Answer:
0.0185 min⁻¹
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:
20.0 % of the initial values is left which means that 0.20 of [tex][A_0][/tex] is left. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 0.20
t = 87.0 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.20=e^{-k\times 87.0}[/tex]
k = 0.0185 min⁻¹
