the rate constant of first order reaction is 6.8 *10-4 s-1 ,if the initial concentration of the reactant is 0.004 M ,what is the molarity after 20 minutes ?How long will it take for 25% of the reactant to react?

Dear Student,
                     we know for the 1st order reaction , we have the integrated rate expression as :

                                                          k=2.303t× log [A]0[A]t
Given,  k = 6.8 x 10-4 s-1
            t =  20 min = 20 x 60 = 1200 s
           [A]0 = 0.004 M

Putting these values in the integrated rate expression we get ,
                               
                                                  6.8 x 10-4   = 2.3031200× log 0.004[A]t

                 This gives ,               0.3543 = log0.004[A]t
                  Taking antilog we get , 100.3543 = 0.004[A]t
                             
                                                       [A]=   0.00177 M

Hence the molarity after 20 minutes is 0.00177 M .

Now, when the 25% of the reactant is over this means we have [A]t = 0.75[A]0

         Again Putting [A]t = 0.75[A]0  in the integrated rate law expression we get ,

                                                                k=2.303t× log [A]00.75 [A]0           
 
                                                 6.8 x 10-4  =  2.303t× log 43

                               This gives t = 423.10 s   = 7.05 min

Hence the Reaction will be 25% completed in 7.05 minutes.

I hope you understood. Keep posting.  

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