If it takes 84.2 min for the concentration of a reactant to drop to 20.0% of its initial value in a first-order reaction, what is the rate constant for the reaction in the units min-1?

Answer 1

ln(No/N) = kt
No = 100 arbitrary number
N = 20
k = unknown
t = 84.2
Solve for k.

Answer 2

ln[80/100]=-k[84.2]
-2.2=-k[84.2]
-2.2/-84.2=k
k=0.0261min^-1

Answer 3

we know that
k=2.303/t log[a/a-x]
here a-x=20
so k=2.303/84.2 log[100/20]