MIT OpenCourseWare
http://ocw.mit.edu
5.62 Physical Chemistry II
Spring 2008
For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.
5.62 Spring 2007
Lecture #33
Page 1
Transition State Theory. I.
Transition State Theory = Activated Complex Theory = Absolute Rate Theory
! [H2F]‡ !k!
" HF + H
H2 + F "
Assume equilibrium between reactants H2 + F and the transition state.
[H 2 F]‡
[H 2 ][F]
Treat the transition state as a molecule with structure that decays unimolecularly with rate
constant k.
d[HF]
‡
= k [ H 2 F ] = kK ‡ [ H 2 ] [ F ]
dt
K‡ =
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]
=ν
dt
!
$ ‡
(q‡* / N)
# *H
& e'E /kT [ H 2 ] [ F ]
F
2
#" ( q
N ) (q / N) &%
! (q‡ / N) $' q‡ * ' q ‡*
&) rot , ) vib
K = # trans
H2
2
N &%( q Hrot2 + )( q*H
#" q trans
vib
‡
(
)
* ' g‡ * ‡
,, ) H 0 F , e-E kT
2
+( g 0 g 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
=
q*asym
vib
1
'
1 ! e!h" kT
kT
h!
Page 2
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!
K‡ =
%( q‡ + ( q‡*. + ( g‡ + ‡
kT "
q‡ N
/E kT
vib
$ H trans F
'* Hrot2 - * *H
-* H20 F - e
2
2
h! $# ( q trans N ) ( q trans N ) '&) q rot , ) q vib , ) g 0 g 0 ,
where
q
‡*!
vib
=
3n"5"1
#
i=1
1
1 " e"h$i /kT
if transition state is linear
or
q‡*!
vib =
3n"6"1
#
i=1
1
1 " e"h$i /kT
if transition state is nonlinear
n is # of atoms in transition state
q‡*!
vib ≡ partition function from which the antisymmetric vibrational mode is
excluded; it has become the reaction coordinate
So
kT ‡"
K‡ = K ‡ =
K
h!
K ‡" = “special” modification of K ‡
that excludes the partition function for the
antisymmetric vibrational mode
What is E‡?
(ZPE)TS
Potential
Energy
E‡
V0
(ZPE)R
H2 + F
HF + H
“Reaction Coordinate”
“Reaction Coordinate”
revised 4/24/08 3:50 PM
5.62 Spring 2007
Lecture #33
Page 3
Since a molecule cannot have a vibrational energy lower than its zero point energy, the
effective barrier along the reaction coordinate is
E‡ = V0 + (ZPE)TS – (ZPE)R
V0 is the potential energy difference between the bare barrier (saddle point) and the reactant
bare minimum.
For linear H2F‡, n = 3, so 3n–5–1 = 3 regular vibrational modes, thus
1
E ‡ = V0 + h #$!‡sym.st. + 2!‡bend " !H2 %&
2 ####"####$
!
Difference in ZPE
FORMULATION of kTST
d[HF]
kT
kT ‡"
= ! K ‡" [ H 2 ] [ F ] =
K [ H 2 ] [ F ] = k TST [ H 2 ] [ F ]
dt
h!
h
so
kTST =
kT ‡′
K
h
but not all reactant molecules make it all the way to products — some are reflected back to
separated reactants.
Thus,
k TST = !
kT ‡"
K
h
where κ ≡ transmission coefficient (a fudge factor)
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
H2 + F → HF + H
Page 4
T = 300K
m H2 = 2
m F = 19
Translational part
!
$
q‡trans N )
(
Nh 3
=
# H
&
2
#" ( q trans
N ) ( q Ftrans N ) &% (2'kT)3/2
6 !10
23
mol
-1
( m‡ +
**
-) m H2 m F ,
3/2
=
$ 6 !10 23 ! 0.021 '
(6.63!10"34 J·s)3
&
)
(2# !1.38 !10"23 J/K !300K)3/2 &% 0.002 ! 0.019kg )(
3/2
= 2.52 × 10–7 m3 mol–1
Rotational part
! H2 = 2
!‡ = 1
8! 2 IkT
"h 2
I‡ = 1.24 !10 "46 m 2 kg
q rot =
IH2 = 4.56 !10 "48 m 2 kg
q‡rot
I‡ ! H2
=
= 54.4
q Hrot2 IH2 ! ‡
(assume linear
transition state)
Vibrational Part
H2F‡ is a linear transition state (assumed)
3n–5–1 = 3 vibrational degrees of freedom (one vibration is reaction coordinate)
h!‡s
h!‡b
= 5771K stretch
= 573K (doubly degenerate) bend
k
k
h!H2 k = 6323K
q‡vib*! (1 " e"h#s kT ) (1 " e"h#b /kT )
=
"1
2
(1 " e"h#H2 /kT )
q*H
vib
‡
"1
‡
"2
= 1.38
revised 4/24/08 3:50 PM
5.62 Spring 2007
Lecture #33
Page 5
Electronic part
g‡0 = 2 (S = 1 / 2 ) g F0 = 6 ( L = 1,S = 1 / 2 )
g H0 2 = 1
(spin orbit splitting of F 2P1/2 – 2P3/2 is 404 cm–1)
g‡0
1
=
F H2
g 0 g0
3
Calculate E‡
V0 = 3.8kJmol!1
"‡s = 1.20 #1014 s!1
"‡b = 1.19 #1013 s!1 (reasonable guesses)
(How do we guess values for !‡s and !‡b ?)
!H2 = 1.32 " 1014 s –1
1
E ‡ = V0 + hN #$!‡s + 2!‡b " !H2 %& = 6.1 kJ mol"1
2
Calculate kT/h
kT/h =
1.38 !10"23 J/K ! 300K
= 6.24 × 1012 s–1
"34
6.63!10 J•s
Putting it all together:
k TST = !
kT ±"
K
h
(
)(
)
1
= 1 6.24 #1012 s$1 2.52 #10 $7 m 3mol$1 ( 54.4 ) (1.38 ) e$6.1 RT
3
7 $6.1 RT
= 3.93 #10 e
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