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P05 Chemical Kinetics

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This is a lecture is a series on combustion chemical kinetics for engineers. The course topics are selections from thermodynamics and kinetics especially geared to the interests of engineers involved ...

This is a lecture is a series on combustion chemical kinetics for engineers. The course topics are selections from thermodynamics and kinetics especially geared to the interests of engineers involved in combusition

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P05 Chemical Kinetics P05 Chemical Kinetics Presentation Transcript

  • Chemical Kinetics A bit more Physical Chemistry for Combustion Kineticists • Arrhenius Equation • Order of Reactions • Fitting the parameters • Steady State • Pressure Dependence • Time Scales
  • Elementary Reaction Rates Bimolecular Reactions and Collision Theory How fast do molecules react? Problem: Given a bunch of hard spheres traveling at certain average velocity, how often to the spheres collide? With the help of classical collision theory, we can derive the basic Arrhenius rate equation that is used throughout chemical kinetics. This tells us how fast species react. The core question is: How often do a bunch of hard spheres travelling through space collide? What do we need to know? - How many there are? - How large the spheres are? - How fast they are traveling?
  • Hard SPhere Collisions Ball Diameter: ! Balls collide if: Between the centers less than σ1 σ2 + =σ 2 2 How often do two spheres collide? The key to answering this is to calculate the volume that the moving spheres sweep out. Then the problem becomes how much overlap is there between these volumes. When the volumes overlap, there is a collision. In a total volume, the probability of overlap is the probability of collision.
  • Transformation Two large spheres equivalent to: One ball diameter 2! running into a point Collisions will be viewed as A large sphere of diameter 2! Moving through a density of points The key to solving this is to do a trick. Two spheres sweeping out a volume is the same as one big sphere (of the combined radiuses) sweeping out a volume, with the other sphere being single points. The probability of collision is related to the density of points (of one sphere) and the volume swept out by the large sphere. So the next question is how fast is this large sphere moving through space?
  • Rate of Collision A single ball (radius !), traveling with speed " sweeps out a cylinder of volume per second 2 πσ ν If the concentration of these balls is CA Then the volume swept out per second per unit volume is: 2 CA πσ ν A large sphere of radius !, traveling with a speed " sweeps out a volume of area times velocity (per second). If we have a concentration of these balls, then each in this concentration is sweeping out this volume. The volume of each sphere time the total concentration yielding volume ver second.
  • Rate of Collision If the concentration of the points is: CB points per unit volume Then the frequency of the two hitting is: 2 CA CB πσ ν (point per unit volume)(volume per second) = points hit per second collisions per second The other balls, which have been transformed to a set of points, have a concentration (or density) of the concentration of B. The frequency of them ‘hitting’ is simply the volume swept out by A multiplied by the density of B, yielding points hit per second, meaning collisions per second
  • Boltzman Statistics relative velocity according to Maxwell-Boltzmann statistics: 8KT ma mb ν= (µ = πµ ma + mb Substituting for collision frequency: 2 8KT CA CB πσ πµ So how fast are these molecules (spheres moving?). For this we can use Maxwell-Boltzmann statistics giving the average speed (by integrating velocity with the distribution) which is essentially a square root dependence on temperature. This expression can be substituted to give the rate of collision dependent on temperature.
  • Dependence 2 8KT CA CB πσ πµ proportional to concentrations proportional to square root of temperature So in the final expression, we see the expected dependence on the concentrations and also a dependence on temperature.
  • Arrhenius Equation Only those molecules over a certain energy will react: Activation Energy: The proportion of molecules Ea at temperature, T, having an energy over the activation energy: Ea e RT So now we know how often two molecules collide. This is the first prerequisite to a reaction taking place. The molecules have to meet somewhere. The second prerequisite is that the bond making and breaking also requires a minimal amount of energy. If the molecule has at least this much energy, then the molecules react to form the products. Otherwise, no. The proportion of molecules at temperature, T, that have an energy over a certain activation energy, Ea, is the exponent of the Ea over RT. Combining with the frequency, the previous derived result, we get a rate of reaction that is proportional to the concentrations and the square root and the inverse exponential of the temperature.
  • Arrhenius Equation Combined with the number of collisions: 2 8KT Ea CA CB πσ e RT πµ The general Arrhenius equation is generalized to: Ea n CA CB AT e RT Three constants are needed: A, n, Ea Combining the prerequisites of collision and activation energy, we have derived an expression close to the expected Arrhenius equation. The exact di!erence is the square root dependence on temperature. But is does at least justify the general Arrhenius equation which has a dependence on three constants, the Arrhenius coe"cient, the temperature exponent and the activation energy.
  • Transition State Theory Reaction occurs only if enough energy Initial energy needed to be able to win the final exothermic total energy gain The transition state represents the state of the reaction at the top of the energy hill. When the molecules have achieved enough energy, the activation energy, they can react. In the transition state, the reactant bonds are starting to break and new product bonds are starting to from. When the transition state is reached, the molecules can either continue reacting to form products, or can break apart and go back to reactants. The transition state is highly unstable and cannot be isolated. It is di!erent from a reactive intermediate which does exist for a length of time, even though it is very short.
  • Potential Energy Surface A + HB −→ AH + B Trough is reaction path between products and reactants In reality, the energy diagram is a multidimensional surface of the potential energy of each geometric configuration. This diagram is a simple potential surface involving three atoms and the reaction (where A is F and B is Cl): A + HB --> AH + B The x and y axis are the AH and BH distances. The single line diagram, as in the previous slide, is drawn through the minimum of the trough.
  • Product Distributions Some surfaces are more complex and impossible to visualize. But if a potential energy line is drawn, along the minimum of the trough, from the reactant to activated complex to products, some useful information about complex configurations can be extracted. In trying to determine the product distributions of complex reactions, it is useful to look at the energies of the reactants, intermediates and transition states. This is an example of the addition of oxygen to cyclohexane radical. From the diagram, it is possible to justify qualitatively and possibly quantitatively the product distributions. The intermediate molecules in the potential energy minima, are reactive intermediates
  • Role of Quantum Mechanics But where do the potential energy diagrams, i.e. the energies of the reactants, products and transition states, values come from? An ever increasing tool for combustion chemists is to use quantum mechanics, QM. Given a particular structure, QM can compute the energy value of the molecule. One of the di!culties is determining the best molecular structures (usually through some optimization process). Another more complicated di!culty is that transition states are often very complicated electronically (for example, bonds are not quite complete and are stretched). This means that higher order, meaning computationally extensive, calculations must be made. In the study where this diagram was taken, the molecular structures and their corresponding potential energies were computed.
  • Role of Quantum Mechanics Interactions are based on springs and classical potentials. Another theoretical field that is coming into play in combustion kinetics is that of molecular mechanics. This uses the potential energy surfaces to the next level beyond using the simple trough to examine the transition state. Here even the full vibrational behavior of the species, treating atoms as connected with springs, are taken into consideration in determining the rate constants.
  • Arrhenius Form −EA −EA k = Ae RT k = AT e n RT •k Rate •A Frequency Factor •n Temperature Coefficient •Ea Activation Energy •T Temperature •R Energy Constant There are basically two common forms of the Arrhenius equation for the rate. One is a simple exponention with three coe!cients, the frequency factor (or Arrhenius coe!cient) and the activation energy . The temperature dependent version has an additional temperature coe!cient.
  • Fitting Coefficients EA logk = logA − RT 1.Measure rates at various temperatures 2.Plot log k versus 1/RT 3.Intercept is log A 4.Slope is Ea The constants to the Arrhenius equation are fitted from experimental data. The rate constant, with or without the temperature coe!cient, is purely a function of temperature. Thus if the rate constant is measured at di"erent temperatures, the coe!cients can be determined. If a plot is made of log k versus 1/RT, then the intercept is the log of the A coe!cient and the slope is the activation energy.
  • Coefficient Fitting With the temperature coefficient it is not as straightforward, but basically the same: EA logk = logA + nlogT − RT Least squares fitting When the expression is to include the temperature coe!cient, then a simple plot is not possible. However, given enough points, a least squares fitting is possible.
  • Simple Example Temperature rate x 10**16 200 2.12 225 6.8 250 17.3 H2 + OH −→ H2 O + H 275 37.1 300 70.1 325 120 Plot and read from axis 350 191 375 284 400 403 3 −17.46 kJ cm mole 425 549 k = 7.7x10 −12 molecule s e RT 450 723 Atkinson, R.; Baulch, D.L.; Cox, R.A.; Crowley, J.N.; Hampson, R.F, Jr.; Kerr, J.A.; Rossi, M.J.; Troe, J. Summary of Evaluated Kinetic and Photochemical Data for Atmospheric Chemistry, 2001 (IUPAC Subcommittee on Gas Kinetic Data Evaluation for Atmospheric ChemistryWeb Version December 2001) Here is a simple example taken from the literature for the reaction of hydrogen and hydroxyl radical forming water and a hydrogen radical. Experimentally, the temperature dependence of the rate constant was measured. The log of the rate versus 1/RT is plotted. The final expression for the rate constant give the frequency factor, read from the intercept and the activation energy, from the slope.
  • Heating Value Heat released when a given amount of fuel is combusted. •Higher Heating Value •Thermodynamic heat of combustion •Enthalpy change for the reaction with the same temperature before and after combustion. •Lower Heating Value •Heat of Vaporization subtracted from higher heating value •water component is in a vapor state after combustion A standard measurement used in industry for the heat content of a species, meaning how much heat is released when the species is combusted. Two values are used. The first is higher heating value. This is the thermodynamic heat of combustion, meaning the enthalpy change of combustion at a constant temperature. The second is the lower heating value. This is where the heat of vaporization is subtracted from the higher heating value.
  • Heating Value Relationship Higher Heating Value = Lower Heating Value + hvapor,H2 O (nH2 O,out /nH2 O,in ) The lower heating value is useful for boilers where in the end the water is evaporators The di!erence between the higher and the lower heating value is basically related to the heat of vaporization of water. It is for this reason that the lower heating value is useful for boilers.
  • NIST Kinetic Data There are published compilations, for example from NIST, of rate constant data extracted from the literature. From this plot, one can see that there is still a degree of experimental error. This can lead to some flexibility in the determination of the rate constants. This in turn can lead to some optimization when putting together the reactions in a larger mechanism.
  • Standards for Combustion Standard (Accepted) Reference Atkinson, R.; Baulch, D.L.; Cox, R.A.; Crowley, J.N.; Hampson, R.F, Jr.; Kerr, J.A.; Rossi, M.J.; Troe, J. Summary of Evaluated Kinetic and Photochemical Data for Atmospheric Chemistry, 2001 (IUPAC Subcommittee on Gas Kinetic Data Evaluation for Atmospheric Chemistry Web Version December 2001) One of the standard references for rate contants is the paper published by Baulch et. al. These are used whenever possible. If these constants are not used in a mechanism, then the modeler must give a very good reason.
  • Subject to Errors •Systematic Errors •Cannot isolate reactions H2 + M --> H + H + M •Assumptions in associated reactions H2 + O2 --> HO_2 + H H+O2 --> O + OH O+H2 --> H + OH H_2+OH --> H_2O + H O+H2O --> OH + OH H+H+M --> H2 + M O+O+M --> O2 + M H+O + M --> OH + M H+OH --> H2O (more involving HO2 and H2O2) The determination of fundamental rate constants can be subject to several sources of errors. Along with systematic errors in measurements, it may be hard to actually isolate the particular reaction out of a system. Often one has to make assumptions about the other reactions in the system in order to derived the isolated reaction. It can happen that as new results are obtained, these assumptions must be modified and the rate recalculated.
  • Arrhenius Equation A + B −→ products Rate of products production: d(products) Ea dt = k[A][B]k = T e RT But what happens with more complicated with sets of reactions? (as in combustion) From the discussion of bimolecular collision theory, we have a rate equation for a single reaction of species A and species B combining to form products. d[products]/dt = k[A][B] where the rate constants has a temperature dependence: k = T**n exp(Ea/RT) But combustion mechanisms consist of many of such reactions, of di!erent orders (number of reactants). The following discussion looks at dealing with the more complicated cases and simplifications that can be made.
  • First Order Unimolecular: A −→ B d[A] d[B] = −k[A] = k[A] dt dt Reversible Unimolecular: A ←→ B d[A] d[B] = −kf [A] + kr [B] = kf [A] − kr [B] dt dt Let us look at the rate equations for unimolecular transformations between species A and species B. For the first case of non-reversible transformation of A to B, the rate of production is simply minus the rate times its concentration, i.e. the concentration is decreasing. The increase of B is also dependent on how much A there is. A reversible reaction of one reactant and one product is also first order. Thought as two independent reactions, one in the forward and one in the reverse, we get the expected equations. But now, for example, the molecule A has two sources and sinks. In the forward direction, A is being converted to B at the rate of -k[A] (note the negative sign). However, at the same time, it is being created by B being converted to A at the rate of - k[B]. Thus the rate of production of A is the sum of the forward and reverse directions: - k_f[A] + k_r[B]
  • Bimolecular Reactions Recombination: A + B −→ C d[A] = −k[A][B] dt d[B] = −k[A][B] dt d[C] = k[A][B] dt Just as in the previous case, the total rate of production is the sum of all the sources and sinks of the molecular species. For the forward reaction, there is only one source of each rate of production, namely plus or minus k[A][B].
  • Bimolecular: Reversible A + B ←→ C d[A] = −kf [A][B] + kr [C] dt d[B] = −kf [A][B] + kr [C] dt d[C] = +kf [A][B] − kr [C] dt However, for the reversible case, once again, there is the e!ect of both sides of the reaction. In general, to determine the rate of production of the species - Collect all the reactions involving that species. - Add all the rates (with concentrations depending on order) producing the species - Subtract all the rates that are consuming the species. For a complex mechanism, one has a set of di!erential equations, one for each species. On the left hand side is the di!erential and on the right hand side a polynomial expression in terms of rates and species concentrations.
  • Rate of Production d[H] dt = +2k1 [H2 ] H2 + M −→ H + H + M +k2 [H2 ][O2 ] H2 + O2 −→ HO2 + H −k3 [H][O2 ] H + O2 −→ O + OH +k4 [O][H2 ] O + H2 −→ H + OH +k5 [H2 ][OH] H2 + OH −→ H2 O + H −2k7 [H] 2 H + H + M −→ H2 + M −k9 [H][O] H + O + M −→ OH + M −k1 0[H][OH] H + OH −→ H2 O A more typical case, for even a simple example like one part of the hydrogen mechanism, the rate of production of just one species can be quite complex. One can see that the set of di!erential equations that must be solved, for even a simple molecule like hydrogen, can be quite complex. Each species di!erential is equal to a complex polynomial. So n complex di!erential equations must be solve to determine the time evolution of a species within a given mechanism. This should give an indication why kinetics is a di"cult computational task when dealing with, for example, reactive flows.
  • Simplifying Assumptions Need Simplifying assumptions not only for solving the system when set up but also, for example, when determining fundamental rate constants. Quasi First Order Steady State Assumption Sometimes, with simplier systems, there are alternatives to having to solve numerically a complex series of di!erential equations. These are often critical assumptions used to estimate fundamental rate constants. Two common methods are - Quasi First Order - Steady State Assumption
  • Quasi-First Order A + B −→ C Assume [B] is constant Often, for example, [B] >> [A] (pool approximation) where: k' = k[B] d[A] dt = −k[A][B] = −k [A] d[B] dt = −k[A][B] = −k [A] d[C] dt = +k[A][B] = +k [A] In a bimolecular reaction, sometimes one of the reactants, for example B, has a comparatively large concentration. The consequence of this is that it total concentration does not change much, the pool approximation. This simplifies the second order equations to a quasi first order equations. For example, suppose that [C] is 1000 moles/liter and that [B] is 1 mole/liter. Even if the reaction, A + B --> C, went to completion meaning all the B would be used up, [C] would still be fairly close to 1000, i.e. with some assumptions, 999 moles/ liter. This means that $k[B]$ is relatively constant and only [A] is changing significantly. Thus the rate equation can be thought of a being basically first order in A.
  • Quasi-Steady State The concentration of a species does not change It reaches an equilibrium. (after a rapid initial buildup) d[A] =0 dt allows algebraic solving reduces the complexity If molecule is in steady state, then that means that its concentration is not changing. This means that the consumption and depletion rates are the same. It is important to note that molecule does indeed have a non-zero steady state concentration. Mathematically, a molecule is in steady state when d[A]/dt = 0. This fact can be very useful in simplifying the set of di!erential equations of a reactive system. For each molecule in steady state, a di!erential equation is eliminated and is replaced by a purely algebraic equation (which is simpler to solve).
  • Steady State: N O + N2 −→ f ast NO + N N + O2 −→ slow NO + O The production of N is slow, but its destruction is fast Low concentrations of N Reaches an equilibrium (of low concentration) A typical two reaction example is one where a fast reaction is followed by a slow reaction. This one is taken from the NOx mechanism. Here we assume that the N has reached an equilibrium state within these two reactions.
  • Steady State: N O + N2 −→ f ast NO + N N + O2 −→ slow NO + O d[N ] = 0 = k1 [O][N2 ] − k2 [N ][O2 ] solving dt k1 [O][N2 ] Note that it is not zero.. [N ]ss = k2 [O2 ] Setting d[N]/dt to zero, leaves an algebraic equation where [N] can be solved for. The computations show that first, the steady state concentration of N is non-zero and is dependent on [O], [N2] and [O2]. More assumptions have to be made to solve the system completely or to be able to make some computations.
  • Equilibrium Constant A + B ←→ C + D Under Equilibrium Conditions d(products) d(reactants) 0= dt = dt = kf [A][B] − kr [C][D] kf [C][D] kf [A][B] = kr [C][D] Kc = kr = [A][B] Consider the reversible bimolecular reaction involving the reactants A and B and the products C and D. Under equilibrium conditions, the rate of production of the reactants is the same as the rate of production of the products. Setting the di!erentials equal, we get an expression relating the forward and reverse rate constants and concentrations of the reactants and products. Collecting the rate terms on one side and the concentration terms on the other side, we have the expression for a defined quantity, the Equilibrium Constant. The equilibrium constant gives the ratio of the reactant concentrations to the product concentrations. For example, a large equilibrium constant means that the reaction goes close to completion and a significant portion of the species went to products.
  • Gibbs Free Energy ∆G = ∆H − T ∆S Useful work from an isothermal, isobaric system The greatest amount of mechanical work which can be obtained from a given quantity of a certain substance in a given initial state, without increasing its total volume or allowing heat to pass to or from external bodies, except such as at the close of the processes are left in their initial condition. By definition it is G=H-TS. In thermodynamic definitions, the term H-TS occurs frequently enough to justify defining a new variable The Gibbs Free Energy is a thermodynamic quantity that represents the useful work from an isothermal, isobaric system. The quote is how Gibbs himself defined it.
  • Connection to Thermodynamics For an ideal gas, there is a connection to equilibrium constant (we will take this as a definition) ∆G = −RT lnKp Significance: We can relate the equilibrium constant, or the forward and reverse rate constants to an equation involving entropy and enthalpy The connection between thermodynamics and kinetics can be found with the Gibbs Free Energy. In this lecture this will be taken as a definition. The relationship holds for ideal gases. Note that this is the equilibrium with respect to partial pressures. The relationship between the equilibrium defined with concentrations has some extra constants and is dependent on the number of reactants and products. The significance of its use is to be able to relate the equilibrium constant, and even the forward and reverse rates to the thermodynamic quantities of entropy and enthalpy. For example, in numeric programs solving kinetic systems, if the thermodynamics of the reaction is known, i.e. enthalpy and entropy, and if the forward reaction rate is known, then the reverse reaction rate can be derived.
  • Conditions For Equilibrium ∆G = −RT lnKp ∆G = ∆ − T ∆S −∆G Kp = e RT −∆H+T ∆S Kp = e RT −∆H T ∆S Kp = e RT e R From the two expressions for the Gibbs Free Energy, we can derive what conditions favor reaction completion to products. Solving for K in terms of the Gibbs Free Energy, we get an exponential. Then substituting the expression for G and separating, we get two exponentials, one with Enthalpy and one with Entropy.
  • Conditions for Equilibrium −∆H T ∆S Kp = e RT e R kf Kp = kr >1 Favors Products Criteria: ∆H < 0 Exothermic Reaction ∆S > 0 Greater molecular chaos If the equilibrium constant is greater than zero, meaning then the reaction favors the products, the forward rate is greater than the reverse state. An exponential is greater than one if the term in the exponential is also greater than zero. So the first term says that if the change in enthalpy is less than zero, meaning that the reaction is exothermic, then the products are favored. The second term says that if the change in entropy is greater than zero, meaning there is greater molecular chaos, then products will also be favored. Note that this is not a proof, but rather an example of why the Gibbs Free Energy was defined in the first place.
  • Pressure Dependence Energy needed for reaction can come from collisions A hitting molecule can just transfer its translational energy without changing Collisions dependent on other molecules represented by M Reactants + M ←→ P roducts + M Reactions depend on collisions. Some of the translational energy involved in the collisions can be transfered to the internal energy of the molecule instigating a bond breaking or making. Remember in the discussion of the bimolecular collision theory, for a reaction to occur, a certain amount of energy must be present. One type of collision is when the colliding molecules rearrange their bonds among themselves. Another type of collision is when energy is transfered to the molecule putting it in an excited state, which leads to a change in bonding. The di!erence is that the molecule that the hitting molecule does not change, it just transfers some of its energy. Since the molecule transferring energy does not change and since, for example, many molecules can have this e!ect, the molecule is represented by a generic M.
  • Lindemann Model Unimolecular decomposition is only possible If enough energy is available to break the bond Energy provided by collision with M A + M −→ A + M ka ∗ A + M −→ ∗ k−a A + M ∗ ku A −→ P roducts Unimolecular decomposition is only possible if enough energy is available to break the bond. This energy is provided by collisions with a generic set of molecules M. The Lindemann model is based on the production of an intermediate excited state of a molecule, here seen as A* produced by a collision with a generic (set) of molecules represented by M. This can be represented by a set of three reactions, production of the activated species, its reverse and the activating species going to products.
  • Lindemann Equations d[P roducts] = ku [A ]∗ dt Quasi-steady state of activated species d[A ] ∗ = 0 = ka [A][M ] − k−a [A ][M ] − ku [A ] ∗ ∗ dt ka [A][M ] [A ] = ∗ k−a [M ] + ku d[P roducts] ku ka [A][M ] = ku [A ] = ∗ dt k−a [M ] + ku From the system of three reactions, the rate of production of the activated species can be written. Since this is an intermediate species, and its decay is very fast, we can assume that the species is in quasi steady state and from that we can derive an algebraic expression for the concentration of this activated species: [A*]= k_a[A][M]/(k(-a)[M] + ku) which can be substituted back in the equation for the production of the product species: d[Products]/dt = k_u [A*] = k_u k_a[A][M]/(k(-a)[M] + ku)
  • Fall Off Limits At high pressure, the rate go asymptotically to a constant, k(inf) At low pressure, the rate has a constant slope, ka[A][M]. In the transition zone the line with the slope bends to the asymptotic constant. The Lindeman model recognizes that some kinetic relationships cannot be represented with a single rate constant. There are several expressions within combustion kinetics which take this e!ect into account.
  • Fitting Low Pressure d[P roducts] ku ka [A][M ] = ku [A ] = ∗ dt k−a [M ] + ku Low Pressure Range: k−a [M ] << ku ∗ ku A −→ P roducts (concentration of collision partners is small) d[P roducts] = ka [A][M ] dt 2nd Order Kinetics At low pressure, the concentration of collision partners is small and the conversion to products dominates. So the term with M disappears in the denominator leaving the expression ka[A][B], meaning second order kinetics. This produces the slope of the line dependent on M in the graph.
  • Fitting High Pressure d[P roducts] ku ka [A][M ] = ku [A ] = ∗ dt k−a [M ] + ku k−a [M ] >> ku A ∗ + M −→ k−a A + M d[P roducts] ku ka [A][M ] = = kinf [A] dt k−a [M ] (first order kinetics) At high pressure, the M term is larger and the reverse dominates. Thus the term with M in the denominator dominates leaving first order kinetics in the final expression. This produces the flat ‘fall o!’ behavior shown in the graph.
  • Fitting Fall off The two fall o! coe"cients can be derived experimentally by plotting the log of the species A versus the log of the rate constant rate constant.
  • Chemical Time Scales How long does a chemical process take? especially in relation to other processes •Mixing •Diffusion •Turbulence Chemical time scales are important for gaining insight into to total combustion process. For example, if the chemical time scale is fast compared to di!usion processes, the chemical process goes to completion (equilibrium) before it di!uses very far. This has important implications for modeling, meaning the chemistry can be simplified if the chemical processes are very fast with respect to all other processes.
  • Combustion time Scales High Temperatures •chemical times can be fast compared to transport •Reaction goes to completion fast •Gas is primarily burn or unburnt Low Temperatures •chemical times can be same order of magnitude as other processes •Resulting in more complex mixing On a computational level, varying time scales of di!erent processes allow simplifying assumptions to be made. For example, fast chemical time scales allow that less detail need be known for the processes reactants and complete combustion products. At high temperatures, the chemical times can be fast with regard to transport processes. The consequence is that the kinetics can be considered as either burnt gases or unburnt gases, meaning the intermediate chemistry does not play an important role. A challenge now for chemical kinetics is the lower temperatures. Not only are these processes important, but since the chemical processes are on the same order of magnitude as other processes, such as transport, they are di"cult to simplify and the mixing of intermediate states dominate.
  • Relation to Rate Constant How is the chemical time scale quantified? d[A] A −→ products = −k[A] dt [A] = [A]0 e −kt It takes an infinite amount of time to reach zero. Not useful How is the chemical time scale quantified? Of course, the chemical time scale is directly influenced by the kinetic rate constant. The higher rate constant, the faster the process and thus the smaller the chemical time scale. Thus there must be an inverse relationship. However, for example, for a first order reaction, the reactant concentration decays exponentially, but it still takes an infinite amount of time to reach zero. This is not a useful measure.
  • Useful Measure By convention: Chemical Time Scale is how long it takes to reach: [A]0 1 [A](τchem ) = = .37 63% of the initial e e [A]0 [A](τchem ) = = [A]0 e−kτchem e 1 τchem = k Since the time scale is a relative concept, it is not entirely important exactly at what point we decide. By convention (this is also used in electronics), the time scale is determined to be the point at which the concentration is 1/e of the original. How much time is needed to consume about 63% of the original substance. Solving the original first order equation we find that the first order chemical time scale is the inverse of the rate constant.