5.62 Physical Chemistry II Spring 2008

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1 MIT OpenCourseWare Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 5.62 Spring 2007 Lecture 33 Page 1 Transition State Theory. I. Transition State Theory Activated Complex Theory Absolute Rate Theory 2 + F! " [2F]! k "! F + Assume equilibrium between reactants 2 + F and the transition state. K [ 2F] [ 2 ][F] Treat the transition state as a molecule with structure that decays unimolecularly with rate constant k. d[f] k [ 2 F] kk [ 2 ][ F] k has units of sec 1 (unimolecular decay). The motion along the reaction coordinate looks like an antisymmetric vibration of 2 F, one-half cycle of this vibration. Therefore k can be approximated by the frequency of the antisymmetric vibration ν[sec 1 ] k ν frequency of antisymmetric vibration (bond formation and cleavage looks like antisymmetric vibration) d[f] ν d[f] νk [2] [F]! (q / N) $ & q 2 " ( N)(q F / N) % [ ] F & e'e /kt 2 [ ]! K (q trans / N) $ ' &) " ( q 2 trans N) %& ( rot, q ' ' vib g 0 ) +,) ( vib + g 2 F ( 0 g 0 +,e -E kt Reaction coordinate is antisymmetric vibrational mode of 2 F. This vibration is fully excited (high T limit) because it leads to the cleavage of the bond and the formation of the F bond. For a fully excited vibration hν kt The vibrational partition function for the antisymmetric mode is

3 5.62 Spring 2007 Lecture 33 Page 2 q asym vib 1 ' 1! e!h" kt kt h! since e hν/kt 1 hν/kt Note that this is an incredibly important simplification. The unknown ν simply disappears! We do not need to estimate it! where K kt " $ h! $ q trans q F ( 2 trans N) q trans N %( ' ( N) &' ) rot + ( - q. + ( vib g,) 2-0 +, g 2 F ) 0 g 0, -e /E kt! 3n"5"1 i1 1 1" e "h$ i /kt if transition state is linear or! 3n"6"1 i1 1 1" e "h$ i /kt if transition state is nonlinear n is of atoms in transition state! partition function from which the antisymmetric vibrational mode is excluded; it has become the reaction coordinate So K K kt h! K " K " special modification of K that excludes the partition function for the antisymmetric vibrational mode What is E? (ZPE) TS Potential Energy V 0 E (ZPE) R 2 + F F + Reaction Coordinate Reaction Coordinate

4 5.62 Spring 2007 Lecture 33 Page 3 Since a molecule cannot have a vibrational energy lower than its zero point energy, the effective barrier along the reaction coordinate is E V 0 + (ZPE) TS (ZPE) R V 0 is the potential energy difference between the bare barrier (saddle point) and the reactant bare minimum. For linear 2 F, n 3, so 3n regular vibrational modes, thus FORMULATION of k TST so d[f] E V h! $ sym.st. + 2! bend "! 2 % &! " $! kt h! K " Difference in ZPE [ 2 ][ F] kt h K " [ 2 ][ F] k TST [ 2 ] F [ ] k TST kt h K but not all reactant molecules make it all the way to products some are reflected back to separated reactants. Thus, k TST! kt h K " where κ transmission coefficient (a fudge factor) EVALUATION OF k TST POTENTIAL ENERGY SURFACE KNOWN: E directly from potential energy surface I (moment of inertia of transition state) calculate from geometric structure of transition state ν analyze shape of potential in saddle point region κ trajectory calculations consider κ 1 for now.

5 5.62 Spring 2007 Lecture 33 Page F F + T 300K m 2 2 m F 19 Translational part! " ( q trans N) $ & F N ( N) %& Nh 3 ( m + (2'kT) 3/2 ) m 2 m - F, ( q 2 trans ) q trans 3/2 6!10 23 mol -1 (6.63!10 "34 J s) 3 $ 6!10 23! ' (2!1.38!10 "23 J/K!300K) 3/2 & % 0.002! 0.019kg ) ( m 3 mol 1 Rotational part 3/2! 2 2! 1 8!2 IkT "h 2 I !10 "48 m 2 kg I 1.24!10 "46 m 2 kg I rot I 2! 2! 54.4 (assume linear transition state) Vibrational Part 2 F is a linear transition state (assumed) 3n vibrational degrees of freedom (one vibration is reaction coordinate) h! s k 5771K stretch h! b k 573K (doubly degenerate) bend h! 2 k 6323K! vib ( 1" kt e"hs ) "1 1" e "h b /kt ( ) " ( ) "1 1" e "h 2 /kt

6 5.62 Spring 2007 Lecture 33 Page 5 Electronic part Calculate E g 0 2( S 1 / 2) g F 0 6( L 1,S 1 / 2) g (spin orbit splitting of F 2 P 1/2 2 P 3/2 is 404 cm 1 ) g 0 g 0 F g V 0 3.8kJmol!1 " s s!1 " b s!1 (reasonable guesses) (ow do we guess values for! s and! b?)! " s 1 Calculate kt/h Putting it all together: E V hn $! s + 2! b "! 2 % & 6.1 kj mol"1 kt/h 1.38!10"23 J/K! 300K 6.63!10 "34 J s s 1 k TST! kt h K ± " 1( s $1 )( $7 m 3 mol $1 )( 54.4) ( 1.38) 1 RT e$ $6.1 RT e k TST m 3 mol 1 s 1 at 300K k EXP 2.70 x 10 6 m 3 mol 1 s 1 acceptable agreement Experimental value is smaller because κ is probably not 1. Sometimes k TST will be smaller than k EXP because of tunneling. This model for k TST does not take the quantum mechanical phenomenon of tunneling into account. Tunneling can make the reaction rate become faster than the k TST prediction. If k TST < k EXP, it may mean that there is some tunneling contribution.