Question

The release of volatile organic compounds into the atmosphere is regulated to limit ozone formation. In a laboratory simulation, \(5 \%\) of the ethanol in a liquid detergent is released. Thus, a "down-the-drain" factor of 0.05 is used to estimate ethanol emissions from the detergent. The \(k_{\mathrm{H}}\) values for ethanol and 2-butoxyethanol \(\left(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) are \(5 \times 10^{-6} \mathrm{~atm} \cdot \mathrm{m}^{3} / \mathrm{mol}\) and \(1.6 \times 10^{-6} \mathrm{~atm} \cdot \mathrm{m}^{3} / \mathrm{mol}\), respectively. (a) Estimate a "down-the-drain" factor for 2 -butoxyethanol in the detergent. (b) What is the \(k_{\mathrm{H}}\) for ethanol in units of \(\mathrm{L} \cdot \mathrm{atm} / \mathrm{mol} ?\) (c) Is the value found in part (b) consistent with a value given as \(0.64 \mathrm{~Pa} \cdot \mathrm{m}^{3} / \mathrm{mol} ?\)

Short answer

a. 0.016, b. 5 x 10^{-3} L·atm/mol, c. Yes

Step-by-step solution

- Understand the Situation

We are given that ethanol in a detergent has a 'down-the-drain' factor of 0.05, meaning that 5% of the ethanol is released. The problem involves comparing this with another compound, 2-butoxyethanol, and converting units for Henry’s law constant, denoted as \( k_{H} \).

- Determine 'Down-the-Drain' Factor for 2-butoxyethanol

Since the 'down-the-drain' factor relates to the volatility and proportion released into the atmosphere, compare the Henry's law constants \( k_{H} \) values of both compounds. Ethanol’s \( k_{H} \) is 5 x 10^{-6} atm·m^3/mol, and 2-butoxyethanol’s is 1.6 x 10^{-6} atm·m^3/mol. The ratio of their \( k_{H} \) values gives the comparative volatility. Thus, the 'down-the-drain' factor for 2-butoxyethanol is calculated as follows: \[ \text{'down-the-drain' factor for 2-butoxyethanol} = \text{'down-the-drain' factor for ethanol} \times \frac{k_{H, \text{2-butoxyethanol}}}{k_{H, \text{ethanol}}} \] \[ = 0.05 \times \frac{1.6 \times 10^{-6}}{5 \times 10^{-6}} \] \[ = 0.05 \times 0.32 = 0.016 \]

- Convert k_H to Different Units

The given \( k_{H} \) for ethanol is in units of atm·m^3/mol. Convert this to L·atm/mol using the conversion factor 1 m^3 = 1000 L: \[ k_{H, \text{ethanol}} = 5 \times 10^{-6} \text{ atm·m}^3/\text{mol} \] \[ = 5 \times 10^{-6} \text{ atm} \times 1000 \text{ L}/\text{mol} \] \[ = 5 \times 10^{-3} \text{ L·atm}/\text{mol} \]

- Compare Values from Different Units

We need to check if \( 5 \times 10^{-6} \text{ atm·m}^3/\text{mol} \) matches the given value of 0.64 Pa·m^3/mol: First, convert 0.64 Pa·m^3/mol to atm·m^3/mol. Since 1 atm = 101325 Pa: \[ 0.64 \text{ Pa·m}^3/\text{mol} \times \frac{1 \text{ atm}}{101325 \text{ Pa}} = \frac{0.64}{101325} \text{ atm·m}^3/\text{mol} \] \[ = 6.31 \times 10^{-6} \text{ atm·m}^3/\text{mol} \] Comparing 6.31 x 10^{-6} atm·m^3/mol to 5 x 10^{-6} atm·m^3/mol shows they are in the same order of magnitude, indicating rough consistency.

Conclusion

Summarize the findings: 'down-the-drain' factor for 2-butoxyethanol is 0.016, converted \( k_{H} \) for ethanol is 5 x 10^{-3} L·atm/mol, and the given value of \( k_{H} \) for ethanol is consistent.

Key concepts

volatile organic compounds

Volatile organic compounds (VOCs) are chemicals that easily evaporate at room temperature and have high vapor pressure. They are often emitted from products like paints, cleaning supplies, pesticides, and even some building materials. When VOCs are released into the atmosphere, they can contribute to air pollution and the formation of ground-level ozone, which is a key component of smog.
In the context of the exercise, ethanol and 2-butoxyethanol are examples of VOCs released from a detergent. It is important to understand how the 'down-the-drain' factor helps estimate the proportion of VOCs entering the environment. VOCs can have various health and environmental impacts, such as respiratory issues and contributing to climate change.

volatility

Volatility refers to the tendency of a substance to vaporize. This property is directly linked to a substance's vapor pressure: the higher the vapor pressure, the more volatile the substance. In the exercise, we compare the volatility of ethanol and 2-butoxyethanol using their Henry's law constants (\(k_H\)).
Henry's law constant expresses the ratio of the concentration of a compound in air to its concentration in water under equilibrium conditions. For volatile substances, a higher \(k_H\) indicates greater volatility. In the exercise, ethanol has a \(k_H\) of 5 x 10^{-6} atm·m^3/mol, and 2-butoxyethanol has a \(k_H\) of 1.6 x 10^{-6} atm·m^3/mol, making ethanol more volatile than 2-butoxyethanol.
The 'down-the-drain' factor is affected by the volatility of the substances because more volatile substances are released into the air more easily when used.

unit conversion

Unit conversion is essential in scientific calculations to ensure consistency and accuracy. In the exercise, we converted the Henry's law constant of ethanol from atm·m^3/mol to L·atm/mol. Understanding how to convert units helps in comparing different values and ensuring the correct application of scientific principles.
For the conversion, we used the fact that 1 m^3 = 1000 L. Hence, the given \(k_H\) of 5 x 10^{-6} atm·m^3/mol for ethanol was converted to:
\(k_{H, \text{ethanol}} = 5 \times 10^{-6} \text{ atm·m}^3/\text{mol} = 5 \times 10^{-3} \text{ L·atm}/\text{mol}\)

Similarly, we compared this value to another given in different units, Pa·m^3/mol. Using the conversion factor where 1 atm = 101325 Pa helps arrive at a consistent check for the data provided:
\(\frac {0.64 \text{ Pa·m}^3/{\text{mol}}} {101325 \text{ Pa}} = 6.31 \times 10^{-6} \text{ atm·m}^3/{\text{mol}} \)
The close match between 6.31 x 10^{-6} and 5 x 10^{-6} atm·m^3/mol values supports the consistency of the given data.

environmental chemistry

Environmental chemistry is a critical field that studies the chemical processes affecting the environment. It involves understanding the sources, reactions, transport, effects, and fate of chemical species in the air, water, and soil.
In the exercise, the focus is on the release of VOCs, such as ethanol and 2-butoxyethanol, into the atmosphere. When these compounds are released into the environment, they undergo various chemical transformations. One crucial effect of these releases is the formation of ground-level ozone, a harmful air pollutant.
By understanding properties like Henry's law constant, scientists can predict the behavior of chemical substances in the environment and devise better regulations and remediation strategies. This helps in protecting public health and preserving environmental quality, emphasizing the importance of studies and exercises such as these in forming a robust scientific understanding of environmental impact and regulatory practices.