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

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

Step-by-step explanation:

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

The rate law is -
`ln[([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[(100.0)/(0.14)]=0.2* t`

or, t = 33

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