MECH 6120: Combustion Notes on Chemical Kinetics and Chain

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MECH 6120: Combustion By reaction rate, we mean Notes on Chemical Kinetics and change in moles of species Rate = Chain Reactions unit time × unit volume Since the number of moles per unit volume is equiv- 1 Chemical Kinetics alently a concentration C of the species, the reaction rate of species i, denoted Ri, corresponds to Thermodynamics is all one needs to predict the equi- dCi 3 librium state that a system reaches following a pro- Ri = , moles/cm · s cess and the heat/work transfers that occur during dt the process. Thermodynamics, however, cannot pre- From the reaction, we can gather how the reaction dict the rate at which a process will occur, i.e., the rates for the various species are related. For example, speed at which a system reaches a new equilibrium consider state. For example, we could have a coffee cup ini- ◦ ◦ tially at 50 C, surrounded by an ambient at 22 C. CH4 + 2O2 −→ CO2 + 2H2O Thermodynamics will tell us that heat will flow from the cup to the environment and that the final state The reaction rates for the products would be positive, has the cup in equilibrium with the environment (2nd and they would be negative for the reactants. And law). It can also tell us how much heat was trans- by the stoichiometry of the reaction; ferred (1st law). It cannot tell us how long it took the coffee to cool down. To predict this, we would 1 1 RCH = RO = −RCO = − RH O need to know how the rate of heat transfer depended 4 2 2 2 2 2 on the temperature difference between the cup and which follows because for every mole of methane, we the environment (i.e., Newton’s law of cooling). consume 2 moles of O2 and create 1 mole of CO2 and Likewise, Thermodynamics can predict the chemi- 2 moles of H O. cal equilibrium state when all reactions (forward and 2 backwards) are in balance. For example, say we have a system of O and O2 in equilibrium. The equilib- 1.1 Elementary vs. Overall Reactions rium condition implies that the rate of O formation Overall (or, equivalently, global) reactions describe by the forward reaction; the overall conversion of a set of reactants to a set of products. These are the type of reactions we have O2 → 2O been dealing with so far, i.e., is balanced by the rate of destruction of O by the reverse reaction; CH4 + 2O2 −→ CO2 + 2H2O 2O → O describes the stoichiometric, complete–combustion Thermodynamics cannot predict how fast either of oxidation of methane. This reaction does not rep- these reactions will occur – it can only predict the resent the actual molecular event that takes place in state when they are in balance. the process. That is, a molecule of methane does not The topic of chemical kinetics deals with reaction run into a molecule of oxygen to form one CO2 and rates. Obviously, we would need to know something two H2O molecules (perhaps this could occur, but it about reaction rates to predict many types of com- would be very, very unlikely). In reality, the oxida- bustion phenomena, i.e., the rate at which gasoline tion process involves a chain of elementary reactions, vapor is consumed in a cylinder, the speed of flame in which an elementary reaction is the actual molec- spread over a combustible surface, etc. ular reaction event.

1 Most elementary reaction can be classiﬁed as uni- where A, B, C, and D correspond to arbitrary molecular, bimolecular, or trimolecular. The molec- species. The overall reaction rate, which is given by ularity of a reaction is deﬁned as the number of dC dC dC dC molecules taking part in each act leading to a chem- R = − A = − B = C = D (6) ical reaction. For example, at high temperatures dt dt dt dt propane can decompose via is by the law of mass action equal to

C3H8 −→ C2H5 + CH3 (1) R = kCACB (7)

In the above, Ci is the molar concentration of species The molecularity of the above reaction is one (uni- 3 3 molecular) because only one molecule is required for i (kmol/m or mol/cm ). Note that Turns uses the the reaction to occur. Examples of bimolecular reac- notation [A], [B], etc. to denote concentration. Since we are assuming ideal gases, the concentration is re- tions are the attack of H2 by the hydroxyl radical; lated to the partial pressure of the species by H2 + OH −→ H2O + H (2) P C = i (8) i R T and the attack of methane by the hydrogen radical; u The quantity k in Eq. (7) is referred to as the rate CH4 + H −→ CH3 + H2 (3) ‘constant’ of the reaction, and it has whatever units are neccessary for the reaction of interest to allow the Finally, an example of a termolecular (molecular- rate to be in units of mole/cm3·s. In the above reac- ity=3) reaction is tion, for example, k would have units of cm3/mole·s. In general, k will be a strong function of the temper- H + OH + M −→ H2O + M (4) ature, but is independent of the concentrations. In general, for the reaction In the above, M refers to an arbitrary third–body molecule, and serves the purpose of either carrying XN XN 0 00 away or providing energy in the reaction. In the νiMi −→ νi Mi (9) above reaction, the reaction of H with OH is highly i=1 i=1 exothermic. If a third body was not present to take we write, by the law of mass action, the reaction rate away the energy released during the reaction, the by, H2O molecule that was formed would simply ﬂy apart N Y ν0 back into H and OH. i R = k Ci (10) i=1 1.2 Law of Mass Action The changes in concentration of the species due to the reaction are A fundamental principle in predicting the reaction dC rate is the law of mass action, which states: i ≡ R = (ν00 − ν0)R (11) dt i i i The rate of an elementary chemical reaction is proportional to the active masses of Molecularity precisely applies only elementary (or the reacting substances. The active masses single-step) reactions. Often, however, the formation are the species concentrations raised to their of a product species from reactants occurs not from stochiometric coeﬃcients. a single, elementary reaction but from a sequence of reactions. For example, in the conversion of hydrogen To illustrate, consider the simple elementary reac- and bromine to hydrogen bromide, the reaction tion A + B −→ C + D (5) H2 + Br2 −→ 2HBr (12)

2 refers to the overall process. The actual process which can also be solved analytically. actually consists of a series of intermediate reac- Most chemical reactions are bimolecular and pro- tions – which ultimately result in the formation of 2 ceed as the result of reactions following binary col- molecules of HBr and the destruction of one molecule lisions. It is therefore not surprising that chemical of Br2 and one molecule of H2. It is therefore not en- reactions frequently follow second-order kinetics. For tirely meaningful to assign a molecularity of 2 to the the general second-order bimolecular reaction, above reaction. It is more convenient and meaningful to classify A + B −→k AB (20) overall reactions by order rather than molecularity. The order of a reaction is the number of atoms or the rate law will be molecules whose concentration determines the rate of dCA dCB dCAB conversion to products. From the law of mass action, = = − = −kCACB (21) one would predict the order to be equal to the molec- dt dt dt ularity assigned to the reaction. This will be true for The initial conditions would have CA = CA,0, CB = elementary reactions, but is not always the case for CB,0, and (most likely) CAB = 0 at time t = 0. So- overall reactions. Most often, the order of a reaction lution of this set of equations is fairly simple. If we is determined from experimental observations. integrate over time the left hand side of the above All unimolecular elementary reactions involving DE, we get decomposition (or dissociation) will have ﬁrst-order rate kinetics. For example, the reaction CA(t) − CA,0 = CB(t) − CB,0 (22)

k1 or A2 −→ 2A (13) CB(t) = CA(t) + CB,0 − CA,0 (23) will have a rate law of Put this back into the DE for A, and dC dC A = −2 A2 = 2k C (14) dC dt dt 1 A2 A = −kC (C + C − C ) dt A A B,0 A,0 with typical initial condition of 2 = −kCA − k(CB,0 − CA,0)CA (24) C (0) = C ,C (0) = 0 (15) A2 A2,0 A There are two special cases in which the solution to the above is easy. First, if we start off with equal Integrating the first two terms in Eq. (14) gives concentrations of A and B, then the second term in the left hand side disappears. Integrating leads to 2(CA2,0 − CA2 (t)) = CA(t) (16) · ¸ and integrating the last two terms (assuming that k 1 −1 CA = + kt (25) is constant) gives a solution for the concentration of CA,0 A2; −k1t The second case is when CB,0 À CA,0, i.e., A is a CA2 = CA2,0 e (17) trace compound in B. The second term in Eq. (24) Likewise, the reaction now dominates, and the reaction becomes effectively k first–order, i.e., AB −→2 A + B (18) dC A ≈ −k(C − C )C ,C À C will have dt B,0 A,0 A B,0 A,0

dCA dCB dCAB under which CA will undergo an exponential decay. = = − = k2CAB (19) dt dt dt When CB,0 and CA,0 are similar yet not equal, we

3 need to integrate the full form of Eq. (24). This can be done analytically, although there is no need to do it here. Suﬃce to say that it can be done. Third order elementary reactions are less common than bimolecular reactions – simply because they re- quire the simultaneous collision of three molecules. Although there are plenty of reactions to which we could assign a molecularity of three, e.g.,

2NO + O2 −→ 2NO2 (26)

3H −→ H2 + H (27)

2CO + O2 −→ 2CO2 (28)

the rate laws will usually be described by orders around 2. Again, this is because the above reactions refer to overall processes consisting of a number of Figure 1: Arrhenius’ plot of ln(k) vs 1/T showing elementary, usually bimolecular reactions. The most the straight–line behavior relevant trimolecular (i.e, third order) elementary reactions for combustion phenomena are the radical recombination reactions, such as H + OH + M −→ vs. time for reactions occurring at constant temperatures, from which the rate constant k for an nth- H2O + M. The importance of these will be discussed n in a subsequent section. order reaction could be obtained. When he plotted Sometimes a reaction does not have a rate law to ln kn vs. 1/T , where T is the temperature at which which an order can be assigned. For example, it is the reaction took place, he obtained the straight-line behavior illustrated in Fig. 1. known that the H2 – Br2 reaction mentioned previ- ously has a rate law of the form, From such plots, Arrhenius deduced that the rate constant could be expressed in the form 1/2 dC kC C HBr = H2 Br2 (29) k = A e−E/RuT (30) dt CHBr 1 + 0 k CBr2 where the constants A and E are referred to as the preexponential factor and the activation energy. where k and k0 are constants. If C ¿ k0C , HBr Br2 The theoretical foundation behind Eq. (30) is the order is 3/2 – whereas mass action would predict fairly straightforward. It is known that the energies an order of 2. If this inequality does not hold, the of molecules (kinetic, rotational, vibrational, elec- reaction has no order – in that the order cannot be tronic) form a distribution – in that some molecules deﬁned. have more energy than others. Arrhenius reasoned that only molecules possessing an energy equal to 1.3 Temperature Dependence of the or exceeding a certain activation energy E would, Rate Constant upon collision with other molecules, react to form a new molecule. From the kinetic theory of gases, The most widely used formulation for the temper- the fraction of molecules having this energy is ature dependency of the rate constant k resulted exp(−E/RuT ). Note that as T → ∞, the fraction from the experimental and theoretical investigations goes to unity – because as temperature increases, of Svante Arrhenius (1859-1927). His experiments so do the energies of the molecules, and more and consisted of measurements of species concentration more molecules would have an energy in excess of E.

4 The second factor that comes into the theory is the In the above, A is a constant, and the exponent n collision rate between the diﬀerent molecules. For a accounts for the temperature dependence in the pre- second-order reaction involving species A and B, the exponential factor. Turns has a listing of the param- collision rate between molecules of A and B will be eters A, n, and E many elementary reactions involv- related to the species concentrations by ing hydrocarbon species (see, e.g., Table 4.1, Tables 5.1—5.3). The data listed in the book was obtained 0 collision rate = A CACB (31) from many sources – experimental and theoretical.

0 1/2 In these tables, some of the listed reactions have E The factor A is a weak (≈ T ) function of tem- (and perhaps n) equal to zero – indicating that the perature. The reaction rate would be expected to reaction rate constant is nearly independent of tem- be simply the fraction of molecules having adequate perature. In reactions of this type, basically there energy to react times the collision rate. However, is no energy ‘barrier’ to overcome to form a prod- not all collisions fulﬁlling the energy requirement will uct. That is, the reactants molecules are so reactive result in a reaction. For many types of molecules, (i.e., unstable), that if they collide at all a reaction the orientation of the colliding molecules with respect is assured. An example is the radical recombination to each other is an important factor in determining reaction of H to form H , in which whether or not a reaction will occur. The orientation 2 ∗ dependency is included in the reaction rate formula- H + H + M −→ H2 + M (34) tion through a so-called steric factor P, which can be The radical species H are very reactive – we will see viewed as a sort of collision eﬃciency. Putting it all in the next set of notes that the formation of these together, the reaction rate can be written species is a key element in combustion reactions. The

0 −E/RuT reaction between H and H is also very exothermic R = A CACB e P = k CACB (32) – and a ‘third body’ – such as a wall or another in which k is given by Eq. (30), with A = A0P molecule – is needed to carry away the energy that Generally, the more exothermic a reaction is, the is released during the reaction. This third body is smaller the activation energy E. And for small E, the represented in the above by M, and M ∗ denotes that reaction is more sensitive to temperature. It is also the third body has been energized by the reaction. If important to note that while the Arrhenius formula- the third body is not present, the H2 molecule would tion accurately predicts the rates of simple reactions receive the energy of reaction, and likely dissociate over a moderate temperature range, it generally can- again into H + H. not describe overall combustion processes over a wide temperature range – because at different tempera- 1.4 Chemical Kinetics and Chemical tures, different reactions involved in the combustion Equilibrium process will become more or less dominant in determining the overall rate. For example, a rate constant Up to now we have examined the rate of a single re- of a reaction could be obtained from a set of mea- action as it proceeds in a single direction (reactants surements at relatively low temperatures. At higher to products). For most combustion situation, there temperatures, however, measurements could result in will be more than one reaction occurring simultane- a completely different rate constant. Therefore, one ously. The topic of chain reactions, which is a key must exercise caution when extrapolating a specific element in initiating and sustaining combustion, will reaction rate to broad temperature ranges. be covered in a subsequent section. What we want to In reporting data on reaction rate constants for el- examine now is a reaction that can go ‘both ways’, ementary reactions, often a modified Arhennius form i.e., forward and backwards, and relate this to what is used, in which we know about chemical equilibrium. Consider the simple situation in which A decom- k = AT ne−E/RuT (33) poses to B and C, and simultaneously B and C

5 recombine back to A. This sequence is known as When the reactions are in balance – leading to the reversible reactions. The equations governing the equilibrium concentration relation in Eq. (41) – we concentrations would be get essentially the same relation that we would pre- k dict from chemical equilibrium. Since the reactions A −→f B + C (35) are balanced, we could view the pair of reactions af- fecting A, B and C as an equilibrium reaction, k B + C −→r A (36) A )* B + C (42) Since A is destroyed in the ﬁrst reaction and created for which in the second, the net rate change of A is PB,ePC,e = KP (T ) (43) dC PA,eP∅ A = −k C + k C C (37) dt f A r B C where the extra e subscript denotes that the partial By writing a rate law for B or C, we’ll get the same pressures are at their equilibrium values (which is thing – except negative, something we always assumed solely in the context of thermodynamics) and P∅ is the standard state pres- dCB dCC dCA sure of 1 atm. This equation looks similar to Eq. (41). = = − (38) dt dt dt In fact, if we use the ideal gas law (our species are all assumed to be ideal gases), then which should make sense because the net rate of B and C formation has to balance the net rate of A P C = A (44) destruction. A R T Suppose that the initial concentrations of B and C u are zero. The above then gives That is, concentration (kmols/m3) is a molar den- sity. Replace the above into Eq. (41) and compare to CB(t) = CC (t) = CA,0 − CA(t) (39) Eq. (43), and we ﬁnd that for this particular reaction

where CA,0 is the initial concentration of A. The rate RuT kf law for A then becomes KP (T ) = (45) P∅ kr dC A = −k C + k (C − C )2 (40) dt f A r A,0 A This provides a nice physical interpretation of the equilibrium constant KP , i.e., it is proportional to which we could (in principle) integrate to obtain the the ratio of the characteristic rates (i.e., the rate con- solution for CA at all times. For now, however, we stants) of the forward and reverse reactions in an want to simply get the value of CA for t −→ ∞, i.e., equilibrium reaction. If we have a situation where the steady–state, or equilibrium, value of the concen- the forward reaction is a lot ‘faster’ than the reverse, tration. Equilibrium occurs when the reactions are we would then have kf À kr and KP À 1. From in balance: the net rate of formation of A (by the re- Eq. (43), we would then expect PB,e and PC,e À PA,e verse reaction) is balanced by the net rate of destruc- because the forward reaction, being so much faster tion (by the forward reaction) so that dCA/dt = 0. than the reverse, would result in little of A being Applying this to Eq. (37) gives, for the equilibrium around – it would react to form B and C. concentrations, For a general equilibrium reaction involving a pair C C k (forward and reverse) of elementary reactions, the B,e C,e = f (41) C k equilibrium constant is related to the rate constants A,e r by µ ¶∆ν in which the subscript e denotes the equilibrium RuT kf value. KP (T ) = (46) P∅ kr

6 in which ∆ν is the change in the number of molecules In the above, M denotes any third body – which in the forward direction. For our example we had can be any other atom or molecule in the system, ∆ν = 1 (i.e., started with 1 (A), ended with 2 (B or even the walls of the system. The asterisk on M and C). This relationship is very useful: if we can indicates that M is an energized atom or molecule use experiments and/or kinetic theory to determine – in that M has received the energy of the reaction. a forward rate constant, then this information, along Note that there are several intermediate species in with thermodynamics, will provide the rate constant the above mechanism: CH3,H2, OH, HCO, H2CO, for the reverse reaction. CO, H, and O. The important intermediate species, as we will see, are the radicals CH3, OH, O, H, and HCO. These species are very unstable and tend to 2 Chain Reactions react with almost anything they collide with. Thus, the formation of these species tends to promote addi- As mentioned in the notes on kinetics, few combus- tional reactions, which results in the propagation of tion processes consist of a single elementary reaction, the chain. and the overall reaction equation seldom represents Although the above chain mechanism may seem the detailed mechanism of the actual chemical reac- complicated, it is actually a greatly–simpliﬁed ap- tion. In reality several reactions may proceed simul- proximation to reality. The current understanding taneously or in sequence. The intermediate species of the methane oxidation mechanism includes 279 (i.e., the species not appearing in the overall reac- elementary reaction steps, which can operate either tion) produced and consumed in such reactions are in the forward or reverse direction (see table 5.3 of important in determining the overall order of the re- Turns). action. A chain reaction occurs, as the name implies, in a chain or sequence of steps. Often the most seem- 2.1 Chain Reaction Mechanisms ingly simple overall reactions can in reality be due We begin our analysis of chain reactions by examining to a complex chain reaction. A good example is the the simplest of chain mechanisms, i.e.,: stoichiometric oxidation of methane: k1 CH4 + 2O2 −→ CO2 + 2H2O (47) A −→ B (R1) k From experimental and theoretical investigations, it B −→2 C (R2) is known that the following reactions are involved in the oxidation process: In the above, B is the intermediate species, and the

CH4 + O2 −→ CH3 + OH + O overall process is the conversion of A to C. CH + OH −→ CH + O + H The law of mass action can be applied to each in- 4 3 2 dividual step in the mechanism. Since species B is CH4 + O −→ CH3 + OH formed in reaction 1 and destroyed in reaction 2, and CH3 + O2 −→ H2CO + OH both reactions are ﬁrst-order, we have H CO + OH −→ HCO + H O 2 2 dC B = k C − k C (48) HCO + OH −→ CO + H2O dt 1 A 2 B CO + OH −→ CO + H 2 Similarly, for A and C we get H + O2 −→ OH + O dCA H + H2O −→ OH + H2 = −k1CA (49) ∗ dt O + O + M −→ O2 + M dCC ∗ = k C (50) 2OH + M −→ H2O + O + M dt 2 B

7 and initial conditions would typically be CA(0) = 1. initiation reaction: mechanistic step that pro- CA,0,CB(0) = CC (0) = 0. The rate laws for the duces the free radicals. problem, in Eqs. (48–50) are a coupled set of first– order differential equations. This particular set can 2. propagation reaction: a reaction that allows the be solved analytically, giving chain to continue by producing at least as many radicals as it consumes. −k1t CA = CA,0 e (51) k C ¡ ¢ 1 A,0 −k1t −k2t 3. inhibition reaction: a reaction in which the final CB = e − e (52) k2 − k1 products are consumed rather than formed. C © ¡ ¢ ¡ ¢ª A,0 −k1t −k2t CC = k2 1 − e − k1 1 − e 4. termination reaction: a reaction that results in k2 − k1 (53) a net destruction of free radicals.

All of the details of the derivation (and there are Using the simple methane oxidation chain given plenty) were omitted. The point of this example above, the reactions can be divided into: was simply to demonstrate that a seemingly simple chain can lead to a complicated mathematical solu- Initiation: tion. Most of the chain mechanisms involved in combustion reactions will have no analytical solution – CH4 + O2 −→ CH3 + OH + O and numerical methods must be used to predict the evolution of the concentrations from the chain. Propagation: An additional point of the example was to demonstrate how we apply the law of mass action to obtain CH4 + OH −→ CH3 + O + H2 rate laws for the species that participate in the chain. CH4 + O −→ CH3 + OH

In more complicated chains (such as the methane CH3 + O2 −→ H2CO + OH oxidation mechanism given above) we can get more CO + OH −→ CO + H than one intermediate. On a chemical basis, these 2 intermediates are usually free radicals. In a precise H + O2 −→ OH + O sense, a free radical is a species that has a spin- H2CO + OH −→ HCO + H2O unpaired electron. It may be thought of as a molecular fragment of a stable molecule. For example, the Inhibition: H atom is a free radical, as is illustrated below, where the dots symbolize electrons H + H2O −→ OH + H2

H:H −→ H· +H· (54) Termination: Similarly, if one H atom is taken away from methane (CH ), two free radicals are formed: 4 HCO + OH −→ CO + H2O ∗ HH O + O + M −→ O2 + M H: C:H¨ −→ H: C¨· +· H (55) ∗ 2OH + M −→ H2O + O + M H¨ H¨ Free radicals are of utmost importance in chain mechanisms because they are very unstable compared to the reactant and product species in the overall re- We can write a general chain reaction mechanism action, and thus tend to promote chemical reactions. involving the four types of chain mechanisms. In The reactions that make up a chain have certain what follows, M denotes a (stable) chemical species, general characteristics that are classiﬁed as follows: R is a radical, P is a product, and α is simply a

8 number. this approximation is that the radicals are so reac- k tive that once they build up to a small level they M −→1 R initiation react quickly, and thus stay at the low and constant k level. Therefore, we can simply set Eq. (60) equal to R + M −→2 αR + M 0 propagation zero and algebraically solve for CR to obtain, k −→3 R + M P termination k C C = 1 M (61) k4 R R −→ destroyed termination (1 − α)k2CM + k3CM + k4 + k5 wall k5 R −→gas destroyed termination Using the above in Eq. (59), we get Note that radicals can be removed in reactions 4 and dC k k C2 P = 1 3 M (62) 5 through interactions with the container walls or dt (1 − α)k2CM + k3CM + k4 + k5 simply with the gas molecules. In reaction 2, if α = 1 the reaction is termed chain carrying – in that the Of utmost interest in combustion reactions is net production of radicals by the reaction is zero. If whether or not the above rate can go to infinity. In α > 1, the reaction is called chain branching – a other words, under what conditions will the rate of net amount (i.e., (α − 1)) of radicals are created by product formation essentially ‘blow up’. By examin- the reaction. In the methane oxidation chain prop- ing the denominator, we see that a critical value of α agation reactions listed above, the first, second, and can be defined, fifth propagation reactions would be chain branch- k3CM + k4 + k5 ing, since they produce two radicals in consuming αcritical = 1 + (63) one. The third, fourth, and sixth propagation reac- k2CM tions are chain carrying – one radical is produced by For α < α , the denominator in Eq. (62) will the consumption of one radical. critical stay positive and finite, and thus so will the rate of We would like to obtain an expression for the rate P production. But for α = α , the denominator change of C in terms of C . Using the law of mass critical P M goes to zero and the rate becomes infinite. We will see action, we set in the following section that an infinite reaction rate dC is referred to as explosive. A necessary criteria for M = −k C − k C C − k C C (58) dt 1 M 2 R M 3 R M the chain mechanism to exhibit explosive behavior is dC the existence of a chain branching step. On the other P = k C C (59) dt 3 R M hand, if all the propagation reactions are chain car- dC rying (i.e., α = 1), then the reaction does not exhibit R = k C + (α − 1)k C C − k C C dt 1 M 2 R M 3 R M explosive behavior. However, it is important to note that for some actual explosion processes, the steady − k4CR − k5CR (60) state approximation may not be valid because the We have three differential equations for the three concentration of R does not remain small and con- unknowns CM , CR, and CP . Assuming that we have stant. Other reaction steps may also be important, values for the rate constants, we could in principle and the postulated reaction mechanism given above integrate these equations to get the concentrations may not be applicable during an explosion. as a function of time – although this would probably have to be done numerically (as in the case above). 2.2 Chain Explosions A commonly used approach to quickly estimate the rate law from the above equations is the steady What we later will call a flame is a reaction that state approximation, for which the rate change of CR propagates through a fuel-oxidizer mixture that is (Eq. (60)) is set equal to zero. The rationale behind explosive. By explosive, we mean that the rate of

9 reaction is fast – so fast that the rate at which prod- What will be the minimum time required for all the ucts are formed becomes nearly infinite. Note that in molecules to react to form products? Since it takes a strict combustion sense, ‘explosive’ does not have one radical to make one product molecule, then we the common meaning of ‘blowing up’ (as in an explo- have to produce 1019 radicals for the reaction to com- sive device, i.e., a bomb), the latter will be referred plete. Since the most efficient process would be if ev- to as a detonation. ery collision resulted in a reaction, then the reaction There are two general ways by which a mixture can time for this case would simply be on the order of the explode. One, as mentioned above, is called a chain time required for the 1019 molecules to collide: explosion. Here the cause of the explosion results 1019 collisions from the existance of chain branching reactions in the t ≈ = 1011 s ≈ 30 years (66) chain mechanism. Because of the branching, more 108 collisions/s and more radicals are formed which continuously ac- Such a slow process cannot be called combustion. celerate the reaction. The other is referred to as a However, let’s now assume that the chain propaga- thermal explosion. Here the explosion results from tion reaction in the container is a chain branching the nature of the Arrhenius rate law. The rate, which reaction, and two free radicals are produced for ev- is proportional to exp(−E/R T ), increases rapidly u ery one consumed during a chain, i.e., with T . Thus a snowball effect can occur – the reaction heats up the reactants, which increases the rate, M + R −→ M 0 + 2R0 (67) which increases the heating of the products, and so on. Eventually, the rate ‘runs away thermally’, and Again, we have to produce 1019 radicals to completely an explosion occurs. convert the reactant molecules to product molecules. Actually, explosions are the result of both effects, After the first chain branching reaction, there will be and any general theory of explosions must incorpo- 2 radicals; after the second, 4; and so on. We then rate both. However, it is instructive to look at both get the series: effects separately, from which the general physical and chemical characteristics of explosions can be ex- 1 + 2 + 4 + 8 + ··· + 2N = 1019 (68) tracted. The profound effect chain branching has on reac- where N is the ‘number of generations’ of the chain tion rate can be dramatically illustrated by the fol- branching reaction – equivalently the number of cy- lowing simple example. Consider a 1 cm3 container cles or loops the chain has to go through. We can which initially contains 1 free radical among other- sum this well-known series to get wise gas reactant molecules. Assume the container is N+1 19 at S.T.P., for which the number of molecules in the 2 − 1 = 10 , or N = 64 generations (69) container will be around 1019 and the average molecu- So we need 64 loops of the chain to complete the lar collision rate will be around 108 collisions/s. (i.e., reaction. The time required to complete each loop a molecule will experience, on the average, 108 col- will still be on the order of the collision rate – thus lisions with other molecules each second). Say the chain propagation reaction in the container is chain t ≈ 64 × 10−8 s ≈ 1 µs (70) carrying, in that This would certainly be considered an explosive pro- 0 0 M + R −→ M + R (64) cess. In actual combustion processes, not all reactions are chain branching. Using the example given in the So one free radical is produced for every one con- previous section, we could have a chain branching re- sumed. Also, the chain termination reaction is of the action of the form, form, R0 + M −→ P (65) M + R −→ M 0 + αR0 (71)

10 where α > 1. After the ﬁrst reaction, we would have Turns gives a good overview of such global formulas α radicals; after the second, α2, and so on. We then on p. 157. Note that the global laws have the look of get the series the law of mass action – yet they apply to the overall reaction and not the elementary steps in the chain. 1 + α + α2 + α3 + ··· + αN = 1019 (72) The formulas are of the form,

or dCCxHy m n N+1 = A exp (−E /R T ) C C (74) α − 1 a u CxHy O2 = 1019 (73) dt α − 1 where A, Ea, m, and n depend on the speciﬁc hy- Say α = 1.01 – which appears to be a small number. drocarbon fuel type (i.e.,x and y). Values of these Solving the above for the number of generations N, constants are listed in Table 5.1 of Turns. we get N = 3934. The reaction time would now be The overall order of the rate law in Eq. (74) is sim- on the order of 40 µs – which is still a very fast (and ply m + n, i.e., the rate is proportional to concentra- explosive) reaction. Consequently, even a relatively tion raised to the m + n power. Since concentration small degree of chain branching can have a profound is proportional to pressure (ideal gas law), the rate eﬀect on the reaction rate. is proportional to P m+n. In addition, Eq. (74) also The explosive behavior of a fuel/air mixture can be shows that the rate is m–order with respect to fuel highly dependent on the temperature and pressure at and n–order with respect to oxygen. which the mixture exists. A good example of this is discussed at the beginning of Ch. 5 in Turns, in which he describes the explosion limit characteristics of the H2/O2 system.

2.3 Global Rate Laws From the previous discussion, an explosive system was defined as that in which the reaction rate could, essentially, go to infinity. Of course, the reaction rate will not be infinite in combustion systems – it will be very fast (compared to most other processes) yet still finite. We need to predict the reaction rate of reactants in order to predict, say, flame speeds or flammabil- ity limits. One route is simply to employ a com- prehensive numerical solution of the detailed chain mechanism for the reaction. This approach may be useful for detailed, scientific–oriented computations (i.e., comparison of theory with experiment), yet it is not practical for many engineering–type calculations. In this respect, there has been much effort to con- struct approximate, yet accurate, correlations for the reaction rates of hydrocarbon/air mixtures. Such correlations are often referred to as global rate laws, and are obtained via a combination of experimental measurements (curve fitting) and theoretical analysis.