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Lecture 2 Chemical Kinetics
1
N�
i=1
ν�iMi →N�
i=1
�i Mi
Example: H2 + O2 → 2 OH
N = 3H2 (i = 1) : ν�1 = 1 ν��1 = 0O2 (i = 2) : ν�2 = 1 ν��2 = 0OH (i = 3) : ν�3 = 0 ν��3 = 2
One (elementary) step reaction
ν�i, ν��i are the stoichiometric coefficients
N is the number of species
ν�i = 0 if i is not a reactantν��i = 0 if i is not a product
Chemical Kinetics
2
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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
N�
i=1
ν�iMi →N�
i=1
�i Mi
there is a relation between the change in the number of moles of each species;i.e., for any two species i and j
dni
ν��i − ν�i=
dnj
ν��j − ν�j
ωi
ν��i − ν�i=
ωj
ν��j − ν�j
If ωi is the time rate of change of the concentration of species i (moles, per unitvolume per second), i.e., ωi = dCi/dt, then
ωi
ν��i − ν�i=
ωj
ν��j − ν�j= ωor
3
and we may the common ratio ω, which is species independent, as the reactionrate (moles per unit volume per second)
ω = k(T )N�
i=1
Cν�i
i
the reaction rate is proportional to the products of the concentrations reactants
Law of Mass Action
CH4 + 2 O2 → CO2 + 2 H2O
ω = k CCH4C2
O2
ωH2O
2=
ωCO2
1= −
ωCH4O
1= −
ωO2
2= ω
phenomenological law that was verified experimentally
(specific) reaction rate constant
the units of k depend on the reaction order n =�
ν�iand is [concentration(n−1)· time]−1
example
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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
N�
i=1
ν�i,jMi →N�
i=1
�i,jMi j = 1, 2, . . . ,M
Chemical reaction involving M elementary steps
ωi =M�
j=1
(ν��i,j−ν�i,j)ωj
net rate of production of species i ωi =M�j=1
ωi,j
where ωj is the reaction rate of the elementary step j, namely
ωj = k(T )N�
i=1
Cν�i,j
i
where forward and backward reactionsare written as separate steps
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Br2 + Mk1→ 2Br + M
Br + H2k2→ HBr + H
H + Br2k3→ HBr + Br
H + HBrk4→ Br + H2
2Br + Mk5→ Br2 + M
Chain reaction - hydrogen-bromine reaction
for example
ωBr =dCBr
dt= 2k1CBr2 − k2CBrCH2 + k3CHCBr2 + k4CHCHBr − 2k5C
2
Br
ωi =5�
j=1
(ν��i,j−ν�i,j)ωj
i = H2, Br2, H, Br, HBr
ωj = k(T )5�
i=1
Cν�i,j
i
⇒ H2 + Br2 → 2HBr
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Two of the important questions in chemical kinetics are
• determine all the elementary steps by which a given chemical reactionactually proceeds
• determine the specific rate constant for each step
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Reaction Mechanisms
8
• First order decomposition reaction
• One step opposing reactions
• Chain reaction
• Reduced mechanisms modeling
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First order decomposition reaction
Ak→ B
dCA
dt= kCA CA(0) given
CA = CA(0)e−kt
t =1
kln
�CA(0)
CA
�⇒ tc ∼
1
k
characteristic time scale
large k corresponds to a small time scale or very fast reactionthis is a source of stiffness in the differential equations
Ca
CA(0)
k
t
9
Akf→ B
Bkb→ A
dCA
dt= −kfCA + kbCB
dCB
dt= kfCA − kbCB
CA(0) specified, CB(0) = 0
M =
�−kf kbkf −kb
�
there are two characteristic times, corresponding to λ−11 and λ−1
2
One step opposing reaction
d
dtC = M ·C C =
�CA
CB
�
C = V1eλ1t +V2e
λ2t
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�CA
CB
�=
CA(0)
kf + kb
�kbkf
�+
kfCA(0)
kf + kb
�1−1
�e−(kf+kb)t
����−kf − λ kb
kf −kb − λ
���� = 0λ1 = 0,
λ2 = −(kf + kb)⇒
�CA
CB
�= Ceq
A
�1
kf/kb
�+ Ceq
B
�1−1
�e−(kf+kb)t
CeqA =
kbkf + kb
CA(0) CeqB =
kfkf + kb
CA(0)
the characteristic times are t1 = ∞ and t2 = (kf + kb)−1 corresponding to thereciprocal of the eigenvalues; i.e., λ−1
1 and λ−12
at equilibrium, kfCeqA − kbC
eqB = 0
11
d
dt[CA + CB ] = 0
In fact, for this simple system, one easily see that
d
dt[kfCA − kbCB ] = −(kf + kb)[kfCA − kbCB ]
d
dt
�z1z2
�=
�0 00 −(kf+kb)
��z1z2
�
z2 = kfCA − kbCB
z1 = CA + CB ⇒
z1 = z1(0) e0
z2 = z2(0) e−(kf+kb)t
CA + CB = CA(0)
kfCA − kbCB = kfCA(0)e−(kf+kb)t
CA
CB
k fCA
− k bCB
when kf + kb � 1, the characteristic times aret1 = 0 (slow time) and t2 = (kf + kb)−1 (fast time)
CA + CB is a conserved quantity,kfCA − kbCB ≈ 0 almost all the time
12
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Chain Reaction
any halogen molecule (F2, CL2 or I2) may replace Br2
the intermediates H, Br are the chain carriers
H2 + Br2 → 2 HBr
Hydrogen-Bromine Reaction
Br2 + Mk1→ 2Br + M
Br + H2k2→ HBr + H
H + Br2k3→ HBr + Br
H + HBrk4→ Br + H2
2Br + Mk5→ Br2 + M
�chain carrying
chain initiating
chain terminating
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Br2 + Mk1→ 2Br + M
Br + H2k2→ HBr + H
H + Br2k3→ HBr + Br
H + HBrk4→ Br + H2
2Br + Mk5→ Br2 + M
⇒ H2 + Br2 → 2HBr
dCHBr
dt= k2CBrCH2 + k3CHCBr2 − k4CHCHBr
dCBr
dt= 2k1CBr2 − k2CBrCH2 + k3CHCBr2 + k4CHCHBr − 2k5C
2
Br
dCH
dt= k2CBrCH2 − k3CHCBr2 − k4CHCHBr
dCH2
dt= −k2CBrCH2 + k4CHCHBr
dCBr2
dt= −k1CBr2 − k3CHCBr2 + k5C
2
Br
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1H.J. Curran, P. Gaffuri, W.J. Pitz and C.K. Westbrook (C&F 2002)
We are faced with a set of N nonlinear coupled differential equations (for spa-tially homogeneous system, as discussed here, these are ODEs), withN 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.
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IllustraLon of the various Lmes scales governing chemical reacLng flows
100 s
10-‐2 s
10-‐4 s
10-‐6 s
10-‐8 s
chemical Lme scales
physical Lme scales
Lme scales of flow, transport, turbulence
slow Lme scales (NO formaLon)
fast Lme scales (steady-‐state, parLal equilibrium)
Intermediate Lme scales
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Radicals form and react very rapidly (at nearly equal rates) such that their concentraLons remains nearly constant
dCBr
dt≈ 0
dCH
dt≈ 0
Rate of formaLon of HBr
dCHBr
dt= k2CBrCH2 + k3CHCBr2 − k4CHCHBr
Br2 + Mk1→ 2Br + M
Br + H2k2→ HBr + H
H + Br2k3→ HBr + Br
H + HBrk4→ Br + H2
2Br + Mk5→ Br2 + M
⇒ H2 + Br2 → 2HBr
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dCH
dt≈ 0 ⇒ k2CBrCH2 − k3CHCBr2 − k4CHCHBr = 0
dCBr
dt≈ 0 ⇒ 2k1CBr2−k2CBrCH2+k3CHCBr2+k4CHCHBr−2k5C
2
Br= 0
CBr =
�k1k5
C1/2Br2 CH =
k2�k1/k5 CH2 C
1/2Br2
k3CBr2 + k4CHBr
dCHBr
dt= 2k0 CH2 CBr2
⇒ dCHBr
dt=
2k2�k1/k5 CH2 C
1/2Br2
1 + (k4/k3)CHBrC−1
Br2
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dCHBr
dt=
2k2�k1/k5 CH2 C
1/2Br2
1 + (k4/k3)CHBrC−1
Br2
dCHBr
dt≈ 2k0 CH2 C
1/2Br2
when k4/k3 � 1; (k0 = k2�k1/k5)
ω = kN�
i=1
Cnii i = reactants only
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21
C1
C3
C2
C = (C1, C2, . . . , CN )T
dC
dt= F(C)
C(0) = C0
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dCdt
= J · C
(J− λI)v = 0 eigenvalues eigenvectors
λnvn
23
dCdt
= VΛV · C
VdCdt
= ΛV · C
z ≡ VCdzdt
= Λz
dzi
dt=λizi i = 1, . . . , n
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25
z ≡ VCdzdt
= Λz
dzi
dt=λizi i = 1, . . . , n
26
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Simple* hydrogen-‐oxygen kineLcs mechanism H2 , H , O , OH , H2O , N2
* RepresentaLon of H2 /O2 chemistry would involve O2 , HO2 , H2O2 as well as NO and related species.
Ren et al. (J. Chem. Phys; 2006)
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Al-‐Khateeb et al. (J. Chem. Phys; 2009)
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Specific Reaction-Rate Constant
The Arrhenius Lawthe pre-exponential factor has a weak temperature dependence,with −1 < α � 2. The coefficient B is the frequency factor,and E is the activation energy.
ReacLon coordinate
Energy
Reactants
Products
E
−∆H
−(∆H) is the heat of reaction
k(T ) = BTαe−E/RT
The probability that a molecule possessesenergy ≥ E is proportional toexp−(E/RT ). The exponential factorin the reaction rate ω representsthe fraction of collisions between reactantmolecules for which products can be formed.
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N�
i=1
ν�iMi →N�
i=1
�i Mi
The reaction rate of a elementary reaction
ω = BTαe−E/RTN�
i=1
Cν�i
i
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