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5.62 Physical Chemistry II
Spring 2008
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5.62 Spring 2008
Lecture #34
Page 1
Transition State Theory. II. E‡ vs. Ea.
Kinetic Isotope Effect.
Want to get kTST into Arrhenius form
k TST = !
kT ‡"
K
h
K ‡" = e#G
but
‡
RT
$$
% #RTln K ‡ = &G‡
because
‡
kT #G‡ /RT
kT ‡
e
= ! eS /R e#H /RT
h
h
‡
‡
‡
G = H # TS
NOW :
H‡ = E ‡ + &nRT
k TST = !
so
where ∆n = (# molecules in TS) – (molecularity of reaction)
(molecularity: e.g., unimolecular, bimolecular, etc.)
e.g.
∆n =
1
–2
= –1
So:
k TST = !
kT S‡ R 1 "E‡ RT
e ee
h
k TST # BTm e"E
‡
RT
where m = 1
Surprisingly, theory predicts a temperature dependence to the pre-exponential factor.
This T-dependence is difficult to observe experimentally unless the rate constant is
measured over a wide temperature range (at least a factor of 5)
Now:
d ln k TST
dT
) " k ‡
‡,
%
d + ln $! eS /R e1 ' + lnT + e(E .
* # h
&
=
dT
=
1 E‡
+
T RT 2
(TST)
Contrast this to Arrhenius model:
d ln k
E
= a2
dT
RT
revised 4/24/08 11:34 AM
5.62 Spring 2008
Lecture #34
Page 2
d ln k d ln k TST
=
dT
dT
Ea
1 E‡
=
+
RT2 T RT2
! E a = RT + E ‡
Again, the experimental Ea is larger than E‡ because Ea is a difference between the
average energy of molecules in the pot and the average energy of molecules that react,
while E‡ is a microscopic quantity, a threshold energy along the PES. Notice that Ea is
not a barrier along PES.
COMPARISON OF TRANSITION STATE THEORY WITH COLLISION THEORY
Calculate kTST in the limit of the assumptions of collision theory (i.e. simplified TST):
1) collisions of hard spheres
2) only translational degrees of freedom
Treat H2 as an atom — a hard sphere of mass 2.
[No rotation, no vibration]
Treat H2F‡ as a diatomic molecule
H 2 + F ! H 2 F ‡ ! HF + H
H2
With these assumptions
k!
+
‡
F
H2
F
% ‡ (E‡ RT
kT "
(q‡ / N)
$ H2 trans F
' q rot e
h # (q trans / N)(q trans / N) &
Note: no vibrational partition function for H2F‡ is included because the one vibrational
mode for the pseudo-diatomic molecule transition state has become the reaction
coordinate. Also, no rotational partition function for H2 is included because we are
treating H2 as an atom.
# (2"(m H + m F )kT)3/2 &
2
( 8" 2 I‡kT ‡
3
kT %%
h
N
(
k!
e*E RT
3/2
2
3/2
h % (2"m H2 kT) (2"m F kT) ( )h
%$
('
h 3N
h 3N
The reason there is no rotational or vibrational partition function for H2 is not that we are
assuming the high-T limit, but rather that we are treating H2 as if it were an atom.
revised 4/24/08 11:34 AM
5.62 Spring 2008
Lecture #34
Page 3
Now: I‡ = µd 2H2 !F
where µ =
m H2 m F
H2
m H2 + m F
F
dH2–F
hard sphere collision diameter
) 8kT # m + m &,
F
%% H2
((.
k ! N+
+* " $ m H2 m F '.-
1/2
"d H2 /F
0
e/E
‡
RT
This looks identical to the collision theory result, and collision theory is not based on
thermodynamics.
" 8kT %
2
(E 0 RT
k =$
' !d ABe
# !µ &
TST
Calculate value for k in limit of collision theory assumptions (i.e. what fraction of
collisions are effective because they have sufficient translational energy along the line of
centers?):
! =1
"d 2H2 #F = 3 $10 #19 m 2
1/2
CT
k % 3.4 $10 8 e#E
‡
m 3 mol & s
RT
Compare to kTST
k TST = 3.9 !10 7 e"E
‡
RT
m 3 mol # s
kTST is smaller because it reflects the more restrictive co-linear steric requirement. kCT is
an upper bound because collision theory treats reactants as spheres with no favored
direction of approach (but with an explicit requirement on the effective collision energy).
revised 4/24/08 11:34 AM
5.62 Spring 2008
Lecture #34
Page 4
TRANSITION STATE THEORY AND KINETIC ISOTOPE EFFECT
Consider the H atom (or proton) transfer reaction
HX + Y → X + HY
(or HX+ + Y → X + HY+)
where the original HX bond is broken and a new HY bond is formed
HX = A,
Y = B, (X + HY) = C
A + B → C
So the key question is how do we know whether breaking and making a bond to H occurs
in the transition state region of the reaction coordinate? The size of the HD kinetic
isotope effect tells us whether the H transfer occurs at or before/after the transition state.
Kinetic isotope effect — reaction rates are slower if deuterium is substituted for
hydrogen and a hydrogen bond is involved in the reaction.
Why? Potential energy of interaction the same for HX and DX (BornOppenheimer appproximation).
1
2 h νHX
1
2 h νDX
Because zero point energy of DX is smaller than for HX
1/2
#k&
Since
!"% (
where k = force constant
$m'
revised 4/24/08 11:34 AM
5.62 Spring 2008
Lecture #34
Page 5
!HX # m DX &
"%
because k HX = k DX
(
!DX $ m HX '
Intramolecular potentials are the same. The “shape”
of potential curve doesn’t change upon isotopic
substitution.
1/2
"m %
!DX = !HX $ DX '
# m HX &
1/2
Since m DX > m HX
1
1
So
h!DX < h!HX
2
2
DX
HX
So
D0 > D0
Kinetic Isotope Effect !
dissociation energy to break D—X bond is
larger than that to break H—X
kH
kD
Calculate this using transition state theory
kT !
h #" (q HA
kH
=
k D kT !
h #" (q DA
$ ' E‡H
q‡H / N
e
B
/ N)(q H / N) &%
$ ' E‡D
q‡D / N
e
B
/ N)(q D / N) &%
RT
RT
k H ! q‡H $ ! q DA q BD $ ('E‡H +E‡D ) RT
= # &#
&e
k D " q‡D % " q HA q BH %
1
1
1
1
!E ‡H + E ‡D = !V0H ! h# "‡i H + h# "Hi + V0D + h# "‡i D ! h# "Di
2 i
2 i
2 i
2 i
But V0H = V0D
!
1
k H ! q‡H $ ! q DA q BD $ ' 2 h #"(i
= # &#
&e
k D " q‡D % " q HA q BH %
D$
)‡i H ' ( )‡i D ' ( )H
i + ( )i & RT
%
i
i
i
Most of the isotope effect is in the difference between the zero point energies of the
deuterated vs. hydrogenated reactants. Can use isotope effect to determine whether a
hydrogen bond was involved in the transition state. Standard diagnostic in kinetics!
revised 4/24/08 11:34 AM