Vietnam part 1 CARL p1 52

This first part of the course in concerned with developing models that can account for chemical reactivity Early models that describe elementary reactions focused on reactions in the gas-phase, since in this environment the time between collisions is relatively large compared to the collision duration itself so that each collision can be considered separately from its environment This is not generally the case for reactions occurring in liquids and on surfaces The following nine pages provide a short illustration of the potential complexity of elementary reactions These points was be discussed later in the course In order to predict the course of a chemical reaction it is useful to know how the total chemical energy of the reaction system changes as the two reactants (AB and CD) approach each other As we will see later, from these energy changes one can gain some understanding of how the products (ABC + D) are formed and also if there are other possible product channels At first sight it might seem reasonable that one could easily predict the change in energy using atomic electronic structure information of the periodic table and estimates of the binding energies of the valence atomic electrons After all, we make good use of this information to predict structures of stable molecules and the presence of double, single, and triple bonds, for example On the figure above, the total chemical energy is the sum of potential energy associated with electron-electron and nuclei-nuclei repulsion and electron-nuclei attraction In fact it would not be so difficult to obtain a rough estimate of this chemical energy as the two molecules AB and CD approach each other using a simple calculator If one did this one would likely obtain the plot above That is, you would not find any significant change within the uncertainties of your calculations - in chemical energy as the reaction proceeded from reactants, AB + CD, to products, ABC + D The time scale for this type of chemical reaction is often in the order of pico-seconds (10-9 s) It is only by ‘zooming in’ by a factor of about ten thousand that one is able to discern the energy associated with the various atomic rearrangements that give a detailed potentialenergy profile of the reaction pathway This potential-energy profile can only be observed when calculating the total chemical energy to the fifth or sixth significant figure As an analogy, if the height of a typical table was total chemical energy of reactants then the energies associated with chemical change describing the reaction would take place within the thickness of the varnish only! Furthermore, in order to make accurate chemical predictions, calculations require at least ten times higher resolution than this, which is quite a challenging task So predicting the rate constant (see later) and products of a chemical reaction from first principles cannot be achieved without very high level computations Chemists rely heavily on of course is insight from similar types of chemical reactions as an aid in predicting reactivity and products Proficiency in this direction is built up over many years of experience, but as we will see later, for most reactions it is simply not possible to predict reactivity with sufficient accuracy to use to model complex chemical systems such as the atmosphere or hydrocarbon combustion To illustrate this point let us consider the reaction of C2H with N2O This page shows all of the possible product channels based on overall standard (298.15 K and atm.) reaction enthalpy fH(N2O )+ fH (C2H) > fH (products) That is, all exothermic channels under standard conditions Here there are nineteen possible (exothermic) product channels From a theoretical standpoint, in order to predict which products are likely, one requires information on the energy change as the C2H radical approaches N2O Here is representation of the energy changes for the C2H + N2O reaction Represented by small yellow circles are local maximum (local transition states – see later) and local minimum (quasi-stable intermediate structures) having their own set of internal energy states The large orange circle on the left represents the energy of the reactants and the green circles on the right represent bi-molecular products having an energy lower than, or close to, that of the reactants It can be seen that the products CCNN + OH (in a slightly endothermic reaction) are not likely to be formed since the reaction path involves transitions over large barriers of more than 50 kcal mol-1 kcal = 4.18 kJ The four lowest-lying product channels, H + CCNNO, N2 + HOCC, CO + HCNN, HCCO + N2 are all accessible from atomic re-arrangements having energies below that of the reactants except for the first step that involves a transition over transition state 1/2c All allowed channels have a common pathway until intermediate structure The most favourable product channel is not necessarily the one having the lowest path (from structure 8) as one must remember that this is a quantized system so in addition to energies one must also take into account the (vibrational and rotational) states that are available to at least some of the local maxima (local transition states) and local minima along the reaction pathways These aspects will be covered later in the course Note also , since total energy of the system is conserved during the passage from reactants to products (which will have a duration of the order of tens to hundreds of pico seconds in this case) the energy of the system does not follow the lines given in this diagram, but remains above the energy of the first transition state (1/2c) The following three pages again illustrates the potential large complexity that can be involved in what might appear to be a fairly simple bi-molecular reaction In this case we list the potential exothermic products of the reaction of nitric acid (HNO3) with the C2H radical Here we find at least 143 exothermic reaction channels! The reaction enthalpies in kJ mol-1 are given in the first column (the figure after the decimal point can be ignored) After these examples one can perhaps appreciate that constructing a chemical kinetic model of a system from first principles using elementary kinetic information can be a very challenging task, but this is the only way that detailed chemical information as a function of time can be ascertained for systems such as the Earth’s atmosphere and combustion It is the aim of this first part of the course to give you some background in how we can theoretically treat and experimentally determine rate constants and product distributions of elementary reactions Later on we will consider real examples of complex chemical systems, particularly photo-chemical processes important in the Earth’s atmosphere There are no notes for this page There are no notes for this page There are no notes for this page Here is an example of a transition state theory calculation for the reaction of OH with acetic acid All the information required to calculate the rate constant (with the exception of Hatom tunneling) at any temperature is contained on this page The vribrational frequencies and moments of intertia come, in this case, from ab initio calculations 39 A visual representation of TST calculations also including the contribution of tunneling Some of the details were mentioned during the lesson 40 This will be derived in the lesson 41 Though potential energy of a system can change in complicated ways as it proceeds from reactants to products, one can identify three general types of potential energy surface The first is a reaction that has a barrier In this case, the rate constant increases as the temperature increases due to the increasing fraction of collisions that have energies above the barrier A reaction between two radicals (and sometimes between a radical and a molecule) often has no barrier and the rate constant is independent of T A more complicated situation arises when the initial complex has a minimum in the potential energy surface For such a case, the initial complex is able to either re-dissociate or proceed to reactants The choice between the two pathways depends on how the energy is distributed in the initial collision complex If the B-C bond receives enough energy then the reaction will proceed to products If the A-B bond receives enough energy then the complex will redissociate to reactants In this latter case no reaction has occurred 42 If the complex lasts long enough to undergo collisions, the propensity for one pathway or the other will change This is because the total vibrational energy of the complex decreases This situation favours formation of reactants rather than re-dissociation As an example, suppose that ABC* complex has 20 units of vibrational energy and that the A-B bond requires of these to dissociate and the B-C bond requires The ratio of A-B dissociations to B-C dissociations can be analysed statistically How many possible ways can 20 units of energy be distributed in two vibrational modes? These are a follows, with the A-B bond energy given first 20:0, 19:1, 18:2, 17:3, 16:4, …….10:10, 9:11, 8:12, ………0:20 The above analysis presumes that a tri-molecular molecule ABC has only two modes of vibration, which is not really the case, the number of vibrational modes is 3N-6 or 3N-5 for a linear or non-linear molecule, respectively, where N is the number of atoms One now looks at how many times A-B has equal to or more than quanta of energy and how many times B-C has energy equal to or more than quanta The ratio of these numbers will give the ratio of dissociation rates If the complex ABC had only 10 units of energy then it will be far more likely that B-C dissociates Thus reduction of energy in the ABC complex is a result of collisions with M (in the atmosphere, M is normally taken as the sum [N2 + O2 + H2O]) Therefore there is a certain range for which the rate constant increases as [M] increases At the point where the total energy of ABC goes below units, A-B can no longer dissociate and therefore the rate constant must be insensitive to any further pressure increase (i.e increase in [M]) One might argue that collisions could also increase the energy of the complex, but one must bear in mind that here we refer to averages over very large numbers of collisions and that the complex has energy above the equilibrium energy so that downward energy collisions only need to be considered In the 1970’s Jurgen Troe considered the form of the pressure dependence of ter-molecular rate constants and arrived at the expression given above which requires knowledge of the 43 rate constant at zero pressure and the rate constants at infinite pressure In practice, the former is derived via a series of low pressure experiments and extrapolated to zero pressure, the latter is derived from high-pressure experiments The effective bi-molecular rate constant (i.e., with pressure factorized out) at intermediate pressures can then be predicted Sometimes, the fall-off parameter Fc is different from 0.6 Note that on the one hand, inter-molecular vibrational re-distribution occurs on the time scales of picoseconds and on the other hand the time between collisions at Bar and 288 K is of the order of a few nanoseconds These two opposing processes are quite competitive under atmospheric conditions if the molecules contain many vibrational degrees of freedom Experiments tell us that a typical value for the energy removed per collision is 30 cm-1 (E=100hcṽ = 6.0e-22 J), where ṽ is the energy expressed in cm-1 This value can be compared the average collision energy at ground level of 1.5kBT of 1.5 * 1.38e-23 * 288 = 6.0e-21 J Note compare this energy with that calculated using the root-mean-square collision velocity of the previous page (0.5 * 20 * 1.66e-27 * 554^2 = 5.0e-21 J It would seen therefore, based on these values, that only about (40/5) collisions, on average, are required to reduced the energy of the formed complex ABC by the initial collision energy These eight collisions will remove sufficient energy that re-dissociation to initial reactants is no longer energetically possible; thus the rate constants reaches the high-pressure regime Since there is a spread in initial collision energies in thermal systems, this the transition from low to high pressure is gradual What now remains to be done is to calculate the expected lifetime of the ABC vibrationally-excited complex 43 The answer to these question form the basis of statistical RRK theory (Rice, Ramsperger and Kassel) The classical version considers the probability that a molecule of s classical identical harmonic oscillators with a total energy E having an energy greater than or equal to the critical energy E0 for dissociation (or isomerization) in a particular bond (mode) This probability is given by the number of ways to attain this (distribution) divided by the total number of ways distribute energy, E , amongst s oscillators 44 Here we look at the number of unique distributions of j objects in s boxes Another approach to that given above that arrives at the same equation is to consider j green objects and j-s empty boxes The total number of objects (including empty boxes) is s The number of ways to arrange j objects is j!, the number of ways to arrange s objects is s! and so on Some arrangements result in the same pattern (distribution) thus we need to divide s! by the number of ways the (green) objects may be distributed amongst themselves (j!) and the number of ways the empty boxes may be distributed amongst themselves (s-j)! Thus one arrives at the total number of unique distributions as s!/(j!(s-j)!) as before The fact that one divides by the total number of ways the boxes can be arranged amongst themselves does not follow that the boxes are indistinguishable Consider one object and three boxes According to the above equation there are 3!/1!(3-1)! = unique distributions This means a distribution comprising an object in the left-hand box, another having the object in the central box, and the last having the object in the right hand-box If the boxes were indistinguishable then there would only be one possible configuration; one full box and two empty boxes Now consider three objects and three boxes According to the equation above there will be 3!/3!(3-3)! = unique distribution This implies that we not note which object is in which box 45 Now consider the number of unique distributions when each available box can accommodate all objects Above shows one example of a distribution that helps us set up the problem to be solved If a box is green it has a least one object in it In the example, box has all six objects in it and boxes to are empty The distribution is now represented differently on the next line Notice that an object in box can change its position to box to (it does not change its position in box 1) Thus there are new choices available to it Now consider a distribution that had one object in each box The same argument holds : each object has new choices available to it This means that the problem as set out above is correct One has effectively j+s-1 virtual objects (importantly not j+s virtual objects ), j of which are green and s-1 are blue (empty boxes) In this case we can proceed as before in finding the number of unique distributions 46 Since we are considering dissociation, one requires that a certain critical number of objects (or number of vibrational quanta) is reached in a particular box (or vibrational mode) Here we consider that the critical value is (shown in the red box above) These objects cannot move Thus we now have j-m+s-1 virtual objects to consider We proceed as before 47 The rate at which the required energy is transferred to a particular degree of freedom is proportional to the probability the distribution given above If we were to label the quanta and label the vibrational modes then each arrangement has equal probability of occurrence, but we are not interested it single arrangements but distributions that put sufficient energy in a particular vibrational mode with no reference to the particular quanta that are present in that mode This means that some arrangements are more probable that others and, in an ensemble, will occur more frequently than others As the critical energy required approaches the total energy of the system, the probability of achieving the critical configuration approaches zero Likewise, for a given total energy and critical energy, as the number of modes, s, increases this probability decreases One now needs to know something about the constant of proportionality, i.e the fundamental rate involved It is reasonable to assume that the rate is closely related to the frequency of vibration of the critical mode Indeed this is the critical result for RRK theory One can equate j (number of quanta) to total energy of the system, E, and m to the critical energy E0 The final form of the RRKM rate constant is given without proof but assumes that j >>s and j-m >>s One can use the first equation though without problem that is the result of a combinatorial problem The only trouble here is one might end up with a computational difficulty if the number of quanta are large Note that the frequency of an typical oscillator is 1013 to 1014 s-1 48 RRK examples will be given in the lesson 49 For RRKM theory we first consider the dissociation or isomerisation of a vibrationally (and rotationally) excited intermediate molecule, which could have been formed in a bimolecular reaction, for example The potential energy surface (PES) follows the reaction coordinate Naturally if the molecule is polyatomic, the PES is multidimensional, nevertheless, all the information required for calculation of the dissociation/isomerisation rate is contained in the slice of PES that follows the reaction co-ordinate Here one is concerned with the distribution of internal energy of the system, thus the bottom of the PES for both the molecule and the transition-state structure is the zero-point vibrational energy Translational energy is not considered here since it is common to both structures and in not redistributable in this system Notice that the (external) rotational energy of the system can change on traversal to the transition state This comes about due to the conservation of angular momentum (thus total angular momentum quantum number J) Rotational states associated with the J quantum number (see later) not mix with vibrational degrees of freedom and are thus considered external for each structure (i.e., for the transition state) The fact that the total rotational energy of the system can change as the reaction proceeds does mean that some rotational energy of the molecule becomes internal vibrational energy of the transition state The active degrees of freedom of the transition state cover the energy range from E0 (the activation energy of the reaction) to E0 + E‡ This energy is denoted in the figure above as vibrational energy, but, in fact this can also be taken up partly by internal rotations and (as seen later) by external rotations associated with the quantum number K The transition state has one less (real) vibrational mode since the one along the reaction coordinate is imaginary and plays no part in the RRKM calculations, though it is used in predicting the width of the barrier should H-atom tunnelling become important 50 51 Examples with be given in the lesson 52 [...]... cartesian expression of molecular structure given below, where all values are in Angstrom N2O4 x y z N 0.00 0.00 0.90 N 0.00 0.00 -0.90 O 0.00 1. 14 1. 39 O 0.00 -1. 14 1. 39 O 0.00 1. 14 -1. 39 O 0.00 -1. 14 -1. 39 For NO2 x y z N 0.00 0.00 0.32 O 0.00 1. 10 -0 .14 O 0.00 -1. 10 -0 .14 32 ... a (bi molecular) rate constant that lies 13 between 1 and 2 x 10 -10 cm3 s -1 There is only a small number of radical-molecule reactions that have rate constants greater than 3 x 10 -10 cm3 s -1 The units of a unimolecular rate constant is s -1 The units of a bi-molecular rate constant is cm3 s -1 (molecule -1) The units of a ter-molecular rate constant is cm6 s -1 (molecule -2) Note that “molecule” is not... by a typical collision cross-sectional area (r2) is of the order of 2 x 10 -10 cm3 If there were two molecules of F (F1 and F2) and three radicals of G (say G1, G2, and G3) per cm3, how many possibilities for collisions (per cm3) are there between F and G? This is simply the product of the concentrations (F1-G1, F1-G2, F1-G3, F2-G1, F2-G2, F2G3) = 6 Thus it is reasonable that collision frequency requires... prior knowledge of similar reactions Radical-molecular reactions and radical-radical reactions dominate atmospheric and combustion chemistry 11 Note that the average thermal energy of a gas is given by (3/2)kBT , where kB is the Boltzmann Constant (1. 38  10 -23 J K -1) 11 Gas-phase reactions are referred to as homogeneous reactions They are, naturally, the most common type of reaction that occurs in the... of a particular degree of freedom For the translational partition function energy spacings are derived from the quantummechanical treatment of the allowed modes in a 3-dimensional box, the derivation of which is not a part of this course The energy spacings associated with translational motion are so small that the translational partition function is by far the largest contributor to the total partition... vibrational partition function is a product of partition functions for individual vibrations 26 This page gives an illustration of the calculation of a vibrational partition function using a spread sheet 27 The page derives the rotational partition function of a diatomic or linear polyatomic molecule having one value of rotational constant, B For a homonuclear diatomic molecule, rotating the molecule by 18 0o... classical approximation for the rotational partition function have been obtained and one of the improved version is given as the last equation where qrot is the classical rotational partition function given in the first equation 30 31 Use the information that you have gained in the first part of this course to attempt the calculation above For the rotational partition functions, instead of using the... are independent and that the individual particles, N, (molecules, atoms) are independent This enables the total partition function of a system, Q, to be expressed in terms of partition functions of individual species and be further factored into the various degrees of freedom 22 The general expression for a partition function is given above In order to calculate a partition function one requires an expression... mass and collision radius, one finds that A(T) remains nearly constant for a very large range of reactants Thus for reactions that occur on every collision, a bi-molecular rate constant of (1 to 3) x 10 -10 cm3 s -1 is expected Notice that the units of a bi-molecular rate constant are volume per unit time Is there a visual representation then of a rate constant in terms of volume? A reasonable representation... take an example of a collision between two species, say OH and CH4 If we also assume the collision radius to be 5 Angstrom then the rate constant, assuming reaction at every collision, should be 1. 5 x 10 -10 cm3 s -1 Naturally, as the reduced mass of the reacting pairs increases so does the collision diameter Curiously though, it turns out that for the vast majority of radical molecule reactions that the