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260    CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION




                                        P
                                      2
                                                     Standard Gibbs Energy
                                                      of mixing of A and B
                                       R

                                                                   a +
                                                                        o
                                                                     o
                                      1           G -μ o c     (1-ε)[μ o μ −2μ ]
                                                                     b
                                                                        c
                                    Normalised Gibbs Energy  0  S    Q



                                                  ε
                                                      ε
                                       T     2RT[(1- )ln((1- )/2)+ε lnε]
                                      -1
                                                             Minimum
                                                            Gibbs Energy
                                                            of mixture


                                     -2
                                        0             0.5            1.0
                                                Fraction of reaction, ε
             FIGURE 12.2
             Variation of the Gibbs energy of a mixture.

                   0
             G   2m with ε is obtained. It can be seen that the larger the difference between the standard chemical
                   c
             potentials (i.e. the steeper the slope of the line PQ), then the larger is the fraction of reaction to achieve
             an equilibrium composition. This is to be expected because the driving force for the reaction has been
             increased. Some examples of dissociation are given later and these can be more readily understood if
             this section is borne in mind.

             12.7 EXAMPLES OF SIGNIFICANCE OF K          p
             12.7.1 EXAMPLE 1
             Consider the reaction in Eqn (12.1).
                                                    1
                                               CO þ O 2 /CO 2
                                                    2

                                                                ¼ 1:
                                        y CO ¼ 1; y O 2  ¼ 1 2; y CO 2
                Hence
                                                      p rCO 2
                                                         ffiffiffiffiffiffiffiffi :                      (12.70)
                                               K p r  ¼
                                                        p
                                                    p rCO p rO 2
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