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)?

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 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.

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.