Answer:
The answer is 5.7 minutes
Explanation:
A first-order reaction follow the law of [tex]Ln [A] = -k.t + Ln [A]_{0}[/tex]. Where [A] is the concentration of the reactant at any t time of the reaction, [tex][A]_{0}[/tex] is the concentration of the reactant at the beginning of the reaction and k is the rate constant.
Dropping the concentration of the reactant to 6.25% means the concentration of A at the end of the reaction has to be [tex][A]=\frac{6.25}{100}.[A]_{0}[/tex]. And the rate constant (k) is 8.10×10−3 s−1
Replacing the equation of the law:
[tex]Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}[/tex]
Clearing the equation:
[tex]Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t[/tex]
Considering the property of logarithms: [tex]Ln A - Ln B = Ln \frac{A}{B}[/tex]
Using the property:
[tex]Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t[/tex]
Clearing t and solving:
[tex]t = \frac{Ln \frac{6.25}{100} }{-8.10x10^{-3}s^{-1} } = 342.3s[/tex]
The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:
[tex]342.3x\frac{1min}{60s} = 5.7min[/tex]
