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                      The solution’s freezing point T is a function of the activity a of A in solution.      Section 12.3
                                                f
                                                                          A
                  Alternatively, we can consider T to be the independent variable and view a as a func-  Freezing-Point Depression and
                                             f
                                                                                 A
                                                                                                       Boiling-Point Elevation
                  tion of T .Wenow differentiate (12.4) with respect to T at constant P.In Chapter 6, we
                         f
                                                                f
                  differentiated ln K°    G°/RT [Eq. (6.14)] with respect to T to get (d/dT)(ln K°)
                                  P
                                                                                       P
                                            2
                  (d/dT)(  G°/RT)   H°/RT [Eq. (6.36)] for a chemical reaction. We can consider
                  the fusion process A(s) → A(l) at pressure P° and temperature T as a reaction with
                                                                          f
                  A(s) as the reactant and A(l) as the product. Therefore the same derivation that gave  A(sln)   B(sln)
                                           2
                  d(  G°/RT)/dT   H°/RT can be applied to the fusion process to give
                                       d    ¢ G   m,A 1T 2  ¢ H  m,A 1T 2
                                                              fus
                                               fus
                                                                     f
                                                      f
                                          a             b                                       A(s)     T   T f
                                      dT f      RT f            RT f 2
                  Taking ( / T ) of (12.4), we thus get
                             f
                                            0  ln  a A  ¢ H  m,A 1T 2                                  A(l)
                                                          fus
                                                                 f
                                           a       b                                 (12.5)
                                              0T f  P      RT f 2
                                                                                                       A(s)
                                                         2

                                    d  ln  a   1¢ H m,A >RT 2 dT      P const.       (12.6)
                                                fus
                                                             f
                                         A
                                                         f
                                                                                                        T   T* f
                  where   H  m,A (T ) is the molar enthalpy of fusion of pure A at T and 1 atm. [Since
                                                                          f
                          fus
                                 f
                  the activity of pure solid A at 1 atm is 1 (Sec. 11.4), a can be viewed as the equilib-  Figure 12.3
                                                                A
                  rium constant K° of Eq. (11.6) for A(s) ∆ A(sln) and (12.5) is the van’t Hoff equa-  The upper figure shows solid A in
                  tion (11.32) for A(s)  ∆ A(sln); the standard state of A(sln) is pure liquid A, so  equilibrium with a solution of
                    H  m,A  is  H° for A(s) ∆ A(sln).]                                       A   B at the solution’s freezing
                    fus
                      Integration of (12.6) from state 1 to state 2 gives                    temperature T . The lower figure
                                                                                                      f
                                                                                             shows solid A in equilibrium with
                                            a A,2    2  ¢ H m,A 1T 2                         pure liquid A at the freezing point
                                                       fus
                                                              f
                                         ln                       dT f                       T* of pure A.
                                                                                              f
                                            a            RT  2
                                             A,1
                                                   1       f
                  Let state 1 be pure A. Then T   T*, the freezing point of pure A, and a A,1    1, since
                                           f,1
                                                f
                  m (which equals m*   RT ln a ) becomes equal to m* when a   1. Let state 2 be a
                                   A
                                                                        A
                    A
                                                                A
                                             A
                  general state with activity a A,2    a and T f,2    T . Using a   g x [Eq. (10.5)],
                                                 A
                                                                      A
                                                                            A A
                                                              f
                  where x and g are the solvent mole fraction and mole-fraction-scale activity coeffi-
                         A
                               A
                  cient in the solution whose freezing point is T , we have
                                                         f
                                       A A    T f  ¢ H m,A 1T2
                                                  fus
                                   ln  g x                    dT   P const.          (12.7)
                                                    RT  2
                                              T * f
                  where the dummy integration variable (Sec. 1.8) was changed from T to T.
                                                                             f
                      If there is only one solute B in the solution, and if B is neither associated nor dis-
                  sociated, then x   1   x and
                                        B
                                A
                                  ln  g x   ln  g   ln  x   ln  g   ln  11   x 2     (12.8)
                                                             A
                                     A A
                                              A
                                                                         B
                                                      A
                                                                            2
                  The Taylor series for ln x is [Eq. (8.36)]: ln x   (x   1)   (x   1) /2       . With
                  x   1   x , this series becomes
                           B
                                         ln  11   x 2   x   x >2    . . .
                                                             2
                                                B
                                                        B
                                                             B
                  Statistical-mechanical theories of solutions and experimental data show that ln g can
                                                                                      A
                  be expanded as (Kirkwood and Oppenheim, pp. 176–177):
                                                3
                                         2
                              ln  g   B x   B x      p     nonelectrolyte solution   (12.9)
                                       2 B
                                              3 B
                                  A
                  where B , B , . . . are functions of T and P. Substitution of these two series into
                          2
                             3
                  (12.8) gives
                                                             1
                                       ln  g x   x   1B   2x       . . .            (12.10)
                                                               2
                                                   B
                                          A A
                                                             2
                                                               B
                                                         2
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