Question

The concentration of a chemical degrades according to first-order kinetics. The degradation constant is 0.2 day-1. If the initial concentration is 100.0 mg L-1, how many days are required for the concentration to reach 0.14 mg L-1?

Answer 1

Answer:

It will require 33 days to reach concentration of the chemical equal to 0.14 mg/ L

Explanation:

For a first order reaction : $A\rightarrow product$

The rate law is - $ln[\frac{[A]_{0}}{[A]_{t}}]=kt$

Where $[A]_{0}$ and  $[A]_{t}$ are initial concentration of A and concentration of A after t time respectively. k is degradation constant or rate constant.

Here k = 0.2 $day^{-1}$ , $[A]_{0}=100.0 mg/L$ and $[A]_{t}=0.14 mg/L$

So plug-in all the given values in the rate equation-

$ln[\frac{100.0}{0.14}]=0.2\times t$

or, t = 33

So, it will require 33 days to reach concentration of the chemical equal to 0.14 mg/ L