The above is a general principle that can be extended to other
concentration units and to liquid solutions, ideal or not. In
non-ideal systems we could replace
and follow the same procedure as above. In stead of an
expression for K involving pressures or concentrations it now read
in activities:
activities of pure condensed
phases
Sometimes one of the reactants or products is a pure solid
(precipitate) or liquid (more solvent e.g.). What activity should
we assign in such a case?
We start by choosing a suitable standard state, say the pure
compound at 1 bar and temperature of interest, we then have:
- μ = μ
o
but also:
- μ = μ
o +RTlna
So a=1 at standard conditions
Any change can be written as
- dμ = RTdlna
We can study the pressure dependence by considering:
- ∂ μ /∂P |T = Vpartial molar (
Vbar)
For a solid or liquid Vbar is a relatively small and
constant value. Thus we can write:
- dμ = VbardP
- RTdlna = VbardP
- dlna = VbardP/RT
Upon integration to a different pressure P' we find
lna' = (P'-1) Vbar/RT
Example 26-12 shows that for graphite the activity is only 1.01
at 100 bars, so the activity is not very pressure dependent.
Mostly if pure condensed compounds are involved in reactions the
activity can taken as unity.
This is also in line with what we said previously about the
solvent following Raoult's law. In the limit of the solvent going
to pure solvent we have that its P goes to P*. As the activity is
defined as P/P* this converges to unity. If a reaction produces
more solvent molecules we can usually consider their activity equal
to one in very good approximation for dilute solutions, even if
they are already non-ideal.
The fact that a=1 for pure condensed phases has an important
consequence for reactions (in general: processes) that only involve
such phases. If all activities are unity, Q=1 and lnQ=0 which means
that ΔrG = ΔrGo + 0. Thus
ΔrG can only be zero -i.e. an equilibrium achieved-
if ΔrGo happens to be
zero, which is generally not the case. In fact there can only be an
equilbrium at one specific temperature:
- ΔrG
o = 0
-
ΔrH
o-TΔrSo
- Tequilibrium=
ΔrH
o/ΔrSo
If the process is the transformation from a solid to a liquid
this is the well-known melting point. At temperature other
than 0oC only one phase can exist: either ice or water.
If the other is present, that is an unstable condition and it will
transform entirely to the stable form. In other words the
process will go to completion, not equilbirium. Only
at 0oC can the two coexist in equilibrium. This
holds for all melting points but it also holds for e.g. a
solid-solid chemical reactions only producing, say, another
solid.
Another way of expressing the above is to say that in order to
have equilibrium at a series of temperatures, one needs at
least one species involved for which activity depends on
composition, e.g. a dilute solute or a gas.