Answer:
The concentration after 115 seconds is 0.0496 M
Explanation:
Applying integrated rate law for a first-order reaction, we will have;
[tex][A]_t = [A]_0e^{-kt}[/tex]
where;
[tex][A]_t[/tex] is the concentration at a time t
[tex][A]_0[/tex] is the initial concentration
k is the rate constant for the reaction
t is the given time
Given;
[tex][A]_0[/tex] = 0.05 M
k = 6.2 x 10⁻⁵ mol L⁻¹s⁻¹
t = 115 seconds
Substitute these values and solve for [tex][A]_t[/tex]
[tex][A]_t = [A]_0e^{-kt} = (0.05)*e^{-(6.2X10^{-5}*115)}\\\\[/tex]
[tex][A]_t = (0.05)(0.9929) =0.0496 M[/tex]
Therefore, the concentration after 115 seconds is 0.0496 M