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Chapter 18: Problem 83

Bioremediation is the process by which bacteria repair their environment in response, for example, to an oil spill. The efficiency of bacteria for "eating" hydrocarbons depends on the amount of oxygen in the system, \(\mathrm{pH}\), temperature, and many other factors. In a certain oil spill, hydrocarbons from the oil disappeared with a first-order rate constant of $2 \times 10^{-6} \mathrm{~s}^{-1}$. At that rate, how many days would it take for the hydrocarbons to decrease to \(10 \%\) of their initial value?

Short Answer

Expert verified
At the given rate constant of \(2 \times 10^{-6}\text{s}^{-1}\), it would take approximately 13.36 days for the hydrocarbons to decrease to 10% of their initial value following a first-order reaction.

Step by step solution

01

Understanding the first-order reaction equation

The formula for a first-order reaction is \(A_t = A_0e^{-kt}\), where \(A_t\) is the amount of the compound at time t, \(A_0\) is the initial amount of the compound, \(k\) is the first-order rate constant, and \(t\) is the time.
02

Substitute given values and solve for time

We want to find the time it takes for the hydrocarbons to decrease to 10% of their initial value. So, \(A_t = 0.1A_0\). Substituting the given and desired values in the first-order reaction equation, we get \(0.1A_0 = A_0e^{-k*t}\), where \(k = 2 \times 10^{-6}\text{s}^{-1}\). Divide both sides by \(A_0\), which gives: \(0.1 = e^{-k*t}\). Now, take the natural logarithm (ln) of both sides to solve for time (t): \(\ln(0.1) = -k*t\). Since we have the value for the rate constant (k), plug it in to find t: \(\ln(0.1) = - (2 \times 10^{-6}\text{s}^{-1}) * t\).
03

Calculate the time (t)

Solving for t, we get: \(t = \dfrac{\ln(0.1)}{-(2 \times 10^{-6}\text{s}^{-1})}\). Now, calculate the value of t: \(t = \dfrac{\ln(0.1)}{-(2 \times 10^{-6}\text{s}^{-1})} = 1154443.305\text{s}\).
04

Convert time from seconds to days

Finally, we need to convert the time from seconds to days. There are 86400 seconds in a day. Therefore, divide t by 86400 to get the number of days: Number of days = \(\dfrac{1154443.305\text{s}}{86400\text{s/day}}\) = 13.36 days. At the given rate, it would take approximately 13.36 days for the hydrocarbons to decrease to 10% of their initial value.

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Most popular questions from this chapter

Copper exposed to air and water may be oxidized. The green oxidized product is referred to as "patina". (a) Write a balanced chemical equation to show the reaction of copper to copper (II) ions with oxygen and protons from acid rain. (b) Would you expect some kind of "patina" on a silver surface? Explain.

A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone: $$ \mathrm{O}(g)+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ At \(298 \mathrm{~K}\) the rate constant for this reaction is $4.8 \times 10^{5}\( \)M^{-1} \mathrm{~s}^{-1}$. (a) Based on the units of the rate constant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not. (c) Use \(\Delta H_{f}^{\circ}\) values from Appendix \(\mathrm{C}\) to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

The degradation of \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of \(k=2.1 \times 10^{8} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(10^{\circ} \mathrm{C}\). If the tropospheric concentrations of \(\mathrm{OH}\) and \(\mathrm{CF}_{3} \mathrm{CH}_{2} \mathrm{~F}\) are \(1.0 \times 10^{12}\) and \(7.5 \times 10^{14}\) molecules \(/ \mathrm{m}^{3}\), respectively, what is the rate of reaction at this temperature in \(M /\) s?

In the following three instances, which choice is greener in each situation? Explain. (a) Petroleum as a raw material or vegetable oil as a raw material. (b) Toluene as a solvent or water as a solvent. (c) Catalyzed reaction at $600 \mathrm{~K}\( or uncatalyzed reaction at \)800 \mathrm{~K}$.

You are working with an artist who has been commissioned to make a sculpture for a big city in the eastern United States. The artist is wondering what material to use to make her sculpture because she has heard that acid rain in the eastern United States might destroy it over time. You take samples of granite, marble, bronze, and other materials, and place them outdoors for a long time in the big city. You periodically examine the appearance and measure the mass of the samples. (a) What observations would lead you to conclude that one or more of the materials are wellsuited for the sculpture? (b) What chemical process (or processes) is (are) the most likely responsible for any observed changes in the materials? [Section 18.2\(]\)
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