Decomposition

Question

When reacted with water, the insecticide DDT decomposes with half life of 10 years.  Approximately how many years will it take for 99% of a given sample to decompose once exposed to water in the environment? 

Answer 1

A = a₀(0.5)^t/10

 

0.01a₀ = a₀(0.5)^t/10

0.01 = (0.5)^t/10

ln 0.01 = (t/10) ln 0.5

-4.6052 ≈ (t/10) (-0.6931)

-46.052 ≈ -0.6931 t

66.4435 ≈ t

Answer 2

we make the insecticide as 1, after the 10 years half life, it will be 50%, and then another 10 years half life, it will be 25%, so we write it like this:

 

1                   10 years

50%              10 years

25%              10 years

12.5%           10 years

6.25%           10 years

3.125%         10 years

1.5625%       10 years

0.78125%

 

that is it will take about 70 years to decompose 99% of the given sample.

Answer 3

This is an exponential decay problem.

 

Let D = amount of DDT remaining after time t has elapsed.

 

D(t) = D(0)e^-kt, k = years^-1

 

Use the half life of 10 years to find k.

 

0.5 = 1e^-10k

 

k = ln(0.5) / (-10) = 0.069315 years^-1

 

D(t) = D(0)e^-0.069315t, t in years

 

After 99% has decomposed, 1% remain, which is 0.01 of the starting amount. 

 

0.01 = 1 e^-0.069315t

 

t = ln(0.01) / (-0.069315) years = 66.438 years

 

It takes 66.438 years for 99% of the DDT to decompose.