Page 83 - Chemical equilibria Volume 4

Thus, we find the sought relation:
∑ i ν Molecular Chemical Equilibria 59
P
P
K () = K (I) ⎛ ⎜ 0 ⎟ i ⎞ [3.10]
⎝ P ⎠
Using expressions [3.2] and [3.10], the equilibrium constant relative to
the partial pressures takes the form:
− ∑ i ν o
⎛ P ⎞ A
P
K () = ⎜ 0 ⎟ i exp r [3.11]
⎝ P ⎠ RT
NOTE 3.3.– If the system is studied at a total pressure equal to the reference
0
P
pressure P (normally 1 bar), the two constants K () and K are identical.
If we now choose to quantify the composition of the gaseous phase in
terms of the concentrations, we write the following for a perfect gas:

P = C k RT [3.12]
k
By substituting this back into relation [3.7], we obtain:

⎛ ⎜∏ C ⎞ i k ν (RT i ν ∑ = K () = (RT i ν ∑ K ( ) [3.13]
c
P
i ⎝ C i ⎠ 0 ⎟ ) i ) i

From relation [3.13], we deduce that the equilibrium constant relative to
the concentrations is linked to the equilibrium constant relative to the molar
fractions by:

⎛ C ⎞ k ν − i ν ⎛∑ P ∑ i ν
i ⎞
c
K () = ⎜∏ i 0 ⎟ = (RT ) i ⎜ 0 ⎟ K (I) [3.14]
i ⎝ C i ⎠ ⎝ P ⎠
Very often, the reference concentration C is taken as 1 mole per liter for
0
i
each component, and if the concentrations are expressed in moles per liter,
the law of mass action takes the form:

C
K () c = ∏ ( ) i ν [3.15]
i
i