Page 88 - Chemical equilibria Volume 4
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64     Chemical Equilibria
                             This gives us the relation between the equilibrium constant K r and those
                           of the combined reactions K ρ:
                                  r ∏
                                 K =    ( )  ρ λ                                         [3.27]
                                         K
                                           ρ
                                      ρ
                             This expression enables  us to calculate the constant of any given
                           equilibrium, if we know the equilibrium constants of the combined reactions,
                           in a chosen convention to describe the solutions.


                           3.1.5. Unit of the equilibrium constants

                             Irrespective of the reaction under study and the expression of the law of
                           mass action used, the corresponding equilibrium constant has no unit,
                           because it is always defined by an exponential. This  means, in  particular,
                           that the law of mass action expressed in partial pressures must contain only
                           the pressure ratios. Similarly, in terms of concentrations, as we have seen,
                           they play a part in the ratios of concentrations. It is true that the denominator
                           of those fractions is often 1 (normal pressure of 1 bar and reference
                           concentration 1 mole/l).

                             However, it must be remembered that the equilibrium constant depends
                           on the convention chosen to define the solution.



                           3.1.6. Variations of the equilibrium constants with temperature

                             By applying relation [A1.22], by deriving relation [3.2] in relation to
                           temperature:

                                          ⎛  0  ⎞ A
                                        d ⎜  r  ⎟
                                 ∂ ln K  =  ⎝  RT  ⎠  =  Δ H 0                           [3.28]
                                                   r
                                  dT       dT     RT 2

                             By deriving relation [3.10], we obtain:

                                 ∂ ln K ()  =  ∂ ln K  (I)  =  Δ r h 0                   [3.29]
                                      P
                                   dT       dT      RT  2
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