F U N D A M E N T A L C H E M I C A L

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1 F U N D A M E N T A L E M I A L P R I N I P L E S E M I A L K I N E T I S OIONI, DYNAMISM OF A SOER PLAYER MUSEUM OF MODERN ART, NEW YORK J. DAVID RAWN (JANUARY 2008) Organic hemistry I and II

2 Practical Kinetics First Order Reactions onsider the reaction A P (1) When A is converted to P ---- v = d[a] = + d[p] = [A] 1 (2) Equation (2), which defines the algebraic relation between reaction velocity, v, and concentration, is called the rate law for the reaction. The exponent in the above equation in the term [A] 1 means that the rate depends upon the concentration of the reactant, [A], raised to the first power; i.e. it means that if the concentration of A is doubled, the rate also doubles. experiment. The exponent, 1, is called the order of the reaction in A. Equation (2) is strictly empirical, which means that in general we do not know the order of a reaction in a given reactant until we have carried out the Thus, when reaction 1 is said to be first order in A. The term k1 is the first order rate constant for the reaction, where the subscript means first order. What are the units of the first order rate constant, k1? To find out, let s look at equation 2 again. The velocity is the change in concentration divided by time; i.e. mol -1 - L -1 -s -1, and the units on the right side of equation 2 are mol-l -1. Thus, mol-1 L-1 s -1 = k mol L-1 (3) The terms for concentration on the right and left hand side of equation (3) cancel, so the units of a first order rate constant are s -1. We can eliminate the term -d[a]/ in equation 2 by integration. v = d[a] [A ] t d[a] = = k [A] [ A] o t o (4) Solving this equation gives the following result. ln[a] + ln[a o ] = kt (5) We can rewrite this as, ln [A o ] [A] = kt (6) Taking the antilog gives, [A] = [A o ]e kt If we make a plot of the natural log (ln) of the concentration of A versus time for a first order reaction, we obtain a straight line whose slope is -k. 2

3 Now let s consider a special case in which the initial concentration of A, [Ao], has decreased by one-half. Substituting these values in equation (6), we obtain ln [A o ] [0.5A] = kt 1/ 2 (7) So that, ln 2 = kt1/2 = (8) The term t1/2 is called the half-life of the reaction. The term t1/2 is often given the symbol τ (tau). Note that for a first order reaction the half-life, τ, is independent of the initial concentration of the reaction, A. Second Order Reactions onsider a reaction in which two reactants are converted to two products. A + + D If this reaction is first order in both A and, then it is second order overall. The rate, or velocity of this reaction is given by v = d[a] = d[] + d[] + d[d] (9) That is, the rate at which either A or disappear is exactly equal to the rate at which and D are produced. We can monitor the rate of reaction experimentally by picking the reactant or product that is easiest to detect. If experimental analysis reveals that the reaction is indeed first order in both A and, then the rate of the reaction is given by, v = k2 [A][] (10) The term k2 in equation (1) is called a second order rate constant (the subscript 2 ) is the moniker for second order. What are the units of k2? As we saw earlier, we want to substitute the units for v, and for the concentrations of A and into equation (10). mol-1 L-1 s -1 = k2 (mol L -1 )(mol-l -1 ) (11) Two concentration terms cancel. Dividing equation (11) by the remaining concentration term, we find that the units of a second order rate constant are L-mol -1 -s -1. If we integrate the equation for a second order reaction, we obtain equation (12). 1 [ o ] [A o ] ln[a ]([ x] o o [ o ]([A o x] = k t 2 (12) where Ao and o are the initial concentrations of A and, respectively, x is the amount of A and have reacted at time t, and k2 is the second order rate constant. This is a bit cumbersome. Suppose we begin the reaction with equal concentrations of A and. the rate law for the reaction then becomes, v = d[a] = k 2 [A][A] = k 2 [A] 2 (13) 3

4 When we integrate equation 13, we obtain, 1 [A] 1 [A o ] = kt (14) It is certainly much easier to work with equation (14) than with equation (13). What is the half-life for the second order reaction defined in equation (14)? To find out, we ll set the concentration of A to 0.5Ao. 1 [0.5A o ] 1 [A o ] = kt = kτ 1/ 2 (15) When we solve equation (15) for τ, we find the relation of the half-life of a second order reaction depends upon the concentration of A, which is quite unlike the behavior of a first order reaction. τ = 1 [A o ]k 2 (16) Pseudo First Order Reactions We have just seen that the rate law for a first order reaction is simpler than the rate law for a second order reaction. It turns out to be quite easy to set up reaction conditions for a second order reaction so that its kinetics will resemble a first order reaction. When this is the case, we say that the reaction is pseudo first order. This situation results if we pick one of the two reactants, say A, and make its initial concentration much greater than the concentration of. Then, we will follow the reaction experimentally only until about 1% of has reacted. At this time the concentration of A will scarcely have changed at all. So the rate law reduces to v = k2[ao]([]-x) (17) Since the initial concentration of A, Ao, is virtually constant, we can rewrite equation (17) as where kapp is the apparent first order rate constant. That is, kapp = k2[ao] (18) Thus, kapp is a pseudo first order rate constant. We can obtain the true second order rate constant by dividing kapp by Ao. We could, of course, have done the same kind of analysis by setting the initial concentration of much greater than that of A. Order and Molecularity When we study reaction mechanisms among the things we would like to discover are the regiospecificty, stereochemistry, and the number of chemical species that are present in the transition state, or activated complex of the of the reaction. The number of chemical species in the transition state is called the molecularity of the reaction. It is easy to fall into the trap of imagining that the molecularity of a reaction is correlated with the order of the reaction. ut we have just seen that the reaction conditions can be manipulated to turn a reaction with second order kinetics into one that is pseudo first order. Thus, molecularity and the order of a given reaction are not deeply related. We can, however, use kinetic data to make inferences about the order of a chemical reaction. onsider the following process in which tert-butyl bromide reacts with water (in a solvent such as tetrahydrofuran) to give tertbutyl alcohol. 4

5 3 r 2 O TF 3 O Kinetic studies show that this reaction is first order in t-butyl bromide and zero order in water. The overall rate is determined by the slow step, so water does not appear in the rate law, which is v = d[tur] = + d[tuo] = [tur] This result might lead us to propose a mechanism in which a carbocation forms in a slow step that s why the reaction is first order in the halide and that the carbocation is captured in a subsequent fast step by water. This process is shown below. 3 r slow 3 fast 2 O 3 O fast O The kinetic result does not prove that a carbocation forms, but it is one piece of evidence supporting the existance of a carbocation. The above reaction is unimolecular because only one molecular species is present in the transition state for the slow, or rate-determining step of the reaction in which the carbocation forms. δ + δ r 3 Unimolecular transition state Next, let s consider a reaction in which bromoethane reacts with water to give ethanol. r 2 O O This reaction is first-order in both in the alkyl halide and in water. The rate law therefore is given by, v = + d[eto] = k 2 [Etr][ 2 0] 5

6 This rate law suggest that the transition state for the reaction has two molecular species, i.e., that it is a bimolecular reaction. δ O δ + r δ 3 imolecular transition state The Steady State Approximation Let us suppose that in the conversion of reactant A to product occurs through the formation of an intermediate, (Scheme I). A k 2 Scheme I The first order rate constants for the two steps are k1 and k2. If the magnitudes of the two rate constants are comparable, this is rather complicated, but if k2 >> k1, then we can write the rate law as shown below in equation (19). v = d[a] = + d[] = [A] (19) Since k2 >> k1, d[]/ = 0. This is the steady state approximation, equation (20). d[] = [A] k 2 [] = 0 (20) Therefore, In other words, k1[a] = k2[] (21) v = d[] = [A] = k 2 [] (22) Now let s make the situation more complex, and more realistic too, by making the first step reversible, Scheme II. A k -1 k 2 Scheme II 6

7 The form of the rate law for Scheme II depends upon the relative magnitudes of k1, k-1, and k2. Suppose that k1 >> k2 and that k-1, >> k2. In this case, the second step, in which is converted to is rate-determining. v = k2[] (23) owever, since k1 >> k2 and that k-1, >> k2, is a transient intermediate (think of the carbocation in the reaction shown above, for instance), whose concentration is constant, i.e. it exists in a steady state. That is, d[] = 0 (24) The intermediate can form by one pathway, but it can disappear by two pathways. The rate law for the formation and disappearance of is given by equation (25). d[] = + [A] k 2 [] k 1 [] = 0 (25) We can rewrite equation (25) as, k1 [A] = (k-1 + k2)[] (26) Solving for yields, [] = [A] (k 1 + k 2 ) (27) This means that the rate law for Scheme II reduces to equation (28). v = k 2 [A] (k 1 + k 2 ) (28) The rate constants in equation (28), which are not observed directly, are called microscopic rate constants. The combination of these constants (k1, k-1, and k2) give an apparent or observed first order rate constant (equation 29). k observed = Interpretation of Rate onstants k 2 (k 1 + k 2 ) (29) What good is a rate constant? That is, once we have obtained the rate constant for a given reaction, how can interpret it? One of the first attempts, to interpret rate constant is the Arrhenius equation (equation 30). k obs = Ae E a / RT (30) The parameter Ea in equation 30 is called the activation energy for the reaction. this is the minimum energy required to allow the reaction to proceed. R is the ideal gas constant, T is the Kelvin temperature, kobs is the observed rate, and the parameter A is called the pre-exponential factor. It is a statistical correction factor. We can rearrange the Arrhenius equation by dividing by A (equation 31). 7

8 k obs A = e E a / RT (31) Now let s take the natural logarithm (equation 32). ln k obs A = E a RT (32) Note that the activation energy, Ea, is independent of temperature. Solving for Ea gives equation 33. E a = RT ln k obs A (33) The term kobs/a is a dimensionless constant. So, we can write, Ea = -RT ln (34) We also recall that the equilibrium constant for a reaction, Keq, is also a dimensionless constant. The form of equation 34 is familiar. We recall from an earlier chapter on thermodynamics that ΔG o = RT ln Keq (35) Equations having similar forms are always intriguing, and in the case we can ask whether we can give the activation energy a thermodynamic significance. The answer is yes, as we shall see in the next section. Transition State Theory We recall that the transition state, or activated complex, for a chemical reaction is a transient configuration of atoms or molecules that lies at the point of maximum energy on the minimum energy pathway leading from reactants to products. y analogy, we can think of this maximum energy position as a saddle point, or the top of a mountain pass. onsider the following equilibrium for the interconversion of a reactant A to a product in a first order reaction. A Scheme III We assert that reactant A is in equilibrium with an activated complex, A, with equilibrium constant, K, as shown in Scheme IV. A K A Scheme IV y the usual conventions for defining an equilibrium constant, K is written as K = [A ]/[A] (36) Since this is a first order reaction, the rate law is 8

9 v = d[a] = [A] (37) We next assert that k1[a] = k [A ] (38) That is, the rate of the reaction depends only on (a) the concentration of the activated complex, A, and (b) on its absolute rate constant, k. The absolute rate constant, k, is the rate of passage over the potential energy barrier separating the reactants and products. k is a constant, k = k b T h (39) where kb is the oltzmann constant, T is the Kelvin temperature, and h is Planck s constant. When we substitute equation 37 into equation 35 we obtain, k1 = k K (40) orresponding to equilibrium constant, K, there is a free energy of activation, ΔG. Therefore, we can write, = k bt h ΔG exp RT (41) where, so that, ΔG = -RT ln K ΔG RT = lnk (42) and, K = exp ΔG RT (43) Furthermore, by analogy with equilibrium thermodynamics we can write a relation between the free energy of activation, ΔG, and the enthalpy, Δ and entropy of activation, ΔS. Thus, ΔG = Δ TΔS (44) Therefore, equation 40 can be rewritten as, 9

10 = k bt h Δ exp exp ΔS RT R (45) It can also be shown that there is a simple relation between the Arrhenius activation energy, Ea, and the enthalpy of activation, Δ. Δ = Ea nrt (46) where n is the order of the reaction. We recall that the Arrhenius equation contains a pre-exponential term called the statistical factor, A. It is aesthetically very pleasing that the statistical term is related to the entropy of activation, (equation 47). A = k b T h ΔS exp e n R (47) where n is again the order of the reaction and ΔS is the order of the reaction. In sum and Ea Δ A ΔS The Theoretical asis of Transition State Theory When we ask what occurs when a reaction occurs, we can consider bond-breaking to results from converting a vibration to a translation. If ν is the vibrational frequency of the bond in its zero point potential energy and ν is the vibrational frequency of the bond that is breaking in the transition state for the reaction, we can write Δε = hν h ν (48) or Δε = h(ν ν ) (49) The reaction coordinate, then, is a normal vibrational mode that is being converted to a translation. Thus, the molecule in the transition state has lost one degree of freedom so that instead of 3N-6 degrees of vibrational freedom, it now has 3N-7 degrees of vibration freedom. Remember that a normal mode of vibration can be represented as a function of one coordinate; let s say the bond is breaking along the x-coordinate. Thus, the potential energy, V, can be represented as a function of only one dimension, r; i.e. V = V(r) (50) All of the other vibrational modes are orthogonal to the reaction coordinate. Thus, changes in the breaking bond do not affect the other vibrational modes. It turns out that the vibrational stretching frequency, ν, is related to the force constant of the bond and the mass of the smallest member of the vibrating bond for instance a - bond in a much larger organic molecule. hanges in rate of reaction that result from isotopic substitution are called kinetic isotope effects. To a first approximation kinetic isotope effects reflect differences in the mass of the isotopes, and to a first approximation bonding forces are constant. The vibrational frequency ν is related to the mass of the atom in question (here deuterium or hydrogen, and the force constant for the bond, k. 10

11 ν = 1 k 2π m 1/ 2 (51) The vibrational energy of a bond, εn, is related to the stretching frequency, v, by the relation εn = (n + 1/2)hν where n = 0, 1, (52) In the zero point vibrational energy level, εn = (1/2)hνo (53) The zero point vibrational energy lis lower bond a bond between and atom, let s say a carbon atom, and deuterium ( 2 ), than between a carbon atom and hydrogen. Therefore, more energy is required to break a carbon-deuterium bond than to break a carbon-hydrogen bond. This is reflected in ΔG and in the rate constants for breaking those bonds, kd and k, respectively. If we call the transition state for breaking these bonds X, then we can write, and Δ = Δ o (X ) Δ o reactants (54) ΔS = ΔS o (X ) ΔS o reactants (55) Remember that in transition state theory the activation energy, Δε = h (v v) (56) Therefore, the change in the activation energy, ΔΔε, give a change in the rate constant and not a change in k. If the rate-determining step in a reaction is cleavage of a - or -D bond, the ratio of the rate constants, k and kd, will be greater than 1.0. k k D >1 (57) On the other hand if k k D 1 then the rate-determining step of the reaction is not cleavage of the - (or -D) bond. (58) The ammond Postulate We are now in a position to discuss the ammond Postulate in a slightly different context. We recall that the ammond Postulate states that the structure of transition state of an exothermic reaction resembles the reactants. This means that very little structural change is required to convert the reactants to the transition state. Therefore, Δ exothermic reaction >> Δ (59) Similarly, the structure of transition state of an endothermic reaction resembles the products, which means that considerable structural change is required to convert the reactants to the transition state. 11