Answer:
[tex]t_{1/2}=19.98\ min[/tex]
[tex][A_t]=0.037\ M[/tex]
Explanation:
(a)
Given that:
Rate constant, k = 0.0347 min⁻¹
The expression for the half-life is:-
[tex]t_{1/2}=\frac{\ln 2}{k}[/tex]
Where, k is rate constant
So,
[tex]t_{1/2}=\frac{\ln 2}{0.0347}\ min[/tex]
[tex]t_{1/2}=19.98\ min[/tex]
(b)
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]0.0347[/tex] min⁻¹
t = 60 min
Initial concentration [tex][A_0][/tex] = 0.300 M
Final concentration [tex][A_t][/tex] =?
Applying in the above equation, we get that:-
[tex][A_t]=0.300\times e^{-0.0347\times 60}\ M[/tex]
[tex][A_t]=0.3\times \frac{1}{e^{2.082}}\ M[/tex]
[tex][A_t]=0.037\ M[/tex]
