Page 63 - Chemical equilibria Volume 4

Let us first write the synthesis reactions for the three components present
in the system: Properties of States of Physico-Chemical Equilibrium 39
=
+
C4H CH 4
=
+
C2O CO
2
+
=
CO CO
Algebraically, by linear combination, we eliminate the monoxygen,
which is absent from the system (whereas the element carbon is present in
the form of graphite). In order to do so, we consider the last two equations,
and by multiplying the last equation by 2, we obtain:
2CO = CO + C
2
We find ourselves with two reactions:

+
=
C4H CH 4
2CO CO + C
=
2
The first reaction involves the element hydrogen, which is neither present
in the system nor algebraically eliminable, so this equation is eliminated. As
no chemical species is present in more than one phase at a time, there is no
phase transfer to be taken into account. Ultimately, the balance equation:

2CO = CO + C
2

constitutes a base for our vector space, which is one-dimensional. There is
only one independent equilibrium in our system.

NOTE 2.2.– dioxygen can be ignored in certain experimental conditions and
in view of certain problems. If experience undermines this hypothesis, then
we must take this into consideration, as well as its synthesis reaction. The
system then becomes two-dimensional with two independent reactions.

Ionic aqueous solutions – in the case of ionic solutions, we first draw up
a list of all the species (ions and neutral molecules) present in the solution.
Then, we write Bronsted’s weak-acid–weak-base equilibria and the redox
reactions, the complexation reactions and finally the reactions of
precipitation of solid phases.