1. 1 Chemical Kinetics Texts: Atkins, 6th edtn., chaps. 25, 26 & 27 “Reaction Kinetics” Pilling & Seakins (1995) l Revision l Photochemical Kinetics l Photolytic activation, flash photolysis l Fast reactions l Theories of reaction rates –Simple collision theory –Transition state theory

2. 2 Overview of kinetics l Qualitative description –rate, order, rate law, rate constant, molecularity, elementary, complex, temperature dependence, steady-state,... l Thermochemical l Reaction dynamics – H ( 2 S) + ICl (v, J)  HI (v ´, J ´ ) + Cl ( 2 P 1/2 ) l Modelling of complex reactions C & E News, 6-Nov-89, pp.25-31 –stratospheric O 3 tropospheric hydrocarbons H 3 CCO 2 ONO 2 –combustion chemical vapour deposition: SiH 4  Si films

3. 3 Photochemical activation l Initiation of rxn by light absorption; very important –photosynthesis; rxns in upper atmosphere l No. of photons absorbed? Einstein-Stark law: 1 photon responsible for primary photochemical act ( untrue ) S 0 + h  S 1 * Jablonski diagram S*  S + h fluorescence, phosphorescence S* + M  S + M collisional deactivation (quenching) S*  P  + Q  photochemical reaction

4. 4 Example & Jablonski diagram l A ruby laser with frequency doubling to 347.2 nm has an output of 100J with pulse widths of 20 ns. l If all the light is absorbed in 10 cm 3 of a 0.10 mol dm -3 solution of perylene, what fraction of the perylene molecules is activated?

5. 5 No. of photons in 20 ns pulse? Energy of 1 photon? use  h & c= so that:  = hc / = (6.626  10 -34 J s)(3.00  10 8 m s-1)/(347.2  10 -9 m) = 5.73  10 -19  J Energy of pulse? is 100J No. of photons = (100 J) / (5.73    J) = 1.75  10 19 l No. of molecules? 1 mol dm -3 contains N A per dm -3 So, 10 cm 3 of a 0.10 mol dm -3 solution contains: (10/1,000)  (1/10)N A = 6.02  10 20 molecules l No. of photons / no. of molecules = 0.29 or 29%

6. 6 Key parameter: quantum yield,   = (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 . Energy of photon?  = hc /  (6.626  10 -34 J s)(3.00  10 8 m s -1 )/(490  10 -9 m) = 4.06  0 -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  0 -19 ) = 4.0  10 23 Mols of photons? (4.0  10 23 ) /(6.023  10 23 ) = 0.66 einsteins  = 0.344 / 0.66 = 0.52

7. 7 Quantum yield Significance?  = 2.0 for H 2 + I 2  2HI reaction HI + h  H + I (i) primary  = 1 H + HI  H 2 + I (p) I + I  I 2 (t) For H 2 + Cl 2  2HCl  > 10 6 Is  constant? No, depends on, T, solvent, time. / nm >430405 400<370  00.360.50 1.0 for NO 2  NO+O

8. 8  l Absolute measurement of F A, etc.? No; use relative method. l Ferrioxalate actinometer: C 2 O 4 2  + 2 Fe 3+  2 Fe 2+ + 2 CO 2  = 1.25 at 334 nm but fairly constant from 254 to 579 nm l For a rxn in an organic solvent the photoreduction of anthraquinone in ethanol has a unit quantum yield in the UV.

9. 9 Rates of photochemical reactions Br 2 + h  Br + Br Definition of rate: where n J is stoichiometric coefficient (+ve for products) Units: mol s -1 So F A is moles of photons absorbed per second l Finally, the reaction rate per unit volume in mol s -1 m -3 l Or mol m -3 s -1

10. 10 Stern-Volmer l Apply SS approx. to M*: d[M*]/dt = (F A /V) - k F [M*] - k Q [M*][Q] l Also (F F / V)= k F [M*] So: (F A / F F ) = 1 + (k Q /k F ) [Q] And hence: Plot reciprocal of fluorescent intensity versus [Q] Intercept is (1/F A ) and slope is = (k Q / k F ) (1/F A ) Measure k F in a separate experiment; eg, measure the half-life of the fluorescence with short light pulse & [Q]=0 since d[M*]/dt = - k F [M*] then [M*]=[M*] 0 exp(-t/  ) M + h M*F A / V M*  M + h F F / V M* + Q  M + Q

11. 11 Problem 26.2 (Atkins) l Benzophenone phosphorescence with triethylamine as quencher in methanol solution. Data: [Q] mol dm -3 1.0E-35.0E-310.0E-3 F F ( arbitrary ) 0.410.250.16 Half-life of benzophenone triplet is 29  s. Calculate k Q. Build table: (1/ F F )2.4 4 4.0 0 6.2 5 Plot; intercept = (1/ F A ) slope = + (1/ F A ) (k Q / k F ) Now  = ln 2 / k so k F = (0.693 / 29  10 -6 ) s -1 Therefore: k Q = 5.1  10 8 dm 3 mol -1 s -1

12. 12 Problem 26.2 (continued) l Intercept  1.96 5 ± 0.11 l Slope  424. 5 ± 17 k Q /k F = 424.5/1.965  = 216 k F = 2.39  10 4 s -1 k Q = 5.16  10 8 dm 3 mol -1 s -1

13. 13 Flash photolysis [RK, Pilling & Seakins, p39 on] l Fast burst of laser light –10 ns, 1 ps down to femtosecond l High concentrations of reactive species instantaneously l Study their fate l Transition state spectroscopy J. Phys. Chem. a 4-6-98

14. 14 Flash photolysis l Adiabatic –Light absorbed => heat => T rise –Low heat capacity of gas => 2,000 K –Pyrolytic not photolytic –Study RH + O 2 spectra of OH, C 2, CH, etc l Isothermal –Reactant ca. 100 Pa, inert gas 100 –T rise ca. 10 K; quantitative study possible –precursor + h  CH subsequent CH + O 2 

15. 15 Example [RK, Pilling & Seakins, p48] CH + O 2  products Excess O 2 present, [O 2 ] 0 = 8.8  10 14 molecules cm -3, 1st order kinetics, follow [ CH ] by LIF t / ms20304060 I F 0.2300.1440.0.880.033 Plot ln(I F ) versus time t Slope = - k 1 k 2 = k 1 / [O 2 ] 0

16. 16 Problem In a flash-photolysis experiment a radical, R , was produced during a 2  s flash of light and its subsequent decay followed by kinetic spectrophotometry: R  + R   R 2 The path-length was 50 cm, the molar absorptivity, , 1.1  10 4 dm 3 /mol/cm. l Calculate the rate constant for recombination. –t /  s 0 10 15 25 40 50 –Absorbance0.750.580.510.410.320.28 How would you determine  ?

17. 17 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:  205   fs; separation  600 pm [C & E News 7-Nov-88]

18. 18 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. 27.9 Atkins 6th ed.

19. 19 Fast flow tubes; 1 m 3 /s, inert coating, t=d/v In a RF discharge: O 2  O + O  or  pass H 2 over heated tungsten filament or O 3 over 1000C quartz, etc. Use non-invasive methods for analysis eg absorption, emission Gas titration : add stable NO 2 (measurable flow rate) Fast O+NO 2  NO+O 2 then O+NO  NO 2   NO 2  h End-point? Lights out when flow( NO 2 ) = flow (O)

20. 20 ClO + NO 3 J. Phys. Chem. 95:7747 (1991) l 1.5 m long, 4 cm od, Pyrex tube with sliding injector to vary reaction time F  + HNO 3   NO 3 + HF [  NO 3 ] monitor at 662 nm F  + HCl   Cl + HFfollowed by Cl  + O 3   ClO + O 2

21. 21 Problem [RK, Pilling & Seakins, p36] HO 2  + C 2 H 4  C 2 H 5  + O 2  C 2 H 5 O 2  MS determines LH channel 11%, RH channel 89% C 2 H 5 signal6.143.952.531.250.700.40 Injector d / cm 3 5 7 10 12 15 Linear flow velocity was 1,080 cm s -1 at 295 K & 263 Pa. Calculate 1 st order rate constant; NB [ O 2 ] 0 >>[ C 2 H 5  ] 0 Either convert d ’s into times via flow rate and then plot ln(signal) versus t Or plot ln(signal) vs d & convert slope

22. 22 Flow tubes; pros & cons l Mixing time restricts timescale to millisecond range l Difficult to work at pressures > (atm/100) l Wall reactions can complicate kinetics –coat with Teflon or halocarbon wax; or vary tube diameter l Cheap to build & operate, sensitive detection available –Resonance fluorescence –Laser induced fluorescence –Mass spectrometry –Laser magnetic resonance

23. 23 Resonance fluorescence l Atomic species (H, N, O, Br, Cl, F) mainly not molecular l Atomic lines are very narrow; chance of absorption by another species is highly unlikely l Resonance lamp: mcwe discharge dissociates H 2 l H atoms formed in electronically excited state; fluoresce, emitting photon which H-atoms in rxn vessel absorb & re- emit them where they can be detected by PMT Lamp: H 2  H  H*  H + h Rxn cell: H + h  H*  H + h

24. 24 LIF; detection of OH l Excitation pulse at 282 nm to upper state of OH with lifetime of ns; fluorescence to ground state at 308 nm I F  n  l relative concentrations not absolute (drawback). l Right angle geometry l Good candidates: –CN, CH, CH 3 O, NH, H, SO

25. 25 Reactions in shock waves Wide range of T ’s & P ’s accessible; 2,000 K, 50 bar routine Thermodynamics of high-T species eg Ar up to 5,000 K Study birth of compounds: C 6 H 5 CHO  CO* + C 6 H 6 Energy transfer rxns.: CO 2 + M  CO 2 * + M l Relative rates, use standard rxn as “clock”

26. 26 Mode of action of shock tube l Fast bunsen-burner (ns) l Shock wave acts as a piston compressing & heating the gas ahead of it Study rxns behind incident shock wave or reflected shock wave (milli-  s times) l Non-invasive techniques T & p by computation from measured shock velocity

27. 27 Problem A single-pulse shock tube used to study 1st order rxn C 2 H 5 I  C 2 H 4 + HI ; to avoid errors in T measurement a comparative study of a standard cpd. S was carried out with C 3 H 7 I  C 3 H 6 + HI for which k S =9.1  10 12 exp(-21,900/T) s -1. For a rxn time of 220  s 5% decomp. of C 3 H 7 I or S occurred. What was the temp. of the shock wave? [  900 K] For C 2 H 5 I 0.90% decomp. occurred; evaluate k If at 800 K (k/k S ) = 0.102 compute the Arrhenius equation for k. [  5.8  10 13 exp(-25,260/T) s -1 ]

28. 28 Partial solution [X] = [X] 0 exp(-kt)  k = (1/t) ln {[X] 0 /[X]} 5% reacted, 95% not reacted so [X]/[X] 0 =0.95 k =(1/t) ln{[X] 0 /[X]} = (10 6 /220) ln(1.0526) s -1 k = 233 s -1 k = A exp(-E/RT)  T= E/{R (ln A - ln k)} T = 21,900 / {ln (9.1  10 12 ) - ln (233)} = 898 K k S ? [S]=[S] 0 exp(-k S t)  k S =(1/t) ln{[S] 0 /[S]} 0.90% reacted, 99.1% not reacted so [S]/[S] 0 =0.991