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1.11 Gibbs Energy of Reaction                  155

             The first four tenus on the right give the standard Gibbs energy change,

                       /)"Go  =lG£ +m~ -aG1-bag =l/),.GPL +m/),.GPM -a/),.GPA -MGPB,   [7.49]
             while the last four tenus combine to give

                                         RTlnJtP~ =RTlnQp.                           [7.50]
                                               PtP~
             Thus, equation (7.48) has the form

                                           /),.G = /)"Go + RTlnQp.                   [7.51 ]
             Expression Qp is called the reaction quotient.
                For a  general homogeneous region,  formula (7.42)  replaces (7.40).  Then equation
             (7.48) is replaced with
                                                   I  m
                                  /),.G = /)"Go  + RTln aLaM  = /)"Go  + RTlnQ       [7.52]
                                                  a~a~
             where
                                                    I  m
                                              Q= aLaM  .                             [7.53]
                                                  a~a~

                Note that in expression Q the activities of products appear in the numerator while
             the activities of reactants appear in the denominator. Each activity is raised to a power
             equal to the coefficient of the constituent in the chemical equation.


             ExampleZ7

                From 1.000 mol H2 at 0.01000 bar, 25° C , and 0.500 mol O2  at 0.1000 bar, 25° C, reac-
             tion produced 1.000 mol water vapor at 0.0500 bar and 25° C.  What was the Gibbs
             energy change?
                For the given reaction




             we have





             From table 7.4
                                            /)"Go  = -228.62 kJ.
                Substituting into equation (7.51) yields

                                                           1
                           /),.G = -228.62 kJ +(8.3145 x 10-3  kJ K- X298.15 K)ln15.81
                              = -228.62 kJ +6.84 kJ = -221.78 kJ.
                The large negative value of /),.G indicates that there is a strong tendency for the reac-
             tion to proceed under the given conditions. However, the thermodynamics does not
             tell us anything about the rate of the process. This will be considered in the chapters
             on kinetics.
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