Lecture 2. Chemical Kinetics. Chemical Kinetics 6/26/11. One (elementary) step reaction

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1 Lecture Chemical Kinetics 1 One (elementary) step reaction im i i M i is the number of species i, i are the stoichiometric coefficients i i Chemical Kinetics =0ifi is not a reactant =0ifi is not a product Example: H +O OH =3 H (i = 1) : 1 =1 1 =0 O (i = ) : =1 =0 OH (i = 3) : 3 =0 3 = permission of the owner, M. Matalon. 1

2 im i i M i there is a relation between the change in the number of moles of each species; i.e., for any two species i and j dn i i i = dn j j j If ˆω i is the time rate of change of the concentration of species i (moles, per unit volume per second), i.e., ˆω i = dc i /, then ˆω i i i = ˆω j j j or ˆω i i i = ˆω j j j = ω and we may the common ratio ω, which is species independent, as the reaction rate (moles per unit volume per second) 3 Law of Mass Action phenomenological law that was verified experimentally the reaction rate is proportional to the products of the concentrations reactants ω = k(t ) C i i (specific) reaction rate constant the units of k depend on the reaction order n = i and is [concentration (n 1) time] 1 example CH 4 +O CO +H O ˆω H O = ˆω CO 1 = ˆω CH 4 O 1 = ˆω O = ω ω = kc CH4 C O 4 permission of the owner, M. Matalon.

3 Chemical reaction involving M elementary steps i,jm i j =1,,...,M i,jm i where forward and backward reactions are written as separate steps net rate of production of species i ˆω i = M ˆω i,j j=1 ˆω i = M j=1 ( i,j i,j) ω j where ω j is the reaction rate of the elementary step j, namely ω j = k(t ) C i,j i 5 Chain reaction - hydrogen-bromine reaction Br +M k 1 Br + H k H + Br k 3 H + HBr k 4 Br + M HBr + H HBr + Br Br + H Br + M k 5 Br +M H +Br HBr ˆω i = 5 j=1 ( i,j i,j) ω j i = ω j = k(t ) 5 C i,j i H,Br, H, Br, HBr for example ˆω Br = dc Br =k 1 C Br k C Br C H + k 3 C H C Br + k 4 C H C HBr k 5 C Br 6 permission of the owner, M. Matalon. 3

4 Two of the important questions in chemical kinetics are determine all the elementary steps by which a given chemical reaction actually proceeds determine the specific rate constant for each step 7 Reaction Mechanisms First order decomposition reaction One step opposing reactions Chain reaction Reduced mechanisms modeling 8 permission of the owner, M. Matalon. 4

5 First order decomposition reaction A k B dc A = kc A C A (0) given Ca CA(0) k C A = C A (0)e kt t t = 1 k ln CA (0) C A t c 1 k characteristic time scale large k corresponds to a small time scale or very fast reaction this is a source of stiffness in the differential equations 9 One step opposing reaction A k f B B k b A dc A = k f C A + k b C B C dc A (0) specified, C B (0) = 0 B = k f C A k b C B d kf k M = b C = M C k f k C = b CA C B C = V 1 e λ1t + V e λt there are two characteristic times, corresponding to λ 1 1 and λ 1 10 permission of the owner, M. Matalon. 5

6 k f λ k f k b k b λ =0 λ 1 =0, λ = (k f + k b ) CA C B = C A(0) k f + k b kb k f + k f C A (0) 1 e (k f +k b )t k f + k b 1 CA C B = C eq 1 A + C eq 1 k f /k B e (k f +k b )t b 1 C eq A = k b k f + k b C A (0) C eq B = k f k f + k b C A (0) the characteristic times are t 1 = and t =(k f + k b ) 1 corresponding to the reciprocal of the eigenvalues; i.e., λ 1 1 and λ 1 at equilibrium, k f C eq A k bc eq B =0 11 In fact, for this simple system, one easily see that d [C A + C B ]=0 d [k f C A k b C B ]= (k f + k b )[k f C A k b C B ] z 1 = C A + C B z = k f C A k b C B d z1 z = 0 0 z1 0 (k f +k b ) z C A + C B = C A (0) k f C A k b C B = k f C A (0)e (k f+k b )t when k f + k b 1, the characteristic times are t 1 = 0 (slow time) and t =(k f + k b ) 1 (fast time) C A + C B is a conserved quantity, k f C A k b C B 0 almost all the time z 1 = z 1 (0) e 0 (kf +kb)t z = z (0) e C B k f C A k bc B C A 1 permission of the owner, M. Matalon. 6

7 Chain Reaction Hydrogen-Bromine Reaction any halogen molecule (F,CL or I ) may replace Br H +Br HBr Br +M k 1 k Br + H Br + M HBr + H chain initiating k H + Br 3 HBr + Br chain carrying H + HBr k 4 Br + H Br + M k 5 Br +M chain terminating the intermediates H, Br are the chain carriers 13 Br +M k 1 Br + H k H + Br k 3 H + HBr k 4 Br + M HBr + H HBr + Br Br + H Br + M k 5 Br +M H +Br HBr dc HBr = k C Br C H + k 3 C H C Br k 4 C H C HBr dc Br dc H = k C Br C H k 3 C H C Br k 4 C H C HBr =k 1 C Br k C Br C H + k 3 C H C Br + k 4 C H C HBr k 5 C Br dc H = k C Br C H + k 4 C H C HBr dc Br = k 1 C Br k 3 C H C Br + k 5 CBr 14 permission of the owner, M. Matalon. 7

8 We are faced with a set of nonlinear coupled differential equations (for spatially homogeneous system, as discussed here, these are ODEs), with generally a large number The description of the combustion of real fuels may involve 1000 species or more, involved in a complex network of elementary steps that add up to few thousands. For example, the detailed kinetic mechanism of the primary reference fuel (PRF) contains 1034 species participating in 436 elementary reactions 1. 1 H.J. Curran, P. Gaffuri, W.J. Pitz and C.K. Westbrook (C&F 00) permission of the owner, M. Matalon. 8

9 chemical Lme scales physical Lme scales slow Lme scales (O formalon) 10 0 s Intermediate Lme scales 10 - s 10-4 s Lme scales of flow, transport, turbulence fast Lme scales (steady- state, parlal equilibrium) 10-6 s 10-8 s IllustraLon of the various Lmes scales governing chemical reaclng flows 17 Br +M k 1 Br + H k H + Br k 3 H + HBr k 4 Br + M HBr + H HBr + Br Br + H Br + M k 5 Br +M H +Br HBr Radicals form and react very rapidly (at nearly equal rates) such that their concentralons remains nearly constant dc Br 0 Rate of formalon of HBr dc H 0 dc HBr = k C Br C H + k 3 C H C Br k 4 C H C HBr 18 permission of the owner, M. Matalon. 9

10 dc Br dc H 0 k C Br C H k 3 C H C Br k 4 C H C HBr =0 0 k 1 C Br k C Br C H +k 3 C H C Br +k 4 C H C HBr k 5 C Br =0 C Br = k1 C 1/ Br k C H = k k1 /k 5 C H C 1/ Br 5 k 3 C Br + k 4 C HBr dc HBr = k k1 /k 5 C H C 1/ Br 1+(k 4 /k 3 )C HBr C 1 Br dc HBr =k 0 C H C Br 19 dc HBr = k k1 /k 5 C H C 1/ Br 1+(k 4 /k 3 )C HBr C 1 Br dc HBr k 0 C H C 1/ Br when k 4 /k 3 1; (k 0 = k k1 /k 5 ) ω = k C n i i i = reactants only 0 permission of the owner, M. Matalon. 10

11 1 dc = F(C) C(0) = C 0 C =(C 1,C,...,C ) T C 1 C C 3 permission of the owner, M. Matalon. 11

12 dc = J C (J λi)v = 0 λ n v n eigenvalues eigenvectors 3 z ṼC dc = VΛṼ C Ṽ dc = ΛṼ C dz = Λz dz i =λ iz i i =1,...,n 4 permission of the owner, M. Matalon. 1

13 5 z ṼC dz i =λ iz i dz = Λz i =1,...,n 6 permission of the owner, M. Matalon. 13

14 Simple * hydrogen- oxygen kinelcs mechanism H, H, O, OH, H O, Ren et al. (J. Chem. Phys; 006) * RepresentaLon of H /O chemistry would involve O, HO, H O as well as O and related species. 7 Al- Khateeb et al. (J. Chem. Phys; 009) 8 permission of the owner, M. Matalon. 14

15 Specific Reaction-Rate Constant The Arrhenius Law k(t )=BT α e E/RT the pre-exponential factor has a weak temperature dependence, with 1 < α. The coefficient B is the frequency factor, and E is the activation energy. The probability that a molecule possesses energy E is proportional to exp (E/RT ). The exponential factor in the reaction rate ω represents the fraction of collisions between reactant molecules for which products can be formed. Energy E Reactants Products H ReacLon coordinate ( H) is the heat of reaction 9 The reaction rate of a elementary reaction im i i M i ω = BT α e E/RT C i i 30 permission of the owner, M. Matalon. 15