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

The answer is 5.7 minutes

Step-by-step 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]=(6.25)/(100).[A]_(0)`. And the rate constant (k) is 8.10×10−3 s−1

Replacing the equation of the law:

`Ln (6.25)/(100).[A]_(0) = -8.10x10^(-3)s^(-1).t + Ln[A]_(0)`

Clearing the equation:

`Ln [A]_(0).(6.25)/(100) - Ln [A]_(0) = -8.10x10^(-3)s^(-1).t`

Considering the property of logarithms:
`Ln A - Ln B = Ln (A)/(B)`

Using the property:

`Ln ([A]_(0))/([A]_(0)).(6.25)/(100) = -8.10x10^(-3)s^(-1).t`

Clearing t and solving:

`t = (Ln (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(1min)/(60s) = 5.7min`