Page 117 - Instant notes
P. 117

Phase equilibria     103


        point, equilibrium is established. Thus the melting and freezing temperatures of a pure
        substance are identical and in this case the terms can be used interchangeably. (However
        this is often not the case for a mixture, as the freezing temperature, where solid first starts
        to appear from a solid mixture, is often not the same as the melting temperature, where
        solid first starts to melt in a solid mixture (see Topic D5).)
           At the melting point, the equilibrium for a pure species A is:


        and as for all equilibria, the change in Gibbs free energy for the forward reaction, ∆G, is
        zero under all conditions (see Topic C1). For  each  phase,  for a small change in free
        energy, dG (see Topic B6), dG=Vdp−SdT and therefore d∆G=∆Vdp−∆SdT, where ∆G,
        ∆V and ∆S are the changes in Gibbs free energy, volume and entropy during the forward
        reaction, so that at equilibrium, ∆G=G (1)−G (s)=0, ∆V=V (1)−V (s) and ∆S=S (1)−S (s), with ∆G,
        ∆V and ∆S being the change in the Gibbs free energy, volume and entropy on melting
        respectively. This means that:



        The change in entropy on melting is always positive, as liquid species have more freedom
        of movement than solid species. The change in volume is also usually positive, as melting
        a solid produces a liquid in which the molecules move around more (have more
        translational energy), and as a consequence occupy more space. In this case, dp/dT is
        positive,  and  increasing  the pressure increases the melting temperature. A notable
        exception to this is water, as solid water (ice) has an open, hydrogen-bonded structure,
        which occupies more volume than liquid water. This is why  icebergs  float,  and  as  a
        consequence ∆V is negative; in this case dp/dT is negative (Fig. 1b).


                                       Boiling point


        At the boiling point of a liquid A, an equilibrium is established for the physical transition
        of A between liquid and vapor:





        Again, the equation        can be applied to this equilibrium, but in this case with
        the changes in volume and entropy being  ∆V=V (g)−V (1) and  ∆S=S (g)−S (1), for the
        transformation between liquid and vapor. ∆S always has a positive value, as molecules
        have more freedom of movement in the vapor than in the liquid.  Furthermore,  ∆V is
        always positive and is usually much larger than that observed in a melting transition, as
        the volume occupied by a mole of gas is much larger than a mole of liquid, whereas the
        difference in the volumes of liquid and solid are comparatively small. As a result, dp/dT
        is positive, but has a smaller value than for the  melting  transition,  which  means  that
   112   113   114   115   116   117   118   119   120   121   122