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270    CHAPTER 12 CHEMICAL EQUILIBRIUM AND DISSOCIATION





             12.9 THE EFFECT OF PRESSURE AND TEMPERATURE ON DEGREE
                    OF DISSOCIATION
             12.9.1 THE EFFECT OF PRESSURE

             The effect of pressure on the degree of dissociation is defined by Eqn (12.62), viz.
                                                      !
                                                 x x y d  ðy c þy d  y a  y b Þ
                                                  y c
                                                  c
                                                    d
                                          K p r  ¼  y a y b  p r
                                                 x a x
                                                    b
                It can be seen that the ratio of the amounts of substance is given by
                                             x x y b  p r ðy c þy d  y a  y b Þ         (12.104)
                                              y a
                                              a
                                                b
                                             x c x  ¼
                                              y c y d
                                                d       K p r
                This can be interpreted in the following way. If y c þ y d   y a   y b ¼ 0 then the mole fractions of the
             products will not be a function of pressure. However, if y c þ y d   y a   y b > 0 then the species on the
             left-hand side of the chemical equation (i.e. the reactants) increase, whereas if y c þ y d   y a   y b < 0
             then the species on the right-hand side of the equation (i.e. the products) increase. The basic rule is
             that the effect of increasing the pressure is to shift the equilibrium to reduce the total amount of
             substance.
                Considering the two reactions introduced previously. The equation for the carbon monoxide re-
             action (12.27) was
                                          1                        a
                                     CO þ O 2 5 ð1   aÞCO 2 þ aCO þ O 2
                                          2                        2
             and the equilibrium Eqn (12.72) was

                                               s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi  s ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
                                  1   a 1 þ a=2  1 þ a=2  1    1   a  1 þ a=2  1
                            K p r  ¼                     p ffiffiffiffi ¼            p ffiffiffiffi
                                 1 þ a=2   a       a=2    p r   a      a=2     p r
                This means that n c þ n d   n a   n b < 0, and this would result in the constituents on the ‘products’
             side of the equation increasing. This is in agreement with the previous statement because the total
             amounts of reactants in Eqn (12.27) is 1.5, whilst the total amounts of products is 1 þ a/2, where a is
             less than 1.0.
                A similar calculation for the combustion and dissociation of a stoichiometric methane (CH 4 ) and
                                                                  1
             air mixture, performed using a computer program entitled EQUIL2 gave the results in Table 12.1. The
             chemical equation for this reaction is

                CH 4 þ 2ðO 2 þ 3:76N 2 Þ
                                                                                        (12.105)
                    /ð1   a 1 ÞCO 2 þ a 1 CO þ 2ð1   a 2 ÞH 2 O þ 2a 2 H 2 þða 1 =2 þ a 2 ÞO 2 þ 7:52N 2


             1
              An executable version of EQUIL2 is available at http://booksite.elsevier.com/9780444633736.
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