1. 1 Questions?

2. 2 Simple collision theory: A + B  products Assumptions?  molecules hard-spheres  every collision is reactive Calculate rate of collision  rate of reaction In 1 s a typical A travels a distance d= v AB t & will collide with all B’s within collision cylinder of volume =  (r A +r B ) 2 v AB m 3 so n B  (r A +r B ) 2 v AB collisions per second  collision frequency z AB Collision density =  (r A +r B ) 2 v AB  n B n A m –3 s –1

3. 3 SCT Reaction rate = k 2 [A][B] = k 2 n A n B Collision rate =  (r A +r B ) 2 v AB  n B n A “collision density” So k 2 =  (r A +r B ) 2 v AB =  AB v AB where v AB ? Maxwell-Boltzmann distrib. of velocity v AB = (8kT/  ) 1/2 where  =(m A m B )/(m A +m B ) collision cross-section,  AB =  (r A +r B ) 2 Practical units: v AB = 4.602 {T (M A +M B )/(M A M B )} 1/2 with M X in kg mol – 1 & T in K

4. 4 Calculation of collision rate Calculate the mean time between collisions experienced by a single Ar atom at 300 K & atmospheric pressure. Collision diameter for Ar is 0.29 nm, the RMM is 0.040 kg mol –1.  AA =  d 2 = 2.64  10 –19 m 2 v AA = 563.7 m s –1 (NB twice the speed of sound) pV=nRTn A =n/V=p/(RT)= 101,325/(8.3143  300) n A =40.1 mol m –3  2.4  10 25 molecules m –3 z AA = n A  AA  v AA = 3.6  10 9 s –1  t C = (1/Z AA ) = 280 ps

5. 5 Test of SCT  CH 3 +  CH 3  C 2 H 6 M (methyl) = 0.015 kg/molT = 300K d (methyl)  0.21 nm  k 2 = (1/2)  AA v AA = 6.6  10 –17 m 3 s –1 molecule –1 convert to cm 3 s –1 mol –1 calculated 4  10 13 experimental 2  10 13 Unique result; usually calculation >> experiment T-dependence all wrong anyway

6. 6 Improvements? Relax assumption that every collision is reactive If  (coll)  critical value  C then rxn occurs otherwise not What fraction, f, of collisions have  C ? M-B f = exp(-  C /kT) = exp(-E c /RT) k 2 =  AB v AB exp(-E c /RT) Much smaller & T-dependence now OK; but how E c ?

7. 7 Comparison with Arrhenius k 2 = A exp(-E A /RT) k 2 =  AB (8kT/  ) 1/2 exp(-E C /RT) Does E C = E A ? d(ln k)/d(1/T) =  E A /R)operational definition k 2 =  AB (8kT/  ) 1/2 exp(-E C /RT) = Y T 1/2 exp(-E C /RT) ln k = (1/2) ln T  (E C /RT) + ln Y ln k =  (1/2) ln (1/T)  (E C /R)(1/T) + ln Y d (ln k)/d(1/T) =  (1/2)T  (E C /R) =  (E A /R) E A = E C + RT/2NB RT is small in cf. with most E A ’s k 2 =  AB v AB exp(-E c /RT) = e 1/2  AB v AB exp(-E A /RT)

8. 8 Summary of SCT So theory computes the A-factor not k 2 Rarely does A(expt) > A(calc) Exception: rxns such as K  + Br 2  KBr + Br  where A(expt)  5 A(calc); harpoon mechanism For simple species & atoms A(expt)  A(calc) As molecular complexity increases A(calc)>>A(expt) Steric factor, p, so that A(expt)=p e 1/2  AB v AB Check on exptal. data for blunders or special effects Upper estimate of rate constant, if data unavailable

9. 9 Problem solving I The reaction of propene gas with ozone has been studied in a very large closed spherical reactor of volume 220 m 3 at a temperature of 337 K. The concentration of ozone was monitored by measuring its absorbance at 254 nm along a pathlength of 17.88 m; preliminary experiments had shown a molar decadic absorption coefficient, , for ozone of 302 m 2 mol –1 under these conditions. The experiment began by introducing a small amount of a mixture of ozone and oxygen into the reactor and measuring the initial absorbance, which was found to be 0.0384 Calculate the initial concentration of ozone.

10. 10 Problem solving II Immediately afterwards 0.798 Pa of propene was added. Calculate the initial concentration of propene in reactor. Measurements of the absorbance, A, & time, t, were made: A 0.02682 0.01890 0.01305 0.00913 0.00634 t 60 120 180 240 300 See next slide for plot Why is ln A = - k t + ln A 0 ?

11. 11

12. 12 AC/TS Theory A + B  C †  P Rate  v [C † ] [C † ] = K [A][B] but rate = k 2 [A][B] so k 2  K = k K K = (Q C /Q A Q B ) exp(-E 0 † /RT) K = q v † K † = (kT/h )  K † k 2  k (kT/h )  K † k 2  k (kT/h)  K † = k (kT/h)  (Q † /Q A Q B ) exp(-E 0 † /RT)

13. 13 Eyring equation ( kT/h ) s –1,  10 13 s –1 ; universal frequency factor Correct T -dependence; Q(T) ’s only weak Q ’s contain molecule specific terms  moments of inertia, force constants, etc. Great improvement in principle on SCT Detailed calculations? Possible but laborious; most difficult? Guess structure of TS Order-of-magnitude calculations?

14. 14 Order of magnitude calcs. Assume all Q X ’s the same; q R (A)=q R (B), etc. atom A + atom B  (AB) †  products A = (kT/h) (q T ) 3 (q R ) 2 (q V ) 0 / (q T ) 3 (q T ) 3 = (kT/h) (q R ) 2 /(q T ) 3 = 10 -9 cm 3 molecule -1 s -1 N-L A + N-L B  (AB) †  products A = (kT/h)(q T ) 3 (q R ) 3 (q V ) ? (q T ) 3 (q R ) 3 (q V ) 3a-6 (q T ) 3 (q R ) 3 (q V ) 3b-6 = (kT/h) (q V ) 5 /(q T ) 3 (q R ) 3 = 10 -14 cm 3 molecule -1 s -1 all based on q T  10 8 cm -1, q R  10, q V  1

15. 15 Sample calculations ReactionA(expt) A(calc) H+H 2  H 2 +H 5.4  10 13 7.4  10 13 CH 3 +H 2  CH 4 +H 2.0  10 12 1.0  10 12 CD 3 +CH 4  CD 3 H + CH 3 1.0  10 11 2.0  10 11 CH 3 +C 4 H 10  CH 4 + C 4 H 10 1.0  10 10 6.0  10 9 Molecular complexity increases the A- factor decreases As before, E 0 & E A are not identical E A = E 0 + nRT n lies between +1/2 and -2

16. 16 Rate constant from 1st principles see J. Phys, Chem. 99:630 (1995) CHF 3 +  OH  F 3 C  + H 2 O Experimental k = 2.1  10 – 16 cm –3 molecule –1 s –1 TST + tunneling correction  Geometry optimisations of reactants, products & TS; MO Theory ab initio at MP2, 6-311G** Calculated k = 1.03  10 – 16 cm –3 molecule –1 s –1 Estimated lifetime for CHF 3 of 65.5 years in the atmosphere

17. 17 Thermodynamic formulation In soln. Q ’s not readily accessible, recast K = exp(-  G 0 /RT)  G 0 =  H 0 - T  S 0 K = exp(+  S 0 /R) exp(-  H 0 /RT) k 2  k (kT/h)  K † improper equilibrium constant k 2  k (kT/h) (c 0 )  n exp(+  S 0 † /R) exp(-  H 0 † /RT) where c 0 is the standard state concentration &  n is the change in the number of species in going from reactants to TS. The extra term arises because of equil. constant defn: improper  K C = K (RT/p 0 )  n  proper

18. 18 Calculation k 2  k (kT/h) (c 0 )  n exp(+  S 0 † /R) exp(-  H 0 † /RT) But in solution E A =  H 0 † + RT k 2  k e (kT/h) (c 0 )  n exp(+  S 0 † /R) exp(-E A /RT) So A = k e (kT/h) (c 0 )  n exp(  S 0 † /R) Example: exptl. A = 1.14  10 7 cm 3 mol –1 s –1 using a standard state of 1 mol m  calculate  S 0 †. Re-arrange:  S 0 † = R ln { A c 0 / e (kT/h)} = 8.3143 ln {1.14  10 7  1 / 2.718 (6.29  10 12 )} = -118 J K -1 mol -1

19. 19 Thermodynamic TST Talk about rxns in solution in familiar terms  Eg  S 0 †,  H 0 †, from T-dependence  and if k as f(P) then  V 0 † (next slide) [V(H 2 O) 6 ] 2+ + [Ru(NH 3 ) 6 ] 3+   S 0 † large & -ve OH - + [Co(NH 3 ) 5 Br] 2+   S 0 † large & +ve Predominant effect? As reactants to  TS  Solvent increasingly ordered (  ) or  disordered (  ) Charged species structure the solvent

20. 20 f(p)?  Dissociation reactions AB  A+B  Di-tertbutyl peroxide decomposition  +13 cm 3 /mol  Diels-Alder? –25 to –50 cm 3 /mol  Dimerisation of cyclopentadiene  –31 cm 3 /mol