Chemical Kinetics Chapter 14 Copyright © The McGraw-Hill Companies, Inc.  Permission required for reproduction or display.

Chemical Kinetics Thermodynamics – does a reaction take place? Kinetics – how fast does a reaction proceed? Reaction rate is the change in the concentration of a reactant or a product with time (M/s). A B rate = - D[A] Dt D[A] = change in concentration of A over time period Dt rate = D[B] Dt D[B] = change in concentration of B over time period Dt Because [A] decreases with time, D[A] is negative.

I- + ClO- -----) Cl- + IO- Rates Example of average rate calculation for: I- + ClO- -----) Cl- + IO- Given the [I-] and time in seconds, then what is the average rate? Time (s) [I-] Molarity 2.00 0.00169 8.00 0.00101 Rate = - ΔM / ΔT Rate = - (0.00101 - 0.00169) M / (8.00 - 2.00) s = 6.8x10-4 M / 6.00s Rate = 1.1 x 10-4 M/s

A B rate = - D[A] Dt rate = D[B] Dt

Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) red-brown Br2 (aq) + HCOOH (aq) 2Br- (aq) + 2H+ (aq) + CO2 (g) time 393 nm light Detector t1< t2 < t3 D[Br2] a D Absorption

Reaction Rates and Stoichiometry 2A B Two moles of A disappear for each mole of B that is formed. rate = - D[A] Dt 1 2 rate = D[B] Dt aA + bB cC + dD rate = - D[A] Dt 1 a = - D[B] Dt 1 b = D[C] Dt 1 c = D[D] Dt 1 d

Expressing reaction rates For a chemical reaction, there are many ways to express the reaction rate. The relationships among expressions depend on the equation. Note the expression and reasons for their relations for the reaction 2 NO + O2 (g) = 2 NO2 (g) [O2] 1 [NO] 1 [NO2] Reaction rate = – ——— = – — ———— = — ———  t 2  t 2  t Make sure you can write expressions for any reaction and figure out the relationships. For example, give the reaction rate expressions for 2 N2O5 = 4 NO2 + O2 How can the rate expression be unique and universal? 15 Chemical Kinetics

Calculating reaction rate The concentrations of N2O5 are 1.24e-2 and 0.93e-2 M at 600 and 1200 sec after the reactants are mixed at the appropriate temperature. Evaluate the reaction rates for 2 N2O5 = 4 NO2 + O2 Solution: (0.93 – 1.24)e-2 – 0.31e-2 M Decomposition rate of N2O5 = – ———————— = – —————— 1200 – 600 600 s = 5.2e-6 M s-1. Note however, rate of formation of NO2 = 1.02e-5 M s-1. rate of formation of O2 = 2.6e-6 M s-1. The reaction rates are expressed in 3 forms Be able to do this type problems

Reaction rates Ex: Balance the following reaction, then determine how the rates of each compound are related: N2O5 (g) → NO2 (g) + O2 (g) 2 4 [] = molarity unless otherwise noted If D[O2]/Dt = 5.0 M/s, what is D[N2O5]/Dt?

The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB cC + dD Rate = k [A]x[B]y Reaction is xth order in A Reaction is yth order in B Reaction is (x +y)th order overall

F2 (g) + 2ClO2 (g) 2FClO2 (g) rate = k [F2]x[ClO2]y Double [F2] with [ClO2] constant (compare exp1 and exp3) Rate doubles x = 1 Quadruple [ClO2] with [F2] constant (compare exp1 and exp2) Rate quadruples y = 1 rate = k [F2][ClO2]

Rate Laws Rate laws are always determined experimentally. Reaction order is always defined in terms of reactant (not product) concentrations. The order of a reactant is not related to the stoichiometric coefficient of the reactant in the balanced chemical equation. F2 (g) + 2ClO2 (g) 2FClO2 (g) 1 rate = k [F2][ClO2]

Ex: Determining the rate law using the following data: Exp # [NH4+] [NO21-] Initial rate (M/s) 1 0.50 0.20 3.0 x 10-3 2 0.50 0.40 6.0 x 10-3 3 1.5 0.40 54 x 10-3 x 1 x 2 x 2 x 3 x 1 x 9 Rate = k [NH4+]m [NO2-]n 3m = rate = 9 , m = 2 2n = rate = 2 , n = 1 Rate = k [NH4+]2 [NO2-]1 3.0 x 10-3 = k [0.50]2 [0.20]1 k = 0.060 Rate = 0.060 [NH4+]2 [NO2-]1

Rate = k [NH4+]2 [NO2-]1

Initial Rate of Formation of Practice Problem: 2 A + 2 B C + D The following data about the reaction above were obtained from three experiments: (a) What is the rate equation for the reaction? (b) What is the numerical value of the rate constant k? What are its dimensions? Experiment [A] [B] Initial Rate of Formation of C in M 1 0.60 0.15 6.3´10-3 2 0.20 2.8´10-3 3 7.0´10-4

rate = k [A]2[B]1 (b) = 0.12 L2mol-2min-1

First-Order Reactions rate = - D[A] Dt A product rate = k [A] D[A] Dt = k [A] - rate [A] M/s M = k = = 1/s or s-1 [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 [A] = [A]0e−kt ln[A] = ln[A]0 - kt

Concentration and time of 1st order reaction Describe the features of plot of [A] vs. t and ln[A] vs. t for 1st order reactions. Apply the technique to evaluate k or [A] at various times. [A] ln[A] ln [A] = ln [A]o – k t [A] = [A]o e – k t t½ t t

13.4 The conversion of cyclopropane to propene in the gas phase is a first-order reaction with a rate constant of 6.7 × 10−4 s−1 at 500°C. If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 min? How long (in minutes) will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? How long (in minutes) will it take to convert 74 percent of the starting material?

13.4 Solution (a) In applying Equation (13.4), we note that because k is given in units of s−1, we must first convert 8.8 min to seconds: We write Hence, Note that in the ln [A]0 term, [A]0 is expressed as a dimensionless quantity (0.25) because we cannot take the logarithm of units. ln[A] = ln[A]0 - kt

13.4 (b) Using Equation (13.3), (c) From Equation (13.3),

13.5 The rate of decomposition of azomethane (C2H6N2) is studied by monitoring the partial pressure of the reactant as a function of time: The data obtained at 300°C are shown in the following table: Are these values consistent with first-order kinetics? If so, determine the rate constant.

13.5 Solution First we construct the following table of t versus ln Pt. Figure 13.11, which is based on the data given in the table, shows that a plot of ln Pt versus t yields a straight line, so the reaction is indeed first order. The slope of the line is given by

13.5 According to Equation (13.4), the slope is equal to −k, so k = 2.55 × 10−3 s−1 . ln[A] = ln[A]0 - kt

First-Order Reactions The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t½ = t when [A] = [A]0/2 ln [A]0 [A]0/2 k = t½ ln 2 k = 0.693 k =

13.6 The decomposition of ethane (C2H6) to methyl radicals is a first-order reaction with a rate constant of 5.36 × 10−4 s−1 at 700°C: Calculate the half-life of the reaction in minutes.

13.6 For a first-order reaction, we only need the rate constant to calculate the half-life of the reaction. From Equation (13.6)

Second-Order Reactions rate = - D[A] Dt A product rate = k [A]2 rate [A]2 M/s M2 = D[A] Dt = k [A]2 - k = = 1/M•s 1 [A] = [A]0 + kt [A] is the concentration of A at any time t [A]0 is the concentration of A at time t=0 t½ = t when [A] = [A]0/2 t½ = 1 k[A]0

13.7 Iodine atoms combine to form molecular iodine in the gas phase This reaction follows second-order kinetics and has the high rate constant 7.0 × 109/M · s at 23°C. If the initial concentration of I was 0.086 M, calculate the concentration after 2.0 min. Calculate the half-life of the reaction if the initial concentration of I is 0.60 M and if it is 0.42 M.

13.7 (a) To calculate the concentration of a species at a later time of a second−order reaction, we need the initial concentration and the rate constant. Applying Equation (13.7) 1 [A] = [A]0 + kt

13.7 where [A]t is the concentration at t = 2.0 min. Solving the equation, we get (b) For [I]0 = 0.60 M t½ = 1 k[A]0

13.7 For [I]0 = 0.42 M

Zero-Order Reactions rate = - D[A] Dt A product rate = k [A]0 = k rate = M/s [A] is the concentration of A at any time t [A] = [A]0 - kt [A]0 is the concentration of A at time t = 0 t½ = t when [A] = [A]0/2 t½ = [A]0 2k

Summary of the Rate Laws

Activation energy ( Ea ): minimum energy required to initiate a chemical reaction Activated complex Ea Reactants Energy DErxn Products Rxn pathway (or rxn coordinate) Note that: DErxn, forward = - DErxn, backward Ea, forward ≠ Ea, backward

Figure 2: Change in Potential Energy

A = frequency factor (related to # of collisions) Arrhenius equation Relationship between rate and T A = frequency factor (related to # of collisions) R = 8.314 J/(mol•K) Swedish, father was a surveryor, taught himself to read at age 3; In 1884, based on this work, he submitted a 150-page dissertation on electrolytic conductivity to Uppsala for the doctorate. It did not impress the professors, and he received the lowest possible passing grade. Later this very work would earn him the Nobel Prize in Chemistry. Worked with Boltzmann and van’t Hoff. Svante Arrhenius (1859-1927)

How to determine Ea: perform rate experiments using various T (and keep concentrations constant.) ln A k Temp (K) ln k 1/T (K-1) http://www.shodor.org/unchem/advanced/kin/arrhenius.html

Ex: Determine the activation energy using the following data: T (K) k (s-1) 190. 2.50 x 10-2 200. 4.50 x 10-2 210. 7.66 x 10-2

Application of Arrhenius Equation From k = A e – Ea / R T, calculate A, Ea, k at a specific temperature and T. The reaction: 2 NO2(g) -----> 2NO(g) + O2(g) The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11? Method: derive various versions of the same formula k = A e – Ea / R T A = k e Ea / R T A / k = e Ea / R T ln (A / k) = Ea / R T Make sure you know how to transform the formula into these forms.

Application of Arrhenius Equation (continue) The reaction: 2 NO2(g) -----> 2NO(g) + O2(g) The rate constant k = 1.0e-10 s-1 at 300 K and the activation energy Ea = 111 kJ mol-1. What are A, k at 273 K and T when k = 1e-11? Use the formula derived earlier: A = k eEa / R T = 1e-10 s-1 exp (111000 J mol-1 / (8.314 J mol-1 K –1*300 K)) = 2.13e9 s-1 k = 2.13e9 s-1 exp (– 111000 J mol-1) / (8.314 J mol-1 K –1*273 K) = 1.23e-12 s-1 T = Ea / [R* ln (A/k)] = 111000 J mol-1 / (8.314*46.8) J mol-1 K-1 = 285 K

Practice Problem: The rate constant for the reaction H2(g)  + I2(g) ---> 2HI(g) is 5.4 x 10-4 M-1s-1 at 326 oC.     At 410 oC the rate constant was found to be 2.8 x 10-2 M-1s-1. Calculate the a) activation energy and b) high temperature limiting rate constant for this reaction.

We know the rate constant for the reaction at two different temperatures and thus we can calculate the activation energy from the above relation. First, and always, convert all temperatures to Kelvin, an absolute temperature scale. Then simply solve for Ea in units of R. ln(5.4 x 10-4 M-1s -1/ 2.8 x 10-2 M-1s-1) = (-Ea /R ){1/599 K - 1/683 K} -3.9484 = - Ea/R {2.053 x 10-4 K-1} Ea= (1.923 x 104 K) (8.314 J/K mol) Ea= 1.60 x 105 J/mol

Now that we know Ea, the pre-exponential factor, A, (which is the largest rate constant that the reaction can possibly have) can be evaluated from any measure of the absolute rate constant of the reaction. so 5.4 x 10-4 M -1s-1 = A exp{-(1.60 x 105 J/mol)/((8.314 J/K mol)(599K))} (5.4 x 10-4 M-1s-1) / (1.141x10-14) = 4.73 x 1010 M-1s-1 The infinite temperature rate constant is 4.73 x 1010 M-1s-1

Alternate Form of the Arrhenius Equation At two temperatures, T1 and T2 or

13.9 The rate constant of a first-order reaction is 3.46 × 10−2 s−1 at 298 K. What is the rate constant at 350 K if the activation energy for the reaction is 50.2 kJ/mol?

13.9 The data are

13.9

N2O2 is detected during the reaction! Reaction Mechanisms The overall progress of a chemical reaction can be represented at the molecular level by a series of simple elementary steps or elementary reactions. The sequence of elementary steps that leads to product formation is the reaction mechanism. 2NO (g) + O2 (g) 2NO2 (g) N2O2 is detected during the reaction! Elementary step: NO + NO N2O2 Overall reaction: 2NO + O2 2NO2 + Elementary step: N2O2 + O2 2NO2

2NO (g) + O2 (g) 2NO2 (g) Mechanism:

Intermediates are species that appear in a reaction mechanism but not in the overall balanced equation. An intermediate is always formed in an early elementary step and consumed in a later elementary step. Elementary step: NO + NO N2O2 N2O2 + O2 2NO2 Overall reaction: 2NO + O2 2NO2 + The molecularity of a reaction is the number of molecules reacting in an elementary step. Unimolecular reaction – elementary step with 1 molecule Bimolecular reaction – elementary step with 2 molecules Termolecular reaction – elementary step with 3 molecules

Rate Laws and Elementary Steps Unimolecular reaction A products rate = k [A] Bimolecular reaction A + B products rate = k [A][B] Bimolecular reaction A + A products rate = k [A]2 Writing plausible reaction mechanisms: The sum of the elementary steps must give the overall balanced equation for the reaction. The rate-determining step should predict the same rate law that is determined experimentally. The rate-determining step is the slowest step in the sequence of steps leading to product formation.

A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. k = A • exp( -Ea / RT ) Ea k Uncatalyzed Catalyzed ratecatalyzed > rateuncatalyzed Ea < Ea ‘

Figure: 14-20

In heterogeneous catalysis, the reactants and the catalysts are in different phases. Haber synthesis of ammonia Ostwald process for the production of nitric acid Catalytic converters In homogeneous catalysis, the reactants and the catalysts are dispersed in a single phase, usually liquid. Acid catalysis Base catalysis

Half-Life Sugar is fermented in a 1st order process (using an enzyme as a catalyst). sugar + enzyme --> products Rate of disappear of sugar = k[sugar] k = 3.3 x 10-4 sec-1 What is the half-life of this reaction?

Half-Life Section 15.4 and Screen 15.8 Rate = k[sugar] and k = 3.3 x 10-4 sec-1. What is the half-life of this reaction? Solution [A] / [A]0 = 1/2 when t = t1/2 Therefore, ln (1/2) = - k • t1/2 - 0.693 = - k • t1/2 t1/2 = 0.693 / k So, for sugar, t1/2 = 0.693 / k = 2100 sec = 35 min

Half-Life Rate = k[sugar] and k = 3.3 x 10-4 sec-1. Half-life is 35 min. Start with 5.00 g sugar. How much is left after 2 hr and 20 min? Solution 2 hr and 20 min = 4 half-lives Half-life Time Elapsed Mass Left 1st 35 min 5.00 g 2nd 70 2.50 g 3rd 105 1.25 g 4th 140 0.625 g

Half-Life Radioactive decay is a first order process. Tritium ---> electron + helium 3H 0-1e 3He If you have 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution

Half-Life Solution ln [A] / [A]0 = -kt Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution ln [A] / [A]0 = -kt [A] = ? [A]0 = 1.50 mg t = 49.2 years Need k, so we calc k from: k = 0.693 / t1/2 Obtain k = 0.0564 y-1 Now ln [A] / [A]0 = -kt = - (0.0564 y-1) • (49.2 y) = - 2.77 Take antilog: [A] / [A]0 = e-2.77 = 0.0627 0.0627 is the fraction remaining!

Half-Life Solution [A] / [A]0 = 0.0627 Start with 1.50 mg of tritium, how much is left after 49.2 years? t1/2 = 12.3 years Solution [A] / [A]0 = 0.0627 0.0627 is the fraction remaining! Because [A]0 = 1.50 mg, [A] = 0.094 mg But notice that 49.2 y = 4.00 half-lives 1.50 mg ---> 0.750 mg after 1 ---> 0.375 mg after 2 ---> 0.188 mg after 3 ---> 0.094 mg after 4

Half-Lives of Radioactive Elements Rate of decay of radioactive isotopes given in terms of 1/2-life. 238U --> 234Th + He 4.5 x 109 y 14C --> 14N + beta 5730 y 131I --> 131Xe + beta 8.05 d Element 106 - seaborgium 263Sg 0.9 s