Absolute Rate Theory

Absolute Rate Theory
Learning Objective
Recognize the role of activated complex in controlling the rate of a
chemical reaction. Use two-dimensional potential energy surfaces to
deduce the expression for the absolute rate constant of a bimolecular
elementary reaction (Eyring equation). Class will estimate the
frequency factor for the reaction of hydrogen and iodine to form
hydrogen iodide using the partition functions and avagadro’s number.
Theory of absolute reaction rates!
Strategy:
(1)  Assume that the reactants are in equilibrium with the activated complex (AC)
(2)  Assume that AC has equal probability of going to products or to reactants
(3)  Use partition functions to calculate the equilibrium constant
(4)  Use the frequency of the fatal vibration as the upper limit for the reaction rate
TS refers to the state of energy, while
AC refers to the mol. structure of TS state
AC Endothermic rxn H
B
A
∆H
From Ken Dill
Equilibrium Approximation
Reactants are assumed to be in equilibrium with the
activated complex! AC can go to products or reactants.
K‡
A+B
[AC]
Equilibrium constant =
k
Products
C‡/Co
= K‡
CA /Co CB /Co
(K should be dimensionless)
∴ [AC] = C ‡ =
K‡
Co
AC H
B
A
C = mol/m3
Co = 1000 mol/m3
CA CB
∂(ξ/v)
K‡
‡
=
ν
C
=
ν
CA CB
Rate =
∂t
C0
ν is the vibration that snaps AC to products!
But d(ξ/v) = k C A C B
dt
∴k=ν
K‡
C0
∆H
Calculate the equilibrium constant and the vibrational frequency
that leads AC to the products!
Molecular Partition Functions and Equilibrium Constant
Partition functions describe how the energy is distributed among different degrees
of freedom of the molecule.
Vibrational partition function:
fv = Σ(e - (s+0.5)hν/kT)
The equilibrium constant in terms of the partition functions is
K = No
∏[qproducts/V]
∏[qreactants/V]
K = equilibrium constant
No = number of molecules per m3
q = the partition function
V = volume of the system
E
q
but v = f e -εo/kT
f = partition function with respect to the zero point energy
ε0 = Zero point energy of vibration along the rxn coordinate
ZPE
Activation Energy and the Fatal Vibration
(q‡/v)
K‡ = N
(qA/v)(qB /v)
o
∆ε‡o
No f‡ - εo - εoA - εo B /kbT
∴K‡ =
e ‡
fA fB
∆E‡o = NA(ε‡ –εA-εB)
f represents partition function (q/V) in units of ε‡o
K‡ = [No f‡ /(fA fB)] e –(∆E‡o/RT)
Replaced ZPE with ∆E‡o, and R= NAkb
Factoring out the vibrational partition function of the AC that
corresponds to the vibration along the reaction coordinate, we have:
f‡ = fv f‡ = (kbT/hν) f‡
(see left for explanations)
K‡ = [No f‡ (kT/hν) /(fA fB)]e –(∆E‡o/RT)
fv = Σ(e - (s+0.5)hν/kT)
= e – 0.5 hv/kT Σ(e - s hν/kT)
= e - hv/2kT (1/ [1- e - hν/kT])
= e - hv/2kT/ [1-(1- hν/kbT)]
= 1* (kbT /hν)
f‡ = fv f‡ = (kbT/hν) f‡
∑ xi = (1/(1-x))
ex = (1+ x)
e - hv/2kT = 1
Eyring Equation, Activation Energy and the Frequency factor
A+BC -> AB+C
K‡ = [(kT/hν) No] [f‡ /(fA fB)] e –(∆E‡o/RT)
The maximum rate for the reaction is when each
vibration along the Rxn coordinate leads to the
product, then
k=ν
K‡
C0
(slide 3)
k = (kT/h)(ν/ν) [No /Co] [f‡ /(fA fB)] e –(∆E‡o/RT)
k = (kT/h) [No /Co] [f‡ /(fA fB)] e –(∆E‡o/RT)
Eyring equation
Frequency factor
Activation energy is the
differences in the zero point
energies of reactants and AC!
From Ken Dill
Summary thus far!
1.  Absolute rate theory incorporates all degrees of freedom (geometries)
2.  Equilibrium is assumed between the reactants and AC, and partition
functions used to calculate the equilibrium constant.
3.  Recognition of the fatal vibration gives the maximum rate constant value.
4.  Factoring out the fatal vibration partition function allows us to calculate
the rate constant of a reaction from first principles.
5.  Frequency factor calculated from first principles.
6.  Activation energy calculated from QM calculations, if structure of the
activated complex is known or postulated.
7.  Physical insight given to describe the activation energy and fatal vibration.
Additional Insight: Entropy of Activation
Let
a + b + c +...
[AC]
Products
∆ν = νproducts -νreactants
k = kbT co Δν K‡
h
k = kbT co Δν e(ΔS‡/R) e(-ΔH‡/RT) Since K ‡ = e -ΔG‡/RT
h
A = kbT co Δν e(ΔS‡/R)
h
A is large when ΔS‡ is > 0
and ΔG = ΔH - TΔS
A is small when ΔS‡ is < 0
If the structure of [AC] resembles products very closely in terms of structure, then
ΔSo = ΔS‡ then we have a late transition state.
E
E
R
p
Early
R
Late TS
P
Comparison with the Collision theory
A+B
[AB]‡
Products
Let A and B be linear molecules with nA and nB atoms
fA = f3t f3r f3VnA - 6
∴ k = kT NA
h
f‡ = f3t f3r f (nA + nB) - 7
V
3
fB = f3t f3r fVnB - 6
3
f3t f3r fV(nA + nB) -7
3
3
f3t f3rfVnA - 6f3t f3r fϑnB - 6
3
e
- (ΔE‡ /RT)
o
5 e - (ΔE‡ /RT)
N
f
= kT A V3 3
h
ft fr
Collision theory considers “molecules” as atoms,
∴ fA = fB = ft3
f‡ = ft3 fr2
kcollision
3 2
f
kT
=
NA t3 fr3
h
ft ft
5
fV
∴k=
k
(fr) collision
‡
e - (ΔEo /RT)
o
In-Class Assignment:
Using collision theory, estimate the frequency factor for the
bimolecular decomposition of HI to H2 and I2.
See answer on the next slide
Example: H2 + I2
2HI
What is the value of the frequency factor ?
fI2 = fH2 = ft3 f2r fϑ
H
I
H
I
3 translational
3 rotational
3n - 6 vibrational modes = 6
[AC] =
(one vibration responsible for product formation is to be excluded)
∴ f‡ = f3t f3r f5v
∴ A = kT
h
NA f3t f3r f5V
f3t f2r f1V f3t f2r f1V
NAf3V
kT
= . 3
h ft fr
At room temperature ft = 1010/m; fr =10; fV= 1; kT/h = 1013/s
∴A=
1013s -1
6 * 1023 mol-1 = 106 m3
mol s
(1010m-1)310 .1
observed values for bimolecular reactions ~ 106 to 8 (m3/mol.s)
considering the approximate values used, the rough description of
[AC], the calculated value is very good!! This theory is valid for all
elementary reactions, unlike the collision theory.

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