1 / 59

Chemical Kinetics

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.

orea
Download Presentation

Chemical Kinetics

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  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.

  42. 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

  43. 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®

  44. 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

  45. 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?

  46. 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

  47. 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.

  48. 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)

  49. 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

More Related