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74                                           3 Basics of Gas Combustion

                    n        1 1 0:5            s                 2        0:5
                           P
                     SO 3
              K P ¼    0:5         ¼                      0:5
                   n  n    n         ð1,000   sÞð1,000   0:5sÞ  4,000   0:5s
                    SO 2 O 2
              At 1,000 K, K P ¼ 1:8, by iteration we can solve this equation and get
                                          s ¼ 515

              So the final mole amount of the species are determined as follows.

                          SO 2      O 2        SO 3      N 2        Total
            n i ðmoleÞ    485       742.5      515       2,000      3742.5



              Since chemical equilibrium constant depends on gas temperature, the tempera-
            ture of the combustion product mixture has to be known in order to determine the
            equilibrium constant. When this information is known, an empirical formula known
            as van’t Hoff’s equation [7] can be used for the computation of the partial pressure
            based equilibrium constant K P :
                                      d         DH R
                                        ð lnK P Þ ¼                      ð3:34Þ
                                      dT         RT 2
            where DH R ¼ the enthalpy of reaction (J/mole) and it generally depends on the
            temperature.
              Assuming there is no phase change, DH R can be calculated using Eq. (3.53).
              When ΔH R is assumed constant over a narrow range of temperature, the inte-
            gration of Eq. (3.34) gives a form that is similar to the Arrhenius equation for a
            narrow range of temperature

                                                DH R
                                    K P ¼ K P0 exp                       ð3:35Þ
                                                 RT
            where K P0 can be determined using a reference temperature such as T 0 ¼ 298K.
            From this analysis it can also be seen that the components in the combustion is
            dependent on the combustion temperature [5]. Therefore, a general form of the
            chemical equilibrium constant is

                                                  B
                                     K P ¼ Aexp                          ð3:36Þ
                                                  T
                                                                   1    1
              For example, the chemical equilibrium constant of the reaction N 2 þ O 2 $
                                                                   2    2
            NO is
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