First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

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First-order reaction that results in the destruction of a pollutant has a rate constant of 0.l/day. (a) how many days will it ta

ke for 90% of the chemical to be destroyed? (b) how long will it take for 99% of the chemical to be destroyed? (c) how long will it take for 99.9% of the chemical to be destroyed?

Chemistry

1 answer:

Alekssandra [2.7K] · Answer 1 · 2026-03-24T09:53:18+03:00

The rate equation for a first order reaction can be expressed as follows:

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}$

In this context, k represents the reaction's rate constant, t denotes the time the reaction takes, $A_{0}$ is the initial concentration, and $A_{t}$ is the concentration at time t.

The rate constant of the reaction is $0.1 day^{-1}$ .

(a) If we start with an initial concentration of 100, when 90% of the substance is eliminated, the remaining quantity at time t will be 100-90=10. By substituting the values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{10}=23.03 days$

The time required to destroy 90% of the substance amounts to 23.03 days.

(b) If the initial concentration is set at 100, when 99% is destroyed, the present amount at time t will be 100-99=1. By substituting the input values,

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{1}=46.06 days$

This results in a duration of 46.06 days required to eradicate 99% of the chemical.

(c) Should the initial concentration be set at 100, with 99.9% of the chemical removed, the remaining quantity at time t will be 100-99.9=0.1. Substituting the values yields

$t=\frac{2.303}{k}log\frac{A_{0}}{A_{t}}=\frac{2.303}{0.1 day^{-1}}log\frac{100}{0.1}=69.09 days$

Thus, the time needed to eliminate 99.9% of the chemical is calculated as 69.09 days.