This document addresses the question of how the Gibbs Free Energy of an equilibrating chemical system approaches its minimum value. This minimum is achieved at constant T and constant P. The extent of reaction adjusts itself until the derivative of the Gibbs Free Energy with respect to the extent of reaction is zero. This defines the position of the minimum in the Gibbs Free Energy (technically an extremum). The example chosen concerns dinitrogentetroxide decomposition, and complements the discussion in J. Chem. Ed., 65,4071988. For the reaction
the molar Gibbs free energy of the two components are
so
, the Gibbs Free Energy of a mixture of
moles of
and
moles of
would be
We define the extent of reaction,
, as
and
where the denominators are the stoichiometric
coëfficients
taken from the balanced chemical equation.
These equations can be inverted to solve for the number of moles of
each component as a function of the starting number of moles of that
component and the extent of reaction.
One has from Equation 4
and, from Equation 5, one has
so, the mixture's Gibbs Free Energy must be (substituting Equations 6 and 7 inot Equation 3)
whicis
We wish to take the derivative of
with respect
to
, with the intent of setting the result equal to zero,
searching for an extremum in
(which we suspect is
a minimum).
Wishing to do this at constant pressure, we need to write each partial
pressure in terms of the total pressure and the mole fraction, using
Dalton's Law.
We have:
where we recognize that each mole fraction is itself a function of
the extent of reaction,
.
From here, the calculus becomes a bit messy, but the result is worth it.
What we know is that
and
which is a little harder than the equimolar cases where
is
zero.
We will do this work in parts, so that the differentiation can be
explicitly followed, line by line.
First, we attempt taking ther derivative of
with
respect to
, i.e.,
where we need to just evaluate the remaining partial derivatives.
and
so, doing the dirty deed, we have
Bringing these two equations separately over a common denominator, one has
which become, again sequentiually,
Expanding into Equation 12
which is, upon substitution of Equations 19 and 20 yields,
which is
Q.E.D.