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## Chapter 26: Chain reactions

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### Chapter 26: Chain reactions

Homework:Exercises 26:1-7 (a only)

Chain Reactions

- In a chain reaction an intermediate produced in one step generates intermediate for next step, etc.
- Intermediates called chain carriers
- Radical chain reactions radicals are chain carriers
- Ions can be chain carriers
- Nuclear reaction neutrons are chain carriers
- Reaction steps
- Initiation step - chain carriers formed e.g., Cl2 -> Cl.
- E.g. Atomic radicals via thermolysis or photolysis
- Propagation step - chain carriers attack other reactants to give a new carrier
- If more than one chain carrier is produced it is a branching step

Reaction Steps (continued)

- Retardation step - chain carrier attacks a product
- Doesn’t end chain, but does deplete the product
- Termination step - chain carriers combine and end the chain
- Inhibition step - chain carriers removed by means other then termination
- Reaction with walls of foreign radicals

Example: Formation of HBr

- Initiation: Br2-> 2Br. , k1
- Propagation : Br + H2-> HBr + H. , k2

H. + Br2-> HBr + Br. , k3

- Inhibition : H. + HBr -> H2 + Br. , k4
- Termination : Br. + Br. + M -> Br2 + M*, k5

Rate Equations for Chain Reactions

1. Write all steps

2. Write net rates of formation for all intermediates

3. Apply steady state approximation to the intermediates

- d[I]/dt = 0

4. Substitute resulting expression for intermediate concentration into the net rate equation

HBr

- Phenomenological rate equation: d[HBr]/dt = {k[H2][Br2]3/2}/{[Br2] + k[HBr]}

Initiation: Br2 +M -> 2Br. + M, ka

Propagation : Br + H2-> HBr + H. , kb

H. + Br2-> HBr + Br. , k’b

Inhibition : H. + HBr -> H2 + Br. , kc

Termination : Br. + Br. + M -> Br2 + M*, kd

- d[Br]/dt = 2ka[Br2][M] - kb[H2][Br] + k’b[Br2][H] + kc[H][HBr] - 2kd[Br]2[M]

2ka[Br2][M] - kb[H2][Br] + k’b[Br2][H] + kc[H][HBr] - 2kd[Br]2[M]=0

- d[H]/dt = kb[H2][Br] - k’b[Br2][H] - k’b[Br2][H]

d[H]/dt = kb[H2][Br] - k’b[Br2][H] - k’b[Br2][H] = 0

- Two equations in two unknowns [H] and [Br]
- [Br] = (ka/kd)1/2 [Br] 1/2
- [H] = {kb(ka/kd)1/2 [H2][Br] 1/2 }/{k’b[Br2] + kc[HBr] }

HBr (continued)

- d[HBr]/dt = kb[Br][H2] + k’b[H][Br2] - kc[H][HBr]
- d[HBr]/dt = kb(ka/kd)1/2 [Br] 1/2[H2] +( k’b[Br2] - kc[Br]) {kb(ka/kd)1/2 [H2][Br] 1/2 }/{k’b[Br2] + kc[HBr] }
- d[HBr]/dt = 2kb(ka/kd)1/2 [Br] 3/2 [H2] /( [Br2] + kck’b[HBr] }
- If k = 2kb(ka/kd)1/2 and k’= kck’b thenabove reduces to phenomenological rate equation
- Notes:
- [M] cancels out
- [HBr] in denominator means it acts as an inhibitor
- [Br2] in denominator since it removes H from chain
- This rate law can be integrated numerically to produce an integrated rate law

Explosions

- Thermal explosion - due to rapid increase in reaction rate with temperature
- Happens when heat can’t escape fast enough
- Chain branching explosion - occurs when reaction has chain branching steps which grow exponentially and rate cascades into an explosion
- Previous chain reaction only one chain carrier
- Br + H2-> HBr + H.
- H. + Br2-> HBr + Br.
- In chain branching, more than one chain carrier is produced in each propagation step
- R + A -> P + eR (. O. + H2-> H. + OH. )
- e is called the branching ratio

Chain Branching Explosive Reactions

- Reaction:
- A -> R, k1
- R + A -> P + eR, k2
- R -> destruction, k3
- At steady state
- d[R]/dt = 0 = k1[A] - k2[R][A] + ek2[R][A] - k3[R]
- 0 = k1[A] -(1-e )k2[R][A] - k3[R] = k1[A] -{k3- ( e -1 )k2[A] }[R]
- [R] = k1[A] / {k3- ( e -1 )k2[A] }
- If the termination reaction is separated into gas phase and wall reaction, + k3= kg+ kw, [R] = k1[A] / {(kg+ kw)- ( e -1 )k2[A] }
- d[P]/dt = k2[R][A] = k2 k1[A]2/ {(kg+ kw)- ( e -1 )k2[A] }

Chain Branching Explosive Reactions - General Comments

- Consider [R] = k1[A] /{(1-e )k2[A] + k3} = k1[A] /{k3-(e - 1 )k2[A] }
- For chain reactions without branching e = 1 so [R] = k1[A] /k3 , i.e. the concentration of the radical is proportional to the rate of formation/rate of destruction
- For chain branching e > 1 - this means amount of radical grows as reaction proceeds
- In particular, if k3 = (e - 1 )k2[A], then [R] goes to infinity and the steady state approximation breaks down
- This leads to upper and lower explosive limits because k3 = kg+ kw
- At lower limit, kw is dominant since it depends on diffusion to the wall which is more favorable at low pressures and radicals are destroyed faster than they are produced - depends on size & type of container
- At upper limit, kg is dominant and destructive gas-phase collisions outweigh branching
- In the middle - BOOM!

Explosive Chain Reactions - H2 + O2

- Book considers the example in which the initiation reaction is photo initiated: H2 -> 2 H rate= v = intensity of radiation = I
- Propagation & Branching steps are of the form X = eX + product where v = ka[X]
- Example: . O2. + H.-> O. + OH. v = k2[ H. ][O2]

. O. + H2-> H. + OH. v = k3[. O. ][H2] {Branching steps}

- Termination steps of the form X-> removed, v = kb[X]
- Example: H. + wall -> 1/2 H2 v = k4[H. ]
- Overall rate of formation of radical formation:

d[X]/dt = I + eka[X] - ka[X]-kb[X] = I + eka[X] - ka[X] - kb[X]

d[X]/dt = I + f [X] {f = (e - 1)ka - kb}

Solving this D.E. : [X](t) = I / f (e-f t - 1)

Consequences of [X](t) = I / f (e-f t - 1)

- kb > (e - 1)ka or f <0, means that termination is dominant
- [X](t) = {I / (kb- ka(e-1))} (1 - e-(e - ka - kb) t)
- as t goes to infinity [X](infinity) = I / (kb- ka(e-1) )
- kb < (e - 1)ka or f >0, means that propagation is dominant and as t increases [X](t) increases without limit
- These two cases defines the transition from combustion to explosion
- Explosive limits hydrogen/oxygen
- Chain reaction
- Lower limit (T,P)
- Upper Limit (T,P)
- Thermal explosive limit above chain reaction temperature

Photochemical Reactions

- Reaction discussed for explosion of a H2/O2 discussed photo-dissociation of H2 initiated the reaction.
- Such reactions are called photo-chemical reactions
- Important in many natural processes
- Photosynthesis
- Environmental chemistry
- Primary Quantum Yield, f, the number of reactant molecules producing specified primary products per photon
- Overall Quantum Yield, F, the number of reactant molecules that react per photon
- Includes all steps (example below F = 2; )
- HI +hn -> H + I
- H + HI -> H2 + I
- I + I ->I2
- In chain reactions can be large (104)

Determining Quantum Yield

- Step 1: Calculate the number of photons:
- Power (W ) x time(s) = Energy (J)
- Energy/hn = # of photons
- Step 2: Calculate number of molecules formed
- Step 3: Divide 3 of molecules/# of photons = F
- Note reverse calculation can also be done to calculate power or time

Self-Test 26.3

- l = 290 nm
- E = (6.626 e-34 j s) x (3 x 8 m/s)/290 e-9 m) = 6.85 e-19 J/photon
- F = 0.30 = molecule /photons
- For 1 mole: 6.02 e 23/.3 photons required = 20.1 e 23 photons
- Total energy required = 20.1 e23 photons x 6.85e-19J/photon = 1.38 e6 J
- Time = Joules/Watts = 1.38 e4 s = 3.8 h

Photochemical Rate Laws

- Photochemical steps in a rate equation affect the rate via an intensity term, Iabs,
- Iabs is the rate ate which photons are absorbed divided by the volume in which absorption occurs
- Example
- H2 + O2
- HBr
- For collisional initiation (Br2 +M -> 2Br. + M, ka ) the rate is

d[HBr]/dt = 2kb(ka/kd)1/2 [Br2] 3/2 [H2] /( [Br2] + kc/k’b[HBr] }

- For photo initiation (Br2 +hn-> 2Br.Iabs ), the rate is

d[HBr]/dt = Iabs 1/2 2kb(1/[M]kd)1/2 [Br2] [H2] /( [Br2] + kc/k’b[HBr] }

- Iab= ka[Br] 3/2 [M]

Photosensitization

- In photosensitization a molecule which cannot directly absorb light can be excited by a collision with one which can
- Hydrogen is a good example
- Hg absorbs photons @ 254 nm Hg + hn -> Hg*
- Hg* + H2 -> Hg + 2H or
- Hg* + H2 -> HgH + H
- Initiation step for reactions
- In solutions carbonyls can be used (benzophenone, C6H5COC6H5 )

Quenching

- A quenching agent removes energy from an excited species
- S + hni -> S* v= I
- S* -> S + + hnf v= kf[S*]
- f = fluoresence
- S* + Q -> S + Q v= kQ[S*][Q]
- At steady state

d[S*]/dt= 0 = I - kf[S*]- kf[S*][Q] = I - (kf+ kQ [Q] )[S*]

Or [S*] = I /(kf+ kQ [Q] )

- Now, the fluoresence intensity , If, is proportional to kf[S*] so
- If a kf[S*] a I kf /(kf+ kQ [Q] )

Quenching - Stern-Volmer Plot

- If I°f is the fluoresence intensity in the absence of a quenching agent then
- I°f/If = I kf /(kf+ kQ [0] )/ I kf /(kf+ kQ [Q] ) = (kf+ kQ [Q] )/ kf
- I°f/If = 1+ kQ [Q] / kf
- A plot of I°f/If versus [Q] is called a Stern-Volmer Plot and the slope is kQ/ kf
- The initial rate after a brief flash d[S]/dt = - (kf+ kQ [Q] )[S*]
- Integrating [S*] = [S*] 0 exp (-t/t) where 1/t= kf+ kQ [Q]
- Plot of 1/t vs [Q] gives kf (intercept) and kQ(slope)

Polymerization Kinetics

- Chain Polymerization
- Process
- Activated monomer, M*, attacks another monomer and adds to it
- Resultant species then attacks new monomer and adds, etc.
- Monomer used up slowly
- High polymers formed rapidly
- Average molar mass increased by long reaction times
- Step Polymerization
- Any two monomers can link at any time
- Monomer used quickly

Chain Polymerization

- Examples ethene, metyl methacrylate and styrene
- -CH2CHXl + CH2=CHX -> -CH2CHXCH2CHXl
- Rate of polymerization, v is proportional to the square root of the initiator concentration, v = k[I]1/2[M]
- Proof:

Steps:

{Initiation} I -> Rl + Rl v = ki[I] (I= initiator, R = radical)

M + Rl -> M1lfast (M=monomer, M1l monmer radical)

{Propagation} M + Mn-1l -> Mnl v = kp[M][Ml]

Rate of monomer radical production is determined by initiation step so

d[Ml]/dt = 2f ki[I] {2 because 2 radicals produced and f is the fraction of radicals which initiate a chain}

{Termination} Mnl + Mml -> Mm+n v = kt[Ml]2 ; d[Ml]/dt = -2kt[Ml]2

Proof (continued)

- Apply steady state approximation
- d[Ml]/dt = 2f ki[I] -2kt[Ml]2 = 0
- 2f ki[I] =2kt[Ml]2 or 2f ki[I] =2kt[Ml] 2
- [Ml]= (2f ki[I]/2kt)0.5 = (f ki[I]/kt)0.5 ([I] )0.5
- Rate of propagation = - rate of monomer consumption = kp[M][Ml]
- Rate of monomer consumption=-v = -kp[M] (f ki[I]/kt)0.5 ([I] )0.5
- This is same as v = k[I]1/2[M] where k = kp(f ki[I]/kt)0.5

Chain Length

- Kinetic chain length, n, ratio of the number of monomer units consumed per active center in the initiation step
- Measure of the efficiency of chain propagation
- n = # of monomer units consumed/#number of active centers
- n = propagation rate/initiation rate
- Since initiation rate = termination rate, n =kp[M][Ml]/ -2kt[Ml]2

or n =kp[M]/ 2kt[Ml]

- But from steady state approximation, [Ml]= (f ki[I]/kt)0.5 ([I] )0.5 so
- n =kp[M][Ml]/ -2kt[Ml]2 becomes n =kp[M]/ -2kt (f ki[I]/kt)0.5 ([I] )0.5
- n =k [M ][I]-0.5 where k = kp/ -2kt (f ki[I]/kt)0.5 ([I] )0.5
- The slower the initiation, the greater the kinetic chain length

Average Number of Monomers in a Chain (Example 26.4 & self test 26.4)

- Average Number of Monomers in a Chain, <n> depends on termination mechanism
- If it is two radicals combining, Mnl + Mml -> Mm+n,<n> is twice the kinetic chain length since two combine to terminate the reaction
- <n> = 2n = 2k [M ][I]-0.5
- If it is disproportionation, Ml + Ml -> M + :M,<n> is the kinetic chain length termination results in two chains
- <n> = n = k [M ][I]-0.5

Stepwise Polymerization

- Any monomer can react at any time
- Proceeds via a condensation reaction in which a small molecule is eliminated in the step
- Usually water
- Example polyesters
- HO-M-COOH + HO-M-COOH -> HO-M-COO-M-COOH
- Rate (A is COOH)
- d[A]/dt = -k[OH][A] = k[A]2 {there is one OH for every A}
- Solution [A] = [A0]/(1 + kt[A0])
- Fraction of groups condensed at t is p

p = [A0]- [A]/ [A0] = kt [A0]/(1+kt [A0])

- <n> = [A0]/[A] = 1/(1-p) = 1 + kt [A0]

Catalysis

- Catalyst accelerates the reaction
- Undergoes no change
- Lowers the activation energy
- Provides alternate pathway for reaction
- Homogeneous catalyst is in the same phase as the reaction mixture
- Heterogeneous catalyst is a different phase
- In auto catalysis, products accelerate the reaction
- Example A-> P v= k[A][P]
- Rate initially slow. As P increases, rate increases. As [A] gets small, reaction slows down
- Demonstrated by integration of rate law:

Oscillating Reactions

- Because of autocatalysis, reactants and products may vary periodically
- Important in biochemistry
- Maintain heart rhythm
- Glycolytic cycle
- Lotka-Volterra mechanism
- Steps
- (1) A + X -> X + X d[A]/dt = -ka[A][X]
- (2) Y + X -> Y + Y d[X]/dt = -kv[X][Y]
- (3) Y -> B d[B]/dt = kc[Y]
- (1) & (2) Autocatalytic
- Conditions: [A] is constant {steady state condition not approximation}
- Numerical solution of [X] and [Y] gives periodic variation

Example: Brusselator

- Steps
- (1) A -> X d[X]/dt = ka[A]
- (2)X+ X+ X -> Y + Y + Y d[Y]/dt = kv[X]2[Y]
- (3) B + X -> Y + C d[Y]/dt = kc[B][X]
- (4) X -> D d[Y]/dt = kc[B][X]
- Conditions: Hold [A] and [B] constant
- Solve numerically
- A plot of [Y] vs [X] will converge to the same periodic variation of X and Y regardless of initial conditions
- Trajectory called a limit cycle

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