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 – 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
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