Page 146 - Essentials of physical chemistry
P. 146

108                                                  Essentials of Physical Chemistry

            Now since we may want to produce NH 3 , we can ask what temperature would shift the equilibrium
            so that the pressure of the desired product NH 3 is doubled (assuming DH 0  remains approximately
                                                                      298
            constant).

                                    0                     0
                                  K (T x )      2     DH  298  1    1
                                    P
                              ln           ¼ ln    ffi                    :
                                  0
                                K (298:15)      1       R    T x  298:15
                                  P
                  R ln(2)    1     1
            Thus,       þ       ¼     and so we find that only a modest 118 increase is needed.
                  DH  0   298:15
                      298          T x
            (8:314 J=mol K) ln(2)                3   1  ) T x ¼ 309:74 K.


              ( 45,900 J=mol)  þ (3:354016435   10 ) ¼  T x
              Actually this process required reaction vessels capable of holding very high pressures and a
            catalyst is also required. Eventually, a temperature of over 6008K was found to work best but we can
            see from this limited example that increasing the temperature shifts the equilibrium toward more
            NH 3 . Here we can also see the first case to comment on the need for care in treating reciprocal
            temperatures. We recommend carrying all digits available in this sort of calculation and then
            rounding to the least number of significant figures only at the end of the calculation.

            VAPOR PRESSURE OF LIQUIDS
            The Gibbs free energy concept can also be used in some cases where DG does not appear in the final
            formula but is still important to analyze the process. An important case is the vapor pressure of
            liquids. A strange concept that is important in forensic applications is that in principle all solids and
            liquids have some small vapor pressure. In everyday experience, we can hold items close to our nose
            and detect faint odors. In some cases like polished granite or other stoneware this slight vapor
            pressure is negligible to human smell detection but we know dogs and other animals can detect
            odors perhaps to a sensitivity more than 1000 times that of the human nose. Even then you say
            granite has no vapor pressure and effectively that is so for the solid but at temperatures where stone
            becomes molten as in lava there will be a vapor pressure. The point is that when a solid changes into
            a gas (sublimation) or a liquid changes into a gas (boiling, vaporization) the atoms=molecules of the
            gas literally ‘‘jump’’ away from the solid or liquid. To motivate this discussion, imagine that when
            water boils some molecules at the surface of the liquid jump out of the container. However, if we
            boil water to make tea and turn off the heat and let the water cool, the gas molecules immediately
            above the surface of the liquid can ‘‘crash’’ into the liquid since their random motion allows some to
            move toward the liquid. We have already used the idea that phase changes can be considered
            reversible in our discussion of entropy when we noted dS ¼ dq rev =T and that DS vap ¼ DH vap =T mp .
            Here we extend that idea to treat an equilibrium condition at the surface of a liquid in boiling or at
            the surface of a subliming solid like CO 2 (dry ice).
              Consider the most common case of boiling. At equilibrium G liq ¼ G vap so dG liq ¼ dG vap . For the
            boiling process the most appropriate variables to use are P, T with the assumption that there is some
            equation of state that will relate P, V, T, and n. Thus, we can relate the general differential of G to
            one of the HUGA equations.

                     qG         qG
                                    dT ¼ VdP   SdT so we can match the liquid and vapor dG values.
                     qP         qT
              dG ¼        dP þ
                         T          P
                                                         dP   (S vap   S liq )
            V liq dP   S liq dT ¼ V vap dP   S vap dT, which leads to  ¼  . Recall (S vap   S liq ) ¼
                                                         dT   (V vap   V liq )
            DH vap      dP     DH vap
                 . Thus,   ¼         . We make a (good) approximation here due to V liq << V vap
             T bp       dT   T bp (DV vap )
            as for water where the liquid volume is less than 19 mL=mol at 1008C while the vapor (steam)
   141   142   143   144   145   146   147   148   149   150   151