WOLFRAM|DEMONSTRATIONS PROJECT

Simplified Statistical Model for Equilibrium Constant

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temperature T (K)
300
energy ΔE (kJ/mol)
-5
energy
Δϵ
A
(kJ/mol)
1
energy
Δϵ
B
(kJ/mol)
0.75
Consider a simple chemical equilibrium
A⇌B
with equilibrium constant
K
eq
=
N
B
N
A
. (This can alternatively be written
K
c
=
[B]
[A]
in terms of the concentrations of
A
and
B
.) The difference in electronic energy for the reaction
A→B
equals
ΔE
, conveniently expressed in kJ/mol. Let the internal structure of each molecule be idealized as a series of equally spaced energy levels (similar to those of a harmonic oscillator), with the energy increments
Δ
ϵ
A
and
Δ
ϵ
B
. The spacings
Δ
ϵ
A
and
Δ
ϵ
B
relative to
ΔE
are exaggerated in the graphic for easier visualization. The sublevels of each molecular species are assumed to occupy a Boltzmann distribution at temperature
T
. Accordingly,
N
A
(n)=
N
A
-nΔ
ϵ
A
/kT
e
q
A
(T)
, where
q
A
(T)=
∞
∑
n=0
-nΔ
ϵ
A
/kT
e
=
-1
(1-
-Δ
ϵ
A
/kT
e
)
, the molecular partition function for
A
, and analogously for
B
. For a mixture of
A
and
B
, a single Boltzmann distribution can be considered to apply for the composite levels of both molecules. This leads to the formula for equilibrium constant in statistical thermodynamics:
K
eq
=
q
B
q
A
-ΔE/RT
e
.
Under constant volume conditions, the equilibrium constant is related to the change in Helmholtz free energy:
ΔA=ΔE-TΔS=-RTln
K
eq
. An exothermic reaction, with
ΔE<0
, tends to give
K
eq
>1
, implying that the forward reaction
AB
is favored. This can be reversed, however, with the endothermic reaction
BA
favored if the entropy change
ΔS
is sufficiently negative. This could result if species
A
has a greater number of thermally accessible levels at the given temperature.