1. Chemical kinetics or dynamics 3 lectures leading to one exam question  Texts: “Elements of Physical Chemistry” Atkins & de Paula  Specialist text in Hardiman Library –“Reaction Kinetics” by Pilling & Seakins, 1995 These notes available On NUI Galway web pages at http://www.nuigalway.ie/chem/degrees.htm What is kinetics all about? 1

2. Academic? Ozone chemistry  Ozone; natural formation (  185  240 nm) –O 2 + h  2O –O + O 2  O 3  Ozone; natural destruction (  280  320 nm) Thomas Midgely –O 3 + h  O + O 2 1922 TEL; 1930 CFCs –O + O 3  2O 2  ‘Man-made’ CCl 2 F 2 + h  Cl + CClF 2 –Cl + O 3  Cl ̶ O + O 2 –Cl ̶ O + O  Cl + O 2 –----------------------------- –Net result is: O + O 3  2 O 2  1995 Nobel for chemistry: Crutzen, Molina & Rowland  1996 CFCs phased out by Montreal protocol of 1987 2

3. Chemical kinetics  Thermodynamics –Direction of change  Kinetics –Rate of change –Key variable: time  What times? –10 18 s age of universe –10 -15 s atomic nuclei –10 8 to 10 -14 s  Ideal theory of kinetics? –structure, energy –calculate fate  Now? –compute rates of elementary reactions –most rxns not elementary –reduce observed rxn. to series of elementary rxns. 3

4. 4 Kinetics determines the rate at which change occurs Thermodynamics vs kinetics

5. Pressure vs CAD in an engine 5

6. ⇌ Kinetics and equilibrium kinetics equilibrium 6

7. Hierarchical structure C.K. Westbrook and F.L. Dryer Prog. Energy Combust. Sci., 10 (1984) 1–57. 7

8. Reaction A f n f Ea f A r n r Ea r k = A T n exp(-E a /RT) Reaction mechanism c2h5oh=c2h4+h2o1.25E+140.12.00E+041.11E+071.778.08E+03 c2h5oh=ch2oh+ch32.00E+23-1.689.64E+048.38E+14-0.227.02E+03 c2h5oh=c2h5+oh2.40E+23-1.629.95E+049.00E+15-0.244.65E+03 c2h5oh=ch3cho+h27.24E+110.19.10E+044.91E+070.997.50E+04 c2h5oh+o2=pc2h4oh+ho22.00E+1305.28E+042.19E+100.284.43E+02 c2h5oh+o2=sc2h4oh+ho21.50E+1305.02E+041.95E+110.094.88E+03 c2h5oh+oh=pc2h4oh+h2o1.81E+110.47.17E+024.02E+080.921.79E+04 c2h5oh+oh=sc2h4oh+h2o6.18E+100.5-3.80E+021.63E+090.832.39E+04 c2h5oh+oh=c2h5o+h2o1.50E+100.82.53E+037.34E+090.911.72E+04 c2h5oh+h=pc2h4oh+h21.88E+033.27.15E+033.93E-013.839.48E+03 c2h5oh+h=sc2h4oh+h28.95E+042.533.42E+032.21E+022.971.28E+04 c2h5oh+h=c2h5o+h25.36E+042.534.41E+032.47E+032.744.19E+03 c2h5oh+ho2=pc2h4oh+h2o22.38E+042.551.65E+042.88E+032.482.83E+03 c2h5oh+ho2=sc2h4oh+h2o26.00E+1201.60E+048.59E+12-0.269.42E+03 c2h5oh+ho2=c2h5o+h2o22.50E+1202.40E+046.66E+13-0.487.78E+03 8

9. 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/m 3 /s, more practically mol dm –3 s –1 9

10. 10 5Br - + BrO 3 - + 6H + = 3Br 2 + 3H 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 Rate = k [Br - ][BrO 3 - ][H + ] 2 where k is the rate constant (variable units) Rate of reaction {symbol: R,n,..}

11. 11 Rate Law  How does the rate depend upon [ ]s?  Find out by experiment The Rate Law equation  R = k n [A]  [B]  … (for many reactions) –order, n =  +  + … (dimensionless) –rate constant, k n (units depend on n) –Rate = k n when each [conc] = unity

12. 12 Experimental rate laws? CO + Cl 2  COCl 2  Rate = k [CO][Cl 2 ] 1/2 –Order = 1.5 or one-and-a-half order H 2 + I 2  2HI  Rate = k [H 2 ][I 2 ] –Order = 2 or second order H 2 + Br 2  2HBr  Rate = k [H 2 ][Br 2 ] / (1 + k’ {[HBr]/[Br 2 ]} ) –Order = undefined or none

13. 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

14. 14 Integration of rate laws  Order of reaction For a reaction aA products, the rate law is: rate of change in the concentration of A

15. 15 First-order reaction

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

17. 17 First-order reaction

18. 18 A  P assume that -(d[A]/dt) = k [A] 1

19. 19 Integrated rate equation ln [A] = -k t + ln [A] 0 Slope = -k 1 st Order reaction

20. 20 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:

21. 21 Half life: first-order reaction

22. 22 When is a reaction over?  [A] = [A] 0 exp{-kt} Technically [A]=0 only after infinite time

23. 23 Second-order reaction

24. 24 Second-order reaction A plot of 1/[A] versus t gives a straight line of slope k A if r = k A [A] 2

25. 25 Second order test: A + A  P Slope = k 2 nd Order reaction

26. 26 Half-life: second-order reaction

27. Kinetics and equilibrium kinetics equilibrium 27

28. Initial Rate Method 5 Br - + BrO 3 - + 6 H +  3 Br 2 + 3 H 2 O General example: A + B +…  P + Q + …  Rate law: rate = k [A]  [B]  …?? log R 0 =  log[A] 0 + (log k+  log[B] 0 +…) y = mx + c  Do series of expts. in which all [B] 0, etc are constant and only [A] 0 is varied; measure R 0  Plot log R 0 (Y-axis) versus log [A] 0 (X-axis)  Slope  28

29. Example: R 0 = k [NO]  [H 2 ]  2 NO + 2H 2  N 2 + 2H 2 O  Expt.[NO] 0 [H 2 ] 0 R 0 –1 25 102.4 × 10 -3 –2 25 51.2 × 10 -3 –3 12.5 100.6 × 10 -3 Deduce orders wrt NO and H 2 and calculate k.  Compare experiments #1 and #2   Compare experiments #1 and #3  Now, solve for k from k = R 0 / ([NO]  [H 2 ]  ) 29

30. How to measure initial rate?  Key: - (d[A]/dt)  - (  [A]/  t)  (  [P]/dt) A + B + …  P + Q + … t=0 100 100  0 0 mol m  3 10  s 99 99  1 1 ditto  Rate?  (100-99)/10 = -0.10 mol m  3 s  1 +(0-1)/10 = -0.10 mol m  3 s  1  Conclusion? Use product analysis for best accuracy. 30

31. Isolation / flooding IO 3 - + 8 I - + 6 H +  3 I 3 - + 3 H 2 O  Rate = k [IO 3 - ]  [I - ]  [H + ]   … –Add excess iodate to reaction mix –Hence [IO 3 - ] is effectively constant –Rate = k [I - ]  [H + ]   … –Add excess acid –Therefore [H + ] is effectively constant  Rate  k  [I - ]   Use integral or initial rate methods as desired 31

32. 32 Rate law for elementary reaction  Law of Mass Action applies: –rate of rxn  product of active masses of reactants –“active mass” molar concentration raised to power of number of species  Examples: – A  P + Qrate = k 1 [A] 1 – A + B  C + Drate = k 2 [A] 1 [B] 1 –2A + B  E + F + Grate = k 3 [A] 2 [B] 1

33. 33 Molecularity of elementary reactions?  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.

34. CO + Cl 2 X  COCl 2 34  Experimental rate law: –(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(1)  Cl + CO  COCl(2)  COCl + Cl 2  COCl 2 + Cl(3)  Cl + Cl  Cl 2 (4) –Steps 1 thru 4 comprise the “mechanism” of the reaction.  decay  collisional

35. 35 - (d[CO]/dt) = k 2 [Cl] [CO] If steps 2 & 3 are slow in comparison to 1 & 4 then, Cl 2 ⇌  2Cl or K = [Cl] 2 / [Cl 2 ] So [Cl] =  K × [Cl 2 ] 1/2 Hence:  - (d[CO] / dt) = k 2 ×  K × [CO][Cl 2 ] 1/2 Predict that: observed k = k 2 ×  K  Therefore mechanism confirmed (?)

36. H 2 + I 2  2 HI  Predict: + (1/2) (d[HI]/dt) = k [H 2 ] [I 2 ]  But if via: – I 2  2 I –I + I + H 2  2 HI rate = k 2 [I] 2 [H 2 ] – I + I  I 2 Assume, as before, that 1 & 3 are fast cf. to 2 Then: I 2 ⇌  2 I or K = [I] 2 / [I 2 ]  Rate = k 2 [I] 2 [H 2 ] = k 2 K [I 2 ] [H 2 ] (identical) Check? I 2 + h  2 I (light of 578 nm) 36

37. 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 90120 [A] / mmol dm  3 8.706.524.893.672.75  Show that the reaction is 1 st order in azomethane & determine the rate constant at this temperature. 37

38. 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 90120 ln [A]2.161.881.591.301.01  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 38

39. More recent questions …  Write down the rate of rxn for the rxn: C 3 H 8 + 5 O 2 = 3 CO 2 + 4 H 2 O  for both products & reactants[8 marks] For a 2 nd 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 NO 2 + NO 2  N 2 O 4 referred to as a bimolecular rxn?[3 marks] 39