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Chapter 18: Problem 85

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, pH, temperature, and many other factors. In a certain oil spill, hydrocarbons from the oil disappeared with a first-order rate constant of \(2 \times 10^{-6} \mathrm{~s}^{-1}\). How many days did it take for the hydrocarbons to decrease to \(10 \%\) of their initial value?

Short Answer

Expert verified
It takes approximately \(13.32\) days for the hydrocarbons to decrease to \(10\%\) of their initial value in the given oil spill scenario.

Step by step solution

01

Understand the first-order rate equation

For a first-order reaction, the rate equation is given by: \[ \ln{\frac{C_0}{C}} = kt, \] where \(C_0\) is the initial concentration of hydrocarbons, \(C\) is the final concentration of hydrocarbons, \(k\) is the rate constant, and \(t\) is the time required for the reaction to occur. Since we are interested in the time required for the concentration of hydrocarbons to decrease to \(10\%\) of their initial value, we can express \(C\) as a fraction of the initial concentration (\(0.1C_0\)).
02

Plug the values into the rate equation

Now, plug in the given rate constant (\(2 \times 10^{-6} \mathrm{~s}^{-1}\)) and the values of \(C_0\) and \(C\) in terms of \(C_0\) into the rate equation: \[ \ln{\frac{C_0}{0.1C_0}} = (2 \times 10^{-6} \mathrm{~s}^{-1})t. \]
03

Simplify the equation and solve for time

The equation can be simplified by canceling out the \(C_0\) terms: \[ \ln{10} = (2 \times 10^{-6} \mathrm{~s}^{-1})t. \] Now, solve for \(t\): \[ t = \frac{\ln{10}}{2 \times 10^{-6} \mathrm{~s}^{-1}}. \] Using the natural logarithm of \(10 (\ln{10} \approx 2.303)\), the equation becomes: \[ t \approx \frac{2.303}{2 \times 10^{-6} \mathrm{~s}^{-1}}. \] Calculate the value of \(t\): \[ t \approx 1.1515 \times 10^6 \mathrm{~s}. \]
04

Convert time from seconds to days

To find the number of days, we need to convert the time from seconds to days: \[ \mathrm{Days} = \frac{t}{60 \times 60 \times 24}. \] Plug in the value of \(t\) obtained in step 3: \[ \mathrm{Days} \approx \frac{1.1515 \times 10^6 \mathrm{~s}}{60 \times 60 \times 24}. \] Perform the calculation: \[ \mathrm{Days} \approx 13.32. \] It takes approximately \(13.32\) days for the hydrocarbons to decrease to \(10\%\) of their initial value.

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Most popular questions from this chapter

The following data were collected for the destruction of \(\mathrm{O}_{3}\) by \(\mathrm{H}\left(\mathrm{O}_{3}+\mathrm{H} \longrightarrow \mathrm{O}_{2}+\mathrm{OH}\right)\) at very low concentrations: $$ \begin{array}{llll} \hline \text { Trial } & {\left[\mathrm{O}_{3}\right](\boldsymbol{M})} & {[\mathrm{H}](\boldsymbol{M})} & \text { Initial Rate }(\boldsymbol{M} / \mathrm{s}) \\ \hline 1 & 5.17 \times 10^{-33} & 3.22 \times 10^{-26} & 1.88 \times 10^{-14} \\\ 2 & 2.59 \times 10^{-33} & 3.25 \times 10^{-26} & 9.44 \times 10^{-15} \\ 3 & 5.19 \times 10^{-33} & 6.46 \times 10^{-26} & 3.77 \times 10^{-14} \\ \hline \end{array} $$ (a) Write the rate law for the reaction. (b) Calculate the rate constant.

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