chapter 3 - chain reaction pnmeela
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Chain Reactions
CPE624 FACULTY OF CHEMICAL ENGINEERING
� Example : Decomposition of acetaldehyde
Initiation step Propagation #1 Propagation #2 Termination step
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CH
COCHCHOCH +→ 43
23623
32233
311343
3
][ 2
][ ]][[
][ ,
•=→•
•=•+→•
•=•+→•+
=•+•→
CHkrHCCH
COCHkrCHCOCOCH
CHAkrCOCHCHCHAAkrCHOCHA
tt
pp
pp
ii
23
AkCr =
� Propagation steps occur faster than initiation and termination steps
� Radical species are very reactive and concentrations are always very low.
� Major products generated by propagation step
� Minor products made by initiation and termination steps.
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Mass balance
� PFTR/Batch : Time derivative,
� CSTR : Concentration gradient,
� Mass balance = - reactant + product
dtdCA
OAA CCr −=τ
Mass balance : PFTR ]][[][][
311 •−−=−−= CHAkAkrrdAd
pipiτ
tppi rrrrdCHd 2][
213 −+−=•
τ2
33231 ][2][]][[][ •−•+•−= CHktCOCHkCHAkAk ppi
][]][[][323121
3 •−•=−=• COCHkCHAkrr
dCOCHd
ppppτ
][][322 •== COCHkr
dCOd
ppτ
]][[][311
4 •== CHAkrdCHd
ppτ
23623
32233
311343
3
][ 2
][ ]][[
][ ,
•=→•
•=•+→•
•=•+→•+
=•+•→
CHkrHCCH
COCHkrCHCOCOCH
CHAkrCOCHCHCHAAkrCHOCHA
tt
pp
pp
ii
Recall:
Activity in Class : Write mass balance for CSTR
When the concentration of the two radicals are low (very reactive) :
� Hence :
Rates of two propagation steps are exactly equal
0][][ 33 =•−•
τoCOCHCOCH
0][]][[ 3231 =•−• COCHkCHAk pp
][]][[ 3231 •=• COCHkCHAk pp
� Hence adding the two propagation steps:
0][][][][ 3333 =•−•
=•−•
ττoo COCHCOCHCHCH
233231 ][2][]][[][ •−•+•− CHktCOCHkCHAkAk ppi
][]][[ 3231 •−•= COCHkCHAk pp
23 ][2][ •= CHktAki
The initiation and termination step are exactly equal
� The major processes are the propagation steps
� Adding p1 and p2 steps yield:
� Autocatalytic – the reaction generates the catalyst that promotes the reaction.
� The overall rate is enhanced by large ki and inhibited by large kt even though ki, kt << kp
� Kinetic chain length – ratio of rp to rt. CPE624 FACULTY OF CHEMICAL ENGINEERING
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•++→•+ 343 CHCOCHCHA
� PSS valid when the concentration of the species is small.
� In batch/PFR – setting time derivative equal to zero
� In CSTR - Mass balance equations are developed by assuming steady state, so that PSS of intermediate species is in steady state and concentration is small.
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� Hence :
0][][][][ 3333 =•−•
=•−•
ττoo COCHCOCHCHCH
233231 ][2][]][[][ •−•+•− CHktCOCHkCHAkAk ppi
][]][[ 3231 •−•= COCHkCHAk pp
23 ][2][ •= CHktAki
t
i
kAkCH2][][ 2
3 =•
2/12/1
3 ][2
][ AkkCHt
i⎟⎟⎠
⎞⎜⎜⎝
⎛=•
τ][][ 44 oCHCH
r−
=
]3][[1 •= CHAkp
2/12/1
1 ][2
][ AkkAkt
ip ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
2/12/3
1 2][ ⎟⎟
⎠
⎞⎜⎜⎝
⎛=
t
ip k
kAk
� The reaction propagates by radical R·.
� ni and nt are the number of molecules react in initiation and termination steps respectively.
� PSS approximation on CR yield the overall rate expression
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CBA +→
XRnRCBRA
RAn
t
i
→•
•++→•+
•→
,, [ ]
[ ][ ][ ] t
i
nRtt
RApp
nAii
Ckr
CCkrCkr
=
=
=
[ ] t
it
nn
Ap
n
tt
i Ckknkr +
⎟⎟⎠
⎞⎜⎜⎝
⎛= 1
1
[ ] effnAeff Ckr =
Generic Chain Reaction
ntni
Ap
nt
tt
inAeff Ck
knkCkr eff
+
⎟⎟⎠
⎞⎜⎜⎝
⎛==
1
1
][][ p
nt
tt
ieff k
knkk
1
⎟⎟⎠
⎞⎜⎜⎝
⎛=
t
ieff n
nn +=1, ,
⎟⎠
⎞⎜⎝
⎛ −=RTEkk o exp
Replacing each k in initiation, propagation and termination with :
Effective activation energy has become :
pt
t
t
ieff E
nE
nEE +−=
Generic Chain Reaction
2][Akkkr pt
i⎟⎟⎠
⎞⎜⎜⎝
⎛=
If ni=nt=1 :
RTE
effot
ieff ekp
kkk
−==
Hence the new effective rate coefficient :
ptieff EEEE +−=
Example 10-1 (Schmidt pg 404) � Consider the chain reaction in a CSTR:
a) Write the mass balance equations for A, B, R and X in a CSTR assuming constant density.
b) What is the overall reaction rate with respect to CA? c) Find τ for 90% conversion of A in CSTR assuming pseudo
steady state if CA0 = 2 moles/liter, ki = 0.001 sec-1, kp = 20 liter/mole sec and kt = 0.1 sec-1.
d) What are CR and Cx for this conversion? CPE624 FACULTY OF CHEMICAL ENGINEERING
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XRRCBRA
RA
→
++→+
→
,, [ ]
[ ][ ][ ]Rtt
RApp
Aii
Ckr
CCkrCkr
=
=
=
Exercise in class
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§ Large temperature dependences.
§ Sensitive to trace impurities that can alter the initiation and termination rates
§ Initiators and Scavengers (promoters and poisons) have large influences.
§ Initiators ú Initiated by adding species I that easily forms radical ú Initiate the reaction faster than reactant.
§ Scavengers ú Termination step - radical species decomposed /reacted with other
radical species to form an inactive species X ú Adding scavengers S ú X is important in determining overall reaction rate
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SRtsts CCkrXSR =→+ ,
Example: Briefly explain the terms initiator and scavenger for a chain reaction � Initiator: Species (I) that can easily form radicals in a
chain reaction and can initiate the reaction faster than the reactant.
� Scavenger: Species (S) added into a chain reaction that can readily scavenge the chain propagator to terminate the reaction.
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� Surface reaction steps important in controlling chain reactions.
� Wall termination reactions introduce a complexity to all chain reactions – the overall reaction rate as a function of the size of reactor.
� In a small reactor, termination reactions on surface keep the radical intermediate small and inhibit chain reaction.
� In large reactor, the termination rate is smaller.
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Wall termination reactions
SRtt CCkrXSRss
=→+ ,
)(, """sbS RRmRtttt CCkCkrr
volarear −==⎟
⎠
⎞⎜⎝
⎛=
DDSkk AhD
mt ==" Mass transfer controlling
mt kvolareak
volareak ** " == Process control by surface
termination
Example Consider the chain reaction in a CSTR:
a) Write the mass balance equations for A, B, R and X in a CSTR
assuming constant density. b) What is the overall reaction rate with respect to CA? c) Find τ for 98% conversion of A in CSTR assuming pseudo
steady state if CA0 = 2 moles/liter, ki = 0.002 sec-1, kp = 10 liter/mole sec and kt = 0.05 sec-1.
d) What are CR and Cx for this conversion?
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XRRCBRA
RA
→
++→+
→ [ ][ ][ ][ ]Rtt
RApp
Aii
Ckr
CCkrCkr
=
=
=
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(a) Initiation: A → R• ; ri = kiCA
Propagation: A + R• → B + C + R•; ri = kpCA CR Termination: R•→ X; rt = ktCR (b) Overall reaction rate: CR/τ = kiCA – ktCR = 0 ∴ CR = ki/kt x CA ∴ r = kpCACR = kikp/kt CA
2 (c) Solving CSTR equation: τ r = CAo – CA τ = (CAo – CA)/r = (CAo – CA )/CA
2 x kt/kikp = [2 moles/L - 2(1-0.98)]/(0.04)2 x 0.05/ s/0.002/s x 10 l/s
= 1225 x 2.5 = 3062.5 s = 51 min
Q#1 Test 2– Oct 2012
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� Fast and exothermic that proceeds by free radical chain reactions
� Chains reactions, once ignited, process proceeds very quickly and becomes very non-isothermal
� Release large amounts of energy
� Applications in the production of power, heat in incineration
� Involve multiphases: oxidants is air, fuel is liquids/solids
� Example is oxidation of H2 and alkanes
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� Autooxidation – autocatalytic process and it is an oxidation that converts alkanes into alkyl peroxides.
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ROOHOHR →+− 2
•+→−+•
•→+•
•+•→−
RROOHHROROOORHRHR
RO
,
2
� Lab safety : � Organic chemicals will react with oxygen
in the air at room temperature in chain reactions
� Hence forming organic peroxides (fuel /C & H atoms in the compounds)
� Consequences : spontaneous react, explosion upon shaking or opening the cap
� Organic peroxide have very exothermic heats of decomposition
� The reactions can only occur if oxygen is present
� The reaction depends crucially on the initiation steps, and different molecules have vastly different capabilities of dissociation
� The initiation step can be photoinduced. If a bottle is sitting in sunlight, UV photons can cause photo dissociation to initiate the chain reaction much faster than in the dark
� Organic molecule (R-H) where R. could be an alkyl or any fragment containing C, H and O atoms
� In the presence of O2, this molecule undergo auto oxidation reaction :
� The chain reactions proceeds at high temperature to break the R-H strong bond (>80 kcal/mole)
ROOHOHR →+− 2
� Consider the chain reaction sequence :
ROOHOHR
RROOHHRROOROOORHCHR
→+−
•+→−+•
•→+•
•+→−
2
2
__________________________,
,, initiation
oxidation
radical transfer
Thermal and chemical autocatalysis
� Chemical autocatalysis – reactions accelerates chemically such as in enzyme-promoted fermentation, or chain branching reaction
� Enzyme reaction nearly isothermal but combustion processes are both chain branching and auto thermal
� Autocatalysis – reactive intermediate or heat can act as catalyst to promote the reaction
� Catalyst is generated by the reaction , by adding promoters or heat to initiate the and accelerate the process
� Produce more than one free radical species in propagation step. Thus, the propagation steps increase the concentration of radical species and destabilize the kinetics.
� Example:
� Rapid rise in the concentration of radical species can accelerate the reaction and possibly a chain-branching explosion.
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•+•+•→ OHORROOH
� Produce two reactive free radicals from one
� Propagation steps produce more radical species that start with. ie:
The hydroperoxy radical .OOH reacts immediately at high temperature
� Most combustion reactions involve chain branching
•+•→•→+• OOHOOHOH 2
� Consider the reaction of :
� The mechanisms are:
� derive
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CBA +→
XRRCBRA
RA
→
++→+
→
,,
α
Rtt
RApp
Aii
Ckr
CCkrCkr
=
=
=
Apt
AiR Ckk
CkC)1( −−
=α
� For α > 1 and CA is large enough and not in SS :
� If the propagation term is large compares to others:
� Sample of chain branching reaction : Combustion of Hydrogen to form water
Apt Ckk )1( −= α
ktRR
RApR
eCC
CCkdtdC
ot=
−+=
)(
)1(α
Exercise in Class : Problem 10-16 (Schmidt) pg 440
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� Chemical industry is the safer industry compare to the risk if you drive on the road
� Some example of disaster
-Texas City disaster I and II
-Flixborough & Philips polyethylene plane explosion
-Bhopal incident
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References
� Schmidt, L.D. (2005). The Engineering of Chemical Reactions, 2nd edition, New York: Oxford University Press.
� Fogler, H.S. (2006). Elements of Chemical Reaction Engineering, 4th Edition, New Jersey: Prentice Hall.
� Levenspiel, O. (1999). Chemical Reaction Engineering, 3rd Edition, New York: John Wiley.
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CBA +→
Aii CkrRA =→ ,A+ R→ B+C +αR , rp = kpC ACR
Rtt CkrXR =→ ,
Apt
AiR Ckk
CkC)1( −−
=α
0)1( =−−+= RtRApAiR CkCCkCkdtdC
α
For linear chain :
t
AiR k
CkC =
Exam Q#1-June 2012
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