Consider the following mechanism  in aqueous solutions. At 298 K the equilibrium constant for the first step is 5.75⋅10−35.75⋅10-3                                                                           k1k1                                Step1:              A + A         −−→←−−←→           A + B                                                                                               k−1k-1                                                                                             k2k2                                Step2:                   B       →→          P     When [A]  is large,  the second step is slow and the first step is

Principles of Modern Chemistry
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Chapter18: Chemical Kinetics
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Consider the following mechanism  in aqueous solutions. At 298 K the equilibrium constant for the first step is 5.75⋅10−35.75⋅10-3  

                                                                        k1k1 

                              Step1:              A + A         −−→←−−←→           A + B

                                                                                              k−1k-1 

 

                                                                                         k2k2 

                              Step2:                   B       →→          P

 

 

When [A]  is large,  the second step is slow and the first step is fast and at equilibrium.   

Under these conditions the following experimental data is gathered.   The third column in the table is a check to make sure [A] is sufficiently large. The ratio should be greater than 100 when the absolute uncertainties are ±0.01±0.01    Note that as time goes on here, the assumption we are making will not work.  

Time  (s) [A]  (mol L-1)  k−1[A]k2k-1[A]k2    (s)
5.00 0.993 1,970
25.0 0.967 1,920

(a) Use the correct integrated rate law obtained by analyzing the mechanism to find a value for the rate constant k2.k2. 

First calculate the effective rate constant. You need to do the proper analysis of the mechanism to find what combinations of k1k1, k−1k-1 and k2k2 give you the effective rate constant.

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