Page 163 - Chemical equilibria Volume 4

Determination of the Values Associated with Reactions – Equilibrium Calculations 139
We can see that if we consider the two reactions:
– synthesis of water, given by:
H 2 + ½ O 2 = H 2O [4R.11]
– and combustion of carbon monoxide, by:

CO + ½ O 2 = CO 2 [4R.12]
the balance equation of the reaction in question is the difference between the
balance equations for reactions [4R.11] and [4R.12]. Thus, we have:

[4R.10]= [4R.11]– [4R.12]

Let us apply the same linear combination to the standard Gibbs energies,
and thus:

Δ 10 0 Δ g = 11 0 Δ g − 12 g 0

The application of relation [4.2] immediately gives us the relation
between the sought equilibrium constant and the known equilibrium
constants:

() I
() I
() I
K 10 = K 11 / K 12
() I
() I
Hence, if K and K are known at the desired temperature, we can
12
11
() I
easily calculate K .
10
Ultimately, the solving of all problems of determination of the
equilibrium constants (or the standard Gibbs energies) relies on our
knowing:
– the standard enthalpy at a chosen temperature;
– the standard entropy at the same temperature;

– the variations of the molar specific heat capacities as a function of the
temperature.

We have seen the advantage of going into greater detail about the
properties and methods for measuring or calculating those values in sections
4.2, 4.3 and 4.4.