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Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097

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BuyFindarrow_forward

Chemistry

9th Edition
Steven S. Zumdahl
ISBN: 9781133611097
Textbook Problem

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 × 10−5 M, how long would it take for the levels of DDT to reach a concentration of 1.41 × 10−7 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.

Interpretation Introduction

Interpretation: The initial concentration, half life and molar mass of DDT is given. It is assumed that decomposition of DDT is first order reaction. By using these values, the time required for the given change in the concentration of DDT is to be calculated.

Concept introduction: The change observed in the concentration of a reactant or a product per unit time is known as the rate of the particular reaction. The differential rate law provides the rate of a reaction at specific reaction concentrations.

To determine: The time taken by 8.75×105M DDT to decrease to a concentration of 1.41×107M .

Explanation

Given

The half life of DDT reaction is 56.0days .

The given equation belongs to first order reaction. Hence,

[DDT]0=8.75×105M[DDT]=1.41×107M

Where,

  • [DDT]0 is initial concentration.
  • [DDT] is concentration at time t .

Formula

The half-life is calculated using the formula,

t12=0.693k

Where,

  • t12 ishalf life.
  • k is rate constant.

Substitute the values of t12 in the above equation.

t12=0.693k56.0days=0.693kk=0

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