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Gibbs’ Free Energy and Equilibria                                           107

            more complicated method because it produces graphical results similar to the treatment of rate
            constants in Chapters 7 and 8, and only depends on DH 0  . The point of this exercise is to
                                                            298
            reinforce the idea that equilibria are dynamic and are the result of a dynamic trade-off of a
            forward reaction and a reverse reaction. We start with the same equation as above,
               0                        	   0
            DG   ¼ RT ln K P ) ln K P ¼  DG   =RT .   Then   the  temperature  derivative  is
               298                          298
                                         0
                           0
                                                            0
            d ln K P   ( DG )     1     d(DG )  þDG 0  1 d(DG )
                                                              . But dG ¼ SdT þ VdP, so that

                   ¼
              dT       RT 2     RT     dT     RT 2   RT   dT
                                           ¼

                                          0                              0
                                        qG        0                  d DG Þ      0
            at constant P in a reactor       ¼ S . Then at 1 atm            ¼ DS , so that
                                                                      ð
                                        qT                             dT
                                             P
                                              0
                                       0
                               0
            d ln K 0  DG 0  ( DS )  (DH   TDS )  DS 0  DH 0   d ln K 0
                 P                                                 P .
              dT   ¼  RT 2     RT  ¼    RT 2    þ  RT  ¼  RT 2  ¼  dT
                                0     0
                           d ln K   DH
                                                          0
              This equation     P  ¼     requires only the DH value, although a way is still needed to
                             dT     RT  2
                   0
            obtain K for some initial condition. This equation can be developed further for graphical analysis
                   P
            by integrating the derivative over a range of temperature.
                    T ð 2      T ð 2             T ð 2
                                                                             0
                                    0
                                 DH (T)dT   DH 0   dT    DH 0  1   1       K (T 2 )
                           0                                                 P
                           P
                      d ln K ¼        2   ffi         2  ¼               ¼ ln  0    :
                                   (RT )     R     T      R   T 2  T 1     K (T 1 )
                                                                             P
                    T 1        T 1              T 1
                                     DH 0     1  1
                     0         0                                                    0
                               P
                    P
                                       R   T 2  T 1
            Thus, ln K (T 2 ) ¼ ln K (T 1 )        , although this equation assumes that DH is con-
            stant over a small temperature range.
            Example 2
            One of the most significant and most highly studied equilibria in the early twentieth century was the
            Haber ammonia synthesis for which Fritz Haber was awarded the Nobel Prize in 1918. In spite of
            the fact that the atmosphere of Earth is over 70% nitrogen, it is chemically difficult to use that
            enormous source of N 2 to produce nitrogen fertilizers because N 2 is quite unreactive. Because of the
            Haber synthesis, as much as one-third of the world food supply is a result of increased agricultural
            yield due to nitrogen fertilizers, such as NH 4 NO 3 , which can be synthesized from NH 3 on a
            commercial scale. Haber might be considered one of the greatest benefactors to humankind but
            the original motivation for developing the process was that Germany needed a way to make nitrates
            for munitions in WWI. In the early 1900s, nitrates were manufactured by acidifying ‘‘guano’’ (bird
            droppings rich in nitrogen compounds) as found in hundred foot layers on islands off the coast of
            Chile to obtain nitric acid and then on to nitrates. However, due to various blockades, Germany was
            cutoff from obtaining guano. Thus, ammonia synthesis was a strategic process, which lengthened
            WWI but later became very beneficial to agricultural yield. Another consideration is that the Nobel
            Prize money itself comes from the earnings of the original patent granted to Alfred Nobel for the
            invention of stabilized tri-nitro-toluene (TNT), so converting atmospheric N 2 to nitrates has great
            significance to humankind whether for war or peace.
                                                                       0
              The 90th Edn. of the CRC Handbook lists values for NH 3 of DH 298  ¼ 45:9 kJ=mol and
            DG 0  ¼ 16:4 kJ=mol. Thus, for the equilibrium N 2 þ 3H 2  ƒ 2NH 3 we can calculate
                                                                  ƒ!
               298
                          0
            DG 0    RT ln K ¼ 2( 16:4 kJ=mol)   0   0 ¼ 32:8 kJ and so we can find the value of K 0  .
               298        P                                                             298
                                32, 000 J=mol                 0           5    P 2
                      0
                  ln K 298  ¼                    ¼ 13:2321;  K 298  ¼ 5:58   10 ¼  NH 3 3  :
                          (8:314 J=mol K)(298:15 K)                              P


                                                                              P N 2
                                                                                  H 2
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