Concise Physical Chemistry

c21 JWBS043-Rogers September 13, 2010 11:30 Printer Name: Yet to Come

THE STEADY-STATE PSEUDO-EQUILIBRIUM 353

21.8 THE STEADY-STATE PSEUDO-EQUILIBRIUM

The enthalpy difference between the reactant state and the activated complex is not

a legitimate equilibrium in the thermodynamic sense; but it is useful to treat it as

though it were. It is best called apseudo-equilibriumorsteady-state equilibrium.

Suppose a second-order reaction proceeds through an activated complex [AB]*:

A+B=[AB]∗→products

The activated complex is controlled by a pseudo-equilibrium constant:

K∗=

[AB]∗

[A][B]

If the reaction to form products is a first-order breakup of the activated complex, we

obtain

rate=−

dA

dt

=k 1 [AB]∗

The concentration of activated complex through its equilibrium constant is

[AB]∗=K∗[A][B]

whence

−

d[A]

dt

=k 1 K∗[A][B]

This equation explains the observation of second-order overall kinetics for the re-

action, even though the rate constant for decomposition of the activated complex is

first order. According to this mechanism,k 1 K* is the observed second-order rate

constant.

The incoming and outgoing species of the activated complex may be weakly

bound. If so, we may regard its breakup as a kind of one-cycle molecular vibration.

The complex stretches along its weakest bond, and then it falls apart to give either

products or the original reactants. The vibrational energy acquired during this one-

dimensional, one-cycle stretch isE=kBTbecause there is one degree of vibrational

freedom contributing^12 kBTto the kinetic energy and one contributing^12 kBTto the

potential energy. It is also true for a vibration thatE=hv, hence

kBT=hv ν=

kBT

h

=

RT

NAh

whereNAis the Avogadro number.

In the limit of an activated complex that always breaks up to give products, the

number of vibrations taking place per second is the number of activated complexes