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158     Chemical Equilibria
                           A1.2.6. Relation between conventions (II) and (III)
                             Let us examine the relation between the activity coefficients expressed on
                           the basis of the dilute solution reference (II), and those stemming from the
                           molar solution (convention (III)).
                             Let us write that the chemical potential does not depend on the chosen
                           reference state, and taking account of relation [A1.5], we find:

                                 μ s  = μ 0(III)  + Rlnγ s (III) C ≅ μ s ∞  + R lnγ (II) x s  [A1.15]
                                             T
                                                               T
                                      s
                                                      s
                                                                    s
                             Thus, we can identify and write:
                                 μ 0(III)  = μ s ∞                                      [A1.16]
                                  s
                             We find:

                                γ s (III) C =γ (II) x s                                 [A1.17]
                                      s
                                          s
                             Thus, the chemical potential in infinite dilution is  equal to that at the
                           concentration of 1 mol/l.

                             The product of the activity coefficient in convention (II) by the  molar
                           fraction (i.e. the activity in reference  (II)) is equal to the product of the
                           activity coefficient in convention (III) by the concentration expressed in
                           moles per liter (i.e. the activity in reference (III)).

                             If the solution is very  dilute, from  relations [A1.17]  and [A1.5], we
                           deduce:

                                      γ (III)
                                γ (II)  =  s                                            [A1.18]
                                  s
                                       v 0 0
                                                            0
                             The molar volume of the solventv  is expressed in liters per mole.
                                                           0
                             If the concentrations are sufficiently low for us to write thatγ s (II)  = 1, then
                           γ s (III)  = v .
                                  0
                                  0
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