Question

A city's water supply is contaminated with a toxin at a concentration of 0.63 mg/L. For the water to be safe for drinking, the concentration of this toxin must be below 1.5 x 10-3 mg/L. Fortunately, this toxin decomposes to a safe mixture of products by first-order kinetics with a rate constant of 0.27 day–1 . How long will it take for the water to be safe to drink?

Answer 1

Answer:

Approximately 22.37 days, will it take for the water to be safe to drink.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

k is rate constant

Given that:- k = 0.27 (day)⁻¹

$[A_0]$ = 0.63 mg/L

$[A_t]=1.5\times 10^{-3}$ mg/L

Applying in the above equation as:-

$1.5\times 10^{-3}=0.63e^{-0.27\times t}$

$63e^{-0.27t}=150\times \:10^{-3}$

$e^{-0.27t}=\frac{1}{420}$

$t=\frac{100\ln \left(420\right)}{27}=22.37$

Approximately 22.37 days, will it take for the water to be safe to drink.