Combustion Chemistry Hai Wang Stanford University 2015 Princeton-CEFRC Summer School On Combustion Course Length: 3 hrs June 22 – 26, 2015 Copyright ©2015 by Hai Wang This material is not to be sold, reproduced or distributed without prior written permission of the owner, Hai Wang. Lecture 4 4. Bimolecular Reaction Rate Coefficients In the last lecture, we learned qualitatively the reaction mechanisms of hydrocarbon combustion. To make this description quantitative, we will need to have a basic knowledge of reaction rate theories. While the rate coefficients of a large number of combustion reactions are experimentally measured, reaction rate theories are often necessary to interpret the experimental data. We shall focus our discussion here to bimolecular reactions of the type A + B → C + D and leave the discussion for unimolecular reactions to a later time. 4.1 Hard Sphere Collision We discussed earlier that an elementary chemical reaction requires molecular collision. The rate coefficient of a bimolecular reaction is essentially the product of reaction probability of two reactants (A and B) upon collision at a given temperature γ(T) and the frequency of collision ZAB, k(T) [A][B] = γ(T) ZAB (4.1) Here we assume that molecules may be described by rigid spheres. Figure 4.1 shows several scenarios of molecular encounter. Starting from the head-on collision, an offset of the two axes of molecular motion leads to off-center collision. Suppose the diameter of A and B are σA and σB, respectively. The limiting off-center collision would have a spacing equal to (σA+σB)/2 between the two axes of molecular motions. It may be concluded from this simple analysis that two molecules with their axes of motions lie within a cylindrical volume of cross section equal to A B σ σA B Head-on collision Off-center collision σAB =(σA+σB)/2 Figure 4.1 Various Limiting off-center collision scenarios of molecule collision. 4-1 Stanford University ©Hai Wang Version 1.2 ! $2 2 σ A +σ B S = πrAB = π # & (4.2) " 2 % would collide with each other. Here rAB is the collision radius. Suppose an NB number of B molecules are at rest and confined to an arbitrary volume V. An A molecule travels through the volume with a mean velocity equal to vA . The 2 cylindrical collision volume that A sweeps through per unit time is πσ ABvA (see, Figure 4.2). The number density of B is NB/V. Therefore the number of collision per unit time is N Z s−1 = πσ 2 v B . (4.3) B ( ) AB A V If we have NA A molecules in the same volume, the collision rate is equal to N N Z cm−3s−1 = πσ 2 v A B . (4.4) AB ( ) AB A V V 2 ps AB vA Figure 4.2 Schematic illustration of the cylindrical collision volume in a total volume V. NB, V The above derivation is simple, but it has one problem. That is, it treats the molecules like “ghost” particles since collision does not lead to changes in the direction of motion of the A molecule. This problem is easily mended by realigning the cylindrical volume with the velocity vector of the A molecule every time it undergo scattering with a B molecule. Since the number densities of A and B are independent of the orientation of the cylindrical volume, the result is not affected by the “ghost” particle assumption. A much more important problem in the derivation comes from the assumption that the B molecules are at rest while A sweeps through the volume. Actually it is the mean, relative velocity between A and B ( vAB ) that determines the collision rate, i.e., 4- 2 Stanford University ©Hai Wang Version 1.2 N N Z cm−3s−1 = πσ 2 v A B . (4.5) AB ( ) AB AB V V In other words, the rate coefficient of bimolecular reaction at the collision limit (i.e., γ(T) ≡ 1) may be given by 3 −1 −1 2 kcoll (cm mol s ) = πσ ABvABNavg . (4.6) The remaining derivation will have to be centered on the relative, mean velocity vAB . 4.2 Mean Molecular Velocity Recall from Lecture 2 that the distribution of translational energy is given by the Maxwell- Boltzmann distribution, E k T N e− i B ρ E = i = . (2.41) ( i ) −E k T N e j B ∑ j We wish to transform the energy distribution in terms of momentum, i.e., 2 2 2 2 p px + p y + pz E = = (4.7) 2m 2m N " p2 + p2 + p2 % i = exp$− x y z ' q . (4.8) $ 2mk T ' trans N # B & Here m is the molecular mass. The ratio Ni/N is proportional to the probability density function of momentum distribution,* i.e., 222 c ⎛⎞pppxyz++ fppp,,= exp− . (4.9) ( xyz) ⎜⎟ qmkTtrans⎝⎠2 B where c is the normalization constant, which can be found by recognizing that the probability of a molecule having any momentum must be unity. Therefore, 2 2 2 ∞ ∞ ∞ ∞ ∞ ∞ $ p + p + p ' c & x y z ) f px , p y , pz dpx dp y dpz = exp − dpx dp y dpz ∫ 0 ∫ 0 ∫ 0 ( ) q ∫ 0 ∫ 0 ∫ 0 & 2mk T ) .(4.10) trans % B ( = 1 * The probability density function describes that the probability to find a molecule having momentum in the ranges of px to px+dpx, py to py+dpy, and pz to pz+dpz is f(px, py, pz) dpx dpy dpz. 4- 3 Stanford University ©Hai Wang Version 1.2 The above integral gives q c = trans (4.11) (2π mkT )32 222 1 ⎛⎞pppxyz++ fppp,,= exp− . (4.12) ( xyz) 32 ⎜⎟ (2πmkT ) ⎝⎠2mkB T Recognizing that the velocity distribution function is proportional to the momentum distribution function, i.e., ⎛⎞dp ⎛⎞⎛⎞dp dp fvvvdvdvdvfppp,,= , , x yz dvdvdv , (4.13) v ( xyz) xyz( x yz)⎜⎟⎜⎟⎜⎟ xyz ⎝⎠dvxyz⎝⎠⎝⎠ dv dv we obtain 32 ⎡⎤222 ⎛⎞m mv( xyz++ v v) fv vxyz, v , v dv xyz dv dv=⎜⎟ exp⎢⎥− dv xyz dv dv . (4.14) ( ) 22πkT k T ⎝⎠⎣⎦⎢⎥B We are, in fact, interested in the total velocity of a molecule rather than the component velocity, 222 vvvv=++xyz . (4.15) In the spherical coordinate, we may write dvxyz dv dv= dv( vdθθφ)( vsin d ) . (4.16) Therefore the probability density function of the total velocity is 222 32 ⎡⎤mv++ v v ππ2 ⎛⎞m ( xyz) 2 f( v) dv=⎜⎟exp⎢⎥− dv v sinθφθ d d 22πkT k T ∫∫00 ⎝⎠⎣⎦⎢⎥B . (4.17) 32 ⎡⎤mv222++ v v 2 ⎛⎞m ( xyz) =4πv⎜⎟ exp⎢⎥− dv 22πkT k T ⎝⎠⎣⎦⎢⎥B The mean velocity of a molecule is ∞ 8kT v== vf( v) dv B . (4.18) ∫0 π m 4- 4 Stanford University ©Hai Wang Version 1.2 The mean, relative velocity of two types of molecules is 88kT⎛⎞11 kT vvv=+=22 BB + = , (4.19) AB A B ⎜⎟ ππ⎝⎠mmAB µ where µ is the reduced mass (see, p. 18 of Lecture 2 notes). It follows that the collision rate constant is 8kT kNcm311 mol-- s = B ps 2 . (4.20) coll( ) pm AB avg Take the example of H• + O2 → products. We have σ = 2.05 (Å); σ = 3.46 (Å); σ = 2.755 (Å); H i O2 H i-O2 "1×32 % 1 −24 µ = $ ' = 1.61×10 (g) #1+ 32 &6.023×1023 In cgs units, the collision rate coefficient is −16 2 3 −1 −1 8×1.3806×10 −8 23 kcoll cm mol s = × π × 2.755×10 ×6.023×10 T ( ) π ×1.61×10−24 ( ) 14 3 –1 –1 14 At 300 K, we have kcoll = 3.7×10 (cm mol s ); and at 1000 K, kcoll = 6.7×10 (cm3mol–1s–1). In other words, the collision rate coefficient is of the order of 1014 cm3mol– 1s–1. The order of magnitude may be compared favorably with the pre-exponential factor of the experimental rate expression for reaction (3.41), H• + O2 → O• + OH• (see Figure 3.7), # & 3 16 −0.671 71.3 (kJ/mol) k(cm mol-s) = 2.65×10 T exp%− ( . $% RuT '( which gives 2.65×1016T −0.671 = 5.8 and 2.6×1014 (cm3mol–1s–1) at 300 and 1000 K, respectively. Therefore, we may conclude that the pre-exponential factor is related to the collision frequency, and the exponential term is related to the reaction probability γ(T). Of course, the reaction probability is roughly determined by the fraction of molecules with combined energy greater than the activation energy Ea. 4- 5 Stanford University ©Hai Wang Version 1.2 4.3 Collision of Non-Rigid Sphere All gas molecules attract each other at a long range of separations, and they repel against each other at short separations. These forces are commonly known as the van der Waals interactions and may be modeled with the Lennard-Jones (L-J) 12-6 potential function, ( 12 6+ !σ $ !σ $ V r = 4ε *# AB & −# AB & - , (4.21) vdW ,AB ( ) AB * r r - )" % " % , where εAB is the collision well depth,σ AB is the collision diameter, and r is the distance of intermolecular separation (see, Figure 4.3). Table 4.1 lists the L-J 12-6 potential parameters for the self collision of some typical species. For dissimilar molecules, the pair potential parameters may be estimated from the self-collision parameters σ +σ σ ≈ A B , (4.22) AB 2 εAB = εAεB , (4.23) Of course, the force on each molecule is related to the potential energy, dV (r ) F r = vdW ,AB . (4.24) ( ) dr Let the positions of two colliding molecules be r1 and r2.