Chlorine is commonly used to disinfect drinking water, and inactivation of pathogens by chlorine follows first-order kinetics. The following data are for \(E\). coli inactivation: $$ \begin{array}{cc} \text { Contact Time (min) } & \text { Percent (\%) Inactivation } \\ \hline 0.00 & 0.0 \\ 0.50 & 68.3 \\ 1.00 & 90.0 \\ 1.50 & 96.8 \\ 2.00 & 99.0 \\ 2.50 & 99.7 \\ 3.00 & 99.9 \end{array} $$ (a) Determine the first-order inactivation constant, \(k\). [Hint: \% inactivation \(\left.=100 \times\left(1-[\mathrm{A}] /[\mathrm{A}]_{0}\right) .\right]\) (b) How much contact time is required for \(95 \%\) inactivation?

Short Answer

Expert verified
The first-order inactivation constant, \(k\), is approximately 2.3. The required contact time for 95% inactivation is approximately 1.3 minutes.

Step by step solution

01

Understand the First-Order Kinetics Formula

First-order kinetics for disinfection can be represented by the formula: \(\frac{d[A]}{dt} = -k[A]\), where \([A]\) = concentration of pathogen at time \(t\), \([A]_0\) = initial concentration of pathogen, \(k\) = first-order inactivation constant.
02

Convert Percent Inactivation to Concentration Ratio

Percent inactivation is given by: \(100 \times \big(1 - \frac{[A]}{[A]_0}\big)\). Rearranging gives: \(\frac{[A]}{[A]_0} = 1 - \frac{\text{Percent Inactivation}}{100}\).
03

Calculate Concentration Ratios

For each contact time in the table, compute \(\frac{[A]}{[A]_0}\): - At 0.50 min: \(\frac{[A]}{[A]_0} = 0.317\)- At 1.00 min: \(\frac{[A]}{[A]_0} = 0.10\)- At 1.50 min: \(\frac{[A]}{[A]_0} = 0.032\)- At 2.00 min: \(\frac{[A]}{[A]_0} = 0.01\)- At 2.50 min: \(\frac{[A]}{[A]_0} = 0.003\)- At 3.00 min: \(\frac{[A]}{[A]_0} = 0.001\)
04

Use the First-Order Kinetics Formula

For first-order reactions, \(\frac{[A]}{[A]_0} = e^{-kt}\). Taking the natural logarithm on both sides, \(\text{ln}\big(\frac{[A]}{[A]_0}\big) = -kt\).
05

Determine the Inactivation Constant \(k\)

For the data points, calculate \(-kt\) and plot \(\text{ln}\big(\frac{[A]}{[A]_0}\big)\) vs. contact time (min). Using linear regression, determine the slope of the line, which represents \(-k\).
06

Calculate Inactivation Constant

Using the data: - At 0.50 min: \(\text{ln}(0.317) = -1.149\)- At 1.00 min: \(\text{ln}(0.10) = -2.303\)- At 1.50 min: \(\text{ln}(0.032) = -3.446\)- At 2.00 min: \(\text{ln}(0.01) = -4.605\)- At 2.50 min: \(\text{ln}(0.003) = -5.811\)- At 3.00 min: \(\text{ln}(0.001) = -6.908\)Plotting these points and finding the slope gives \(k \approx 2.3\).
07

Determine Contact Time for 95% Inactivation

For 95% inactivation, \(\frac{[A]}{[A]_0} = 0.05\). Use \(\text{ln}(0.05) = -kt\) and solve for \(t\): \[-2.996 = -2.3t\] \[t = \frac{2.996}{2.3} \approx 1.3 \text{ min}\].

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Key Concepts

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

first-order kinetics
Understanding first-order kinetics is key for disinfection processes. In first-order kinetics, the rate of reaction is directly proportional to the concentration of the reacting species. Mathematically, this can be expressed as \(\frac{d[A]}{dt} = -k[A]\), where \([A]\) represents the concentration of pathogens and \({k}\) is the kinetic constant. This relationship means if you have a higher concentration of pathogens, the rate of disinfection will be faster.

The equation also allows us to calculate how the concentration of a pathogen decreases over time. By solving the equation, we get \([A] = [A]_0 e^{-kt}\), demonstrating an exponential decay. This tells us that, in a given time period, the fraction of pathogens not yet disinfected diminishes exponentially.
chlorine disinfection
Chlorine is widely used for the disinfection of drinking water because of its effectiveness against a variety of pathogens. When chlorine is added to water, it forms hypochlorous acid and hypochlorite ion, both of which are powerful disinfectants.

The effectiveness of chlorine disinfection depends on several factors:
  • Contact time: Longer contact time usually increases the level of disinfection.
  • Concentration: The level of chlorine available in the water to inactivate the pathogens.
  • Water quality: Factors like pH, temperature, and presence of organic matter can affect chlorine's efficacy.
Typically, chlorine inactivates pathogens following first-order kinetics. This means that the effectiveness of chlorine at killing pathogens decreases exponentially over time, making it highly reliable for ensuring water is safe to drink.
pathogen inactivation
Pathogen inactivation is a critical step in ensuring safe drinking water. By inactivating pathogens, we mean reducing their viability to cause disease. Chlorine, for instance, targets cell walls and disrupts the metabolism of pathogens, effectively neutralizing them.

The inactivation process can be quantified by the inactivation percentage, which can be calculated using the formula \(Percent \, Inactivation = 100 \times (1 - \frac{[A]}{[A]_0})\). For instance, if 90% of the pathogens are inactivated, only 10% remain viable.

Higher levels of inactivation, such as 99.9%, imply a very small fraction of pathogens survive, which is crucial for public health because even a small number of pathogens can cause illness. By understanding and manipulating the inactivation process, we can ensure that water treatment processes are effective and reliable.
kinetic constant calculation
The kinetic constant \({k}\) is a crucial parameter in first-order kinetics because it indicates the disinfection rate. To calculate \({k}\) for chlorine disinfection, you need data on contact time and the percentage of pathogen inactivation.

Here’s how to determine the kinetic constant:
  • Convert the percentage inactivation into a concentration ratio, \(\frac{[A]}{[A]_0} = 1 - \frac{\text{Percent Inactivation}}{100}\).
  • Apply the first-order kinetics formula where \( \frac{[A]}{[A]_0} = e^{-kt} \).
  • Take the natural logarithm on both sides to get \(\text{ln}(\frac{[A]}{[A]_0}) = -kt\).

Using plotted points for different contact times, you can derive the slope of the line, which will be \(-k\). For example, data from the exercise indicate \(\text{ln}(0.317) = -1.149\) at 0.50 minutes and \(\text{ln}(0.10) = -2.303\) at 1.00 minute. Plot these data points and apply linear regression to find the slope. The slope is \(-k\) and thus, \({k}\) can be approximated. For the given data, \({k} \) was found to be approximately 2.3.

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Most popular questions from this chapter

In a kinctics experiment, a chemist places crystals of iodine in a closed reaction vessel, introduces a given quantity of \(\mathrm{H}_{2}\) gas, and obtains data to calculate the rate of HI formation. In a second experiment, she uses the same amounts of iodine and hydrogen but first warms the flask to \(130^{\circ} \mathrm{C}\), a temperature above the sublimation point of iodine. In which of these experiments does the reaction proceed at a higher rate? Explain.

Many drugs decompose in blood by a first-order process. (a) Two tablets of aspirin supply \(0.60 \mathrm{~g}\) of the active compound. After 30 min, this compound reaches a maximum concentration of \(2 \mathrm{mg} / 100 \mathrm{~mL}\) of blood. If the half-life for its breakdown is \(90 \mathrm{~min},\) what is its concentration (in \(\mathrm{mg} / 100 \mathrm{~mL}\) ) \(2.5 \mathrm{~h}\) after it reaches its maximum concentration? (b) For the decomposition of an antibiotic in a person with a normal temperature \(\left(98.6^{\circ} \mathrm{F}\right)\), \(k=3.1 \times 10^{-5} \mathrm{~s}^{-1} ;\) for a person with a fever (temperature of \(\left.101.9^{\circ} \mathrm{F}\right), k=3.9 \times 10^{-5} \mathrm{~s}^{-1}\). If the person with the fever must take another pill when \(\frac{2}{3}\) of the first pill has decomposed, how many hours should she wait to take a second pill? A third pill? (Assume that the pill is effective immediately.) (c) Calculate \(E_{\mathrm{a}}\) for decomposition of the antibiotic in part (b).

What is the difference between a reaction intermediate and a transition state?

For the reaction \(4 \mathrm{~A}(g)+3 \mathrm{~B}(g) \longrightarrow 2 \mathrm{C}(g)\) the following data were obtained at constant temperature: $$ \begin{array}{cccc} \text { Experiment } & \begin{array}{c} \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \end{array} & \begin{array}{c} \text { Initial [A] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} & \begin{array}{c} \text { Initial [B] } \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 1 & 5.00 & 0.100 & 0.100 \\ 2 & 45.0 & 0.300 & 0.100 \\ 3 & 10.0 & 0.100 & 0.200 \\ 4 & 90.0 & 0.300 & 0.200 \end{array} $$ (a) What is the order with respect to each reactant? (b) Write the rate law. (c) Calculate \(k\) (using the data from Expt 1 ). (d) Using the value of \(k\) calculated in part (c), calculate the rate when \([\mathrm{A}]=\) \([\mathrm{B}]=0.400 \mathrm{~mol} / \mathrm{L}\).

Insulin is a polypeptide hormone that is released into the blood from the pancreas and stimulates fat and muscle to take up glucose; the insulin is used up in a first-order process. In a certain patient, this process has a half- life of 3.5 min. To maintain an adequate blood concentration of insulin, it must be replenished in a time interval equal to \(1 / k\). How long is the time interval for this patient?

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