1. Ch.22 Reaction dynamics Collision theory
Diffusion-controlled reactions
Transition state theory
Thermodynamic aspect
Potential energy surfaces
Electron transfer reactions

2. Collision theory Energy requirement
Theory of reaction rates based on reactants collisions.
Consider a bimolecular reaction
A + B → P; v = kr [A][B] or A + A → P; v = kr [A]2
Collision frequency z = σ<crel> N, where
σ is the collision cross-section and N is the # density.
σ = πd2 and d = ½ (dA + dB). <crel> = (8RT/πμ)1/2
Collision number = # of collision /unit volume s
ZAA = ½ σ <crel> NA2
ZAB = σ <crel> NANB A B
collision cross-section σ
Energy requirement
Impact parameter b;
collision occurs only if b ≤ d. If b = 0, head-on collision.
Relative kinetic energy ε = ½ μvrel,
but the collision energy = ½ μvrel2 cos2 θ.
Reaction occurs, only if ε cos2 θ ≥ εa (activation energy).
Since b = d sin θ, the above condition equivalent to
cos2 θ = {1- (b/d)2} ≥ εa/ε, namely 1- εa/ε ≥ (b/d)2
For a collision with ε ≥ εa, the reactive collision cross-section
σr(ε) = πb2 = π(d sinθ)2 = πd2 (1- εa/ε) = σ(1- εa/ε), which is shown
in the bottom figure.
H+D2→ HD+D

3. Averaging over the speed distribution

5. Steric factor P transition state k2 = σ (8kT/πμ)1/2 exp(- εa /kT)
Collision theory overestimate k2.
Steric factor P = k2(exp) /k2(theory),
k2 = Pσ (8RT/πμ)1/2 exp(- εa /kT)
Example: H2+ C2H4 → C2H6 at 628 K
Experimental Pre-exponential factor; A exp = 1.24 x106 l mol-1 s-1
μ = (m1 m2)/ (m1+ m2) = (2 x 28)/(2+28) g/ NA = 3.12 x10-27 g
<crel> = (8kT/πμ)1/2 = 2.66 x 103 m/s
σ(H2) = 0.27 nm2, σ(C2H4) = 0.64 nm2 → σ = 0.46 nm2
A theory = 7.37 x1011 l mol-1s-1
P = A exp / A theory = 1.24 x106 / 7.37 x1011 = 1.7 x10-6
Among energetic collisions with ε(collision) > εa,
the reaction probability is only 1.7 x10-6.
Most collisions do not have the proper orientation
to form the transition state.
The collision theory does not tell how to calculate P; a weak point.
transition state

6. Why do alkali atom + halogen reactions have P >1?
Ex: K + Br2 → KBr + Br; P = 4.8
Harpoon mechanism
The potential energy curves of
K + Br2 and K+ + Br2- cross at R*.
At R* electron transfer from K to Br2 occurs:
curve crossing
Estimation of Rx
The energy difference ΔE0 between
K+ + Br2- and K + Br2 at R = ∞ is
ΔE0 = IP(K) - EA(Br2) = (420 – 250) = 170 kJ/mol
K → K+ ; IP = 420 kJ/mol
Br2- → Br2 ; EA(electron affinity) =
At R*, IP - EA - (e2/4πε0R*) = 0
σ* = πR*2
P = σ*/σ = πd2/ πR*2 = [e2/4πε0d(IP-EA)]2
Using the IP an EA values and d = 400 pm,
P = 4.8 is estimated.
This is in good agreement with the experimental value. e- K Br Br Br R*

7. Reactions in solutions
A liquid does not always mix with another liquid (example: oil + water), but
these liquids may dissolve in a proper common solvent.
Liquid reactions are usually carried in such a solvent.
A reaction A + B → P in solution occurs in the following steps
1. A + B → AB; formation of an encounter pair (EP).
EP is surrounded by solvent molecules; cage effect
2. AB → A + B; separation
EP has a finite lifetime.
3. AB → AB* → P; activation & reaction
Applying a steady-state approximation for AB,
with
encounter pair →

8. kd’<< ka : diffusion-controlled reaction
+ OH-
Two limiting cases
kd’<< ka : diffusion-controlled reaction
Activation is much faster than the separation of EP,
This is the case when the activation energy Ea is not much larger than kT.
Then, diffusion is the rate-determining step.
2) ka << kd’ : activation-controlled reaction
Activation is a slow process, so it is the rate-determining step.
This is the case when the activation energy Ea is much larger than kT.

9. Diffusion-controlled reactions
A diffuses towards and B diffuses towards A.
Consider a particular reactant A
Let [B]r the concentration of B at r. [B]r = [B] (bulk concentration) at r =∞.
In a diffusion-controlled reaction, A acts as a sink for B, and
therefore [B]r = 0 at r = R*.
When the system has reached a steady-state, and
because of the spherical symmetry of the system.
The solution is [B]r = a + b/r. From the boundary condition above,
[B]r = [B] (1- R*/r).
Rate of reaction = 4πR*2J, and
Rate of reaction = 4πR* DB[B],
Considering the mutual diffusion, D = DA + DB
= kd [A][B]
A little error may be introduced by letting RA = RB = ½ R*
kd = 8RT/3η
Ex: I + I → I2 in hexane (η = cP, 1cP = 0.1 kg m K.
The calculated kd = 2.0 x1010 l mol-1s-1 is in good agreement with kexp = 1.3 x 1010. ,
y’’ + 2y’/x = 0
xy’’ + 2y’ = 0
(xy’)’ + y’ = 0
xy’ + y = a
(xy)’ = a
xy = ax + b
y = a + b/x

10. Spherical coordinate system

11. Transition state theory
An activated complex AB≠ generated in an energetic collision between reactants
A and B can form a product if it passes through the transition state via a proper
vibrational mototion.
A + B ↔ C ≠ → P; pre-equilibrium assumed.
K ≠ = (pC≠/po)/ (pA/po)(pB/po)= po pC≠ /pApB ; dimensionless
The partial pressure of J’s species in terms of the molar concentration is pJ = RT[J].
K ≠ = RT[C ≠] po/ {RT[A] RT[B]} = (po/ RT) [C ≠]/ [A][B] = (RT/po) K ≠ [A][B]
C ≠ → P; v = k ≠ [C ≠] = (RT/po) k ≠K ≠ [A][B]
Then, v = kr [A][B] and kr = (RT/po) k≠ K ≠
K ≠ can be calculated based on statistical thermodynamics and k ≠ obtained
by considering the vibrational motions of the TS.
The steric factor is naturally included in the theory.

12. Statistical Thermodynamic
Energy partitioning & Boltzmann distribution law
Microcanonical ensemble
Configuration & the number of possible events
Maximum term method: Lagrange undetermined multiplier
Molecular partition function q
Canonical Ensenble & Canonical partition function Q
All thermodynamic quantities from Q
Equilibrium constant K from Q

13. Energy Partitioning E = ∑ ½ mvi2 : fixedNA
E = ∑ ½ mvi2 : fixed
1 mole Ar
i =1
Internal energy U
N,V,E constant
Q: How is the energy distributed among
NA Ar atoms ?
Q: How is the thermal energy absorbed by
each mode of N2 motions (translation,
rotation, vibration)?
V fixed
1 mole N2 q

14. Microcanonical ensemble
N,V,E fixed
As N → ∞, for any thermodynamic quantity
Time average = ensemble average
Equal a priori probability. …… …… ……
N = The # of systems
in the ensemble

15. Statistical approach ni : population ∑ = N , ∑ ni εi = E
Configuration {n0,n1,n2,n3,n4,n5 …}
Number of possible ways W
W = N! / n0! n1! n2! ….
i = 0
i = 0
Ex: N = 5, E = 5
εi ε 2ε 3ε 4ε 5ε W
{ni} {4, 0, 0, 0, 0, 1} 5!/4! =
{3, 1, 0, 0, 1, 0} 5!/3! =
{3, 0, 1, 1, 0, 0} 5!/3! =
{2, 2, 0, 1, 0, 0} 5!/2! 2! = 30
{2, 1, 2, 0, 0, 0} 5!/2! 2! = 30
{1, 3, 1, 0, 0, 0} 5!/3! =
{0, 5, 0, 0, 0, 0} 5!/5! =
total
Absolute
monarchism
Communism

16. W becomes a highly peaked function as N→NA ∑ Wi ~ Wmax = W*
How can one maximize under two constraints
∑ ni = N and ∑ ni εi = E ?
Lagrange undetermined multiplier method.
(∂ lnW/∂ni) = 0 ; I =1,2 …..m-2
From these m equations one can determine
{ni*} = (n1*, n2*, n3*, … nm*)
The probability for an Ar atom to have energy εi is
pi = ni* / N = exp (-εi /kT) / q
q = ∑ exp (-εi /kT) : Molecular partition function
Boltzmann Distribution Law i
β = 1/kT

17. } Lagrange undetermined multiplier method
1. Maximization under no constraint
f = f (x1, x2,….. , xn)
δf = (∂f /∂x1)dx1 +(∂f /∂x1)dx1 + ….. + (∂f /∂xn)dxn = 0 is always true at the max.
If there is no constraint among the variables, namely all xi’s are independent,
(∂f/∂x1) = 0, (∂f/∂x2) = 0 …. , (∂f/∂xn) = 0.
2. Maximization under constraints
f = f (x1, x2,….. , xn),
g = ∑ xi - N = 0 ; ∑ xi = N
h = ∑ xi εi - E = 0 ; ∑ xi εi = E
Let x1, x2, ….., xn-2 are independent variables and xn-1 and xn are not.
δf = (∂f /∂x1)dx1 + ….. + (∂f /∂xn) dxn = 0 is still true at the max. (1)
δg = (∂g /∂x1)dx1 +(∂g /∂x1)dx2 + ….. + (∂g /∂xn)dxn = (2)
δh = (∂h /∂x1)dx1 +(∂h /∂x1)dx2 + ….. + (∂h /∂xn)dxn = (3)
(1) + α (2) + β (3) = 0, where α and β are undetermined constants
Σ1,n-2 {(∂f /∂xi) + α (∂g /∂xi) + β (∂h /∂xi)} dxi +
+ {(∂f /∂xn-1) + α (∂g /∂xn-1) + β (∂h /∂xn-1)} dxn-1
+ {(∂f /∂xn) + α (∂g /∂xn)dxn + β (∂h /∂xn)} dxn = (4)
When x1, x2,….. , and xn-2 ) vary independently, the conditions for satisfying the eq. (4) are
{(∂f /∂xi) + α (∂g /∂xi) + β (∂h /∂xi)} = 0 (i = 1, 2, ….. , n-2) and simultaneously
one has to choose α and β such that {(∂f /∂xi) + α (∂g /∂xi) + β (∂h /∂xi)} = 0 for i = n-1, n.
In conclusion, {(∂f /∂xi) + α (∂g /∂xi) + β (∂h /∂xi)} = 0 for all i. }
f = f(x,y)
y = ax + b; constraint
g(x,y) = y- ax+b = 0
two constraints

18. Example
Q: When a rectangular parallelpiped box having the sides x, y, and z is made with a
cardboard of a fixed area S, what are the ratios xy and z/y which maximize
the volume V?
V = xyz
S = 2 (xy + yz +zx) → g = 2 (xy + yz + zx) – S = 0 ; constraint
At the maximum,
δV = (∂V/∂x)dx + (∂V/∂y)dy + (∂V/∂z)dz = 0, (1)
δg = (∂S/∂x)dx + (∂S/∂y)dy + (∂S/∂z)dz = (2)
(1) + α (2) = 0, which is
{(∂V/∂x)+α(∂g/∂x)}dx + {(∂V/∂y)+α (∂g/∂y)}dy = 0
The maximization condition is that
(∂V/∂x) + α (∂g/∂x) = yz + α 2(y+z) = 0, (3)
(∂V/∂y) + α (∂g/∂y) = zx + α 2(z+x) = 0, (4)
(∂V/∂z) + α (∂g/∂z) = xy + α 2(x+y) = (5)
(3) - (4) = (y-x) z = 2α(y-x), where z ≠ → 2α
One gets from these conditions that x = y, and similarly y = z.
A cube has a maximum volume for a fixed surface area S !!

19. Maximization of ln W W = N! / n0! n1! n2! …..
ln W = ln N! - ∑ ln ni ! ~ (N ln N –N) - ∑ (ni ln ni - ni ) (Stirling’s formula)
∂ ln W/ ∂ni = {(ln N +1) -1} - {(ln ni +1) -1} = ln (N/ ni) (1)
g = ∑ ni - N → (∂g/ ∂ni) = (2)
h = ∑ ni εi - E → (∂h/ ∂ni) = εi (3)
(1) + α x (2) + β x (3) = 0 ; α , β constants (undetermined multiplier)
∂ ln W/ ∂ni + α(∂g/ ∂ni)+ β(∂h/ ∂ni) = 0 for all i
- ln ni /N + α - β εi = 0, and therefore
The probability Pi for a molecule to populate an energy level εi is
Pi = ni / N = eα exp (-βεi).
Since ∑i Pi = ∑i eα exp (-βεi) = 1 → eα = 1/ ∑i exp (-βεi).
Pi = exp (-βεi) / ∑i exp (-βεi).
It will be shown later that β = 1/kT. So,
Pi = exp (- εi /kT)/∑i exp (- εi /kT) = exp (- εi /kT)/q; Boltzmann distribution law
q ≡ ∑ exp (-εi /kT) : Molecular partition function

20. Boltzmann energy distribution law
Pi = Ni /N= exp (-εi/kT)/ ∑ exp (-εi/kT); discrete ε state
q =∑ exp (-εi /kT); molecular partition function
P = exp (-ε/kT) dε / ∫ exp(-ε/kT) dε ; continuous ε state
q = ∫ exp (-ε/kT); molecular partition function
degenerate states
Degenerate states: states with the same energy
but with different motions. gi = degeneracy
Pi = ni / N = gi exp (-εi/kT)/q , gi = degeneracy
The probability for a molecule to have energy of ε~ ε+ dε
dP = g(ε) exp(-ε/kT) dε / ∫ g(ε) exp(-ε/kT) dε
Example: vibrational partition function qv
εv = hν (v+½), where the vibrational quantum number v =0,1,2 ….
The lowest level can be conveniently taken as the 0 of energy.
qv = ∑0,∞ exp (-εv/kT) = ∑0,∞ exp (- vε/kT) = 1/(1- e- ε/kT)
ν (Cl2) = 1/λ = cm-1 → ε = hν = hc/ λ = 69.6 meV
At 298 K kT = 26 meV and → ε/kT = 2.68
qv (298 K) = 1 + e e e …. =
…. = 1.074
qv (1000 K) = 1.811
qv (∞ K) = ∞
q is the # of energy levels populated at T. ε

21. Molecular partition function q
The energy of a molecule is the sum of contributions from its different modes of motion:
εi = εiT + εiR + εiV + εiE ;( translation, rotation, vibration, electronic contribution).
Translational partition function qT
1-D particle in a box of size X
The translational energy levels can be treated as continuous levels.
, where β = 1/kT.

22. 3-D particle in a box of size X, Y, and Z
εT = εXT+ εYT+ εZT
qT = qXT+ qYT+ qZT = (2πm/h2β)3/2 XYZ = (2πm/h2β)3/2 V,
where V is the volume of the box.
qT = V/Λ3, Λ = h(β/2πm) 1/2 = h/(2πmkT)1/2
Ex: N2 at 298 K in a cube of the side L = 1 m.
Λ = h/(2πmkT)1/2 = x10-11 m and
qT = V/Λ3 = 1.42 x1032 >>>> 1
This tells that the translational energy levels are indeed almost continuous
regardless of the mass. → Classical mechanics can be applied.
Rotational partition function qR
A rigid rotor model
For a linear rotor with a moment of inertia I
εR = L2/2I = J(J+1)ħ2/2I = hcBJ(J+1),
where rotational quantum number J = 0,1,2, …. , ħ = h/2π, and B = h/8π2cI (cm-1)
Degeneracy gJ = 2J+1 (MJ = m ħ; m = 0, ±1, 2… , ±J)
qR = ∑J=0,∞ (2J+1)[exp{- hcBJ(J+1)/kT}]
Since the rotational energy gap is much smaller than kT,

23. B = 10.591 cm-1. hc/kT ≈ 200 cm-1at RT. hcB/kT = 0.05111. qR = 19.9
1H35Cl at 273 K
= kT/hcB
Example: qR of 1H35Cl at 273 K.
B = cm-1. hc/kT ≈ 200 cm-1at RT. hcB/kT =
qR = 19.9

24. Origin of the symmetry number,
;characteristic rotational temperature.
Origin of the symmetry number
Some rotational states are forbidden due to
the Pauli principle.
All particles have an intrinsic spin.
The spin quantum number is either half-integral,
Fermion, (I = 1/2, 3/2, 5/2, ……) or integral number,
Boson, (I = 0, 1, 2, 3 ……).
The wavefunction should be antisymmetric for Fermion
and symmetric for Boson with respect to the
permutation of two identical particles. (see the next 4
pages below) .

29. Symmetry number of H2, HD
ψrot = ψJ ; even (when J = even) or odd (when J = even) function. and
Rotation by 180o is equivalent to the exchange of H1 and H2.
H is a Fermion with a nuclear spin of I =1/2.
ψtotal = ψrot ψspin
The total wavefunction should be antisymetric with respect
to exchange of H1 and H2.
ψspin (ortho-H2) = α(1)α(2), β(1)β(2), or 1/√2 {α(1)β(2)+β(1)α(2)} ; symmetric
ψspin (para-H2) = 1/√2 {α(1)β(2)- β(1)α(2)} ; antisymmetric
P1,2 ψtotal = P1,2 ψrot P1,2 ψrot = antisymmetric
Therefore, the rotation of ortho-H2 can only have odd J states, i.e., J = 1, 3, 5, ….
The rotation of para-H2 can only have even J states, i.e., J = 0, 2, 4, ….
The number of allowed states are diminished by half. → symmetry number σ = 2.
Normal H2 is a mixture of ortho + para (3:1) H2. Therefore,

30. Electronic partition function qE
qE = Σi exp(- βεi) = … = 1 because Δε/kT <<< 1.
In special cases, qE ≠ 1.
1) Atoms or radicals with unpaired electrons: qE = g Σi giexp(- βεi) = 2
ex: H atom (1s1)
2) O2 or …
; S = s1 + s2 = 1 … …
electronic energy
levels
Degeneracy = 2S +1 = 3
qE = 3

31. Relation between U and q Justification of β = 1/ kT
From thermodynamics we know that for a monoatomic gas
E =(3/2) nRT = (3/2) NkT (1)
q T = (2πm/h2β)3/2 V
(∂ln q/∂β)V = - 3/2 (∂ln β/∂β)V = -3/2β
E = - N (∂ ln q/∂β)V = 3N/2β (2)
Comparing the eq.’s (1) and (2),one gets
β = 1/ kT →
= - N (∂ ln q/∂β)V
U =U(0) + E
U(T) - U(0) = - N (∂ln q/∂β)V

32. Entropy : 3rd Law of Thermodynamics
dU = dq + dw = ∑ ( εi dni + ni dεi )
dS ≡ dqrev/ T = kβ∑ εi dni = k ∑ βεi dni (β = 1/kT)
From the maximization condition (∂ lnW/∂ni) = - α + βεi
dS = k ∑ βεi dni = k ∑ {(∂ lnW/∂ni)V +α)} dni
= k d lnW + k α ∑ dni = k d lnW
because ∑ dni = d∑ ni = dN (fixed) = 0
S = k ln W : absolute entropy (In 1877 by Boltzmann)
S = k ln W = k(N ln N - N) - k Σ(ni ln ni -ni) = k (N ln N - Σ ni ln ni)
= k (Σni ln N - Σ ni ln ni ) = - kN Σ(ni/N) ln (ni /N) = - kN Σpi ln pi
= - kN Σpi (-βεi – ln q) = kβE + kN ln q = kβ{U-U(0)} + kN ln q
S = {U-U(0)}/T+ kN ln q
: 3rd Law of Thermodynamics

33. Canonical ensemble Ei (1) (2) Heat bath at T , where
A system with fixed N,V, and T.
N replicated systems immersed in a heat bath.
The total ensemble energy has a fixed energy E.
Each system has a different energy Ei.
Configuration {N1,N2 …… Nn}
Constraints: Σ Ni = N, Σ Ni Ei = E. (1)
Weight (# of the possible events)
W (N1,N2 …… Nn) = N !/N1! N2! …… Nn!
Maximization of W under the constraints as before;
∂ lnW/∂Ni + α + β Ei = 0 for all i. So,
pi = exp(- βEi) /Q, where
Q = Σi exp(-βEi) ; canonical partition function.
Ei = εia + εib + εi(c) + ……
Q = Σi exp [-β {εia + εib + εic ……}]
Since the number of available quantum states >>> # of particles,
εi’s in eq.(1) never be the same.
Q (N,V,T) = [q(V,T)]N ; distinguishable particles.(solid)
Q (N,V,T) = [q(V,T)]N /N! ; indistinguishable particles (gas)
because the permutation of N particles with a particular set of
energy states is indistinguishable. Ei
(1)
(2)
Heat bath at T
, where c a b
N particles in
each system

34. Q = Σi exp (-βEi) = exp (-βE1) + exp (-βE2) + exp (-βE3) + ….
= exp {- β(ε1a + ε1b + ε1c + … + ε1N )} + exp {- β(ε2a + ε2b + ε2c + … + ε1N)} +
= exp(-βε1a) exp(-βε1b) exp(-βε1c) …. exp(-βε1N)
+ exp(-βε2a) exp(-βε2b) exp(-βε2c) …. exp(-βε2N)
+ exp(-βε3a) exp(-βε3b) exp(-βε3c) …. exp(-βε3N) + … upto N terms
= [exp(-βε1a) + exp(-βε2a) + exp(-βε3a) + …. + exp(-βεN a)] x
[exp(-βε1b) + exp(-βε2b) + exp(-βε3b) + …. + exp(-βεN b)] x
[exp(-βε1c) + exp(-βε2c) + exp(-βε3c) + …. + exp(-βεN c)] x
: : : :
[exp(-βε1N) + exp(-βε2N) + exp(-βε3N) + …. + exp(-βεN N)]
= q N
upto N terms
molecules
states
states
molecules
x x x x
x x x x

35. Q = Σ i exp (-βEi) = exp (-βE1) + exp (-βE2) + exp (-βE3) + ….
= exp {- β(ε1a + ε1b + ε1c + … + ε1N )} + exp {- β(ε2a + ε2b + ε2c + … + ε1N)} + ….
pi a = exp(-βεia)/q, exp(-βεia) = q pi a
exp {- β(ε1a + ε1b + ε1c + … + ε1N )} = qN (p1a p1b p1c …. p1N)
Q/qN = (p1a p1b p1c …. p1N) + (p2a p2b p2c …. p2N) + …. + (pN a pN b pN c …. pN N)
= (p1a + p2a + p3a + …. + pN a) x
(p1b + p2b + p3b + …. + pN b) x
…………………………… x
(p1N + p2N + p3N + …. + pN N) = 1N =
molecules
states
states
molecules
x x x x
x x x x

36. Thermodynamic functions from Q
For an ideal gas,
← A = E -TS = U - U(0) -TS
← dA = - pdV - SdT, p = - (∂A/∂V)T
← G = A + pV

37. Equilibrium constant K
a A + b B → c C + d D

39. Calculation of equilibrium constant K
Consider a simplest case of R ↔ P
In thermal equilibrium at T the thermal energy is partitioned in A and B according
to the Boltzmann distribution law.
Boltzmann distribution law
Let N = NR + NP and ∆E0 = εP0 – εR0
For the combined system
q = ∑i=0,∞ [exp(- εRi/kT) + exp{- (εPi+ ∆E0} /kT]
= qR + qP exp(- ∆E0/kT)
NR/N = qR /q and NP/N = qP exp(- ∆E0/kT) /q
K = [P]/[R] = NP/ NR = qP exp(- ∆E0/kT) /qR
K can be estimated, if the levels of R and P are known.
The energy levels, in principle can be calculated by
solving the Schrödinger equation Hψ = Eψ or
experimentally (spectroscopically).
Case 1: both R and P have the same energies levels.
Then, K = qP exp(- ∆E0/kT) /qR = exp(- ∆E0/kT) <1.
At equilibrium, R predominates over P.
Case 2: the energy levels of P much narrower than
that of R.
Then, K = qP exp(- ∆E0/kT) /qR = exp(- ∆E0/kT) >1.
At equilibrium, P predominates over R.
case 1
case 2

40. Transition state theory (continued)
A + B ↔ C ≠ → P
v = k ≠ [C ≠] = (RT/po) k ≠K ≠ [A][B] = kr [A][B]
kr = (RT/po) k≠ K ≠
One particular normal mode of the TS leads to the product.
The vibrational frequency ν≠ of that mode is quite low
( ) because the potential well is shallow.
The partition function of the mode is
k≠ = κ ν≠ , where κ is the transmission coefficient which is assumed to be about 1.
The unknown frequency ν≠ is cancelled .
transition state , , .
Eyring equation

41. Reaction between structureless particles.
A + B ↔ C ≠ → P
Lets assume that both A and B molecules do not have any structure like an atom.
Then, the activated complex resembles a diatomic molecule with mass of
mC≠ = mA + mB and moment of inertia I = μr2.
The transitional partition function is
qR = (2IkT/ħ2)
With σ* = κπr2 and ∆rE0 = Ea,
we arrive at precisely the same expression obtained from the simple collision theory.
Since the reactant particles are structureless, there is no steric factor, i.e. P =1.
= NA

42. Reaction between nonlinear molecules, ,

43. Observation and manipulation of activated complex
1. NaI + hν → Na + I
Pulsed laser, Femtosecond laser
Curve-crossing, Adiabatic potential
Laser-induced fluorescence
Low frequency ν ≠ = 27 cm-1
Oscillation in the excited state: T = 1.25 ps
2. H + O=C=O→ [H...O…C=O] ≠ → HO + CO
Molecular beam
van der Waals molecule IH…OCO
Femtosecond transition state spectroscopy

44. Thermodynamic aspect Activaton parameters
For the reaction A + B ↔ C ≠ at equilibrium, the Gibbs energy of activation is
∆≠G = - RT ln K≠
From the TS theory (standard state notation not used for simplicity).
∆≠H = enthalpy of activation, ∆≠S = entropy of activation
Substituting with ∆≠G = ∆≠H – T∆≠S,
and B = (kT/h)(RT/po),
From the formal definition of activation energy Ea = RT2(∂ lnk/∂T),
we get Ea = ∆≠H + 2RT.
and A =
Steric factor P = e- ∆≠S/R

45. Linear Free Energy Relationship (LFER)
Hammett equation(1937) describes correlation between the equilibrium constant K and
rate constant k for reactions ( ionization in water, alkaline hydrolisis, etc.) involving benzoic acid
with different substituents.
ln K/ ln K0 = σρ, ln k/ ln k0 = σρ; σ = substituent constant, ρ = reaction constant.
The reference (K0 and k0) is referred to H-substitutied compound.
ln K (or k) - σ (or ρ) plot is linear.

46. Reactions between ions
For a reaction A + B ↔ C≠ → P in solution,
d[P]/dt = k≠ C≠ with K≠ = aC≠ /aAaB = Kγ[C≠]co / [A][B], where Kγ = γC≠ /γAγB.
Then, d[P]/dt = kr[A][B] with kr = k≠ K≠.
kr = kr0 /Kγ , where kr0 is the rate constant when all γ’s are 1.
Using the Debye-Hückel limiting law, log γ± = - A(= 0.529)│z+ z-│ I1/2,
where I = ½ ∑ zi2 (bi/bo).
I = ½ (z+2 b+ + z-2 b-)/bo for solutions consisting of two types of ions.
Considering I is contributed by two types of ions, we may write
log γA = - A zA2 I1/2 and log γB = - A zB2 I1/2
For a reaction between ions, log γC≠ = - A (zA2 + zB2)I1/2
log kr = log k0r – A [zA2+zB2-(zA+zB)2]I1/2
= log k0r + 2A zAzBI1/2
Kinetic salt effect
The effect of ionic strength on the rate of reaction between ions.
1. If zAzB > 0 (same charge), kr ↑ with increasing I
→ Coulomb repulsion is more screened at higher I
→ Eais lowered.
2. If zAzB < 0 (opposite charge), kr↓ with increasing I
→ Coulomb attraction is more screened, at higher I
→ Ea is lowered.

47. Dynamics of molecular collisions
Reactive collisions
State-to-state dynamics
A + B → P carried at in thermal equilibrim at T.
A and B have Boltzmann population of their quantum states.
kr = <σ vrel> NA : < > denotes Boltzmann average.
kn,n’ = <σn,n’ vrel> NA
n and n’ are the quantum states of reactant and product , respectively.
kn,n’ = state-to-state rate constant
kr = Σ kn,n’ (T) fn(T); fn(T) is the Boltzmann factor at T.
Experimental approach
Reactanst as a molecular beam
Pulsed, collimated, super-sonic beam: mono-energetic beam
Time-of-flight technique + Mass spectrometer
Detector
Chemiluminescence
Laser-induced fluorescence
The Chemiluminescence spectrum of CO* produced in O + CS → CO* +S. CO*(v’>0) → CO (v’-1)+ hν (IR)

48. Supersonic Molecular Beam
I. λ > d at low P
Molecular flow → effusion
II. λ << d at high P
Many collisions during adiabatic expansion.
Tg ↓ by cooling in expansion.
Energy conservation: ΔH = Cp ΔT = ½ Muz2, where
uz = flow velocity along the beam axis. z
f(v)
cosθ (molecular flow)
supersonic beam v uz
cosn θ (n>>1)
Intense, forward-directed, monoenergetic beam.
Super-sonic beam: vs = (γRT/M)1/2, Mn = uz / vs> 1
λ = h/p = h/mv ; all emerging atoms have the same λ.
He diffraction from H-adsorbed Ni(110) surface
can be observed.
Extensively applied to the studies of molecular reaction
dynamics.
Scattered intensity θ

49. Potential energy surfaces (PES)
The reaction AB + C→ A + BC involves 3 atoms.
A full description of the motions of the system requires 9 coordinates.
A simple case: co-linear approach of C along the nuclear axis of AB.
V = V (RAB,RBC)
V obtained by quantum mechanical calculation.
3D representation of V.
Trajectory on the PES; reaction coordinate
Minimum energy path
TS = Saddle point
Ex: HA + HBHC → HAHB + HC
F + HCl → Cl + HF
RAB
RBC
A B C

50. Various trajectories in HA + HBHC → HAHB + HC
a) C1*: the minimum potential path. The relative translational energy is large enough to overcome the E-barrier. HBHC is in the vibrational ground state (v =0) → successful reaction.
b) C2*: The same total energy as in a) but HBHC is in the vibrationally excited state (v = 1) → successful reaction, producing HAHB (v=1).
c) C3*: not enough total energy to overcome the E-barrier. HBHC is in its vibrational ground state (v =0) → unsuccessful reaction.
d) C4*: not enough total energy even though HBHC is in its vibrational ly excited state (v = 1)
→ unsuccessful reaction.
The co-linear approach is the minimum energy path in the reaction A+ BC → AB + C.
Potential energy for
non-colinear approach

51. Attractive and repulsive surfaces
In A+ BC → AB + C reaction, both cases have an asymmetrical TC like A..B….C
or A....B..C.
Attractive surface
Early transition state: R ≠AB > R ≠BC.
This is the case when the bond length rAB > rBC
Ex: Cl + ClH → Cl2 + H, in which the TS is Cl….Cl..H
Endothermic reaction
A trajectory with high translational energy is attracted to the product side
A vibrationally excited Cl2 is formed.
Translational energy is more effective in driving the reaction.
RBC
RAB
attractive surface
Repulsive surface
Late transition state: R ≠AB < R ≠BC
This is the case when the bond length rAB < rBC.
Ex: H + Cl2 → HCl + Cl (the reverse reaction of the above), in which
the same TS (H..Cl….Cl) is involved.
Exothermic reaction
A trajectory with high translational energy is repelled to the reactant side.
but a trajectory with vibrational energy is attracted to the product side.
No vibrational excitation of the Cl2 product .
∆E is mainly released in the form of translational energy.
RBC
RAB
repulsive surface

52. Classical trajectories
HA + HBHC → HAHB + HC
Classical trajectory can be calculated on the PES
for a given initial conditions ( relative velocity, vibration
and quantum number).
The reaction HA + HBHC → HAHB + HC (Fig.) takes place
rapidly in less than 10 fs: direct mode process
On the other hand, in the reaction
KCl + NaBr → KBr + NaCl
the tetroatomic activated complex survives for ~ 5 ps,
during which time atoms makes 15 oscillations before
dissociate into products: complex mode process
KCl + NaBr → KBr + NaCl

53. Electron transfer in homogeneous systems
D+ A ↔ D+ + A- ; v = kr[D][A]
D = e- donor, A = e- acceptor
Many reactions in solution occurs via electron transfer; red-ox reaction.
Electrochemical reactions are e- transfer reactions in heterogeneous systems.
3 steps are involved;
1) : encounter pair formation (association)
2) : e- transfer
3) : dissociation
Applying a steady-state approximation to [DA] and [D+A-], , →
Usually, back e- transfer is a slow process; kd>> k’et
When ket >> k’a , kr ≈ ka : diffusion-controlled reaction
When ket << k’a , kr ≈ ka ket /k’a = ket KDA : e- transfer -controlled reaction →

54. From the transition state theory,
ket = (RT/po) k≠ K ≠ = κ ν≠ exp(-∆≠G/RT)
Theoretical expressions needed for ν≠ and ∆≠G.
Molecular motions in e- transfer reaction
DA (or D+A-) is surrounded by solvent molecules.
Since DA is neutral and D+A- is ion pair, the orientation of
the solvent molecules around them are different.
In e- transfer, DA and D+A- as well as the solvent molecules
have to reorient (reorganize) to a more stable new configuration.
Model for estimating ∆≠G
Solvent molecules undergo thermal motions, which leads to
a configuration suitable for e- transfer reaction.
The potential energy associated with such a change in solvent
configuration is modeled as a harmonic oscillator.
, where
R = reactant, P = product
V = ½ kq2 and ω = (k/μ)1/2, so V = ½ μω2 (q-q0)
Gm(q) = ½ NA μω2 (q-q0)

55. e- transfer by tunneling through a potential barrier,
where the parameter α is the fractional change in q.
Let ; reorganization energy L
e- transfer by tunneling through a potential barrier
Tunneling probability
; ε = E/V < 1, κ = {2m(V-E)}1/2/ħ
When κL >> 1,
ket ∝ e-βr ; 28 nm-1< β < 35 nm-1 in vacuum and β ≈ 9 nm-1
when the intervening medium is a molecular link like in liquid.
ket ∝ e-βr e-∆≠G/RT

56. Full expression for the rate constant
, where
If the coupling is weak, , where r is the edge-to-edge distance,
β is a parameter, and <HoDA> is the value of <HDA> at r = 0, i.e., A and A are in contact.
The theory assumes a large β (weak coupling, large D-A distance, no overlap
between the molecular orbitals of D and A).
Applicable to a system with a large D-A distance.
Not applicable to a system in which the D-A distance is relatively short
Ex: binuclear d-metal complexes LmM+n -B- MP+Lm
The theory also assumes that ∆≠G is surmounted only by thermal energy, so it
can be applied at high temperature.
However, observed e- transfer reactions occur at ordinary T, often via nuclear tunneling.

57. Experimental results 1. Distance dependence
Experimental test with a given DA pairs with covalently-
bonded, variable length linkage molecules.
∆≠G is constant because only the linkage molecule is changed.
ln ket = - βr + constant →
linear ln ket vs. r plot with the slope β.
2. ∆rGo dependence
Experimental test with different sets of given DA pairs
and fixed linkage ,
ket is maximum when -∆rGo = λ.
A parabolic curve in ln ket -∆rGo plot observed.
ket increases with ↓ ∆≠G but deceases in the region
(inverted region) where ∆rGo < -λ.