Answer:
[tex]t=0.012\ min[/tex]
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 = 0.0642 s⁻¹
Initial concentration [tex][A_0][/tex] = 0.502 mol/L
Final concentration [tex][A_t][/tex] = 0.479 mol/L
Time = ?
Applying in the above equation, we get that:-
[tex]0.479=0.502e^{-0.0642\times t}[/tex]
[tex]e^{-0.0642t}=\frac{0.479}{0.502}[/tex]
[tex]t=-\frac{\ln \left(\frac{0.479}{0.502}\right)}{0.0642}[/tex]
[tex]t=0.731\ sec[/tex]
Also, 1 sec = 1/60 min
So, [tex]t=0.012\ min[/tex]
