# 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 :

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   =

This gives ,               0.3543 = $\mathrm{log}\frac{0.004}{\left[A\right]t}$
Taking antilog we get , 100.3543 = $\frac{0.004}{\left[A\right]t}$

[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 ,

6.8 x 10-4  =

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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