Introduction to Chain ReactionsBGH/UCDavis/April_08

IntroductionIn these notes we will consider chain reactions in a homogeneous system. Chain reactions can be broken downinto the following steps: initiation, chain propagation, inhibition, branching, and termination steps. The reactivespecies in a chain reaction are called the chain carriers which are reactive intermediates (normally free radicals)that are generated in an initiation step. In the propagation step the chain carriers react with reactants to produceproducts and regenerate the chain carriers. Chain carriers can also react with products to reform the reactants andthe chain carrier. When this step occurs it is called the inhibition step which involves no reduction in the numberof chain carriers. If there is a step in which two or more carriers are produced from a single chain carrier, the stepis called chain branching. In the termination step chain carriers are consumed.

Free RadicalsFree radicals are typically produced by the homolytic cleavage of covalent bonds. There are numerous methodsfor producing free radicals. As an example consider the dissociation of chlorine gas Cl2 which has an absorptionband in the visible spectrum. The bond dissociation energy for Cl2 is about 58 kcal/mol. Thus sunlight canprovide sufficient energy for the cleavage of the chlorine-chlorine bond:

(1)Sunlight : Cl - Cl öhn

ClË + ClË

The bond cleavage is called homolytic because each chlorine atom takes one electron from the bonding electronpair. The product from the bond cleavage ClË is called a neutral radical, or free radical, that has an unpairedelectron in its outer shell. Recall each chlorine atom has 7 electrons in its outer shell.

Free radicals can also be formed by thermal cracking, an important process in the refining of petroleum crude. Attemperatures in excess of 500°C, high molecular weight alkanes break down into smaller alkane and alkenefragments through the homolysis of the C-C bond.

Peroxides are another source of free radicals. The weak O-O bond of peroxides are cleaved at temperaturesaround 80-150°C to produce oxy radicals, as shown in this example involving the decomposition of tert-bytylperoxide

(2)HCH3L3 C - 0 - 0 - C HCH3L3 öHCH3L3 C - 0Ë + 0Ë - C HCH3L3Azobisisobutyronitrile (AIBN) is a common radical initiator from the azo compound family (R|N=N|R) thatdecomposes at 70 to 80 ºC to produce 2 isobutyronitrile radicals and nitrogen:

(3)H3 C - C»CN

»CH3

- N N - C»CN

»CH3

- CH3 ö2 H3 C - CË»CN

»CH3

+ N2

Example 1: Decomposition of AcetaldehydeThe classic example of a chain reaction is the decomposition of acetaldehyde to methane and carbon monoxide.The overall reaction is given by

(4)CH3 CHO ö CH4 + CO

In a chain reaction extremely reactive species (called chain carriers) are initially produced by thermal effects(from collisions) or photochemically in what is called an initiation step. The chain carriers are regenerated in aseries of propagation steps and if the chain is not terminated one can achieve a run-away reaction leading toexplosions. Chain carriers are normally neutral radicals such as HË, CH3

Ë with unpaired electrons that make themhighly reactive. Several mechanisms for the decomposition of acetaldehyde have been proposed. We will startour discussion with the following mechanism (see Schmidt, 1998)

(5)

Initiation Step : CH3 CHO ök1 CH3

Ë + CHOË

Propagation Step I : CH3Ë + CH3 CHO ö

k4 CH4 + CH3 COË

Propagation Step II : CH3 COË ök5

CH3Ë + CO

Termination Step : 2 CH3Ë ö

k6C2 H6

The above mechanism does not predict the formation of a minor product H2 which is known to occur. But forour initial discussion the above mechanism will suffice. We will examine a more complicated mechanism in thenext section. First, note that the two propagations steps yield the overall decomposition reaction. Physically whatthis means is that the propagation steps occur much faster than the initiation and termination steps. Note theformation of ethane in the termination step. This is called a minor product of the chain reaction as it is notproduced during the propagation steps. The free radicals involved in the propagation step are highly unstable andreact rapidly. The notation Ë means that these species have an unpaired electron - the reason for their reactivity.We say the chain is fed by the initiation step (where the chain carriers are produced) and destroyed by thetermination step which consumes the radical CH3

Ë. One can think of the propagation steps as a kinetic chainreaction : carriers are feed into the reaction sequence, and products are produced.

Note if we add the two propagation steps we get

(6)CH3 CHO + CH3Ë ö CO + CH4 + CH3

Ë

Thus we can think of CH3Ë as the catalyst for the reaction. In fact mechanism (6) is an autocatalytic reaction,

because the initiation step generates the catalyst ICH3ËM, which is needed for the overall reaction. The following is

a sketch of the chain reaction and the catalytic loop (taken from Schmidt, 1998):

2 ChainReactions.nb

Figure 1

In the above sketch, RI and RT denote the rates of reaction for the initiator and termination steps

(7)

RI = kI CCH3 CHO

RT = kT CCH3Ë2

The reaction rates for the propagations steps are

(8)

RP1 = kP1 CCH3Ë CCH3 CHO

RP2 = kP2 CCH3 COË

The above figure show clearly how the chain carrier is fed into the autocatalytic loop by the initiation step andremove from the autocatalytic loop by the termination step. Note that the figure does not account for the radicalspecies CHOË. Thus the diagram of the mechanism is incomplete. The reason is because we have not accountedfor the minor products H2 . This is done in the next section.

Rice-Hersfeld MechanismIn this section examine the free radical chain mechanism for the thermal decomposition of acetaldehyde toCH4and CO, that also produces the minor products H2 and C2 H6 The mechanism involves what is calledtransfer steps. The mechanism is due to Rice-Hersfeld. There are 6 elementary steps

(9)

1 - Initiation Step I : CH3 CHO ök1 CH3

Ë + CHOË

2 - Chain Transfer Step I : CHOË ök2 CO + HË

3 - Chain Transfer Step II : HË + CH3 CHO ök3

CH3 COË + H2

4 - Propagation Step I : CH3Ë + CH3 CHO ö

k4 CH4 + CH3 COË

ChainReactions.nb 3

(9)

4 - Propagation Step I : CH3Ë + CH3 CHO ö

k4 CH4 + CH3 COË

5 - Propagation Step II : CH3 COË ök5

CH3Ë + CO

6 - Termination Step I : 2 CH3Ë ö

k6C2 H6

In this reaction network CH3Ë and CH3 COË are the chain carriers. The first reaction step is the chain initiation

step for one of the chain carriers: CH3Ë. This is followed by two chain transfer steps, in which H2is produced, as

well as the second chain carrier CH3 COË. In the chain propagation steps CH3Ë is reproduced. The chain is

terminated in the last step by the consumption of CH3Ë.

Thus in Step I, the radicals CH3Ë and CHOË are formed by the rupture of a C-C bond. Only CH3

Ë is believed to bea chain carrier. The radical CHOË is involved in a chain transfer step (II) in which HËis formed. The HË radicalis then involved in a chain transfer reaction (step II) in which the other chain carrier CH3 COË is produced. InStep IV one of the products CH4 is formed and the chain carrier CH3 COË is reproduced, while in Step V thechain carrier CH3

Ë is regenerated. Step VI consumes CH3 which in turn terminates the reaction.

The overall reaction in a chain reaction is usually given by the propagation steps, which in this case are Steps IVand V

(10)Overall Reaction : CH3 CHO ö CH4 + CO

In this particular reaction sequence side reactions takes place involving the chain transfer and termination stepsthat produce C2 H6 and H2. We can determine a rate law for the overall reaction by finding a rate expression forRCH4 . A schematic of the various pathways for the Rice-Hersfeld mechanism is shown below

4 ChainReactions.nb

Figure 2

Let us construct the stoichiometric matrix for this system

(11) =

-1 0 -1 -1 0 01 0 0 -1 1 -21 -1 0 0 0 00 1 0 0 1 00 1 -1 0 0 00 0 1 1 -1 00 0 1 0 0 00 0 0 1 0 00 0 0 0 0 1

Here the rows represent the species {CH3 CHO, CH3Ë,CHOË,CO, HË, CH3 COË, H2, CH4, CH3 COË, C2 H6} andthe columns the reaction steps 1, 2,…, 6. It is a simple matter to check that the rankHL = 6. Thus the reac-tion steps are linearly independent and hence the dimension of the null space of is zero. Thus if this reactionwere undertaken in a batch reactor there would be no cycles at steady state. The reaction rate of each step wouldbe zero. Note if we add up the two propagation steps we see that this system is autocatalytic

(12)CH3Ë + CH3 CHO Ø CH3

Ë + CO

where the radical CH3Ë plays the role of a catalyst.

ChainReactions.nb 5

Mathematica Assisted AnalysisFor each elementary step we write down a rate expression based on mass action kinetics. We will use Mathemat-ica to assist us with the algebra. To keep the Mathematica code readable, we will use the molecular formulas assubscripts on the production rate expressions and concentrations. Since we will not use the notations package, itis important to use appropriate symbols. To designate a radical we use the “FilledSmallCircle“ symbol Ë. Thiscan be entered from the CompleteCharacters palette, or by the shortcut using the escape key: ÂfsciÂ. Notealso We use RX@iD, where X is a chemical formula for an individual species to denote the rate of production ofspecies X in the ithstep.

We define a function ReactionRates that is a list of all the production rates for the species in the individual steps. The mechanism involves 6 steps, see Eq. 9

ReactionRates = :RCH3 CHO@1D = -k1 CCH3 CHO, RCH3Ë@1D = k1 CCH3 CHO,

RCHOË@1D = k1 CCH3 CHO, RCHOË@2D = -k2 CCHOË, RCO@2D = k2 CCHOË,

RHË@2D = k2 CCHOË, RHË@3D = -k3 CHË CCH3 CHO, RCH3 CHO@3D = -k3 CHË CCH3 CHO,

RH2@3D = k3 CHË CCH3 CHO, RCH3 COË@3D = k3 CHË CCH3 CHO, RCH3Ë@4D = -k4 CCH3Ë CCH3 CHO,

RCH3 CHO@4D = -k4 CCH3Ë CCH3 CHO, RCH4@4D = k4 CCH3Ë CCH3 CHO,

RCH3 COË@4D = k4 CCH3Ë CCH3 CHO, RCH3 COË@5D = -k5 CCH3 COË, RCH3Ë@5D = k5 CCH3 COË,

RCO@5D = k5 CCH3 COË, RCH3Ë@6D = -2 k6 CCH3Ë2 , RC2 H6@6D = k6 CCH3Ë

2 >

:-CCHO CH3 k1, CCHO CH3 k1, CCHO CH3 k1, -CCHOË k2, CCHOË k2,

CCHOË k2, -CHË CCHO CH3 k3, -CHË CCHO CH3 k3, CHË CCHO CH3 k3,

CHË CCHO CH3 k3, -CCHO CH3 CCH3Ë k4, -CCHO CH3 CCH3Ë k4, CCHO CH3 CCH3Ë k4,

CCHO CH3 CCH3Ë k4, -CCOË CH3 k5, CCOË CH3 k5, CCOË CH3 k5, -2 CCH3Ë2 k6, CCH3Ë

2 k6>

Here is the rate of production for CH3Ë in step 4

RCH3Ë@4D

-CCHO CH3 CCH3Ë k4

We will make the Bodenstein approximation (or pseudo-steady state assumption) that the overall rate of produc-tion of the intermediates species (the radicals CHOË, CH3

Ë, CH3 COË , HË) is approximately zero. This assump-tion is expressed by the following equations

eqns = 9RCHOË@1D + RCHOË@2D ã 0, RCH3Ë@6D + RCH3Ë@5D + RCH3Ë@4D + RCH3Ë@1D ã 0,

RHË@3D + RHË@2D ã 0, RCH3 COË@5D + RCH3 COË@4D + RCH3 COË@3D ã 0=

:CCHO CH3 k1 - CCHOË k2 ã 0, CCHO CH3 k1 - CCHO CH3 CCH3Ë k4 + CCOË CH3 k5 - 2 CCH3Ë2 k6 ã 0,

CCHOË k2 - CHË CCHO CH3 k3 ã 0, CHË CCHO CH3 k3 + CCHO CH3 CCH3Ë k4 - CCOË CH3 k5 ã 0>

6 ChainReactions.nb

Note that we have four algebraic equations for the 4 radical species (chain carriers). We can use Mathematica‘sSolve routine to determine a solution

sol1 = SolveAeqns, 9CCHOË, CCH3Ë, CHË, CCOË CH3=E

::CCHOË ØCCHO CH3 k1

k2, CCOË CH3 Ø

CCHO CH3 k1 -CCHO CH33ê2 k1 k4

k6

k5,

CHË Øk1

k3, CCH3Ë Ø -

CCHO CH3 k1

k6

>, :CCHOË ØCCHO CH3 k1

k2,

CCOË CH3 Ø

CCHO CH3 k1 +CCHO CH33ê2 k1 k4

k6

k5, CHË Ø

k1

k3, CCH3Ë Ø

CCHO CH3 k1

k6

>>

Note that we get two solutions. Since concentrations cannot be negative, we can discard the first solution. If weapply the appropriate solution to the expression for the rate of production RCH4 @4D we get

RCH4@4D ê. sol1@@2DD

CCHO CH33ê2 k1 k4

k6

Thus the rate of production of CH4 is given by

(13)RCH4 =CCHOCH32ê3 k4 k1

k6

Note further that the overall reaction is based on steps IV and V ( see eq. 10) and thus RCH3 CHO is determined by

RCH3 CHO@4D ê. sol1@@2DD

-CCHO CH33ê2 k1 k4

k6

This result is consistent with Step 4 in Eq. 9. Finally, the overall rate of consumption of CH3 CHO (includingside reactions) is given by

IRCH3 CHO@1D + RCH3 CHO@3D + RCH3 CHO@4DM ê. sol1@@2DD

-2 CCHO CH3 k1 -CCHO CH33ê2 k1 k4

k6

Thus if we represent the reaction as the overall schema

ChainReactions.nb 7

Thus if we represent the reaction as the overall schema

(14)Overall Reaction : CH3 CHO ö CH4 + CO

then we can write for this schema, the overall rate of reaction as

(15)roverall = -2 CCHO CH3 k1 -CCHO CH33ê2 k1 k4

k6

Analysis of Composition MatrixFor this system we have the following species:

CH3 CHO, CH3Ë, CHOË, CO, HË, CH3 COË, H2, CH4, C2 H6

The composition matrix is

(16) =2 1 1 1 0 2 0 1 24 3 1 0 1 3 2 4 61 0 1 1 0 1 0 0 0

Let us compute the Null Space and rank of this matrix

=2 1 1 1 0 2 0 1 24 3 1 0 1 3 2 4 61 0 1 1 0 1 0 0 0

;

RowReduce@D êê MatrixForm

1 0 1 0 1 0 2 1 00 1 -1 0 -1 1 -2 0 20 0 0 1 -1 1 -2 -1 0

Thus the rank is 3, and hence we have - 3 = 6 basis vectors for the null space. These are

NullSpace@D êê MatrixForm

0 -2 0 0 0 0 0 0 1-1 0 0 1 0 0 0 1 0-2 2 0 2 0 0 1 0 00 -1 0 -1 0 1 0 0 0-1 1 0 1 1 0 0 0 0-1 1 1 0 0 0 0 0 0

The stoichiometric schema for this null space is

(17)

Schema I : 2 CH3Ë öC2 H6

Schema II : CH3 CHO ö CO + CH4

Schema III : 2 CH3 CHO ö 2 CH3Ë + 2 CO + H2

8 ChainReactions.nb

(17)

Schema IV : CH3Ë + CO öCH3 COË

Schema V : CH3 CHO öCH3Ë + CO + HË

Schema VI : CH3 CHO öCH3Ë + CHOË

These schema can be used to construct the elementary steps described in the previous section .

Example 2: The Hydrogen Bromide ReactionThe mechanism suggested for the reaction of hydrogen with bromine to produce hydrogen bromide is

(18)

Initiation Step : Br2 + M ök1 2 BrË + M q1 = k1 CBr2 CM

Propagation Step I : BrË + H2ök2 HBr + HË q2 = k2 CBrË CH2

Propagation Step II : HË + Br2ök3

HBr + BrË q3 = k3 CHË CBr2

Inhibition Step : HË + HBr ök4 H2 + BrË q4 = k4 CHË CHBr

Termination Step : 2 BrË + M ök5

Br2 + MË q5 = k5 CBrË2 CM

Here qi are progress (reaction) rates for the individual steps. The quantity M is called a collision partner thatprovides collision energy to cause bond cleavage. The concentration of M depends on the concentration of inerts(such as N2), as well as Br2, H2 and HBr. Note that in this chain reaction there is an inhibition step. A schematicof the hydrogen bromide cycle is shown below

Figure 3

The reaction rate constants ki at 500 K are estimated to be

(19)

k1 = 3.8ä10-8 CM dm3 mol-1 s-1

k2 = 380 dm3 mol-1 s-1

ChainReactions.nb 9

(19)

k2 = 380 dm3 mol s

k3 = 9.6 ä1010 dm3 mol-1 s-1

k4 = 7.2ä109 dm3 mol-1 s-1

k5 = 4.2ä10-13 CM dm3 mol-1 s-1

An estimate of the reaction rates qi are given below

(20)q1 º 1 mol ê s, q2 º 100 mol ê s,q3 º 100 mol ê s, q4 º 0.1 mol ê s, q5 º 1 mol ê s

Now the rates of production of the species are

(21)

RHBr = q2 + q3 - q4 º 200 - 0.1 º 200 mol ê sRHË = q2 - q3 - q4 º -0.1 mol ê sRBrË = 2 q1 - q2 + q3 + q4 - 2 q5 = 0.1 mol ê sRBr2 = -q1 + q5 º 0

RH2 = -q2 + q4 º -100 mol ê sApplying the PSS assumption to the radical species gives

(22)q2 - q3 - q4 = 0, 2 q1 - q2 + q3 + q4 - 2 q5 = 0

We can use these two equations to solve for the concentrations of the intermediate species CHËand CBrË . Here isthe Mathematica code to do these calculations

ReactionRates =

9q1 = k1 CBr2 CM, q2 = k2 CBrË CH2, q3 = k3 CHË CBr2, q4 = k4 CHË CHBr, q5 = k5 CBrË2 CM=

9CM CBr2 k1, CBrË CH2 k2, CHË CBr2 k3, CHË CHBr k4, CBrË2 CM k5=

ProductionRates = 8RHBr = q2 + q3 - q4, RHË = q2 - q3 - q4,RBrË = 2 q1 - q2 + q3 + q4 - 2 q5, RBr2 = -q1 + q5, RH2 = -q2 + q4<

9CBrË CH2 k2 + CHË CBr2 k3 - CHË CHBr k4, CBrË CH2 k2 - CHË CBr2 k3 - CHË CHBr k4,

2 CM CBr2 k1 - CBrË CH2 k2 + CHË CBr2 k3 + CHË CHBr k4 - 2 CBrË2 CM k5,

-CM CBr2 k1 + CBrË2 CM k5, -CBrË CH2 k2 + CHË CHBr k4=

sol2 = Solve@8RHË ã 0, RBrË ã 0<, 8CHË, CBrË<D

::CHË Ø -CBr2 CH2 k1 k2

HCBr2 k3 + CHBr k4L k5

, CBrË Ø -CBr2 k1

k5

>,

:CHË ØCBr2 CH2 k1 k2

HCBr2 k3 + CHBr k4L k5

, CBrË ØCBr2 k1

k5

>>

We have two solutions but the last set is relevant as the concentrations must be positive. The rate of productionof HBr is then

10 ChainReactions.nb

RHBr ê. sol2@@2DD êê Simplify

2 CBr23ê2 CH2 k1 k2 k3

HCBr2 k3 + CHBr k4L k5

Dividing through by k3CBr2 gives

(23)RHBr =2 k1 k2 CBr2

1ê2 CH2

J1 + CHBr k4CBr2 k3

N k5

=K CBr2

1ê2 CH2

1 + K£ CHBrCBr2

which is identical to what Bodenstein and Lind determined empirically.

ReferencesThe following texts were a helpful in compiling these notes.1. J. J. Carberry, Chemical Catalytic Reaction Engineering, Dover Publications, 2001

2. H. S. Fogler, Elements of Chemical Reaction Engineering, Third Edition, Prentice Hall, 1999

3. F.G. Helffrich, Kinetics of Homogeneous Multistep Reactions, Comprehensive Chemical Kinetics, Vol 38, Elsevier, 2001

4. M. J. Pillings and P. W. Seakins, Reaction Kinetics, Oxford University Press, 2003

5. L.D. Schmidt, The Engineering of Chemical Reactions, Oxford University Press, 1998

ChainReactions.nb 11