Page 55 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:36      9in x 6in     b985-ch05     Elementary Physical Chemistry





                               40                   Elementary Physical Chemistry



                                 Comment: It should be emphasized that the chemical potential is
                                 an intensive property, which depends on T and P as well as on the
                                 composition but not on the amount.



                                  We have shown in the last section that for one mole of a one-component
                               system at pressure P,

                                                    G = G  o∗  + RT ln P f /P o           (5.30)
                                                      ∗
                               In terms of activities [Eq. (5.27)]

                                                      G = G  o∗  + RT ln a                (5.31)
                                                        ∗
                               or
                                                             o
                                                        µ = µ + RT ln a                   (5.32)
                                  In a mixture, we can write for component i,

                                                             o
                                                       µ i = µ + RT ln a i                (5.33)
                                                             i
                               5.7. Relation of ∆G o  to the Equilibrium Constant, K
                                                   ¯
                               Consider the reaction,

                                                      aA + bB   cC + dD                   (5.34)

                               Denoting the molar quantities of A as m and of B as m , we can express
                                                                                ∗
                                                                  ∗
                                                                  A             B
                               the free energy change of the reaction as
                                                   o      o     o     o
                                         ∆ rG = {cG +dG − aG − bG
                                                   C      D     A     B
                                                +RT [c lna C +d ln a D − aln a A − bln a B ]}  (5.35a)
                                                             c
                                                   o
                                                                d
                                                                     b
                                                                   a
                                              =∆G + RT ln(a a /a a )                     (5.35b)
                                                                   A B
                                                             C D
                                                   o
                                              =∆G + RT ln K                              (5.35c)
                                  At equilibrium, ∆G =0 and so
                                                         ln K = −∆G o                    (5.36a)
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