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Chapter 26: Chain reactions

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

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Chapter 26: Chain reactions

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  1. Chapter 26: Chain reactions Homework:Exercises 26:1-7 (a only)

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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] }

  7. 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

  8. 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

  9. 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] }

  10. 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!

  11. 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)

  12. 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

  13. 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)

  14. 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

  15. 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

  16. 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]

  17. 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 )

  18. 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] )

  19. 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)

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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]

  26. 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:

  27. 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

  28. 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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