Foundations of Chemical Kinetics Lecture 17: Unimolecular reactions - - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 17: Unimolecular reactions in the gas phase: Lindemann-Hinshelwood theory

Marc R. Roussel Department of Chemistry and Biochemistry

The factorial

◮ The number n(n − 1)(n − 2) . . . 1 is called the factorial of n.

Notation: n! (read “n factorial”)

◮ By convention, 0! = 1.

Marc’s notation vs the textbook’s

◮ Albert Goldbeter once told me that you knew that a student

was taking ownership of their project when they wanted to change the notation. . . Quantity Textbook Marc Sum (number) of states G G Density of states N g

Density of states for s harmonic oscillators

◮ In lecture 6, we derived the following expressions for the sum

and density of states of a harmonic oscillator: G(ǫ) = ǫ(ω0)−1 g(ǫ) = (ω0)−1

◮ Recall: Roughly speaking, the partition function counts the

number of states with energies below kBT. Therefore, Q ≈ kBT ω0 Note: You can also derive this equation from the harmonic

• scillator partition function by assuming that ω0/kBT is

small, as we did in lecture 12.

Density of states for s harmonic oscillators

(continued)

◮ If we have s distinguishable, independent harmonic oscillators

whose natural frequencies are ωi, the partition function should therefore be Qs ≈

s

• i=1

kBT ωi

◮ From the definition of the classical partition function, we have

Qs = ∞ gs(E) exp

• − E

kBT

• dE

Density of states for s harmonic oscillators

(continued) Qs = ∞ gs(E) exp

• − E

kBT

• dE ≈

s

• i=1

kBT ωi

◮ The problem now is to find the density of states corresponding

to our partition function Qs. This problem turns out to be solved by taking a mathematical

• peration called an inverse Laplace transform of Qs.

The answer is gs(E) = E s−1 (s − 1)! s

i=1 ωi

Hinshelwood theory

◮ Recall that a collision-theory treatment badly underestimates

the Lindemann rate constant k1.

◮ Hinshelwood’s idea was that the energy acquired in a collision

can be stored in any of the bonds in a molecule, and that this therefore introduces a statistical factor (the degeneracy of the corresponding total vibrational energy) into the calculation of the rate constant.

◮ A classical treatment, assuming that the temperature is

sufficiently high that we can treat the vibrational levels as continuous, will use the density of states rather than the degeneracy.

Hinshelwood theory (continued)

◮ For simplicity, Hinshelwood assumed that the s vibrational

modes of a molecule had a common vibrational frequency ω0. Then, Qs ≈ kBT ω0 s gs(E) ≈ E s−1 (s − 1)! (ω0)s

◮ The probability that a molecule has vibrational energy

between E and E + dE is thus gs(E) Qs exp

• − E

kBT

• dE =

E s−1 (kBT)s(s − 1)! exp

• − E

kBT

• dE

Hinshelwood theory (continued)

◮ The probability that a molecule has energy greater than Ea is

therefore ∞

Ea

E s−1 (kBT)s(s − 1)! exp

• − E

kBT

• dE

◮ This integral gives Γ(s, Ea/kBT)/(s − 1)!, where Γ() is the

incomplete gamma function.

◮ Typically, Ea ≫ kBT. In this case, the integral is well

approximated by 1 (s − 1)! Ea kBT s−1 exp

• − Ea

kBT

Hinshelwood theory (continued)

◮ Assuming a collision-limited rate, the rate constant k1 is

therefore k1 = Act Pr(E > Ea) = Act (s − 1)! Ea kBT s−1 exp

• − Ea

kBT

• ◮ Since Ea/kBT ≫ 1,

1 (s − 1)! Ea kBT s−1 ≫ 1, which explains why collision theory fails so badly for some unimolecular reactions.

Hinshelwood theory

Summary and comparison to experiment

◮ Explains why k1 is larger than the collision-limited value: the

vibrational degeneracy allows molecules to store the same amount of energy in many different ways, introducing a statistical factor into the theory.

◮ We assume s oscillators with equal vibrational frequencies.

In practice, we treat s as a parameter which we choose to get the best fit to the data. Typically we find that s is about half of the normal modes of the reactant.

◮ Hinshelwood theory fits the pressure dependence of the

• bserved rate constant better than plain Lindemann theory.

However, there are still deviations at low pressures.

◮ Because of the strongly T-dependent preexponential factor,

Arrhenius plots for k1 (or k∞) should be curved. They are not.