Lecture 2 Chemical Kinetics

6/26/11

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

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)

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

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

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

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 •

Reaction Mechanisms First order decomposition reaction • One step opposing reactions • Chain reaction • Reduced mechanisms modeling •

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−

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

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−

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

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

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)

chemical physical me scales me scales

slow me scales 100 s (NO formaon)

10-2 s

Intermediate me scales of ﬂow, me scales 10-4 s transport, turbulence

10-6 s fast me scales (steady-state, -8 paral equilibrium) 10 s

Illustraon of the various mes scales governing chemical reacng ﬂows

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

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

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

dC = F(C) dt

C(0) = C0

C T 1 C =(C1,C2,...,CN )

 C 2

dC = J C dt ·

(J λI)v = 0 λn eigenvalues − vn eigenvectors

dC = VΛV˜ C dt · dC V˜ = ΛV˜ C dt · dz z VC˜ = Λz ≡ dt dz i =λ z i =1,...,n dt i i

dz z VC˜ = Λz ≡ dt dz i =λ z i =1,...,n dt i i

Simple* hydrogen-oxygen kinecs mechanism

H2 , H , O , OH , H2O , N2

Ren et al. (J. Chem. Phys; 2006)

* Representaon of H2 /O2 chemistry would involve O2 , HO2 , H2O2 as well as NO and related species. 27

Al-Khateeb et al. (J. Chem. Phys; 2009)

Speciﬁc Reaction-Rate Constant

α E/ T The Arrhenius Law k(T )= T e− R B the pre-exponential factor has a weak temperature dependence, with 1 < α 2. The coeﬃcient is the frequency factor, and E−is the activation energy. B

E Reactants The probability that a molecule possesses Energy energy E is proportional to Products exp (E/≥ T ). The exponential factor ∆H in the− reactionR rate ω represents − the fraction of collisions between reactant molecules for which products can be formed. Reacon coordinate (∆H) is the heat of reaction − 29

The reaction rate of a elementary reaction N N ν ν iMi → i Mi i=1 i=1

N α E/ T νi ω = T e− R C B i i=1