K, If R you = 8.34 get a J/K mol, smaller and k at a k = 2. 0 higher -2 temperature, s -. check your solution for mathematical errors. Solution k 2 = J/mol 8.34 J/K mol s - The rate constant of 300 C is 0 s -. e k 573 K 423 K =.0 0 s - 36