I don't remember doing this proof
Mao, but I do remember using a really cumbersome method to prove some acidity constant question. (Then finding an elegant method
)
But yes, using the theory of equilibrium, we can prove this with mathematics
(which you don't need to do or know for VCE Chemistry!):
Take a reaction:
with
Suppose that the initial concentrations (at equilibrium) are:
,
, and
, where
.
Now, half the volume so that every concentration is doubled. Now,
The system is now thrown into dis-equilibrum, and
must decrease in order to return to
.
Thus,
and
must decrease and
must increase (via the backward reaction of the chemical equation above).
Following the reaction, the system will re-establish the equilibrium as follows:
,
,
,
where
(2c-x)}{2a+x})
Since
:
(2c-x)}{2a+x} = \frac{bc}{a})
(2c-x)}{bc} = \frac{2a+x}{a})
(***)
 \left(\frac{c}{a}\right))
Since
:
 \left(\frac{c}{a}\right) > \left(\frac{2a}{2c}\right) \left(\frac{c}{a}\right) = 1)
Similar working from line
(***) can be done to conclude
as well.
This proves that the final equilibrium
and
is greater than the initial equilibrium
and
, despite the principles of equilibrium working to reduce both
and
from
and
respectively.
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