Respuesta :
Answer:
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
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:
Concentration is decreased to 1.56 % which means that 0.0156 of [tex][A_0][/tex] is decomposed. So,
[tex]\frac {[A_t]}{[A_0]}[/tex] = 0.0156
Thus,
[tex]\frac {[A_t]}{[A_0]}=e^{-k\times t}[/tex]
[tex]0.0156=e^{-k\times t}[/tex]
kt = 4.1604
The expression for the half life is:-
Half life = 15.0 hours
[tex]t_{1/2}=\frac {ln\ 2}{k}[/tex]
Where, k is rate constant
So,
[tex]k=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]\frac{4.1604}{t}=\frac {ln\ 2}{t_{1/2}}[/tex]
[tex]t = 6\times t_{1/2}[/tex]
6 half-lives are required for the concentration of reactant to decrease to 1.56% of its original value.
