Foundations of Chemical Kinetics Lecture 6: Further studies of the - - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 6: Further studies of the Boltzmann distribution

Marc R. Roussel Department of Chemistry and Biochemistry

The Boltzmann distribution in classical mechanics

◮ One of the main differences between classical and quantum

mechanics is that energy is not quantized in the former theory.

◮ Accordingly, in classical mechanics, it doesn’t make sense to

ask for the probability that a molecule has exactly energy ǫ. The answer to this question is zero.

◮ The correct question to ask is what is the probability that a

molecule has energy between two specified limits, i.e. what is P(ǫlow ≤ ǫ ≤ ǫhigh) for some specified values of the two limits?

The Boltzmann distribution in classical mechanics

(continued)

◮ Define the density of states g(ǫ) such that g(ǫ) dǫ is the

number of states between energies ǫ and ǫ + dǫ.

◮ Let ǫo be the result of an observation of ǫ. Then,

P(ǫ ≤ ǫo ≤ ǫ + dǫ) = g(ǫ) exp

• −

ǫ kBT

• dǫ/Q
• r

P(ǫlow ≤ ǫo ≤ ǫhigh) = 1 Q ǫhigh

ǫlow

g(ǫ) exp

• −

ǫ kBT

• dǫ

Classical partition function

◮ Let A be the set of allowed energies. This will normally be an

interval of energies, possibly semi-infinite.

◮ Using very similar reasoning to that used to obtain the

quantum partition function, we get Q =

• A

g(ǫ) exp

• −

ǫ kBT

• dǫ

Example: harmonic oscillator density of states

A “calculation” based on the quantum-mechanical energy

◮ For the quantum-mechanical harmonic oscillator, the energy

levels are equally spaced by ω0.

◮ Provided ∆ǫ ≫ ω0, the number (∆G) of energy levels in an

interval of size ∆ǫ is therefore ∆G ≈ ∆ǫ ω0

◮ By definition, the density of states is g(ǫ) = dG/dǫ.

Rearrange the last equation and take a “physicist’s limit” ∆ǫ → 0: ∆G ∆ǫ = 1 ω0 ∴ g(ǫ) = lim

∆ǫ→0

∆G ∆ǫ = dG dǫ = 1 ω0

The Arrhenius equation

reactant product εa V x) x (

specific rate of reaction = (probability that ǫ > ǫa) × (specific rate of crossing if ǫ > ǫa)

The Arrhenius equation (continued)

◮ If the barrier is high enough, there will be many states below

Ea, so energy can be treated approximately as a continuous variable. P(ǫ > ǫa) = ∞

ǫa g(ǫ) exp(−ǫ/kBT) dǫ

∞

0 g(ǫ) exp(−ǫ/kBT) dǫ

◮ Suppose that there is one, roughly constant, density of states

below the top of the barrier and another above, i.e. that g(ǫ) = gb for ǫ < ǫa ga for ǫ > ǫa

The Arrhenius equation (continued)

◮ Then

P(ǫ > ǫa) = ga ∞

ǫa exp(−ǫ/kBT) dǫ

gb ǫa

0 exp(−ǫ/kBT) dǫ + ga

∞

ǫa exp(−ǫ/kBT) dǫ

◮ For a high barrier, very few states above the barrier will be

populated compared to the number of states in the reactant

• well. Thus,

P(ǫ > ǫa) ≈ ga ∞

ǫa exp(−ǫ/kBT) dǫ

gb ǫa

0 exp(−ǫ/kBT) dǫ

◮ For the same reason, we make only a small error by extending

the range of integration in the denominator to infinity: P(ǫ > ǫa) ≈ ga ∞

ǫa exp(−ǫ/kBT) dǫ

gb ∞

0 exp(−ǫ/kBT) dǫ

The Arrhenius equation (continued)

P(ǫ > ǫa) ≈ ga ∞

ǫa exp(−ǫ/kBT) dǫ

gb ∞

0 exp(−ǫ/kBT) dǫ

= − gakBT exp(−ǫ/kBT)|∞

ǫa

− gbkBT exp(−ǫ/kBT)|∞ = ga gb exp(−ǫa/kBT) = ga gb exp(−Ea/RT)

◮ Suppose that the specific rate at which molecules with

sufficient energy cross the barrier is r. Then k = r ga gb exp(−Ea/RT) = A exp(−Ea/RT)

The distribution of velocities

◮ The Boltzmann distribution ought to apply to kinetic energy:

K = 1 2mv2 = m 2

• v2

x + v2 y + v2 z

• ◮ Sum of terms: apply Boltzmann distribution to each term

independently. p(vx) dvx = 1 Q exp −mv2

x

2kBT

• dvx

∴ Q = ∞

−∞

exp −mv2

x

2kBT

• dvx

=

• 2πkBT/m

∴ p(vx) dvx =

• m

2πkBT exp −mv2

x

2kBT

• dvx

The distribution of velocities (continued)

p(vx, vy, vz) dvx dvy dvz =

• m

2πkBT 3/2 exp

• −m(v2

x + v2 y + v2 z )

2kBT

• dvx dvy dvz

This gives the probability density at a point in the velocity space, i.e. near particular values of (vx, vy, vz).

The distribution of molecular speeds

◮ Speed v related to velocity components by v2 = v2

x + v2 y + v2 z

◮ The same speed is obtained at every point on a sphere

satisfying this equation.

◮ dvx dvy dvz is a volume element. Integrated over the surface

• f a sphere of radius v, it gives 4πv2 dv.

◮ Distribution of speeds:

p(v) dv = 4πv2

• m

2πkBT 3/2 exp −mv2 2kBT

• dv
• r

p(v) dv = 4πv2

• M

2πRT 3/2 exp −Mv2 2RT

• dv

The distribution of molecular speeds

0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 200 400 600 800 1000 p(v)/s m-1 v/m s-1 T = 100 K T = 200 K T = 300 K

Average speed

◮ Given a probability density p(x), the average of f (x) is given

by

• A

f (x) p(x) dx where A is the region over which x is defined.

◮ For example, the average speed is

¯ v = ∞ v p(v) dv = ∞ 4πv3

• m

2πkBT 3/2 exp −mv2 2kBT

• dv

=

• 8kBT

πm