Foundations of Chemical Kinetics Lecture 7: Statistical treatment - - PowerPoint PPT Presentation

Foundations of Chemical Kinetics Lecture 7: Statistical treatment of equilibrium

Marc R. Roussel Department of Chemistry and Biochemistry

Equilibrium: A statistical picture

For a reaction A ⇋ B,

both together ε B A ∆ε

Equilibrium: A statistical picture (continued)

◮ There is a Boltzmann distribution for both sets of molecular

states together.

◮ By summing the probabilities of the states belonging to one of

the two chemical identities and multiplying by the total number of molecules (N), we get the average number of molecules which are of the corresponding type. P(A) = 1 Q

• a

exp

• − ǫa

kBT

• ∴ NA = N

Q

• a

exp

• − ǫa

kBT

• and

NB = N Q

• b

exp

• − ǫb

kBT

Equilibrium: A statistical picture (continued)

◮ It is tempting to conclude that the sums

a exp(−ǫa/kBT)

and

b exp(−ǫb/kBT) are the partition functions of A and B.

◮ Normally, the partition function of a molecule is computed by

setting the zero of energy at the ground state. (Think about our computation of the partition function of the harmonic oscillator.)

◮ To deal with both molecular states together, we need to add

∆ǫ0, the difference between the ground-state energies of A and B, to the energies of B. This has the effect of multiplying the partition function of B by exp(−∆ǫ0/kBT).

◮ Thus,

a exp(−ǫa/kBT) = QA and

• b exp(−ǫb/kBT) = QB exp(−∆ǫ0/kBT).

Equilibrium: A statistical picture (continued)

◮ The expected numbers of molecules of A and B are therefore

NA = NQA/Q NB = (NQB/Q) exp(−∆ǫ0/kBT)

◮ The equilibrium constant is K = NB/NA, or

K = QB QA exp

• − ∆ǫ0

kBT

General case: aA + bB ⇋ cC + dD

K = Qc

CQd D

Qa

AQb B

N−∆n exp

• −∆E0

kBT

• where

◮ ∆ǫ0 = cǫ(C)

+ dǫ(D) −

• aǫ(A)

+ bǫ(B)

• ◮ ∆n = c + d − (a + b) (difference of stoichiometric coefficients)

◮ N is the number of molecules (dimensionless). ◮ The translational partition function depends on V .

N and V are chosen to be consistent with the standard state. Example: For p◦ = 1 bar at 25 ◦C, p/RT = 40.34 mol m−3, so we could pick V = 1 m3 and N = 40.34 × 6.022 × 1023 = 2.429 × 1025.