Page 239 - Physical chemistry understanding our chemical world

206 PHASE EQUILIBRIA

The value of G is only ever negative, as required by a thermodynamically sponta-
neous process, if the initial pressure p (initial) is greater than the final pressure p (final) ,
i.e. the fraction is less than one. In other words, Equation (5.9) shows why G is
negative only if the pressure of the CO 2 in the space above the liquid has a pressure
O
that is greater than p .
We disrupted the equilibrium in the bottle when we allowed out much of the CO 2
gas that formerly resided within the space above the liquid; conversely, the CO 2
dissolved in the liquid remains in solution.
After drinking a mouthful of the drink, we screw on the bottle top to stop any
more CO 2 being lost, and come back to the bottle later when a thirst returns. The
CO 2 re-equilibrates rapidly, with some of the CO 2 in the liquid phase passing to the
gaseous phase. Movement of CO 2 occurs in order to maintain the constant value of
K (partition) : we call it ‘re-partitioning’.
Although the value of K (partition) does not alter, the amount of
A fizzy drink goes ‘flat’ CO 2 in each of the phases has decreased because some of the CO 2
after opening it sev- was lost on opening the bottle. The liquid, therefore, contains less
eral times because the CO 2 than before, which is why it is perceived to be less fizzy. And
water is depleted of after opening the bottle several times, and losing gaseous CO 2 each
CO 2 . time, the overall amount of CO 2 in the liquid is so depleted that the
drink no longer sparkles, which is when we say it has ‘gone flat’.

Worked Example 5.4 A bottle of fizzy pop contains CO 2 . What are the relative amounts
of CO 2 in the water and air if K (partition) = 4?

Firstly, we need to note that stating a value of K (partition) is useless unless we know how
the equilibrium constant K (partition) was written, i.e. which of the phases ‘1’ and ‘2’ in
Equation (5.8) is the air and which is the liquid?
In fact, most of the CO 2 resides in the liquid, so Equation (5.8) would be written as

concentration of CO 2 in the drink
K (partition) =
concentration of CO 2 in the air above the liquid

This partition constant has a value of 4, which means that four times
as much CO 2 resides in the drink as in the liquid of the space above
A bottle of fizzy drink the drink. Stated another way, four-fifths of the CO 2 is in the gas
going flat is a fairly triv- phase and one-fifth is in solution (in the drink).
ial example of partition,
but the principle is vital
to processes such as SAQ 5.5 An aqueous solution of sucrose is prepared. It
reactions in two-phase is shaken with an equal volume of pure chloroform. The
media or the operation two solutions do not mix. The sucrose partitions between
of a high-performance the two solutions, and is more soluble in the water. The
liquid chromatography value of K (partition) for this water–chloroform system is 5.3.
column.
What percentage of the sucrose resides in the chloro-
form?