Page 280 - Advanced thermodynamics for engineers
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268    CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION




                Thus
                                 d         d     1     0    0     0     0
                                   ln K p r  ¼     y a m þ y b m   y c m   y d m d
                                                      a
                                                                  c
                                                            b
                                dT        dT   <T
                                                                                         (12.96)
                                           d     1
                                         ¼          ym 0    ym 0
                                           dT  <T       R       P
                                    0
                Consider definition of m
                                          0
                                         m ¼ h 0 þ hðTÞ  Tfs 0 þ sðTÞg
                                                              Z                          (12.97)
                                                                 dh
                                           ¼ h 0 þ hðTÞ  Ts 0   T  :
                                                                 T
                Consider the terms

                                               Z               Z
                                                 dh      hðTÞ    dh
                                       hðTÞ  T      ¼ T                                  (12.98)
                                                  T       T       T
                Let
                                             v ¼ hðTÞ dv ¼ dh
                                                 1           dT
                                             u ¼      du ¼
                                                 T           T 2
                Integrating by parts gives

                                               Z             Z
                                         hðTÞ    dhðTÞ          hðTÞdT
                                      T                 ¼ T           :                  (12.99)
                                          T        T              T 2
                Thus
                                      d          X  y hðTÞ     X  yhðTÞ
                                             ¼             ¼                            (12.100)
                                        lnK p r         2            2
                                      dT            < T           <T
                Although S has been used as a shorthand form it does include both positive and negative signs;
             these must be taken into account when evaluating the significance of the term.
                Thus
                                                 1
                                   d
                                     lnKp r  ¼    2  ðy a h a þ y b h b   y c h c   y d y d Þ
                                  dT           <T
                                                                                        (12.101)
                                                 1
                                            ¼      2  ½yhŠ  ½yhŠ P
                                                        R
                                                <T
                But, by definition

                                       Q p ¼ y c h c þ y d h d   y a h a þ y b h b      (12.102)
                Hence
                                               d 	        Q p
                                                       ¼     :                          (12.103)
                                                  lnK p r   2
                                              dT         <T
                Equation (12.103) is known as the Van’t Hoff equation. It is useful for evaluating the heat of
             reaction for any particular reaction since

                                                     <d lnK p r
                                              Q p ¼          :
                                                      dð1=TÞ
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