Answer:
28.43 min
Explanation:
Using integrated rate law for first order kinetics as:
[tex][A_t]=[A_0]e^{-kt}[/tex]
Where,
[tex][A_t][/tex] is the concentration at time t
[tex][A_0][/tex] is the initial concentration
Given that:
The rate constant, k = [tex]8.1\times 10^{-2}[/tex] min⁻¹
Initial concentration [tex][A_0][/tex] = 0.1 M
Final concentration [tex][A_t][/tex] = [tex]1.0\times 10^{-2}[/tex] M
Time = ?
Applying in the above equation, we get that:-
[tex]1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}[/tex]
[tex]0.1e^{-8.1\times \:10^{-2}t}=10^{-2}[/tex]
[tex]e^{-8.1\times \:10^{-2}t}=\frac{1}{10}[/tex]
[tex]\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)[/tex]
[tex]t=28.43\ min[/tex]
