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Lecture 2 Chemical Kinetics

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Lecture 2 Chemical Kinetics 6/26/11 Lecture 2 Chemical Kinetics 1 Chemical Kinetics One (elementary) step reaction N N ν ν iMi → i Mi i=1 i=1 N is the number of species νi, νi are the stoichiometric coefficients νi =0ifi is not a reactant νi =0ifi is not a product Example: H +O 2OH 2 2 → N =3 H2 (i = 1) : ν1 =1 ν1 =0 O2 (i = 2) : ν2 =1 ν2 =0 OH (i = 3) : ν3 =0 ν3 =2 2 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 1 6/26/11 N N ν ν iMi → i Mi i=1 i=1 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 dn i = j ν ν ν ν i − i j − j Ifω ˆi is the time rate of change of the concentration of species i (moles, per unit volume per second), i.e.,ω ˆi = dCi/dt,then ωˆ ωˆ i = j ν ν ν ν i − i j − j ωˆ ωˆ or i = j = ω ν ν ν ν i − i j − j 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 N νi ω = k(T ) Ci i=1 (specific) reaction rate constant the units of k depend on the reaction order n = νi (n 1) 1 and is [concentration − time]− · example CH +2O CO +2HO 4 2 → 2 2 ωˆ ωˆ ωˆ ωˆ H2O = CO2 = CH4O = O2 = ω 2 1 − 1 − 2 ω = kC C2 CH4 O2 4 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 2 6/26/11 Chemical reaction involving M elementary steps N N ν ν j =1, 2,...,M i,jMi → i,jMi i=1 i=1 where forward and backward reactions are written as separate steps M net rate of production of species i ωˆi = ωˆi,j j=1 M ωˆ = (ν ν ) ω i i,j − i,j j j=1 where ωj is the reaction rate of the elementary step j, namely N νi,j ωj = k(T ) Ci i=1 5 Chain reaction - hydrogen-bromine reaction k1 Br2 +M 2Br + M k→2 Br + H2 HBr + H →k3 H + Br HBr + Br H2 +Br2 2HBr 2 ⇒ → →k4 H + HBr Br + H2 →k 2Br + M 5 Br +M → 2 5 5 νi,j ωˆ = (ν ν ) ω ωj = k(T ) C i i,j − i,j j i j=1 i=1 i =H2,Br2, H, Br, HBr for example dC ωˆ = Br =2k C k C C + k C C + k C C 2k C2 Br dt 1 Br2 − 2 Br H2 3 H Br2 4 H HBr − 5 Br 6 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 3 6/26/11 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 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 4 6/26/11 First order decomposition reaction A k B → Ca dC CA(0) A = kC C (0) given dt A A k kt CA = CA(0)e− t 1 C (0) t = ln A 1 k C tc A ⇒ ∼ 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 k A f B →k B b A → dC A = k C + k C dt f A b B − C (0) specified, C (0) = 0 dC A B B = k C k C dt f A − b B d k k C C = M C M = − f b C = A dt · kf kb CB − λ1t λ2t C = V1e + V2e 1 1 there are two characteristic times, corresponding to λ1− and λ2− 10 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 5 6/26/11 k λ k λ1 =0, − f − b =0 k k λ λ = (k + k ) f − b − ⇒ 2 f b − CA CA(0) kb kf CA(0) 1 (k +k )t = + e− f b CB k + k kf k + k 1 f b f b − C 1 1 A = Ceq + Ceq e (kf +kb)t C A k /k B 1 − B f b − eq kb eq kf CA = CA(0) CB = CA(0) kf + kb kf + kb 1 the characteristic times are t1 = and t2 =(kf + kb)− corresponding to the ∞ 1 1 reciprocal of the eigenvalues; i.e., λ1− and λ2− at equilibrium, k Ceq k Ceq =0 f A − b B 11 In fact, for this simple system, one easily see that d [C + C ]=0 dt A B d [k C k C ]= (k + k )[k C k C ] dt f A − b B − f b f A − b B z1 = CA + CB d z 00 z 1 = 1 z2 = kf CA kbCB ⇒ dt z2 0 (kf +kb) z2 − − 0 z1 = z1(0) e (kf +kb)t z2 = z2(0) e− CA + CB = CA(0) (kf+kb)t kf CA kbCB = kf CA(0)e− CB − C B k b − C A f when kf + kb 1, the characteristic times are k 1 t1 = 0 (slow time) and t2 =(kf + kb)− (fast time) CA CA + CB is a conserved quantity, kf CA kbCB 0 almost all the time − ≈ 12 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 6 6/26/11 Chain Reaction Hydrogen-Bromine Reaction any halogen molecule (F2,CL2 or I2) may replace Br2 H +Br 2 HBr 2 2 → k Br +M 1 2Br + M chain initiating 2 → k Br + H 2 HBr + H 2 → k H + Br 3 HBr + Br chain carrying 2 → k H + HBr 4 Br + H → 2 k 2Br + M 5 Br +M chain terminating → 2 the intermediates H, Br are the chain carriers 13 k1 Br2 +M 2Br + M k→2 Br + H2 HBr + H →k3 H + Br HBr + Br H2 +Br2 2HBr 2 ⇒ → →k4 H + HBr Br + H2 →k 2Br + M 5 Br +M → 2 dC HBr = k C C + k C C k C C dt 2 Br H2 3 H Br2 − 4 H HBr dC H = k C C k C C k C C dt 2 Br H2 − 3 H Br2 − 4 H HBr dC Br =2k C k C C + k C C + k C C 2k C2 dt 1 Br2 − 2 Br H2 3 H Br2 4 H HBr − 5 Br dC H2 = k C C + k C C dt − 2 Br H2 4 H HBr dC Br2 = k C k C C + k C2 dt − 1 Br2 − 3 H Br2 5 Br 14 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 7 6/26/11 We are faced with a set of N nonlinear coupled differential equations (for spa- tially homogeneous system, as discussed here, these are ODEs), with N 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 4236 elementary reactions1. 1 H.J. Curran, P. Gaffuri, W.J. Pitz and C.K. Westbrook (C&F 2002) 15 16 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 8 6/26/11 chemical physical Lme scales Lme scales slow Lme scales 100 s (NO formaon) 10-2 s Intermediate Lme scales of flow, Lme scales 10-4 s transport, turbulence 10-6 s fast Lme scales (steady-state, -8 parLal equilibrium) 10 s Illustraon of the various Lmes scales governing chemical reacLng flows 17 k1 Br2 +M 2Br + M k→ Br + H 2 HBr + H 2 → k3 H +Br 2HBr H + Br2 HBr + Br ⇒ 2 2 → →k4 H + HBr Br + H2 →k 2Br + M 5 Br +M → 2 Radicals form and react very rapidly (at nearly equal rates) such that their concentraons remains nearly constant dC dCH Br 0 0 dt ≈ dt ≈ Rate of formaon of HBr dC HBr = k C C + k C C k C C dt 2 Br H2 3 H Br2 − 4 H HBr 18 Copyright ©2011 by Moshe Matalon. This material is not to be sold, reproduced or distributed without the prior wri@en permission of the owner, M. Matalon. 9 6/26/11 dC H 0 k C C k C C k C C =0 dt ≈ ⇒ 2 Br H2 − 3 H Br2 − 4 H HBr dC Br 0 2k C k C C +k C C +k C C 2k C2 =0 dt ≈ ⇒ 1 Br2 − 2 Br H2 3 H Br2 4 H HBr − 5 Br 1/2 k1 1/2 k2 k1/k5 CH C C = C 2 Br2 Br Br2 CH = k5 k C + k C 3 Br2 4 HBr 2k k /k C C1/2 dCHBr 2 1 5 H2 Br2 = 1 ⇒ dt 1+(k /k )C C− 4 3 HBr Br2 dC HBr =2k C C dt 0 H2 Br2 19 2k k /k C C1/2 dCHBr 2 1 5 H2 Br2 = 1 dt 1+(k /k )C C− 4 3 HBr Br2 dCHBr 1/2 2k0 CH C when k4/k3 1; (k0 = k2 k1/k5) dt ≈ 2 Br2 N ni ω = k Ci i = reactants only i=1 20 Copyright ©2011 by Moshe Matalon.
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