Answer:
0.01034 moles of reactant are left after 5 minutes.
101.63 sec is the half-life.
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 = [tex]8.00\times 10^{-2}[/tex] mol
k is the rate constant = [tex]6.82\times 10^{-3}[/tex] s⁻¹
Time = 5 minutes = 5*60 seconds = 300 seconds ( 1 min = 60 sec)
Thus,
[tex][A_t]=8.00\times 10^{-2}e^{-6.82\times 10^{-3}\times 300}\ mol=0.01034\ mol[/tex]
0.01034 moles of reactant are left after 5 minutes.
Given that:
k = [tex]6.82\times 10^{-3}[/tex] s⁻¹
The expression for half life is:-
[tex]t_{1/2}=\frac{\ln2}{k}[/tex]
Thus,
[tex]t_{1/2}=\frac{\ln2}{6.82\times 10^{-3}}\ s=101.63\ s[/tex]
101.63 sec is the half-life.
