Page 96 - Bruno Linder Elementary Physical Chemistry
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August 18, 2010 11:36      9in x 6in     b985-ch07     Elementary Physical Chemistry





                                                    Chemical Kinetics                      81

                            Solutions

                            a) Equilibrium Approximation:
                                                                          2
                                              2
                                                         2



                                   k 1 [I 2 ]= k [I ] ;  K =[I ] /[I 2 ]= k 1 /k ;  [I ] =(k 1 /k )[I 2 ]
                                                                    1
                                                                                  1
                                           1
                               The rate-determining step is Step (2);
                                                      1
                                                                    2
                                                  r = d[HI ]dt = k 2 [I ] [H 2 ]
                                                      2

                                            d[HI ]/dt =2k 2 [I 2 ]= (2k 2 k 1 /k )[I 2 ][H 2 ]
                                                                       1
                            b) Steady-State Approximation:
                               The intermediate substance, I, is
                                                                1
                               formed in Step (1) forward direction:  d[I ]/dt = k 1[I 2 ]
                                                                2
                                                                 1
                               reduced in Step (1) reverse direction: − d[I ]/dt = k [I ] 2

                                                                 2           1
                                                  1
                                                                2
                               reduced in Step (2) − d[I ]/dt = k 2 [I ] [H 2]
                                                  2
                              Since the intermediate substance must disappear, we have
                                                           2
                                                                  2
                                               k 1 [I 2 ] − k [I ] − k 2[I ] [H 2]= 0

                                                        1
                                       2

                              and so [I ] = k 1 [I 2]/(k + k 2 [H 2]). Thus, using Step 2 gives
                                                  1
                                      d[HI ]/dt =2k 2[I 2 ][H 2 ]= 2k 2 k 1[I 2 ][H 2 ]/(k + k 2[H 2 ])

                                                                          1
                               Realizing that Step (2) is the slow step, it means that k 2   k and so

                                                                                     1
                               neglecting the k 2 term in the denominator gives

                                                d[HI ]/dt ≈ (2k 2 k 1 /k )[I 2][H 2]
                                                                  1
                              which is the same as in Part (a).
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