Question

A contaminant degrades at a rate o f0.0035 min-1. The initial concentration is 1.5 mg/L. The regulatory limit is 1ppb.

a. Find the half-life of the contaminant.

b. Find the concentration of the contaminant one hour after the reaction started.

c. How long will it take until the contaminant reaches the regulatory limit?

Answer 1

a) The unit of rate constant is min-1. So, the reaction is 1st order.

For 1st order reaction, half life = 0.693/rate constant = 0.693/0.0035 = 198 min

b) For 1st order reaction,

ln[A] = ln[Ao] - kt

[A] = final concentration (to be found out)

[Ao] = initial concentration = 1.5 mg/L

k = rate constant = 0.0035 min^-1

t = time = 1 hour = 60 min

So, ln[A] = ln(1.5) - 0.0035*60

ln [A] =0.41 - 0.21 = 0.20

[A] = 1.2 mg/L

So, concentration of contaminant left after one hour = 1.2 mg/L

c) regulatory limit = 1 ppb = 1 microgram/L = 10^-3 mg/L

We will use the same formula here again.

[A] = 10^-3 mg/L

[Ao] = 1.5 mg/L

k = 0.0035 min^-1

t = ?

ln[A] = ln[Ao] - kt

ln(10^-3) = ln(1.5) - 0.0035*t

-6.91 = 0.41 - 0.0035*t

0.0035*t = 7.32

t = 2091 min

It will take 2091 min to reach regulatory limit.

[A] = 1.2 mg/L