Answer:
1.32 seconds would take for the concentration of A to decrease from 0.680 M to 0.370 M.
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]0.460[/tex] s⁻¹
Initial concentration [tex][A_0][/tex] = 0.680 M
Final concentration [tex][A_t][/tex] = 0.370 M
Time = ?
Applying in the above equation, we get that:-
[tex]0.370=0.680\times e^{-0.460\times t}[/tex]
[tex]0.68e^{-0.46t}=0.37[/tex]
[tex]68e^{-0.46t}=37[/tex]
[tex]-0.46t=\ln \left(\frac{37}{68}\right)[/tex]
[tex]t=1.32\ s[/tex]
1.32 seconds would take for the concentration of A to decrease from 0.680 M to 0.370 M.
