2. 1 Chemical kinetics or dynamics 3 lectures leading to one exam question l Texts: “Elements of Physical Chemistry” Atkins & de Paula l 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?

3. 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 l 1995 Nobel for chemistry: Crutzen, Molina & Rowland l 1996 CFCs phased out by Montreal protocol of 1987

4. 3 Chemical kinetics l Thermodynamics –Direction of change l Kinetics –Rate of change –Key variable: time l What times? –10 18 s age of universe –10 -15 s atomic nuclei –10 8 to 10 -14 s lIlIdeal theory of kinetics? –s–structure, energy –c–calculate fate lNlNow? –c–compute rates of elementary reactions –m–most rxns not elementary –r–reduce observed rxn. to series of elementary rxns.

5. 4 Thermodynamics vs kinetics Kinetics determines the rate at which change occurs

6. 5 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

7. 6 Rate Law l How does the rate depend upon [ ]s? l 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

8. 7 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

9. 8 Determining the Rate Law l Integration –Trial & error approach –Not suitable for multi-reactant systems –Most accurate l Initial rates –Best for multi-reactant reactions –Lower accuracy l Flooding or Isolation –Composite technique –Uses integration or initial rates methods

10. 9 Integration of rate laws l Order of reaction For a reaction aA products, the rate law is: rate of change in the concentration of A

11. 10 First-order reaction

12. 11 First-order reaction A plot of ln[A] versus t gives a straight line of slope -k A if r = k A [A] 1

13. 12 First-order reaction

14. 13 A  P assume that -(d[A]/dt) = k [A] 1

15. 14 Integrated rate equation ln [A] = -k t + ln [A] 0

16. 15 Half life: first-order reaction l 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:

17. 16 Half life: first-order reaction

18. 17 When is a reaction over? [A] = [A] 0 exp{-kt} Technically [A]=0 only after infinite time

19. 18 Second-order reaction

20. 19 Second-order reaction A plot of 1/[A] versus t gives a straight line of slope k A if r = k A [A] 2

21. 20 Second order test: A + A  P

22. 21 Half-life: second-order reaction

23. 22 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 l Plot log R 0 (Y-axis) versus log [A] 0 (X-axis) Slope 

24. 23 Example: R 0 = k [NO]  [H 2 ]  2 NO + 2 H 2  N 2 + 2 H 2 O Expt. [NO] 0 [H2] 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 ]  )

25. 24 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  s 99 99  1 1 ditto l 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.

26. 25 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 - ]  l Use integral or initial rate methods as desired

27. 26 Rate law for elementary reaction l 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 l 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

28. 27 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] l No other are feasible! Statistically highly unlikely.

29. 28 CO + Cl 2   COCl 2 Exptal 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. u decay u collisional

30. 29 - (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 l Therefore mechanism confirmed (?)

31. 30 H 2 + I 2  2 HI Predict: + (1/2) (d[HI]/dt) = k [H 2 ] [I 2 ] l 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)

32. 31 Problem l 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 l Show that the reaction is 1 st order in azomethane & determine the rate constant at this temperature.

33. 32 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 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

34. 33 More recent questions … l Write down the rate of rxn for the rxn: C 3 H 8 + 5 O 2 = 3 CO 2 + 4 H 2 O l 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] l Why is the elementary rxn NO 2 + NO 2  N 2 O 4 referred to as a bimolecular rxn?[3 marks]

35. 34 Rate constant expression