chapter 26 chain reactions l.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 26: Chain reactions PowerPoint Presentation
Download Presentation
Chapter 26: Chain reactions

Loading in 2 Seconds...

play fullscreen
1 / 28

Chapter 26: Chain reactions - PowerPoint PPT Presentation


  • 411 Views
  • Uploaded on

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

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter 26: Chain reactions' - paul


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
chapter 26 chain reactions

Chapter 26: Chain reactions

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

chain reactions
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
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
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
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

slide6
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
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
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
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
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 h 2 o 2
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
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
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
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
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 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
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
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
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
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
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
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
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 (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
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
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
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
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