Question

For the reaction A → B + C, when the natural log of [A] is plotted versus the time in seconds a straight line is obtained whose slope is -0.027 s^-1. What is the concentration of A (in M) after 60.0 s if [A]_o = 0.10 M?

Answer 1

Answer – Given, slope for the natural log of [A] is plotted versus the time in seconds = -0.027 s^-1 . time, t = 60.0 s , [A]o = 0.10 M

We know for the natural log of [A] is plotted versus the time in seconds slope is negative for the first order of reaction

And the slope = -k

So,m k = 0.027 s^-1

We know the integrate first order rate law

ln [A] = -kt + ln [A]o

ln [A] = - 0.027 s^-1 * 60.0 s + ln 0.10 M

           = -3.92

Taking antiln form both side

[A] = 0.020 M

So the concentration of A after 60.0 s is 0.02 M