Chemical Kinetics

1. Chemical Kinetics Texts: Atkins, 8th edtn., chaps. 22, 23 & 24 Specialist: “Reaction Kinetics” Pilling & Seakins (1995) • Revision • Photochemical Kinetics • Photolytic activation, flash photolysis • Fast reactions • Theories of reaction rates • Simple collision theory • Transition state theory

2. Overview of kinetics • Qualitative description • rate, order, rate law, rate constant, molecularity, elementary, complex, temperature dependence, steady-state, ... • Reaction dynamics • H (2S) + ICl (v, J)® HI (v´, J´) + Cl (2P1/2) • Modelling of complex reactions C & E News, 6-Nov-89, pp.25-31 • stratospheric O3 tropospheric hydrocarbons H3CCO2ONO2 • combustion chemical vapour deposition: SiH4® Si films

3. Rate of reaction {symbol:R,v,…} Stoichiometric equation • m 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/m3/s, more practically mol dm–3 s–1

4. Rate Law • How does the rate depend upon [ ]s? • Find out by experiment The Rate Law equation • 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

5. Experimental rate laws? CO + Cl2® COCl2 • Rate = k [CO][Cl2]1/2 • Order = 1.5 or one-and-a-half order H2 + I2® 2HI • Rate = k [H2][I2] • Order = 2 or second order H2 + Br2® 2HBr • Rate = k [H2][Br2] / (1 + k’ {[HBr]/[Br2]} ) • Order = undefined or none

6. Determining the Rate Law • Integration • Trial & error approach • Not suitable for multi-reactant systems • Most accurate • Initial rates • Best for multi-reactant reactions • Lower accuracy • Flooding or Isolation • Composite technique • Uses integration or initial rates methods

7. Integration of rate laws • Order of reaction For a reaction aA products the rate law is: rate of change in the concentration of A

8. First-order reaction

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

10. First-order reaction

11. A ® Passume that -(d[A]/dt) = k [A]1

12. Integrated rate equationln [A] = -k t + ln [A]0

13. Half life: first-order reaction • The time taken for [A] to drop to half its original value is called the reaction’s half-life, t1/2. Setting [A] = ½[A]0and t = t1/2 in:

14. Half life: first-order reaction

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

16. Second-order reaction

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

18. Second order test: A + A ® P

19. Half-life: second-order reaction

20. Rate law for elementary reaction • Law of Mass Action applies: • rate of rxnµproduct of active masses of reactants • “active mass” molar concentration raised to power of number of species • Examples: • A ® P + Q rate = k1 [A]1 • A + B ® C + D rate = k2 [A]1 [B]1 • 2A + B ® E + F + G rate = k3 [A]2 [B]1

21. Molecularity of elementary reactions? • Unimolecular (decay) A ® P - (d[A]/dt) = k1 [A] • Bimolecular (collision) A + B ® P - (d[A]/dt) = k2 [A] [B] • Termolecular (collision) A + B + C ® P - (d[A]/dt) = k3 [A] [B] [C] • No other are feasible! Statistically highly unlikely.

22. Exptal rate law: - (d[CO]/dt) = k [CO] [Cl2]1/2 Conclusion?: reaction does not proceed as written “Elementary” reactions; rxns. that proceed as written at the molecular level. Cl2®Cl + Cl(1) ● Decay Cl + CO®COCl (2) ● Collisional COCl + Cl2®COCl2 + Cl(3) ● Collisional Cl + Cl®Cl2 (4) ● Collisional Steps 1 thru 4 comprise the “mechanism” of the reaction. CO + Cl2´COCl2

23. - (d[CO]/dt) = k2 [Cl] [CO] If steps 2 & 3 are slow in comparison to 1 & 4 then,Cl2⇌2Cl or K = [Cl]2 / [Cl2] So [Cl] = ÖK × [Cl2]1/2 Hence: • -(d[CO] / dt)= k2 × ÖK × [CO][Cl2]1/2 Predict that: observedk = k2 × ÖK • Therefore mechanism confirmed (?)

24. H2 + I2® 2 HI • Predict: + (1/2) (d[HI]/dt) = k [H2] [I2] • But if via: • I2® 2 I • I + I + H2® 2 HI rate = k2 [I]2 [H2] • I + I ® I2 Assume, as before, that 1 & 3 are fast cf. to 2 Then: I2 ⇌2 I or K = [I]2 / [I2] • Rate =k2 [I]2 [H2] = k2 K [I2] [H2] (identical) Check? I2 + hn® 2 I (light of 578 nm)

25. Problem • In the decomposition of azomethane, A, at a pressure of 21.8 kPa & 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 • Show that the reaction is 1st order in azomethane & determine the rate constant at this temperature.

26. 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) • 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 • Plot ln [A] along y-axis; t along x-axis • Is it linear? Yes. Conclusion follows. Calc. slope as: -0.00959 so k = + 9.6´10-3 min-1

27. More recent questions … • Write down the rate of rxn for the rxn: C3H8 + 5 O2 = 3 CO2 + 4 H2O • for both products & reactants [8 marks] For a 2nd order rxn the rate law can be written: - (d[A]/dt) = k [A]2 What are the units of k ? [5 marks] • Why is the elementary rxn NO2 + NO2 N2O4 referred to as a bimolecular rxn? [3 marks]

28. Temperature dependence? • C2H5Cl  C2H4 + HCl k/s-1 T/K 6.1 ´ 10-5 700 30 ´ 10-5 727 242 ´ 10-5 765 • Conclusion: very sensitive to temperature • Rule of thumb: rate » doubles for a 10 K rise

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

30. Arrhenius eqn. k=A exp{-E/RT} Useful linear form: ln k = -(E/R)(1/T) + ln A • Plot ln k along Y-axis vs(1/T) along X-axis Slope is negative -(E/R); intercept = ln A • Experimental Es range from 0 to +400 kJ mol-1 Examples: • H· + HCl ® H2 + Cl·19 kJ mol-1 • H· + HF ® H2 + F· 139 kJ mol-1 • C2H5I ® C2H4 + HI 209 kJ mol-1 • C2H6® 2 CH3368 kJ mol-1

31. Practical Arrhenius plot, origin not included

32. Rate constant expression

33. Photochemical activation • Initiation of reaction by light absorption; very important • photosynthesis; reactions in upper atmosphere • No. of photons absorbed? Einstein-Stark law: 1 photon responsible for primary photochemical act (untrue) S0 + hn® S1* Jablonski diagram S* ® S0 + hnfluorescence, phosphorescence S* + M ® S0 + M collisional deactivation (quenching) S* ® P· + Q·photochemical reaction

34. Example & Jablonski diagram • A ruby laser with frequency doubling to 347.2 nm has an output of 100J with pulse widths of 20 ns. • If all the light is absorbed in 10 cm3 of a 0.10 mol dm-3 solution of perylene, what fraction of the perylene molecules are activated?

35. # of photons = total energy / energy of 1 photon • Energy of photon? # of photons = 100 / 5.725  10−19 = 1.7467  1020 • # of molecules: 0.1 mol in 1000 cm3, => 1  10−3 mol in 10 cm3 => 6.022  1020 molecules fraction activated: 1.7467  1020 / 6.022  1020 = 0.29

36. Key parameter: quantum yield, F F = (no. of molecules reacted)/(no. of photons absorbed) • Example: 40% of 490 nm radiation from 100 W source transmitted thru a sample for 45 minutes; 344 mmol of absorbing compound decomposed. Find F. Energy of photon? e = hc / l Þ(6.626´10−34 J s)(3.00´108 m s−1)/(490´10−9 m) = 4.06´10−19 J Power: 100 Watts = 100 J s-1 • Total energy into sample = (100 J s−1)(45´60 s)(0.60)= 162 kJ • Photons absorbed = (162,000)/(4.06´10−19) = 4.0´1023 • Molecules reacted? (6.023´1023) ´ (0.344) = 2.07 ´1023 Þ F = 2.07 ´1023 /4.0´1023 = 0.52

37. Quantum yield Significance? F = 2.0 for 2HI® H2 + I2reaction HI + hn® H• + I• (i) primaryf = 1 H• + HI ® H2 + I•(p) I• + I•® I2(t) • For H2 + Cl2® 2HClF > 106 Is F constant? No, depends on l, T, solvent, time. • l / nm >430 405 400 <370 • F 0 0.36 0.50 1.0 for NO2®NO+O

38. F? • Absolute measurement of FA, etc.? No; use relative method. • Ferrioxalate actinometer: C2O42- + 2 Fe3+® 2 Fe2+ + 2 CO2 • F = 1.25 at 334 nm but fairly constant from 254 to 579 nm • For a reaction in an organic solvent the photo-reduction of anthraquinone in ethanol has a unit quantum yield in the UV.

39. Rates of photochemical reactions • Br2 +hn®Br + Br Definition of rate: • where nJ is stoichiometric coefficient (+ve for products) Units: mol s-1 • So FA is moles of photons absorbed per second • Finally, the reaction rate per unit volume in mol s-1 m-3 • or mol m-3 s-1

40. Apply SS approx. to M*: d[M*]/dt = (FA/V) - kF[M*] - kQ[M*][Q] Also (FF / V)= kF[M*] So: (FA / FF ) = 1 + (kQ /kF) [Q] And hence: Plot reciprocal of fluorescent intensity versus [Q] Intercept is (1/FA) and slope is = (kQ / kF) (1/FA) Measure kF in a separate experiment; e.g. measure the half-life of the fluorescence with short light pulse & [Q]=0 since d[M*]/dt = - kF[M*] then [M*]=[M*]0exp(-t/t) M + hn ® M* FA / V M* ® M + hn FF / V M* + Q ® M + Q Stern-Volmer

41. Problem 23.8 (Atkins) Benzophenone phosphorescence with triethylamine as quencher in methanol solution. Data is: [Q] / mol dm-31.0E-3 5.0E-3 10.0E-3 FF /(arbitrary) 0.41 0.25 0.16 Half-life of benzophenone triplet is 29 ms. Calculate kQ.

43. Flash photolysis [RK,Pilling & Seakins, p39 on] • Fast burst of laser light • 10 ns, 1 ps down to femtosecond • High concentrations of reactive species instantaneously • Study their fate • Transition state spectroscopy J. Phys. Chem. a 4-6-98

44. Flash photolysis • Adiabatic • Light absorbed => heat => T rise • Low heat capacity of gas => 2,000 K • Pyrolytic not photolytic • Study RH + O2 spectra of OH•, C2, CH, etc • Isothermal • Reactant ca. 100 Pa, inert gas 100 kPa • T rise ca. 10 K; quantitative study possible • precursor +hn®CHsubsequent CH + O2®

45. Example [RK,Pilling & Seakins, p48] CH + O2® products Excess O2 present [O2]0 = 8.8´1014 molecules cm-3 1st order kinetics Follow [CH] by LIF t / ms 20 30 40 60 IF 0.230 0.144 0.088 0.033 Calculate k1 and k2

46. Problem • In a flash-photolysis experiment a radical, R·, was produced during a 2 ms flash of light and its subsequent decay followed by kinetic spectrophotometry: R· + R·® R2 • The path-length was 50 cm, the molar absorptivity, e, 1.1´104 dm3/mol/cm. • Calculate the rate constant for recombination. • t / ms 0 10 15 25 40 50 • Absorbance 0.75 0.58 0.51 0.41 0.32 0.28 How would you determine e?

47. Photodissociation [RK, p. 288] • Same laser dissociates ICN at 306 nm & is used to measure [CN] by LIF at 388.5 nm • Aim: measure time delay between photolysis pulse and appearance of CN by changing the timing of the two pulses. Experimentally: t » 205±30 fs; separation » 600 pm [C & E News 7-Nov-88] Beam Splitter

48. TS spectroscopy; Atkins p. 834 • Changing the wavelength of the probing pulse can allow not just the final product, free CN, to be determined but the intermediates along the reaction path including the transition state. • For NaI one can see the activated complex vibrate at (27 cm-1) 1.25 ps intervals surviving for »10 oscillations • see fig. 24.75 Atkins 8th ed.

49. Fast flow tubes; 1 m3/s, inert coating, t=d/v • In a RF discharge: O2 ® O + Oorpass H2 over heated tungsten filament or O3 over 1000oC quartz, etc. Use non-invasive methods for analysis e.g. absorption, emission Gas titration: add stable NO2 (measurable flow rate) • Fast O+NO2 ®NO+O2 then O+NO®NO2*® NO2 +hn End-point? Lights out when flow(NO2) = flow(O)

50. ClO + NO3J. Phys. Chem. 95:7747 (1991) • 1.5 m long, 4 cm od, Pyrex tube with sliding injector to vary reaction time • F· + HNO3®·NO3 + HF [·NO3] monitor at 662 nm • F· + HCl ®·Cl + HF followed by Cl· + O3®·ClO + O2