Advanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS

Experimental Kinetics and Gas Phase Reactions Advanced Physical Chemistry CHAPTER 18 ELEMENTARY CHEMICAL KINETICS Professor Angelo R. Rossi http://homepages.uconn.edu/rossi Department of Chemistry, Room CHMT214 The University of Connecticut Fall Semester 2013 angelo.rossi@uconn.edu Suppose you have a substance, or several substances, in the gas or liquid phase under a certain set of conditions, and you find that when you add another substance, add a catalyst, irradiate the system, or change the temperature and pressure, chemical changes take place in the system. From a purely experimental point of view, there is a question as to how rapidly these changes take place and how their rates depend on independent variables such as concentrations of reactants and catalysts, temperature, and pressure. However, chemists are not satisfied with simply representing experimental results but want to understand what is going on in molecular terms. They want to understand why the steps in a mechanism have the rates they do and how the time course of the reaction can be predicted in advance. 2 Rate of Reaction Rate of Reaction In discussing the rate of a reaction, the first point to be clear about is the stoichiometry because the rates of consumption of reactants and rates of production of products are in the ratios of their stoichiometric numbers. Given the reaction Rearrange to give A + 2 B C 0 = A 2 B + C A chemical reaction can be represented by N s 0 = ν J J where ν J is the stoichiometric number of reactant J. For example, in the equation above, J = A, B, C and ν A = 1, ν B = 2, and ν C = +1 J 3 The extent of reaction ξ is the same for each reactant and product and n J = n J0 + ν J ξ where n J0 is the initial amount of reactant or product, J. Taking the time derivative of n J yields dn J = ν J dξ where the quantity dξ is called the rate of conversion. By convention the rate of reaction (v) is defined in terms of the rate of change of the concentration of a reactant or product, so that v = 1 dξ V = 1 dn J = 1 ( d nj ) V = 1 d[conc J ] ν J V ν J ν J This has the advantage that the same rate of reaction v is obtained for either the reactant or the product. 4

Rate of Reaction Rate of Reaction Conversion and Rate of Reaction The rate of reaction depends on how the stoichiometric equation is written. A + 2 B C 0 = A 2 B + C and ν A = 1, ν B = 2, and ν C = +1. The rate of the reaction for the above chemical equation then becomes v = 1 = 1 d[b] = d[c] 1 2 If the reaction goes in the forward direction, the rate of reaction is positive. If the reaction is in the backward direction, the reaction is negative. If the reaction is at equilibrium, the rate of reaction v is 0. 5 The reaction H 2 + Br 2 2 HBr is carried out in a 0.250-L reaction vessel. The change in the amount of Br 2 is -0.001 mol in 0.01 s. (a) What is the rate of conversion dξ dξ 0.001 mol = = 0.1 mol s 1 0.01 s? (b) What is the rate of reaction v? v = ( ) ( 1 dξ ) V = 0.1 mol s 1 = 0.40 mol L 1 s 1 0.25 L (c) What are the values of d[h2] = -0.40 mol L 1 s 1 d[h 2] d[br 2] = -0.40 mol L 1 s 1, d[br2], and d[hbr]? d[hbr] = 0.80 mol L 1 s 1 Fall 2013 Last Updated: October 6, 2013 at 8:55pm 6 Dependence of Reaction Rate on the Chemical Equation Measuring Rates of Chemical Reactions In discussing the rate v of a chemical reaction, it is important to know how the stoichiometric equation is written because the rate may depend on that. Different rates are obtained when a reaction is written differently. but 2 A + B 2 C v = 1 = d[b] = 1 d[c] 2 2 A + 1 2 B C v = 1 = 2 d[b] = d[c] 1 The above problem is largely avoided by not using fractional stoichiometric numbers in stoichiometric equations used to interpret rates of reactions. The rates of chemical reactions are obtained from measurements of concentration as a function of time. Chemical analytical methods may be used when the reaction can be stopped suddenly, e.g. by rapid cooling for high-temperature reactions or by catalyst inactivation for a catalyzed reaction. Physical methods are especially useful for determining the rate of a chemical reaction because they offer the possibility of continuous measurement of the extent of reaction. Spectroscopic methods are the most useful. An important characteristic of any measurement is its response time. The measuring device must respond more rapidly than the concentration is changing. For example, pulsed lasers may be used to study reactions occurring in picoseconds (10 12 ) seconds. High-temperature gas reactions may be carried out using a shock tube, and photochemical reactions may be initiated by flash photolysis. 7 8

Order of Reaction Integrated Rate Laws At constant temperature, the rate of a reaction v depends on the concentrations of reactants and products For example, the reaction goes essentially to completion to the right. A + 2 B C The rate of reaction is found experimentally and given by a rate equation v = k[a] α [B] β k is the rate constant. α and β are independent of concentration and time. They are not the stoichiometric numbers in the balanced chemical equation and must be determined by experiment. The exponents α and β are referred to as the order of the reaction with respect to reactant A and B, respectively and are not necessarily integers. The sum of the orders for the reactants is referred to as the overall order of the reaction. In this case the overall order is α + β Although the focus in kinetics is on rates, integrated rate equations can be used so that concentrations measured at various times can be used directly in the quantitative representation of kinetic data. When rate laws are simple, they can be integrated to give the concentration of reactant as a function of time. Integrated forms will be derived for first-order, second-order, zero-order, and higher-order reactions for certain special cases. All of the integrated equations for simple rate laws apply only at constant temperature and volume for reactions that go to completion. 9 10 Integrated Rate Laws - First Order Reaction Integrated Rate Laws - First Order Reaction The rate equation for a first order reaction A products = k[a] may be integrated after it is written in the form [A] = k If the concentration of A is [A] 1 at t 1 and [A] 2 at t 2 then, [A]2 [A]1 = k t2 t1 ln [A] 1 [A] 2 = k(t 2 t 1 ) An especially useful form of the first order rate law is obtained if t 1 is taken to be zero, and the initial concentration is represented by [A] 0 is given as or and ln [A] 0 [A] = kt [A] = [A] 0 e kt ln[a] = ln[a] 0 kt The last form indicates that the rate constant k may be calculated from a plot of ln[a] versus t, and the slope of the line in such a plot is k. The rate constant for a first-order reaction has units of reciprocal time. 11 12

Integrated Rate Laws - First Order Reaction Integrated Rate Laws - First Order Reaction For a first-order reaction so that where It is evident from aa products v = 1 = k[a] a = ak[a] = k A [A] k A = ak ln [A] 1 [A] 2 = k(t 2 t 1 ) that it is necessary to only determine the ratio of concentrations at two times to determine the rate constant for a first-order reaction. 13 The half-life t 1 2 of a reaction is the time required for half of the reactant to disappear. For the first-order reaction A products k = 1 ln 1 = 0.693 t 1 1 t 1 2 2 2 For a first-order reaction the half-life is independent of the initial concentration. Thus 50% of the substance remains after one half life, 25% after two half-lives, 12.5% after three... The relaxation time τ for a first-order reaction is equal to the reciprocal of the first-order rate constant: τ = 1 leading to the equation k [A] = [A] 0e t τ 14 Integrated Rate Laws - Second-Order Reaction Integrated Rate Laws - Second-Order Reaction A reaction is second order is the rate is proportional to the square of the concentration or is proportional to the product of concentrations of two reactants. If the rate is proportional to the square of the concentration of A in the reaction A products, the rate law = k[a] 2 may be integrated after rearranging it in the form [A] 2 = k If the concentration is [A] 0 at t = 0 and [A] at time t, integration yields kt = 1 [A] 1 [A] 0 1 Thus, a plot of [A] versus t is linear for a second-order reaction, and the slope is equal to the second-order rate constant. The rate constant for a second-order reaction has the units (concentration) 1, i.e. L mol 1 s 1. 15 The half-life for a second-order reaction is given by t 1 = 1 2 k[a] 0 Thus, the half-life of a second-order reaction is inversely proportional to the initial concentration. If the reaction aa products is second order, then which gives with a half-life given by v = 1 = k[a] 2 a akt = 1 [A] 1 [A] 0 t 1 = 1 2 ak[a] 0 16

Integrated Rate Laws - Second-Order Reaction Integrated Rate Laws - Second-Order Reaction The integrated rate equation for k[a] 2 also applies if the rate is given by k[a][b] where the stoichiometry is represented by A + B products and A and B are initially the same concentration. For a second order reaction A + B products with [A] 0 = [B] 0, the half-life is inversely proportional to [A] 0, i.e. t 1/2 = 1 k[a] 0 Thus, when [A] 0 is reduced by a factor of 2, the half-life doubles. A different rate law for a second-order reaction is obtained if the rate is given by [A][B], and the stoichiometry is given by aa + bb products The rate is defined by v = 1 = 1 d[b] = k[a][b] a b If the initial reactants are not in stoichiometric proportions (i.e b[a] 0 a[b] 0 ), then the integrated rate equation is kt = 1 ln [A] [B] 0 b[a] 0 a[b] 0 [A] 0 [B] If the stoichiometric numbers of A and B are unity, the above equation becomes kt = 1 ln [A] [B] 0 [A] 0 [B] 0 [A] 0 [B] 17 18 Integrated Rate Laws - Zero-Order Reaction A reaction is zero order if the rate is independent of the concentration of the reactant: This can occur if the the rate is limited by the concentration of the catalyst or if the rate is determined by the light intensity in a photochemical reaction. Then, k may be proportional to the concentration of catalyst or to the light intensity. Integration of the above equation yields 19 = k [A] 0 [A] = kt The zero-order rate constant has the units mol L 1 s 1. Reversible First-Order Reactions For a system in which both the forward and backward reactions are important, the net rate of reaction can generally be expressed as the difference between the rate in forward direction and that in the backward direction. Consider the reversible reaction A k1 B k2 The rate law for this reversible reaction is = k 1 [A] + k 2 [B] If initially only A is present, then [B] = [A] 0 [A] = k 1 [A] + k 2 ([A] 0 [A]) = k 2 [A] 0 (k 1 + k 2 )[A] 20

Reversible First-Order Reactions ( = (k 1 + k 2) [A] k1 k 1 + k 2 where the expression for [A] is obtained as follows: [B] eq [A]0 [A]eq = = k1 = K [A] eq [A] eq k 2 ) [A] 0 = (k 1 + k 2)([A] [A] eq) where K is the equilibrium constant for A B. This equation can be solved for [A] eq: Reversible First-Order Reactions Integrating yields = (k 1 + k 2)([A [A] eq) [A]0 [A]eq ln = (k 1 + k 2)t [A] [A] eq For this reaction, the concentrations of [A] and [B] as a function of time are illustrated below. [A] eq = k2 k 1 + k 2 [A] 0 The expression for the equilibrium constant in terms of the forward and backward rate constants can be used to eliminate k 2 ( = k 1[A] 1 [B] ) [A]K Thus, it can be seen that the reaction has an initial rate of k 1[A] and that is slows down as [B] accumulates. When [B] = K, the reaction is at equilibrium, and the rate = 0. [A] It should be noted that the rate of approach to equilibrium in this reaction is determined by the sum of the rate constants of the forward and reverse reactions, not by the rate constant for the reverse reaction. 21 22 Consecutive First-Order Reactions Two consecutive irreversible first-order reactions can be represented by k A 1 k 1 2 A2 A3 To determine the way in which the concentrations of the substances in such a mechanism depend on time, the rate equations are first written down for each substance: d[a 1] = k 1 [A 1 ] d[a 2] = k 1 [A 1 ] k 2 [A 2 ] d[a 3] = k 2 [A 2 ] It is then necessary to obtain the solution of these simultaneous differential equations. It will be assumed that, at t = 0, [A 1 ] = [A 1 ] 0, [A 2 ] = 0, [A 3 ] = 0 23 Consecutive First-Order Reactions The rate for [A 1] is readily integrated to obtain Substitution of this expression into d[a2] which may be integrated to obtain d[a 2] [A] 1 = [A 1] 0e k1t = k 1[A 1] k 2[A 2] yields = k 1[A 1] 0e k1t k 2[A 2] [A 2] = k1[a1]0 k 2 k 1 ( e k1t e k2t) Because of the conservation of the number of moles, [A 1] 0 = [A 1] + [A 2] + [A 3] at any time, so the concentration of [A 3] is given by ] 1 [A 3] = [A 1] 0 [1 + (k 2e k1t k 1e k2t ) k 1 k 2 24

Consecutive First-Order Reactions The Steady-State Approximation The adjacent figure shows the concentrations of A 1, A 2 and A 3 as a function of time when k 1 = 1s 1, and k 2 = 1 s 1 (a), 5 s 1 (b), 25 s 1 (c) Note the induction period in (a) for the appearance of A 3. It is not formed initially because A 2 has to be formed first. As k 2 is increased, the induction period becomes less important. Also note that as k 2 becomes larger than k 1, less A 2 is formed after that induction period, i.e. d[a 2] 0 This is the basis for the steady-state approximation that is often useful in deriving rate equations for systems of reactions. When k 2 >> k 1, the steady-state approximation can be used to treat the kinetics of the system of reactions k1 k2 A 1 A 2 A 3. In the steady state (ss), the rate of change of [A 2] = 0 so that d[a 2] = k 2[A 2] + k 1[A 1] = 0 The concentration of A 2 in the steady state is given by [A 2] ss = ( k1 )[A1], and the steady-state k2 concentration of A 2 is given by [A 2] ss = k1 [A 1] 0e k1t k 2 The steady state-concentration of A 3 is given by ( ) ] [A 3] ss = [A 1] 0 [A 1] [A 2] ss = [A 1] 0 [1 1 + k1 e k1t k 2 which agrees with the previous equation for [A 3] when the term k 1e k2t is neglected. Thus, as k 2 becomes larger and larger with respect to k 1, the behavior of this system approaches that of A 1 A 3 25 26 The Steady-State Approximation Consider the following sequential reaction scheme A k A k I 1 k 1 2 I2 P In this reaction, product formation results from the formation and decay of two intermediate species, I 1 and I 2. The differential rate expressions for this scheme are given below: = k A [A] d[i 1] = k A [A] k 1 [I 1 ] d[i 2] = k 1 [I 1 ] k 2 [I 2 ] d[p] = k 2 [I 2 ] The Steady-State Approximation One approach in determining the time-dependent concentrations for the species involved in this reaction is the use of Euler s method for solving differential equations to determine numerically the concentrations as a function of time. The result of this approach for k A = 0.02 s 1, k 1 = k 2 = 0.2 s 1 is provided in the adjacent figure. The relative magnitude of the rate constants results in only modest intermediate concentrations. The figure also illustrates that [I 1] and [I 2] change very little with time such that the time derivative of these concentrations can be set approximately equal to zero. d[i] = 0 which is known as the steady-state approximation. 27 28

The Steady-State Approximation The Steady-State Approximation The steady-state approximation is used to evaluate the differential rate expressions by simply setting the time derivative of all intermediates to zero. This approximation is particularly good when the decay rate of the intermediates is greater than the rate of production so that the intermediates are present at very small concentrations during the reaction. Applying the steady-state approximation to [I 1 ] results in the following expression: d[i 1 ] SS = 0 = k A [A] k 1 [I 1 ] SS [I 1 ] SS = k A k 1 [A] = k A k 1 [A] 0 e kat where the subscript SS indicates that the concentration is spredicted using the steady-atate approximation. The corresponding expression for [I 2 ] under the steady-state approximation is d[i 2 ] SS Finally, the differential expression for P is = 0 = k 1 [I 1 ] SS k 2 [I 2 ] SS [I 2 ] SS = k 1 k 2 [I 1 ] SS = k A k 2 [A] 0 e kat d[p] SS Integration of the above equation results in = 0 = k 2 [I 2 ] SS = k A [A] 0 e kat [P] = [A] 0 (1 e kat ) which shows that the formation of P is consistent with the first order decay of A withing the steady-state approximation. 29 30 The Steady-State Approximation Reaction Mechanisms The steady state approximation requires that the concentration of intermediate be constant as a function of time. It is valid when k 1 and k 2 are sufficiently large that [I 1] and [I 2] are small at all times. A comparison between the numerically determined concentrations and those predicted for the steadystate approximation for the two-intermediates where k A = 0.02 s 1, k 1 = k 2 = 0.2 s 1 is provided in the adjacent figure. Even for those conditions where the steady-state approximation is expected to be valid, the discrepancy between [P] and [P] ss is evident. The steady-state approximation is difficult to implement if the intermediate concentrations are not isolated to one or two of the differential rate expressions derived from the mechanism of interest. As discussed previously, the order of a reaction with respect to a given reactant is not determined by the stoichiometry of the reaction. The reason for the inequivalence of the stoichiometric coefficient and reaction order is that the balanced chemical reaction provides no information with respect to the mechanism of a reaction. A reaction mechanism is defined as the collection of individual kinetic processes or elementary steps involved in the transformation of reactants to products. The rate law expression for a chemical reaction, including the order of a reaction is entirely dependent on the reaction mechanism. In contrast, the Gibbs free energy for a reaction is dependent on the equilibrium concentration of reactants and products. 31 32

Reaction Mechanisms All reaction mechanisms consist of elementary reaction steps or chemical processes that occur in a single step. The molecularity of a reaction step is the stoichiometric quantity of reactants involved in the reaction step. For example, a unimolecular reaction involves a single reactant species as in the decomposition of a diatomic molecule into its atomic fragments: k d 2I I 2 Bimolecular reaction steps involve the interaction of two reactants, e.g. the reaction of nitric acid with ozone: NO + O 3 k r NO2 + O 2 33 Reaction Mechanisms The importance of elementary reaction steps is that the corresponding rate law expression for the reaction can be expressed in terms of the molecularity of the reaction. For the unimolecular decomposition of I 2, the rate law expression is that of a first-order reaction and is given as v = d[i 2] = k d [I 2 ] and the rate law expression for the bimolecular reaction of NO and O 3 is v = d[no] = k r [NO][O 3 ] The order of a reaction can be inferred from the molecularity of a reaction. The equivalence of order and molecularity is true for only elementary reaction steps. 34 Reaction Mechanisms Microscopic Reversibility and Detailed Balance A common problem in kinetics is identifying which of several proposed reaction mechanisms is the correct mechanism. A general rule of kinetics is that although it may be possible to rule out a proposed mechanism, it is never possible to prove unequivocally that a given mechanism is correct. The following example illustrates this rule. Consider the reaction A P As written, this reaction is a simple first-order transformation of reactant A into product B, and it may occur through a single elementary step. However, what if the reaction were to occur through two elementary steps as follows: A k1 I k2 I P In order for a reaction mechanism to be valid, the order of the reaction predicted by the mechanism must be in agreement with the experimentally determined rate law. In evaluating a reaction mechanism, one must express the mechanism in terms of elementary steps. The principle of microscopic reversibility is a consequence of invariance of the classical mechanics when time is reversed. Consider a particle that moves under the action of a force that is a function of position only: The particle moves from position r(0) at t = 0 to r(t 1) at t = t 1 The initial velocity of the particle is v(0), and its velocity at t 1 is v(t 1). Now suppose that at time t 1 all the components of velocity v(t 1) could be instantaneously reversed and allow the particle to move for another time period t 1. The particle would retrace its path to its initial position r(0), but the velocity components would be reversed. The reversed trajectory can be thought of as beginning at time t 1 and evolving as time goes forward to t = 0. 35 36

Microscopic Reversibility and Detailed Balance Rate-Determining Steps A system is said to be invariant under time reversal if the equation is invariant under the following transformation t = t r(t) = r( t) v(t) = v( t) Starting with the principle of microscopic reversibility, it can be shown that, for an elementary reaction, the ratio of the forward and backward rate constants is equal to the equilibrium constant. For an elementary reaction at equilibrium, A + B C + D [C] eq[d] eq = k f [A] eq[b] eq k b the rate of the forward reaction equals the rate of the backward reaction. In a system with many elementary reactions, the principle of detailed balance applies to each elementary reaction individually. 37 Consider the following sequential reaction scheme which involves a series of elementary reactions: A ka ki I P in this scheme, the reactant A decays to form intermediate I, and this intermediate undergoes subsequent decay resulting in the formation of product P The differential rate expressions can be written as follows d[i] d[p] = k A [A] = k A[A] k I [I] = k I [I] The above expressions follow naturally from the elementary reaction steps in which a given species participates. The rate of product formation for a sequential reaction was found to depend on the timescale for the production and decay of the intermediate species. There are two limiting situations discussed on the next two slides. 38 Rate-Determining Steps Rate-Determining Steps The rate constant for the intermediate decay is much greater than the rate constant for the production: k I >> k A. In this limit, any intermediate formed will rapidly go on to product, and the rate depends on the rate of reactant decay. The rate constant for the intermediate production is significantly greater than the intermediate decay: k A >> k I. In this limit, the reactants quickly produce intermediate, but the rate depends on the rate of intermediate decay. 39 40

Parallel Reactions In many situations, a single reactant can become a variety of products, and these reactions are called parallel reactions. Consider the following reaction in which the reactant A can form one of two products, B or C: Parallel Reactions The differential rate expressions for the reactant and products are: = k B [A] k C [A] = (k B + k C )[A] d[b] d[c] = k B [A] = k C [A] Integration of the above equation involving [A] with initial conditions [A] 0 0, [B] = [C] = 0 yields [A] = [A] 0 e (k B+k C )t The product concentrations can be determined by substituting the expression for [A] into the differential rate expressions and integrating [B] = k B [A] k B +k 0 (1 e (k B+k C )t ) C [C] = k C k B +k C [A] 0 (1 e (k B+k C )t ) 41 42 Parallel Reactions Parallel Reactions The adjacent figure shows the reactant and product concentrations for a branching reaction when k B = 2k C = 0.1 s 1. 1. The decay of A occurs with an apparent rate constant equal to k A + k C, the sum of the rate constants for each reaction branch. 2. The ratio of product concentrations is independent of time. At any time point, the ratio [B] [C] is identical. 3. This behavior is consistent with the last two equations given above where the ratio of the concentrations is predicted to be [B] [C] = k B k C 4. The above equation states that as the rate constant for one on the reaction branches increases relative to the other, the greater the concentration of the corresponding product will be. The equation [B] [C] = kb demonstrates that the extent of product formation is dependent kc on the rate constants. On can view this in terms of probability where the larger the rate constant for a given process, the more likely the product will be formed The yield Φ is defined that a given product will be formed from by decay of the reactant: Φ i = k i n k n where k i is the rate constant for the reaction leading to formation of the product of interest. The denominator is the sum of the rate constants for the reaction branches. The total yield is the sum of the yields for forming each product, and it is normalized such that Φ i = 1 i 43 44

Parallel Reactions - Example Problem Temperature Dependence of Rate Constants Under acidic conditions, benzyl penicillin (BP) undergoes the following parallel reaction: where R 1 and R 2 indicate alkyl substitutents. In a solution where the ph = 3, the rate constants for the processes at 22 C are k 1 = 7.0 x 10 4 s 1, k 1 = 4.1 x 10 3 s 1, k 1 = 5.7 x 10 3 s 1. What is the yield for P 1 formation? Φ P1 = k 1 7.0 x 10 4 s 1 = k 1 + k 2 + k 3 7.0 x 10 4 s 1 + 4.1 x 10 3 s 1 + 5.7 x 10 3 = 0.067 s 1 Of the BP that undergoes acid-catalyzed dissociation, 6.7% will result in the formation of P 1. The dependence of rate constants on temperature over a limited range can usually be represented by an empirical equation proposed by Arrhenius in 1889: k = Ae Ea RT where A is the pre-exponential factor and E a is the activation energy. The pre-exponential factor A has the same units as the rate constant. The above equation may be written in logarithmic form ln k = ln A E a RT giving a straight line when the logarithm of the rate constant is plotted against the absolute temperature and is called an Arrhenius plot. 45 46 Temperature Dependence of Rate Constants Differentiating the equation for an Arrhenius with respect to temperature plot yields E a = RT 2 d ln k dt which may be regarded as the definition of activation energy. This equation can be integrated to obtain ln k 2 = E ( ) a T2 T 1 k 1 R T 2 T 1 Temperature Dependence of Rate Constants The activation energy, E a, in the Arrhenius expression, corresponds to the energy needed for the chemical reaction to occur. One can envision a chemical reaction occurring along an energy profile illustrated in the adjacent figure. If the reactants have an energy greater than the activation energy, E a, then the reaction will occur. The exponential dependence on the activation energy is consistent with Boltzmann statistics with e Ea RT representing the fraction of molecules with sufficient kinetic energy to get over the energy barrier given by E a. As the activation energy increases, the fraction of molecules with sufficient energy to react will decrease, as will the reaction rate. 47 48

Distribution of Kinetic Energies as a Function of Temperature Lower Temperature Temperature Dependence of Rate Constants Example Problem The temperature dependence of the acid-catalyzed hydrolysis of penicillin is investigated, Fraction of Molecules Minimum Kinetic Energy Needed to Overcome Ea Higher Temperature Kinetic Energy Distribution of energies of molecules is a gaseous reaction mixture. Only the those molecules have sufficient kinetic energy to overcome the activation energy barrier (E a ) will react. The higher the temperature, the larger the fraction of molecules that are able to react. 49 Temperature Dependence of Rate Constants Example Problem A plot of ln k 1 versus 1 T figure. is shown in the adjacent The data are indicated by points, and the solid line corresponds to the linear least-squares fit to the data. The equation for the line is and the dependence of k 1 on temperature is given in the following table. Temperature ( C) k 1 (s 1 ) 22.2 7.0 x 10 4 27.2 9.8 x 10 4 33.7 1.6 x 10 3 38.0 2.0 x 10 3 What is the activation energy E a and the Arrhenius pre-exponential factor for this branch of the hydrolysis reaction? 50 Activated Complex Theory The activated complex theory or transition state theory depends on the concept of equilibrium for a theoretical description of reaction rates. Consider the reaction A + B k P where the reaction coordinated for this process is illustrated in the figure. ln k = (-6306.3 K) 1 T + 14.1 The slope of the line is equal to Ea R : 6306.3 K = Ea R ; The intercept is equal to ln A: Ea = 52.4 kj mol 1 A = e 14.1 = 1.33 x 10 6 51 A and B react to form an activated complex that undergoes decay, resulting in product formation. The activated complex represents the system at the transition state, but this complex is not stable and has a lifetime on the order of one or a few vibrational periods ( 10 14 s). Activated complex theory involves the following major assumptions: 1. An equilibrium exists between the reactants and the activated complex. 2. The reaction coordinate describing the reaction can be mapped onto a single degree of freedom for the activated complex. For example, if product formation involves bond breaking, the vibrational degree of freedom corresponding to bond breaking is taken to be the reaction coordinate. 52

Activated Complex Theory For a biomolecular reaction, the kinetic mechanism corresponding to the activated complex model is A + B k 1 AB k 1 AB k 2 P The top equation represents an equilibrium between the reactants and the activated complex, while the bottom reaction corresponds to the decay of the activated complex to form product. The central result of activated complex theory is given by: k2 = κ k BT hc K c The Eyring Equation It is important to connect the above equation to the earlier Arrhenius description of chemical reactions. G = RT ln K and G = H T S leads to k2 = k BT hc e G RT = k BT hc e S R Making the appropriate comparison with the Arrhenius pre-exponential factor gives e H RT solutions, unimolecular Ea = H + RT A = ek B T e S R h solutions, bimolecular Ea = H + RT A = ek B T hc e S R gas, unimolecular Ea = H + RT A = ek B T e S R h gas, bimolecular Ea = H + 2RT A = e2 k B T hc e S R 53