Page 200 - Basic physical chemistry for the atmospheric sciences
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1 8 6               Basic physical chemistry

                                n
            7.9.     For nitroge ,   Amax = 0.08  JLm ;  for  oxygen,  Amax = 0.099  JLm.
                     Both these wavelengths lie in the ultraviolet region .
                          I
                                   s
            7 . I O .    2 . 9   x  0 7 photon .
            7 . 1 1 .   34 %  .
                     [N0 2 (g)]
            7 . 1 2 .          2.5
                      [NO(g)]
                     Solution: From  Eq. (7.23)

                                   [N02(g)]  1 . 0 x i o - 2 0  [0 (g)]
                                   [NO(g)]  4.0 x 1 0 -  3   3
                                                                 3
                     where [OJ(g)]  must  be in units of molecules m - .  The  con­
                     centration  of 0 is  given  in  ppmv ;  to  change  this to mole­
                                   3
                               ,   we must first apply the gas equation in the form
                     cules m  3 -
                                                          (
                     of Eq.  (l  8 g) to air at a pressure of I  atm  1 0 1 3   x  0 2 Pa) and
                             .
                                                                 1
                                                                         2
                     a  temperature  of 20°C or  293K  ,   which  gives n = 2.5 x  I 0 5
                                                                0
                                   ,   where  n 0   is  the  concentration  of  all  the
                      molecules  m - 3
                      molecules  in  air  at  1  atm  and  20°C.  Since  03(g)  occupies
                                                                       6
                     0 . 0 40  ppmv  of  air,  [OJ(g)] = (2.5 x  1 025)(0.040 x  l 0 - ) =
                             8
                          I
                      1 . 0 x  0 1 molecules m  -  3 .  Hence,
                               [NOi(g)]   1 . 0 x 1 0 - 2  0
                                                     O
                                                          1 8
                                                            -

                               [NO(g)]  4.0 x 1 0 - 3   ( I . x   1 0   ) - 2 . 5
                     dn 1
            7 . 1 3 .    dt = 2jan2 - kbn 1 n2nM +  jcn3 - kdn n3
                                                  1
                      dn3
                     -;Jf = kbn 1 n2nM -jcn 3  k dn 1n3
                                        -
            7 . 1 4 .    Result  could  have  been  predicted  because  n + 2 n + 3n3  is
                                                                   2
                                                              1
                                               s
                      the  sum  of the  oxygen  atom ,   which  cannot change.  Hint:
                     To  obtain  n 1  + 2n2 + 3n3 =  c onstant,  add  the  left  and  right
                      sides of the three differential equations given  above for the
                      solution of Exercise 7  1 3 , and then integrate.
                                         .
            7 . 1 5 .    When X  =  H
                            H +  0 3 � OH + 0 2
                           OH + O  � H + 0 2
                      Net:  0  + 0 3 � 02 +  0
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