Question

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

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

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.

Convert Percent Inactivation to Concentration Ratio

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

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: \(\frac{[A]}{[A]_0} = 0.001\)

Use the First-Order Kinetics Formula

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

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 \(-k\).

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 \(k \approx 2.3\).

Determine Contact Time for 95% Inactivation

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

Key concepts

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.