Foundations of Chemical Kinetics Lecture 18: Unimolecular reactions in the gas phase: RRK theory Marc R. Roussel Department of Chemistry and Biochemistry
Frequentist interpretation of probability and chemical probability ◮ When we say that the probability of an event E is P , what do we mean? ◮ One of the earliest concepts of probability is the frequentist interpretation: If we repeat the observation infinitely many times for identically prepared systems, P is the fraction of times that E will occurred. ◮ In chemistry, we often apply this reasoning to molecules, and in fact it is implicit in most applications of the Boltzmann distribution.
Frequentist interpretation of probability and chemical probability (continued) ◮ There is a nuance to our application of this interpretation of probability. ◮ Typically, we consider a system containing many identical molecules which may however be in different states. The states of these molecules were not “identically prepared” in quite the sense envisaged in probability theory, but we assume that our method of preparation and the subsequent equilibration introduces no bias, i.e. that they have, if not identical histories, at least equivalent histories.
Frequentist interpretation of probability and chemical probability (continued) ◮ We imagine picking a molecule at random out of this system and ask “What is the probability that a randomly selected molecule is in state S ?” ◮ Because we generally have very large numbers of molecules, the probability can then literally be interpreted as a frequency, i.e. P ( S ) = N S / N total ◮ Moreover, because all the molecules are in a common container and c = N / V , P ( S ) = [ S ] / c total
Review of combinatorics Permutations ◮ Suppose that we have n distinguishable objects, of which we will pick m , keeping track of the order in which they were picked. This is called a permutation of m objects chosen from n . The number of different permutations is symbolized n P m .
Review of combinatorics Permutations (continued) ◮ I’m going to pick objects in order from my set of n . ◮ For the first one, I have n choices. ◮ For the second one, I have n − 1 choices. The total number of different ways I could pick two objects is therefore n ( n − 1). ◮ For the third one, I have n − 2 choices, so I could pick the first three objects in any of n ( n − 1)( n − 2) ways. . . . ◮ In general, the number of ways I could pick m objects from a set of n objects is n ! n P m = n ( n − 1)( n − 2) . . . ( n − m + 1) = ( n − m )! ◮ Special case: If m = n , we have n P n = n ! / ( n − n )! = n ! / 0! = n !
Review of combinatorics Combinations ◮ Suppose that we have n distinguishable objects of which we will pick m , but we don’t care about the order in which we picked them. The number of combinations (unordered subsets) is denoted � n � n C m or . m ◮ The number of permutations n P m is n ! / ( n − m )!. ◮ For each subset of m objects, there are m ! permutations. The number of permutations therefore counts each combination m ! times. Thus, � n � n ! = n P m m ! = m !( n − m )! m
An important problem: Balls and walls ◮ Suppose that we have j indistinguishable balls that we want to place in s distinguishable rooms. We want to know how many different ways there are to do this. ◮ Rather than placing the balls in rooms with fixed walls, think about a situation with movable walls: | • •|| • | • • • | . . . To make s rooms, we need s − 1 movable walls. (The outer walls don’t move.) The movable walls are indistinguishable objects.
An important problem: Balls and walls ◮ If the balls and walls were distinguishable, there would be ( j + s − 1)! arrangements of the balls and walls. ◮ However, the j balls are indistinguishable, so we are overcounting the number of arrangements by the number of permutations of j objects. Similarly, we are overcounting the number of arrangements by the number of permutations of the s − 1 walls. ◮ The actual number of different arrangements of j balls in s urns is therefore W = ( j + s − 1)! j !( s − 1)!
Stirling’s approximation ◮ In chemistry, we often want to evaluate N ! for large values of N . ◮ In that case, the factorial can be approximated as follows: ln N ! ≈ N ln N − N
RRK theory ◮ Developed independently by Rice and Ramsperger and by Kassel in the late 1920s, hence the name. ◮ Modify the Lindemann mechanism to take into account the formation of the transition state: k 1 A ∗ + M , − − ⇀ A + M ↽ − − k − 1 → A ‡ k ‡ A ∗ k 2 K − − − → P . ◮ Why? A ∗ represents an energized molecule, but getting to the transition state requires that the energy stored in its vibrations move to the correct bond(s). The step with rate constant k 2 K represents this process, which is known in the literature as intramolecular vibrational relaxation (IVR).
RRK theory (continued) ◮ A ‡ should be the fastest decaying species in this mechanism since it isn’t even a stable molecule. ◮ Apply the steady-state approximation for [A ‡ ], and solve for k 2 K . (You’ll see why later.) d [A ‡ ] = k 2 K [A ∗ ] − k ‡ [A ‡ ] ≈ 0 dt ∴ k 2 K = k ‡ [A ‡ ] [A ∗ ] Note: p ‡ = [A ‡ ] / [A ∗ ] is the probability that the energy stored in an energized molecule is in the reactive mode, so k 2 K = k ‡ p ‡ .
RRK theory (continued) ◮ Applying the steady-state approximation the normal way, we get [A ‡ ] = k 2 K k ‡ [A ∗ ] ◮ Since v = k ‡ [A ‡ ], we get v = k 2 K [A ∗ ]. ◮ In the original Lindemann mechanism, v = k 2 [A ∗ ]. Comparing the two, we conclude that k 2 from the Lindemann mechanism is k 2 K in the more detailed RRK mechanism.
RRK theory (continued) ◮ As in Lindemann-Hinshelwood theory, we assume that all normal modes have the same frequency. ◮ Suppose that a particular energized molecule has energy E = j � ω 0 spread over s normal modes. The degeneracy of energy level E is just the number of different ways of storing j quanta in s modes: G ∗ = ( j + s − 1)! j !( s − 1)!
RRK theory (continued) ◮ Suppose that we need at least m quanta in the reactive mode in order for the reaction to occur, with E ‡ = m � ω 0 . The degeneracy of the set of molecules that have at least m quanta in the reactive mode is the number of ways of storing j − m quanta in the s modes (which allows for some of the extra quanta to also be in the reactive mode): G ‡ = ( j − m + s − 1)! ( j − m )!( s − 1)! ◮ The probability that a molecule with j quanta has at least m of them in the reactive mode is therefore p ‡ = G ‡ G ∗ = j !( j − m + s − 1)! ( j + s − 1)!( j − m )!
RRK theory (continued) p ‡ = G ‡ G ∗ = j !( j − m + s − 1)! ( j + s − 1)!( j − m )! ◮ Usually, the transition state corresponds to a large m ≫ s . Since j > m , j is also large. ◮ Apply Stirling’s approximation: ln p ‡ = ln j ! + ln( j − m + s − 1)! − ln( j + s − 1)! − ln( j − m )! ≈ j ln j − j + ( j − m + s − 1) ln( j − m + s − 1) − ( j − m + s − 1) − [( j + s − 1) ln( j + s − 1) − ( j + s − 1)] − [( j − m ) ln( j − m ) − ( j − m )] = j ln j + ( j − m + s − 1) ln( j − m + s − 1) − ( j + s − 1) ln( j + s − 1) − ( j − m ) ln( j − m )
RRK theory (continued) ◮ Two of the terms involve s − 1 ≪ m < j . Use a Taylor expansion for ( a + x ) ln( a + x ) with x = s − 1: ( a + x ) ln( a + x ) ≈ a ln a + x (ln a + 1) ◮ Therefore ln p ‡ ≈ j ln j + ( j − m ) ln( j − m ) + ( s − 1) [ln( j − m ) + 1] − [ j ln j + ( s − 1) (ln j + 1)] − ( j − m ) ln( j − m ) = ( s − 1) [ln( j − m ) − ln j ] � s − 1 � j − m � � j − m = ( s − 1) ln = ln j j � s − 1 � j − m ∴ p ‡ = j
RRK theory (continued) � s − 1 � j − m p ‡ = j � ω 0 and m = E ‡ ◮ Since j = E � ω 0 , we have � s − 1 � E − E ‡ p ‡ = E ◮ Since k 2 K = k ‡ p ‡ , � s − 1 � E − E ‡ k 2 K = k ‡ E ◮ If we think of the reactive mode as a vibration, we can replace k ‡ by the frequency of that mode, ν ‡ .
RRK theory (continued) ◮ In our study of Lindemann-Hinshelwood theory, we found that the probability that s oscillators have a total energy between E and E + dE is E s − 1 � � − E ( k B T ) s ( s − 1)! exp dE k B T ◮ If we assume that the first step is in quasiequilibrium, we have [A ∗ ] [A] = k 1 k − 1 ◮ A ∗ represents molecules with a range of different energies above E ‡ . If we, instead, think of A ∗ as representing molecules with energy between E and E + dE for any E > E ‡ , then E s − 1 [A ∗ ] � � [A] = k 1 − E ( k B T ) s ( s − 1)! exp dE k − 1 k B T
RRK theory (continued) ◮ The Lindemann rate constant is k 1 k 2 [M] = ( k 1 / k − 1 ) k 2 [M] k L = k − 1 [M] + k 2 [M] + k 2 / k − 1 ◮ We now have equations for k 1 / k − 1 and for k 2 = k 2 K . Substituting them in, we get � s − 1 � � � E − E ‡ E s − 1 ν ‡ [M] − E ( k B T ) s ( s − 1)! exp E k B T dk RRK = dE � s − 1 � [M] + ν ‡ E − E ‡ k − 1 E where we write dk L since this represents the rate constant only for reactants with energies between E and E + dE .
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