Lecture 2 - Chemical Kinetics - Princeton University · Two of the important questions in chemical...

6/26/11

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

Lecture 2 Chemical Kinetics

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N�

i=1

ν�iMi →N�

i=1

ν��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

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N�

i=1

ν�iMi →N�

i=1

ν��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

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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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N�

i=1

ν�i,jMi →N�

i=1

ν��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

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• 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

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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

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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

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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




2Br + Mk5→ Br2 + M

�chain carrying

chain initiating

chain terminating

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Br2 + Mk1→ 2Br + M




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




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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z ≡ VCdzdt

= Λz

dzi

dt=λizi i = 1, . . . , n

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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

ν��i Mi

The reaction rate of a elementary reaction

ω = BTαe−E/RTN�

i=1

Cν�i

i

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Lecture 2 - Chemical Kinetics - Princeton University · Two of the important questions in chemical...

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