Problem 37 - Chemistry An Atoms First Approach

Question

Ethanethiol \((\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH} ;\) also called ethyl mercaptan) is commonly added to natural gas to provide the "rotten egg" smell of a gas leak. The boiling point of ethanethiol is \(35^{\circ} \mathrm{C}\) and its heat of vaporization is \(27.5 \mathrm{kJ} / \mathrm{mol} .\) What is the entropy of vaporization for this substance?

Step-by-Step Solution

Answer

The entropy of vaporization for ethanethiol can be calculated using the formula ΔS_vap = ΔH_vap / bP. First, convert the boiling point to Kelvin: bP (K) = 35°C + 273.15 = 308.15K. Then, convert the heat of vaporization to J/mol: ΔH_vap (J/mol) = 27.5 kJ/mol * (1000 J/1 kJ) = 27500 J/mol. Finally, calculate the entropy of vaporization: ΔS_vap = 27500 J/mol / 308.15K ≈ 89.25 J/mol K.

Step 1: Convert the boiling point from Celsius to Kelvin

Formula to convert Celsius to Kelvin: Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15 Given boiling point, bP = 35°C Now, convert the boiling point to Kelvin. bP (K) = 35°C + 273.15 = 308.15K

Step 2: Calculate entropy of vaporization

The entropy of vaporization can be calculated using the following formula: Entropy of vaporization (ΔS_vap) = Heat of vaporization (ΔH_vap) / Boiling point (bP) Given the heat of vaporization, ΔH_vap = 27.5 kJ/mol. Now, we need to convert the ΔH_vap from kJ/mol to J/mol: ΔH_vap (J/mol) = 27.5 kJ/mol * (1000 J/1 kJ) = 27500 J/mol Now, calculate the entropy of vaporization, ΔS_vap: ΔS_vap = ΔH_vap (J/mol) / bP (K) = 27500 J/mol / 308.15K ΔS_vap ≈ 89.25 J/mol K (rounded to two decimal places) The entropy of vaporization for ethanethiol is approximately 89.25 J/mol K.

Key Concepts

Ethanethiol

Ethanethiol, also known as ethyl mercaptan, is a colorless liquid with a distinct feature; it is notable for its strong and pungent odor similar to that of rotten eggs. This scent makes it particularly useful as an additive in natural gas to alert users of potential gas leaks, as natural gas is otherwise odorless. The chemical formula for ethanethiol is \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH}\), which classifies it as a thiol, an organosulfur compound similar to alcohols but with sulfur replacing the oxygen atom in its hydroxyl group.

Some key properties of ethanethiol include:

• Its boiling point is relatively low at 35°C, allowing it to vaporize easily at room temperature.
• The presence of sulfur in its structure is responsible for its characteristic odor.
• Ethanethiol's utility in safety applications, owing to its odor, makes it an essential component in the energy sector.

Understanding the properties of chemicals like ethanethiol can help in applying them appropriately in industrial and safety contexts.

Boiling Point Conversion

Converting the boiling point of a substance from Celsius to Kelvin is essential in thermodynamics calculations as it facilitates the use of absolute temperature scale, necessary for accurate thermal assessments. The conversion formula is straightforward: add 273.15 to the Celsius value.

For ethanethiol, with a boiling point of 35°C, the conversion is calculated as follows:
\[ 35°C + 273.15 = 308.15 \text{ K} \]

This conversion is vital when using formulas that integrate temperature, such as those for entropy and energy calculations in scientific contexts. Absolute temperature scales avoid negative values, simplifying calculations in physics and chemistry by maintaining uniformity in energy exchange processes.

Heat of Vaporization

The heat of vaporization is a critical physical property that signifies the amount of energy required to convert a unit amount of liquid into vapor at its boiling point, without changing its temperature. For ethanethiol, this value is given as 27.5 kJ/mol.

This energy is necessary to overcome the intermolecular forces holding the liquid molecules together. The higher the heat of vaporization, the stronger the intermolecular forces. In calculations, it is often essential to convert heat of vaporization from kJ/mol to J/mol, as seen here:
\[ 27.5 \text{ kJ/mol} \times 1000 = 27500 \text{ J/mol} \]

Understanding the heat of vaporization helps predict how much energy is needed in processes like distillation or evaporation, crucial in various industrial applications.

Thermodynamics Calculations

Thermodynamics involves the study of energy transformations and is fundamental in understanding processes like vaporization. One of the key calculations here is determining the entropy of vaporization, which measures the increase in disorder as a liquid becomes a gas.

This can be calculated using the formula:
\[ \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{\text{bP}} \]

Here, \(\Delta S_{\text{vap}}\) represents the entropy change, \(\Delta H_{\text{vap}}\) is the heat of vaporization converted to Joules, and bP is the boiling point in Kelvin. Using the values for ethanethiol, \(27500 \text{ J/mol}\) and \(308.15 \text{ K}\), we calculate:
\[ \Delta S_{\text{vap}} = \frac{27500}{308.15} \approx 89.25 \text{ J/mol K} \]

This increase in entropy signifies the greater freedom of gaseous molecules compared to those in the liquid state. Such calculations are pivotal in understanding the energetic and dynamic behavior of substances during phase transitions.