Question

A certain first-order reaction (A→products) has a rate constant of 8.10×10−3 s−1 at 45 ∘C. How many minutes does it take for the concentration of the reactant, [A], to drop to 6.25% of the original concentration? Express your answer with the appropriate units.

Answer 1

Answer:

The answer is 5.7 minutes

Explanation:

A first-order reaction follow the law of $Ln [A] = -k.t + Ln [A]_{0}$. Where [A] is the concentration of the reactant at any t time of the reaction, $[A]_{0}$ 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 $[A]=\frac{6.25}{100}.[A]_{0}$. And the rate constant (k) is 8.10×10−3 s−1

Replacing the equation of the law:

$Ln \frac{6.25}{100}.[A]_{0} = -8.10x10^{-3}s^{-1}.t + Ln[A]_{0}$

Clearing the equation:

$Ln [A]_{0}.\frac{6.25}{100} - Ln [A]_{0} = -8.10x10^{-3}s^{-1}.t$

Considering the property of logarithms: $Ln A - Ln B = Ln \frac{A}{B}$

Using the property:

$Ln \frac{[A]_{0}}{[A]_{0}}.\frac{6.25}{100} = -8.10x10^{-3}s^{-1}.t$

Clearing t and solving:

$t = \frac{Ln \frac{6.25}{100} }{-8.10x10^{-3}s^{-1} } = 342.3s$

The answer is in the unit of seconds, but every minute contains 60 seconds, converting the units:

$342.3x\frac{1min}{60s} = 5.7min$