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5.62 Physical Chemistry II Spring 2008

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Transition State Theory. I.

Transition State Theory = Activated Complex Theory = Absolute Rate Theory

‡ k H2 + F "! [H2F] !!" HF + H

Assume equilibrium between reactants H2 + F and the transition state.

[H F]‡ K‡ = 2 [H2 ][F] Treat the transition state as a molecule with structure that decays unimolecularly with rate constant k.

d[HF] ‡ ‡ = k[H2F] = kK [H2 ][F] dt k has units of sec–1 (unimolecular decay). The motion along the reaction coordinate looks ‡ like an antisymmetric vibration of H2F , 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[HF] = νK‡ [H2] [F] dt

‡* d[HF] ! (q / N) $ ‡ 'E /kT [ ] = ν # *H F &e [H2 ] F dt "#(q 2 N)(q / N)%&

! ‡ $ ‡ ‡* ‡ (q / N) ' q *' q *' g * ‡ K‡ = # trans &) rot ,) vib ,) 0 ,e-E kT H2 qH2 ) q*H2 , gH2 gF "# (qtrans N) %&( rot +( vib +( 0 0 +

‡ Reaction coordinate is antisymmetric vibrational mode of H2F . This vibration is fully excited (high T limit) because it leads to the cleavage of the H–H bond and the formation of the H–F bond. For a fully excited vibration

hν  kT

The vibrational partition function for the antisymmetric mode is

revised 4/24/08 3:50 PM 5.62 Spring 2007 Lecture #33 Page 2

1 kT q*asym = ' since e–hν/kT ≈ 1 – hν/kT vib 1! e!h" kT h!

Note that this is an incredibly important simplification. The unknown ν simply disappears! We do not need to estimate it!

‡ ‡ ‡*. ‡ kT" q N %( q +( q +( g + ‡ K‡ = $ trans '* rot -* vib -* 0 -e /E kT H2 F H2 *H2 H2 F h! #$(qtrans N)(qtrans N)&') qrot ,) qvib ,)g0 g0 , where

3n"5"1 ‡*! 1 qvib = if transition state is linear # "h$i /kT i=1 1" e or

3n"6"1 ‡*! 1 qvib = if transition state is nonlinear # "h$i /kT i=1 1" e n is # of atoms in transition state

‡*! qvib ≡ partition function from which the antisymmetric vibrational mode is excluded; it has become the reaction coordinate

kT So K‡ = K‡ = K‡" K‡" = “special” modification of K‡ h! that excludes the partition function for the antisymmetric vibrational mode

What is E‡?

(ZPE)TS

‡ Potential E V0 Energy (ZPE)R

H2 + F

HF + H

“Reaction Coordinate” “Reaction Coordinate”

revised 4/24/08 3:50 PM 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 = V0 + (ZPE)TS – (ZPE)R

V0 is the potential energy difference between the bare barrier (saddle point) and the reactant bare minimum.

‡ For linear H2F , n = 3, so 3n–5–1 = 3 regular vibrational modes, thus

1 E‡ = V + h#!‡ + 2!‡ " ! % 0 $ sym.st. bend H2 & !2 ####"####$ Difference in ZPE FORMULATION of kTST

d[HF] kT ‡" kT ‡" TST = ! K [H2 ][F] = K [H2 ][F] = k [H2 ][F] dt h! h so

kT kTST = K‡′ h but not all reactant molecules make it all the way to products — some are reflected back to separated reactants.

Thus,

kT k TST = ! K‡" where κ ≡ transmission coefficient (a fudge factor) h

EVALUATION OF kTST

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.

revised 4/24/08 3:50 PM 5.62 Spring 2007 Lecture #33 Page 4

H2 + F → HF + H T = 300K

m 2 m 19 H2 = F =

Translational part

3/2 ! ‡ $ 3 ( ‡ + (qtrans N) Nh m # & = 3/2 * - = qH2 N qF N (2'kT) * m m - "#( trans )( trans )%& ) H2 F ,

3/2 "34 3 $ 23 ' 23 -1 (6.63!10 J·s) 6!10 ! 0.021 mol 6!10 "23 3/2 & ) (2#!1.38!10 J/K !300K) %0.002! 0.019kg (

= 2.52 × 10–7 m3 mol–1

Rotational part

! = 2 !‡ = 1 8!2IkT H2 q = rot "h2 I = 4.56 !10"48 m2kg I‡ = 1.24 !10"46 m2kg (assume linear H2 transition state) q‡ I‡ ! rot = H2 = 54.4 qH2 I ‡ rot H2 !

Vibrational Part

‡ H2F is a linear transition state (assumed)

3n–5–1 = 3 vibrational degrees of freedom (one vibration is reaction coordinate)

h!‡ h!‡ s = 5771K stretch b = 573K (doubly degenerate) bend k k

h k = 6323K !H2

"1 "2 ‡ "h#‡ /kT q‡*! (1" e"h#s kT ) (1" e b ) vib = = 1.38 *H "1 2 "h#H2 /kT qvib (1" e )

revised 4/24/08 3:50 PM 5.62 Spring 2007 Lecture #33 Page 5

Electronic part

‡ F H2 g0 = 2(S = 1 / 2) g 0 = 6(L = 1,S = 1 / 2) g0 = 1 2 2 –1 (spin orbit splitting of F P1/2 – P3/2 is 404 cm ) g‡ 1 0 = F H2 g 0 g0 3

Calculate E‡

!1 ‡ 14 !1 ‡ 13 !1 V0 = 3.8kJmol "s = 1.20 #10 s "b = 1.19 #10 s (reasonable guesses) ‡ ‡ (How do we guess values for !s and !b ?) = 1.32 1014 s–1 !H2 "

1 E‡ = V + hN#!‡ + 2!‡ " ! % = 6.1 kJ mol"1 0 2 $ s b H2 &

Calculate kT/h

1.38!10"23 J/K ! 300K kT/h = = 6.24 × 1012 s– 1 6.63!10"34 J•s

Putting it all together:

kT k TST = ! K±" h 1 = 1 6.24 #1012 s$1 2.52 #10$7 m3mol$1 (54.4)(1.38) e$6.1 RT ( )( ) 3 = 3.93#107 e$6.1 RT

kTST = 3.40 × 106m3mol–1s–1 at 300K

kEXP = 2.70 x 106 m3 mol–1 s–1 acceptable agreement

Experimental value is smaller because κ is probably not 1. Sometimes kTST will be smaller than kEXP because of tunneling. This model for kTST does not take the quantum mechanical phenomenon of tunneling into account. Tunneling can make the reaction rate become faster than the kTST prediction.

If kTST < kEXP, it may mean that there is some tunneling contribution.

revised 4/24/08 3:50 PM