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Chapter 16: Problem 97

A biochemist studying the breakdown of the insecticide DDT finds that it decomposes by a first-order reaction with a halflife of 12 yr. How long does it take DDT in a soil sample to decrease from 275 ppbm to \(10 .\) ppbm (parts per billion by mass)?

Short Answer

Expert verified
It takes approximately 46.13 years for the concentration of DDT to decrease from 275 ppbm to 10 ppbm in a soil sample.

Step by step solution

01

- Understand the Problem

The problem involves the decomposition of DDT by a first-order reaction with a known half-life. We need to calculate the time it takes for the concentration to drop from 275 ppbm to 10 ppbm.
02

- Apply First-Order Kinetics Formula

For a first-order reaction, the relationship between concentration and time can be described by the formula
03

- Calculate the Rate Constant

The half-life formula for a first-order reaction is given by
04

- Use the Natural Logarithm to Isolate Time

Rearrange the integrated first-order rate equation to solve for time:
05

- Perform the Calculation

Substitute the known values (initial concentration, final concentration, and rate constant) into the rearranged equation to compute the time.
06

- Interpret the Result

Solve the equation and interpret the result in the context of the problem to determine the time required for the concentration decrease.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

DDT decomposition
DDT, short for dichlorodiphenyltrichloroethane, is a synthetic pesticide widely used in agriculture. Unfortunately, DDT decomposes slowly in the environment.Decomposition of DDT follows first-order reaction kinetics. What does this mean? In simple terms, the rate at which DDT breaks down depends on its current concentration.This property helps scientists predict how long DDT will persist in environments like soil. Through understanding first-order kinetics, you can estimate the remaining amount of DDT after a given period, which is crucial for environmental monitoring and public health.
half-life calculation
Half-life is the time required for the concentration of a substance to decrease by half.For DDT, the half-life is given as 12 years. Mathematically, for first-order reactions, the half-life (\t_{1/2} ) formula is written as:\[ t_{1/2} = \frac{0.693}{k} \]Where k is the rate constant of the reaction. Knowing the half-life helps to determine the rate constant k, as we can rearrange the formula to:\[ k = \frac{0.693}{t_{1/2}} \]Substituting the given half-life (12 years) into this formula, we get:\[ k = \frac{0.693}{12} \approx 0.0578 \text{ yr}^{-1} \]This rate constant is essential for further calculations.
rate constant
The rate constant (k) is a crucial parameter in first-order reactions. It provides the speed at which the substance is decomposing.With the rate constant calculated as approximately 0.0578 yr^{-1}, we can find out how long it takes for the concentration of DDT to decrease from 275 ppbm to 10 ppbm. Use the first-order kinetics formula:\[ \text{ln} \frac{[A]_t}{[A]_0} = -kt \]Where [*][A]_t is the final concentration (10 ppbm),[*][A]_0 is the initial concentration (275 ppbm), and[*]t is the time we need to find.Substitute the known values:\[ \text{ln} \frac{10}{275} = -0.0578 \times t \]By solving this, we rearrange to find t:\[ t = \frac{\text{ln} \frac{10}{275}}{-0.0578} \]On calculating, this results in approximately 67.6 years.Thus, it takes around 67.6 years for DDT to decompose from 275 ppbm to 10 ppbm in the soil.

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Most popular questions from this chapter

The mathematics of the first-order rate law can be applied to any situation in which a quantity decreases by a constant fraction per unit of time (or unit of any other variable). (a) As light moves through a solution, its intensity decreases per unit distance traveled in the solution. Show that \(\ln \left(\frac{\text { intensity of light leaving the solution }}{\text { intensity of light entering the solution }}\right)\) \(=-\) fraction of light removed per unit of length \(\times\) distance traveled in solution (b) The value of your savings declines under conditions of constant inflation. Show that \(\ln \left(\frac{\text { value remaining }}{\text { initial value }}\right)\) \(=-\) fraction lost per unit of time \(\times\) savings time interval

The effect of substrate concentration on the first-order growth rate of a microbial population follows the Monod equation: \(\mu=\frac{\mu_{\max } S}{K_{\mathrm{s}}+S}\) where \(\mu\) is the first-order growth rate \(\left(\mathrm{s}^{-1}\right), \mu_{\max }\) is the maximum growth rate \(\left(\mathrm{s}^{-1}\right), S\) is the substrate concentration \(\left(\mathrm{kg} / \mathrm{m}^{3}\right),\) and \(K_{\mathrm{s}}\) is the value of \(S\) that gives one-half of the maximum growth rate (in \(\mathrm{kg} / \mathrm{m}^{3}\) ). For \(\mu_{\max }=1.5 \times 10^{-4} \mathrm{~s}^{-1}\) and \(K_{\mathrm{s}}=0.03 \mathrm{~kg} / \mathrm{m}^{3}\). (a) Plot \(\mu\) vs. \(S\) for \(S\) between 0.0 and \(1.0 \mathrm{~kg} / \mathrm{m}^{3}\). (b) The initial population density is \(5.0 \times 10^{3}\) cells \(/ \mathrm{m}^{3}\). What is the density after \(1.0 \mathrm{~h}\), if the initial \(S\) is \(0.30 \mathrm{~kg} / \mathrm{m}^{3} ?\) (c) What is it if the initial \(S\) is \(0.70 \mathrm{~kg} / \mathrm{m}^{3}\) ?

A gas reacts with a solid that is present in large chunks. Then the reaction is run again with the solid pulverized. How does the increase in the surface area of the solid affect the rate of its reaction with the gas? Explain.

Nitrification is a biological process in which \(\mathrm{NH}_{3}\) in wastewater is converted to \(\mathrm{NH}_{4}^{+}\) and then removed according to the following reaction: \(\mathrm{NH}_{4}^{+}+2 \mathrm{O}_{2} \longrightarrow \mathrm{NO}_{3}^{-}+2 \mathrm{H}^{+}+\mathrm{H}_{2} \mathrm{O}\) The first-order rate constant is given as \(k_{1}=0.47 e^{0.095\left(T-15^{\circ} \mathrm{C}\right)}\) where \(k_{1}\) is in day \(^{-1}\) and \(T\) is in \({ }^{\circ} \mathrm{C}\). (a) If the initial concentration of \(\mathrm{NH}_{3}\) is \(3.0 \mathrm{~mol} / \mathrm{m}^{3},\) how long will it take to reduce the concentration to \(0.35 \mathrm{~mol} / \mathrm{m}^{3}\) in the spring \(\left(T=20^{\circ} \mathrm{C}\right) ?\) (b) In the winter \(\left(T=10^{\circ} \mathrm{C}\right) ?\) (c) Using your answer to part (a), what is the rate of \(\mathrm{O}_{2}\) consumption?

At body temperature \(\left(37^{\circ} \mathrm{C}\right),\) the rate constant of an enzyme-catalyzed decomposition is \(2.3 \times 10^{14}\) times that of the uncatalyzed reaction. If the frequency factor, \(A,\) is the same for both processes, by how much does the enzyme lower the \(E_{\mathrm{a}}\) ?
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