## Related questions with answers

DDT (molar mass = 354.49 g/mol) was a widely used insecticide that was banned from use in the United States in 1973. This ban was brought about due to the persistence of DDT in many different ecosystems, leading to high accumulations of the substance in many birds of prey. The insecticide was shown to cause a thinning of egg shells, pushing many birds toward extinction. If a 20-L drum of DDT was spilled into a pond, resulting in a DDT concentration of $8.75 \times 10 ^ { - 5 }\ \mathrm { M } ,$ how long would it take for the levels of DDT to reach a concentration of $1.41 \times 10 ^ { - 7 }\ \mathrm { M }$ (a level that is generally assumed safe in mammals)? Assume the decomposition of DDT is a first-order process with a half-life of 56.0 days.

Solution

VerifiedFrom the data we have we can determine reaction rate constant k:

$\text{t}_{1/2} = \dfrac{\ln2}{\text{k}}$

$\text{k} = \dfrac{\ln2}{\text{t}_{1/2}} = \dfrac{\ln2}{56.0 \ \text{days}}$

$\boxed{\text{k} = 0.012 \ \text{days}^{-1}}$

Now we can, using integrated rate law, calculate time needed for DDT levels to reach a concentration which is generally assumed to be safe for mammals:

$\ln\left(\text{DDT}\right) = - \text{kt} + \ln\left[\text{DDT}\right]_0$

Solve for t:

$\text{t} = \dfrac{\ln\left[\text{DDT}\right]_0 - \ln\left[\text{DDT}\right]}{\text{k}} = \dfrac{\ln\left(8.75 \times 10^{-5}\right) - \ln\left(1.41 \times 10^{-7}\right)}{0.012 \ \text{days}^{-1}}$

$\boxed{\text{t} = 519\ \text{days}}$

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