Answer:
t = 71.47 min
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:
20.8 % is decomposed which means that 0.208 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.208 = 0.792
t = 7.8 min
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.792=e^{-k\times 7.8}[/tex]
k = 0.0299 min⁻¹
Also,
Given:
88.2 % is decomposed which means that 0.882 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 1 - 0.882 = 0.118
t = ?
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.118=e^{-0.0299\times t}[/tex]
t = 71.47 min
