ABSTRACT
This appendix shows how the various activity coefficients are related. Imagine that the chemical potential for a compound in two different states is denoted by m and m0. The
difference mm0 should be independent of the choice of the reference function, i.e., whether it is mole fraction, molarity, or molality. If we denote the chemical potential on the basis of mole fraction as mX and the activity as aX the following relations hold:
m ¼ moX þ RT ln aX (C:1) m0 ¼ moX þ RT ln a0X (C:2)
So that:
m m0 ¼ RT ln aX a0X
(C:3)
The same exercise for molality expressed as mm and am gives:
m ¼ mom þ RT ln am (C:4) m0 ¼ mom þ RT ln a0m (C:5)
m m0 ¼ RT ln am a0m
(C:6)
aX a0X ¼ am
a0m (C:7)
This last equation shows that the activity ratios do not depend on the chosen reference function. Now, we choose as the state indicated by the prime (0) the state at infinite dilution. Using Equations 3.62 and 3.65 it follows that:
fX f 0X0
¼ gm g0m0
(C:8)
Rearrangement leads to:
g ¼ f g 0
f 0 m0
X0 X m
(C:9)
Of course, the mole fraction X and molality m are related in a straightforward way (see also Appendix B):
X ¼ m mþ 1000
Ms
¼ Msm Msmþ 1000 (C:10)
X m ¼ 1
mþ 1000 Ms
(C:11)
In this equation,Ms is the molecular weight of the solvent. In the state of infinite dilution when m! 0, it follows that:
X0
m0 ¼ 11000
Ms
¼ Ms 1000
(C:12)
Combining this with Equation C.9 and remembering that in the state of infinite dilution g 0 ! 1 and f! 1 the result is
g ¼ f 1000 Ms
mþ 1000 Ms
¼ f 1000 Ms mþ 1000Ms
¼ f 1000 Msmþ 1000 (C:13)
This can be rearranged into:
f g ¼ 1þ 0:001Msm (C:14)
This equation thus shows how the two activity coefficients relate to each other. A similar expression can be derived for the relation between f and y, the activity coefficient based on concentration, but now there is an extra dependency on density of the solution (see also Appendix B):
f y ¼ 1 0:001ci Mi Ms
rsln (C:15)
ci and Mi are the concentration in mol dm
3 and the molecular weight of the solute, respectively rsln is the density of the solution in g cm
The relation between y and g can be derived as follows. The chemical potential expressed for molarity is
m ¼ m0c þ RT ln yc (C:16)
and the one expressed for molality is
m ¼ m0m þ RT ln gm (C:17)
Since the chemical potential does not depend on the choice of concentration unit, it follows that:
m0c m0m ¼ RT ln gm RT ln yc ¼ RT ln g RT ln y þ RT ln m c
(C:18)
At infinite dilution, g! 1 and y! 1 while m=c! 1=rs (where rs is the density of the pure solvent in kg dm3). From this behavior at infinite dilution, it follows that the difference in standard states for the two scales is
m0c m0m ¼ RT ln 1 rs
(C:19)
Combining this with Equations C.16 and C.17 results in the relation we are looking for:
g
y ¼ c rsm
(C:20)
Gibbs-Duhem equation to relate activity coefficients For a two-component system the Gibbs-Duhem equation is
n1dm1 þ n2dm2 ¼ 0 X1dm1 ¼ X2dm2
(C:21)
X1 and X2 are the mole fractions, and X1þX2¼ 1. The chemical potential can be written as
m1 ¼ m01 þ RT ln a1 ¼ m01 þ RT lnX1 þ ln g1 (C:22)
A similar equation can be written for component 2. Differentiating gives:
dm1 ¼ dm01 þ RT(d lnX1 þ d ln g1) ¼ RT dX1 X1
þ RTd lng1 (C:23)
realizing that dm0¼ 0 because it is a constant.
