Question

Naphthalene, better known as "mothballs," has bp \(=218^{\circ} \mathrm{C}\) and \(\Delta H_{\text {vap }}=43.3 \mathrm{~kJ} / \mathrm{mol}\). What is the entropy of vaporization, \(\Delta S_{\text {vap }}\) in \(\mathrm{J} /(\mathrm{K} \cdot \mathrm{mol})\) for naphthalene?

Short answer

The entropy of vaporization \( \Delta S_{\text{vap}} \) for naphthalene is approximately \( 88.18 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \).

Step-by-step solution

Understanding the Parameters

We are tasked to find the entropy of vaporization \( \Delta S_{\text{vap}} \) for naphthalene. The given parameters are the boiling point \( bp = 218^{\circ} \mathrm{C} \) and the enthalpy of vaporization \( \Delta H_{\text{vap}} = 43.3 \mathrm{~kJ} / \mathrm{mol} \). We'll need to convert the boiling point into Kelvin and use the Clausius-Clapeyron equation to determine the answer.

Convert Boiling Point to Kelvin

The boiling point is given in degrees Celsius. To use it in our calculations, we must convert it to Kelvin using the formula \( K = ^{\circ}C + 273.15 \). Thus, \( K = 218 + 273.15 = 491.15 \, \text{K} \).

Apply the Clausius-Clapeyron Equation

The Clausius-Clapeyron equation relates the enthalpy of vaporization to the entropy of vaporization at the boiling point: \( \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_{\text{b}}} \). Here, \( \Delta H_{\text{vap}} = 43.3 \, \mathrm{kJ/mol} \) and \( T_{\text{b}} = 491.15 \, \text{K} \).

Convert Units of Enthalpy

Since \( \Delta H_{\text{vap}} \) is in \( \mathrm{kJ/mol} \), but we need \( \Delta S_{\text{vap}} \) in \( \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \), we convert \( \Delta H_{\text{vap}} \) to \( \mathrm{J/mol} \). \( 43.3 \, \mathrm{kJ/mol} \) is equivalent to \( 43300 \, \mathrm{J/mol} \).

Calculate the Entropy of Vaporization

Substitute the converted values into the equation: \( \Delta S_{\text{vap}} = \frac{43300}{491.15} \approx 88.18 \, \mathrm{J}/(\mathrm{K} \cdot \mathrm{mol}) \). This is the entropy of vaporization for naphthalene.

Key concepts

Clausius-Clapeyron equation

The Clausius-Clapeyron equation is a powerful tool in thermodynamics that allows us to relate the enthalpy of phase transitions, such as vaporization, to temperature and pressure changes. This equation is especially useful for understanding how substances change their state from liquid to vapor. The formula derived from the equation is:

• \( \Delta S_{\text{vap}} = \frac{\Delta H_{\text{vap}}}{T_{\text{b}}} \)

Where \( \Delta S_{\text{vap}} \) is the entropy of vaporization, \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization, and \( T_{\text{b}} \) is the boiling temperature in Kelvin.
This relationship shows us that entropy, a measure of disorder, increases as liquids become gases. This is due to the increased freedom of movement for particles in the gas phase.
Therefore, applying the Clausius-Clapeyron equation helps us calculate the change in entropy when a substance transitions from liquid to vapor, as seen with naphthalene in our exercise.

boiling point conversion

Boiling point conversion is a necessary step when working with thermodynamic equations, especially when using the Clausius-Clapeyron equation. This is because thermodynamic calculations typically require temperature in Kelvin, unlike everyday use where Celsius or Fahrenheit might be more common.
To convert a temperature from degrees Celsius to Kelvin, simply add 273.15 to the Celsius temperature:

• \( K = ^{\circ}C + 273.15 \)

This simple conversion is crucial because Kelvin is an absolute scale starting from zero at absolute zero, where all molecular motion ceases.
For example, naphthalene's boiling point of 218°C is converted to 491.15 K, preparing it for calculations in the Clausius-Clapeyron equation.
Using Kelvin ensures uniformity and accuracy in scientific calculations.

enthalpy of vaporization

Enthalpy of vaporization is the amount of energy required to turn a liquid into a gas at its boiling point under constant pressure. This energy is required to overcome the intermolecular forces in the liquid phase to allow molecules to enter the vapor phase, a state of higher energy and greater disorder.
In our exercise, naphthalene has an enthalpy of vaporization of 43.3 kJ/mol, which signifies the amount of energy needed per mole to vaporize.
To use this in our calculations with entropy, we convert it to Joules per mole, since the standard measure for entropy uses Joules. This results in 43300 J/mol, making it compatible with the units used in thermodynamic equations.
Understanding the enthalpy of vaporization is key to mastering concepts related to phase transitions, such as boiling and condensation.