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 1^st order reaction , we have the integrated rate expression as :

                                                          $k=\frac{2.303}{t}\times \log \frac{\left[A\right]0}{\left[A\right]t}$
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   = $\frac{2.303}{1200}\times \log \frac{0.004}{\left[A\right]t}$

                 This gives ,               0.3543 = $\log \frac{0.004}{\left[A\right]t}$
                  Taking antilog we get , 10^0.3543 = $\frac{0.004}{\left[A\right]t}$
                             
                                                       [A]_t  =   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 ,

                                                                $k=\frac{2.303}{t}\times \log \frac{\left[A\right]0}{0.75 \left[A\right]0}$           
 
                                                 6.8 x 10^-4  =  $\frac{2.303}{t}\times \log \frac{4}{3}$

                               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.