Correct Symmetry Treatment for X+X Reactions Prevents Large Errors

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Correct Symmetry Treatment for X+X Reactions Prevents Large Errors in Predicted Isotope Enrichment Mark Jacob Goldman, Shuhei Ono, and William H. Green J. Phys. Chem. A, Just Accepted Manuscript • DOI: 10.1021/acs.jpca.8b09024 • Publication Date (Web): 01 Mar 2019 Downloaded from http://pubs.acs.org on March 1, 2019

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The Journal of Physical Chemistry

Correct Symmetry Treatment for X+X Reactions Prevents Large Errors in Predicted Isotope Enrichment Mark Jacob Goldman,† Shuhei Ono,‡ and William H. Green∗,† †Department of Chemical Engineering, Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139, USA ‡Department of Earth, Atmospheric, and Planetary Sciences Massachusetts Institute of Technology, 77 Massachusetts Avenue, Cambridge, MA 02139, USA E-mail: [email protected]

Abstract Confusion over how to account for symmetry numbers when reactants are identical can cause significant errors in isotopic studies. An extraneous factor of two in the reaction symmetry number, as proposed in literature, violates reaction equilibrium and causes huge enrichment errors in isotopic analysis. In actuality, no extra symmetry factor is needed with identical reactants.

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Introduction

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The rate of consumption of molecule A, rA , in a bimolecular reaction can be calculated with

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rA = −k[A][B]. If the two reactants are identical, a stoichiometric coefficient of two appears

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in front of the equation for the consumption of A: rA = −2k[A]2 . The rate coefficient k is

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often computed using transition state theory (TST) at the high pressure limit: 1 qT‡†S /V mT S kB T e k(T ) = κσrxn † mA mB h q † /V q /V A B

−Eo kB T

(1)

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where κ is the tunneling factor, σrxn is the reaction symmetry number, m is the number

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of accessible optical, or ‘mirror image’, isomers of a molecule or transition state, kB is

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Boltzmann’s constant, h is Planck’s constant, Eo represents the difference in zero point

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energy between the transition state and reactants, q represents the partition function of

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molecule or transition state with E = 0 at its zero point energy, † indicates removal of the

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symmetry in the rotational partition function to form σrxn , and ‡ indicates the removal of

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one degree of freedom corresponding to the reaction coordinate.

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This article focuses on a discrepancy in defining the reaction symmetry number, σrxn ,

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when used in TST with two identical reactants. With different reactants, σrxn is defined

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according to eq 2 with σX being the rotational symmetry number of X.

σrxn =

σB σA σT S

(2)

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Competing theories exist for the correct value of σrxn when the two reactants are identical.

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A highly-cited article by Pollak and Pechukas in 1978 argues that σrxn should be increased

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by a factor of two when A and B are identical, since the reactants could be translationally

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interchanged. 2 This assertion conflicted with the previous recommendation by Bishop and

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Laidler who used an alternative method to correct for reaction symmetry, reaction path de-

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generacy (RPD). 3 RPD is the number of ways a reaction can proceed with different atoms

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and is equivalent to the combination of symmetry and optical isomers terms in eq 1. When

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reactants are identical, the RPD algorithm will interchange the two molecules and count

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the number of possible pathways twice. To prevent double-counting due to molecule inter-

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changeability, Bishop and Laidler recommend dividing the RPD by two when the reactants

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are identical. 3 These two competing perspectives on how to deal with identical reactants

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have persisted in the literature. A widely-cited review 4 recommends multiplying the reac-

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tant symmetry numbers by a factor of two, referencing Pollak and Pechukas, though many

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studies involving identical reactants do not mention, and likely do not include, this factor of

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two. 5–9

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Though a factor of two mistake can impact many phenomena, it causes huge errors in

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isotopic enrichment since the disputed factor of two is much larger than most non-hydrogen

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kinetic isotope effects. 10

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This paper uses two methods to explain proper symmetry treatment in reaction kinetics.

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We first show that the factor of two proposed by Pollak and Pechukas causes thermodynamic

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inconsistencies in the equilibrium constant and then describe how this error creates huge

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discrepancies in isotopic enrichment, using Cl + Cl −−→ Cl2 as an example.

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Methods

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When estimating rate coefficients, we use TST, shown in eq 1. Reactions in this paper

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are analyzed at the high pressure limit, since errors in the reaction symmetry number would

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impact reaction rates at both the low and high pressure limit. When reactions are barrierless,

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the transition state structure is defined along the reaction path by variational TST. 11 This

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approach leads to the same relative rate coefficients for O + O2 −−→ O3 as an alternative

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approach which involved analysis of metastable O3 . 12

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To evaluate the correct method for treating symmetry numbers, this work varies the

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reaction symmetry number, shown in eq 2, by either adding or not adding a factor of two.

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The two different rate coefficients derived from the different treatments of eq 2 are evaluated

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for thermodynamic equilibrium and for determining isotopic enrichment. To isolate the

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effect of symmetry numbers, examples are chosen which highlight the variation introduced

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in reaction symmetry number.

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The usage of symmetry numbers in this work leads to the same rates as the full treatment

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of rotation with nuclear spin degeneracy at the classical limit. 4,13

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Results and Discussion

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Thermodynamic Derivation

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The extra factor of two recommended in Ref. 2 leads to inconsistencies in equilibria. To show

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this, we plug eq 1 into the definition of equilibrium constant for the reaction 2 B −−→ BB

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and compare that to the equilibrium constant for this reaction derived from the Gibbs free

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energy. We can then determine if an extra factor of two appears when evaluating the forward

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reaction symmetry number, σf,rxn .

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The equilibrium constant, Kc , is the ratio of forward, kf , and reverse, kr , rate coefficients. Substituting kf and kr obtained from eq 1 results in † σf,rxn mBB qBB /V kf e = Kc = 2 kr σr,rxn mB q † /V q † /V B

−Erxn kB T

(3)

B

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where σf,rxn is the reaction symmetry number in the forward direction, σr,rxn is the

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equivalent in the reverse direction, and Erxn is the difference between the zero point energies

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of the product and reactants. The tunneling factors and transition state partition function

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cancel out since neither affects equilibrium.

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To determine whether σf,rxn contains the factor of two recommended by Pollak and

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Pechukas, 2 we derive the equilibrium constant in a different way based on the definition

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that at equilibrium the Gibbs free energy, G, is at a minimum in a constant temperature

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and pressure system. The change in G can be represented by a constant temperature and

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pressure chemical potential, µ.

dG = −2µB + µBB = 0

(4)

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The chemical potential of a pure species j, µj , can be derived from its partition function,

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Qj . Neglecting intermolecular forces with the ideal gas assumption, the partition function

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can be put in terms of a single molecule’s partition function, qj , and the number of those

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molecules, Nj . Since the molecules are also indistinguishable, a factor of Nj ! appears, which Nj N , as shown in eq 5. 14 can be simplified by Stirling’s approximation, Nj ! ≈ ej qj Nj ≈ Qj = Nj !

eqj Nj

Nj (5)

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If species j represents a mixture with mj optical isomers of equal concentration, energy

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and reactivity, then the partition function (Qj ) can be separated into the product of each

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isomer i, Qj = Πi j Qj,i = (Qj,i )mj , which is simplified with Stirling’s approximation in eq 6.

m

Qj = 79

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qj Nj /mj (Nj /mj )!

≈

eqj Nj /mj

Nj /mj !mj =

δA δNj

= −RT

T,V,Ni

δ ln Qj δNj

mj eqj Nj

≈ −RT ln T,V,Ni

Nj (6)

m j qj Nj

(7)

Substituting the chemical potentials for B and BB from eq 7 into the equilibrium relationship in eq 4 results in NBB mBB qBB 2 = mB 2 q B 2 NB

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found using Qj and eq 6.

µj =

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Since µj can equivalently be defined using Gibbs or Helmholtz free energy, 14 µj can be

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mj

(8)

Since the partition function of an ideal gas molecule is proportional to volume, each term 5



Page 6 of 17

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in eq 8 can be divided by volume to obtain concentrations, Cj =

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constant, Kc . 14 mBB (qBB /V ) CBB = Kc 2 = CB mB 2 (qB /V )2

Nj , V

and the equilibrium

(9)

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There is no special factor of two in eq 9 due to translational indistinguishability, and this

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equation is analogous to the equilibrium constant for a reaction with two different reactants. We then extract the symmetry and zero point energy from the partition functions, qj =

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qj† σj

eEo,j /kB T , to get eq 9 into the same form to compare with eq 3. † /V ) σB 2 mBB (qBB Kc = 2 e 2 σBB mB † qB /V

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−Erxn kB T

(10)

Equating eq 10 with eq 3 results in σB 2 /σT S σf,rxn σB 2 = = σBB σBB /σT S σr,rxn

(11)

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By equating these two equilibrium constants, no factor of two appears in the definition

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of σf,rxn even if A = B. Ref. 14 includes a derivation for Kc when A 6= B, which results in

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the same conclusion.

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Interchange of the identical reactants, postulated by Pollack and Pechukas as the reason

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for the factor of two, 2 is already accounted for when going from an ensemble partition

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function to a molecular partition function in eq 5. Since RPD algorithm accounts for the

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interchange, it must be removed by halving the RPD for reactions with identical reactants,

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as recommended by Bishop and Laidler. 3

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Due to this thermodynamic inconsistency, kinetic simulators that estimate reverse ki-

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netics using equilibrium and forward kinetics will also incorrectly double the reverse rate of

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reaction if the factor of two recommended by Pollack and Pechukas is used for the forward

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reaction. Section S2 of the Supporting Information shows how including an erroneous factor

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of two affects reverse reaction rates in ethane pyrolysis.

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Isotopic Enrichment

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In many fields a factor of two error in a computed rate coefficient can have modest effects,

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but when studying isotopic enrichment, it dominates any realistic enrichment. Increasing

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the rate coefficients of reactions with identical reactants by the extra factor of two proposed

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in some literature 2,4 leads to unphysical behavior which is significantly inconsistent with ex-

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perimental isotopic enrichment values of identical reactant reactions. 15 As a simple example,

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consider chlorine atom recombination involving

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tions depending on the isotopes involved. By analyzing the rates of these three reactions,

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we show that adding the factor of two to eq 2 creates unrealistic isotopic enrichment.

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114

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Cl and

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Cl which has three distinct reac-

To isolate the effect of the factor of two, this example makes several simplifying assumptions:

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1. The reactions occur at the high pressure limit, so variable lifetimes of excited states,

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like in atmospheric ozone, 16 would not be expected, and we can use canonical TST (eq

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1) to estimate the rate coefficients.

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2. Mass-dependent kinetic isotope effects for isotopic carbon and chlorine are negligible (

Correct Symmetry Treatment for X+X Reactions Prevents Large Errors

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