Chemical Kinetics Elements of Physical Chemistry Atkins de

Chemical Kinetics – ‘Elements of Physical Chemistry’, Atkins & de Paula – “Reaction Kinetics” , Pilling & Seakins – “An Examples Course in CK” , Campbell MRapid revision of second year MKinetics in flow reactors; plug vs stirred MTemperature dependence; Arrhenius equation MComplex reactions – chain reactions – polymerisations – explosions; branching chain reactions 1

Revision of 2 nd year l 5 Br- + Br. O 3 - + 6 H+ = 3 Br 2 + 3 H 2 O Rate? conc/time or in SI mol dm-3 s-1 – (1/5)(d[Br-]/dt) = – (1/6) (d[H+]/dt) = (1/3)(d[Br 2]/dt) = (1/3)(d[H 2 O]/dt) Rate law? Comes from experiment l Rate = k [Br-][Br. O 3 -][H+]2 where k is the rate constant (variable units) 2

Rate of reaction {symbol: R, v, . . } Stoichiometric equation lm. A+n. B=p. X+q. Y § Rate = - (1/m) d[A]/dt § = - (1/n) d[B]/dt § = + (1/p) d[X]/dt § = + (1/q) d[Y]/dt – Units: (concentration/time) – in SI mol/m 3/s, more practically mol dm– 3 s– 1 3

Rate Law How does the rate depend upon [ ]s? l Find out by experiment l The Rate Law equation l R = kn [A]a [B]b … (for many reactions) – order, n = a + b + … (dimensionless) – rate constant, kn (units depend on n) – Rate = kn when each [conc] = unity 4

Experimental rate laws? l Rate = k [CO][Cl 2]1/2 CO + Cl 2 ® COCl 2 – Order = 1. 5 or one-and-a-half order l Rate = k [H 2][I 2] H 2 + I 2 ® 2 HI – Order = 2 or second order l H 2 + Br 2 ® 2 HBr Rate = k [H 2][Br 2] / (1 + k’ {[HBr]/[Br 2]} ) – Order = undefined or none 5

Integration of rate laws l Order of reaction For a reaction a. A products, the rate law is: rate of change in the concentration of A 6

First-order reaction 7

First-order reaction A plot of ln[A] versus t gives a straight line of slope -k. A if r = k. A[A]1 8

First-order reaction 9

A ® P assume that -(d[A]/dt) = k [A]1 10

Integrated rate equation ln [A] = -k t + ln [A]0 11

l Half life: first-order reaction The time taken for [A] to drop to half its original value is called the reaction’s half-life, t 1/2. Setting [A] = ½[A]0 and t = t 1/2 in: 12

Half life: first-order reaction 13

When is a reaction over? l [A] = [A]0 exp{-kt} Technically [A]=0 only after infinite time 14

Second-order reaction 15

Second-order reaction A plot of 1/[A] versus t gives a straight line of slope k. A if r = k. A[A]2 16

Second order test: A + A ® P 17

Half-life: second-order reaction 18

Rate law for elementary reaction l Law of Mass Action applies: – rate of rxn µ product of active masses of reactants – “active mass” molar concentration raised to power of number of species l Examples: – – A P + Q A + B C + D rate = k 1 [A]1 rate = k 2 [A]1 [B]1 – 2 A + B E + F + G rate = k 3 [A]2 [B]1 19

Molecularity of elementary reactions? l l 20 Unimolecular (decay) A ® P - (d[A]/dt) = k 1 [A] Bimolecular (collision) A + B ® P - (d[A]/dt) = k 2 [A] [B] Termolecular (collision) A + B + C ® P - (d[A]/dt) = k 3 [A] [B] [C] No other are feasible! Statistically highly unlikely.

CO + Cl 2 l ´ Exptal rate law: COCl 2 - (d[CO]/dt) = k [CO] [Cl 2]1/2 – Conclusion? : reaction does not proceed as written – “Elementary” reactions; rxns. that proceed as written at the molecular level. Cl 2 Cl + Cl Cl + CO COCl + Cl 2 COCl 2 + Cl (3) Cl + Cl 2 l l (1) (2) (4) u decay u collisional – Steps 1 thru 4 comprise the “mechanism” of the reaction. 21

- (d[CO]/dt) = k 2 [Cl] [CO] If steps 2 & 3 are slow in comparison to 1 & 4 then, Cl 2 ⇌ 2 Cl or K = [Cl]2 / [Cl 2] So [Cl] = ÖK × [Cl 2]1/2 Hence: l - (d[CO] / dt) = k 2 × ÖK × [CO][Cl 2]1/2 Predict that: observed k = k 2 × ÖK l 22 Therefore mechanism confirmed (? )

H 2 + I 2 2 HI l l Predict: + (1/2) (d[HI]/dt) = k [H 2] [I 2] But if via: – I 2 2 I – I + H 2 2 HI – I + I I 2 rate = k 2 [I]2 [H 2] Assume, as before, that 1 & 3 are fast cf. to 2 Then: I 2 ⇌ 2 I or K = [I]2 / [I 2] l Rate = k 2 [I]2 [H 2] = k 2 K [I 2] [H 2] (identical) Check? I 2 + hn ® 2 I (light of 578 nm) 23

Problem In the decomposition of azomethane, A, at a pressure of 21. 8 k. Pa & a temperature of 576 K the following concentrations were recorded as a function of time, t: Time, t /mins 0 30 60 90 120 [A] / mmol dm-3 8. 70 6. 52 4. 89 3. 67 2. 75 l Show that the reaction is 1 st order in azomethane & determine the rate constant at this temperature. l 24

Recognise that this is a rate law question dealing with the integral method. - (d[A]/dt) = k [A]? = k [A]1 Re-arrange & integrate (bookwork) l Test: ln [A] = - k t + ln [A]0 Complete table: Time, t /mins 0 30 60 90 120 ln [A] 2. 16 1. 88 1. 59 1. 30 1. 01 l Plot ln [A] along y-axis; t along x-axis l Is it linear? Yes. Conclusion follows. Calc. slope as: -0. 00959 so k = + 9. 6´ 10 -3 min-1 25

Proper rate of reaction l Define as: (1/ni)(dni/dt) mol s-1 where ni is a stoichiometric coefficient +ve for products, – ve for reactants Example: 2 H 2 + O 2 = 2 H 2 O l n(H 2) = – 2, n(O 2)= – 1 and n(H 2 O)=+2 l At constant volume: [A] = n /V or n = V × [A] So: (dn/dt) = V. (d[A]/dt) Old rate of reaction (was rxn rate per unit volume) l (d[A]/dt) = (1 / V) (dn. A/dt) 26

Flowvs batch reactors [A]=[A]0 exp{-kt} Stirred flow; perfect mixing l spherical shape of volume V [A]outlet = [A]inlet / { 1 + k (V/n)} ‘Plug’ flow; no mixing at all l cylindrical shape of volume V [A]outlet = [A]inlet exp{ - k (V/n)} NB (V/n) has dimensions (m 3 / m 3 s-1) = s A ‘residence’ or ‘contact’ time 27

Stirred flow; 1 st order -(dn. A/dt) = k n. A = k V [A] l inflow = outflow + reaction Assume system at constant volume so that inlet flow rate is equal to outlet flow rate l n[A]i = n[A]o - (dn. A/dt) l n[A]i = n[A]o + k V [A]o l [A]o = [A]i / {1 + k(V/n)} l Test? vary v, measure [A]o & [A]i, is k the same? 28

Plug flow; 1 st order rxn -(dn. A/dt) = k n. A = k V [A] inflow = outflow + reaction n[A] = n ([A]+d[A]) – (dn. A/dt) –nd[A] = k d. V [A] l òd [A]/[A] = – (k/n) òd. V At inlet [A] = [A]i ; at outlet [A] = [A]o l or l ln [A]o = – k (V/n) + ln [A]i [A]o = [A]i exp{– k (V/n)} 29

Flow reactor problem l Peracetic acid vapour at 2% by volume in nitrogen carrier gas at a total pressure of 101 k. Pa and 490 K flows thru 1. 4 mm radius teflon tubing and is sampled 1. 2 m downstream with the following results: ([A]o/[A]i) l 0. 346 0. 578 0. 718 n / m 3 s-1 3. 71´ 10 -5 7. 20´ 10 -5 11. 9´ 10 -5 Show that rxn follows 1 st order kinetics & calculate k. Eqn? [A]o=[A]i exp{- k (V/n)} or ln{[A]o/[A]i}= - k(V/n) l Reactor vol. V = pr 2 h = 3. 14 ´ (0. 0014)2 ´ 1. 2 = 7. 39 ´ 10 -6 m 3 l k = - (n / V) ´ ln{[A]o/[A]i} = 5. 3208, 5. 3410, 5. 3349 s-1 30

Temperature dependence? l C 2 H 5 Cl C 2 H 4 + HCl k/s-1 6. 1 ´ 10 -5 30 ´ 10 -5 242 ´ 10 -5 T/K 700 727 765 Conclusion: very sensitive to temperature l Rule of thumb: rate » doubles for a 10 K rise l 31

Details of T dependence Hood l k = A exp{ -B/T } Arrhenius l k = A exp{ - E / RT } A A-factor or pre-exponential factor º k at T ¥ E activation energy (energy barrier) J mol R gas constant. 32 -1 or k. J mol-1

Arrhenius eqn. k=A exp{-E/RT} Useful linear form: ln k = -(E/R)(1/T) + ln A l Plot ln k along Y-axis vs (1/T) along X-axis Slope is negative -(E/R); intercept = ln A l Experimental Es range from 0 to +400 k. J mol-1 Examples: – – 33 H· + HCl ® H 2 + Cl· H· + HF ® H 2 + F· C 2 H 5 I ® C 2 H 4 + HI C 2 H 6 2 CH 3 19 k. J mol-1 139 k. J mol-1 209 k. J mol-1 368 k. J mol-1

Practical Arrhenius plot, 34 origin not included

Problem l In consecutive rxns the slower step usually determines the overall rate of rxn. Diethyl adipate (DA) is hydrolysed in 2 steps as: DA ® B ® P l 35 The Arrhenius equations are: first step 1. 23´ 106 exp{-5, 080/T} s-1 second step 8. 80´ 105 exp{-3, 010/T} s-1 Which step is rate determining if the rxn is carried out in aqueous solution under atmospheric pressure?

36

Rxn. at equilibrium: AÛB forward rate constant k. F, reverse k. R Kinetics: l k. F[A]e = k. R[B]e so [B]e/[A]e = k. F/k. R = K l k. F = AF exp{-EF/RT} k. R = AR exp{-ER/RT} l (k. F/k. R) = (AF/AR) exp{-(EF- ER)/RT} Thermodynamics? l DG 0 = - RT ln(K) or K = exp{-DG 0/RT} DG 0 = DH 0 - TDS 0 K = exp{+DS 0/R} exp{-DH 0/RT} l (AF/AR) = exp{+DS 0/R} EF - ER = DH 0 37

Reaction profile diagram l Y-axis? – Energy – common path via TS X-axis? Time? Reactant-product identity Multi-dimensional l Reaction co-ordinate l Schematic diagram l – Concept of energy barrier to rxn 38

Reaction profile diagram CH 3 CHO CH 4 + CO l EF = 198 k. J/mol l Add some iodine as catalyst EF = 134 k. J/mol – Modest reduction in barrier – Massive increase in rate l 39 At 773 K catalysed rxn is 20, 000 times faster – D(C-C)=340, D(C-H)=420 – But D(I-I)=153 k. J/mol

Thermochemistry from kinetics C 2 H 6 2 CH 3· 368 k. J mol-1 EF = 368 k. J mol-1 ER = 0 k. J mol-1 DHRXN = EF-ER = + 368 k. J mol-1 = 2 DHf(CH 3·) - DHf(C 2 H 6) But DHf(C 2 H 6) = - 85 k. J mol-1 l DHf(CH 3·) = (368 -85)/2 = +141 k. J mol-1 l DHRXN = Energy of bond broken = D(H 3 C-CH 3) l 40 Þ Tables of bond dissociation energies

Problem For the rxn trans ® cis perfluorobut-2 -ene l measured k. F=3. 16´ 1013 exp{-31, 030/T} s-1 l l l Given that DH 0=+3. 42 k. J mol-1 and DS 0=-2. 03 J K-1 mol-1 Calculate k. R at 750 K. (AF/AR) = exp{+DS 0/R} Þ AR = AF exp{-DS 0/R} Þ ER = EF - DH 0 -(EF/R) = -31, 030 Þ EF = 8. 3143´ 31, 030 J mol-1 EF = 258. 0 k. J mol-1 Þ ER = 254. 6 k. J mol-1 EF - ER = DH 0 41

Common mathematical functions Kinetics, Arrhenius l k = A exp{-E/RT} Vapour pressure, Clausius-Clayperon l p = p¥ exp{-DHV/RT} Viscosity, Andrade l h = A exp{+E/RT} l 42 But re-define as inverse: F = F¥ exp{-E/RT}

Data from Keith Laidler, J Chem Educ 49, 343 (1972) T 17. 3— 25. 0 C, chirp rates of 100 to 178 min-1 43

Complex (non-elementary) rxns Chain reactions H 2 + I 2 do not react; I 2 2 I then I+H 2+I 2 HI CH 3 CHO ® CH 4 + CO in over 90% yield Traces of C 2 H 6, H 2, CH 3 COCH 3, etc. Rate law rate = k[CH 3 CHO]3/2 l Possible mechanism? l CH 3 CHO CH 3· + HCO· initiation l CH 3· + CH 3 CHO CH 4 + CH 3 CO· propagation l CH 3 CO· CH 3· + CO propagation l CH 3· + CH 3· C 2 H 6 termination 44

Kinetic analysis? Mathematically difficult Steady-state approximation l l l – For reactive intermediates assume that rate of formation » rate of destruction ie (d[R]/dt) » 0 (d[CH 3·]/dt)= v 1 + v 3 - v 2 - v 4 » 0 (d[CH 3 CO·]/dt)= v 2 - v 3 » 0 Þ v 2= v 3 and v 1 = v 4 k 1[CH 3 CHO] = 2 k 4[CH 3·]2 S-S conc. [CH 3·]=(2 k 1/k 4)1/2 [CH 3 CHO]1/2 (d[CH 4]/dt) = k 2[CH 3·][CH 3 CHO] = k 2(2 k 1/k 4)1/2 [CH 3 CHO]3/2 l So kobs = k 2(2 k 1/k 4)1/2 45

Relationships for A and Es l kobs = k 2(2 k 1/k 4)1/2 Since the A-factor is essentially a rate constant: l Aobs = A 2(2 A 1/A 4)1/2 The activation energies are related as the logs of the rate constants: l log kobs = log k 2 + (1/2){log(k 1) - log(k 4)} Eobs = E 2 + (1/2){E 1 - E 4} composite quantity = 35 + (1/2){340 - 0} = 205 k. J mol-1 l 205 « 340; Note how in a chain rxn. the high-energy barrier of 340 k. J mol-1 is circumvented. 46

Chain polymerisation I + M ® M· M · + M ® M 2· + M ® M 3· Mn· + Mm· ® Mn+m l i p Let [R]=total radical conc. p t (d[R]/dt)=-2 k. T [R]2 + k. I[M][I] (neglect r. P) SS-approx. : (d[R]/dt)=0 l So [R] = (k. I/2 k. T)1/2 [M]1/2 [I]1/2 Rate of propagation of chains: l l 47 - (d[M]/dt = k. P [R][M] - (d[M]/dt) = k. P (k. I/2 k. T)1/2 [I]1/2 [M]3/2 (neglect r. I)

Branching chain reactions l “On a windy Thursday in the first week of May in 1937, the Hindenberg airship exploded as it approached its mooring in New Jersey, USA. Newsreel film of the disaster, in which 35 passengers & 1 ground crew died, shows that all 200, 000 m 3 of H 2 in the ship burned in under 45 s. ” – Physical Chemistry, Winn, p. 1016 – http: //www. otr. com/hindenburg. html 48

Branching chain reactions O· + H 2 ® HO· + H· The power of two l 1® 2® 4® 8® 16® 32® 64® 128® 256® 512 etc l Fifty doublings or 250 » 1015 € 1015 for the 380 million citizens of the EU l – € 2. 6 million each l 49 So reactions in which branching is important can proceed explosively

Branching chains M®R i R + M®R + P p R®P t R + M®R + R b l (d[R]/dt)=vi+k. B[R]-k. T[R] If termination>branching l [R] µ {1 -exp[-(k. T-k. B)t]} If branching>termination l [R] µ {exp[-(k. B-k. T)t]-1} 50

2 H 2 + O 2 = 2 H 2 O l H 2 + O 2 HO 2· + H· i H· + O 2 + M HO 2· + M p HO 2· + H 2 O + HO· p l H· + O 2 HO· + O· b l H· Þ diffuses to wall l l 51 t

2 H 2 + O 2 = 2 H 2 O l 1 st limit (sensitive to surface, vessel shape & added inert gas) – Competition between branching and diffusion to wall of H · – no explosion wall ¬ H· + O 2 ® HO· + O· explosion l 2 nd limit (not) – – l Competition: H· +O 2 + M ® HO 2· + M no explosion third order H· +O 2 ® HO· + O· explosion second order hydroperoxyl radical HO 2· much less reactive than H·, HO· 3 rd limit (sensitive to vessel shape/size) – rate very fast, high rate of heat release, faster than can be conducted away Þ thermal explosion – added gases have effect µ to their heat transport properties 52