Question

The molar heat of vaporization of ethanol is \(39.3 \mathrm{~kJ} / \mathrm{mol}\), and the boiling point of ethanol is \(78.3^{\circ} \mathrm{C}\). Calculate \(\Delta S\) for the vaporization of 0.50 mole of ethanol.

Short answer

\( \Delta S = 55.95 \text{ J/K} \).

Step-by-step solution

Identify Given Values

We are given the molar heat of vaporization of ethanol, which is \(39.3 \text{ kJ/mol}\), and the boiling point, which is \(78.3^\circ \text{C}\). We are asked to find \( \Delta S \) for 0.50 moles of ethanol.

Convert Temperature to Kelvin

The boiling point temperature is given in degrees Celsius. To use it in thermodynamic equations, we need to convert it to Kelvin. \(78.3^\circ \text{C} = 78.3 + 273.15 = 351.45 \text{ K}\).

Relate \(\Delta S\) to \(\Delta H_{vap}\)

The change in entropy for the vaporization, \( \Delta S \), is calculated using the formula: \( \Delta S = \frac{\Delta H_{vap}}{T} \), where \( \Delta H_{vap} \) is the enthalpy change and \( T \) is the temperature in Kelvin.

Calculate \(\Delta S\) for One Mole

Using the formula \( \Delta S = \frac{\Delta H_{vap}}{T} \), and inserting the given values: \( \Delta S = \frac{39.3 \text{ kJ/mol}}{351.45 \text{ K}} \approx 0.1119 \text{ kJ/mol} \cdot \text{K}\).

Adjust \(\Delta S\) for 0.50 Moles

Since the problem asks for \( \Delta S \) for 0.50 moles, multiply the entropy change for one mole by 0.50: \( \Delta S = 0.1119 \text{ kJ/mol} \cdot \text{K} \times 0.50 = 0.05595 \text{ kJ/K}\).

Convert to Joules

To express the answer in Joules per Kelvin, convert \( \Delta S \) from \( \text{kJ/K} \) to \( \text{J/K} \): \( \Delta S = 0.05595 \text{ kJ/K} \times 1000 \text{ J/kJ} = 55.95 \text{ J/K}\).

Key concepts

Molar Heat of Vaporization

Molar heat of vaporization is a fundamental concept in thermodynamics. It refers to the amount of heat energy required to turn one mole of a substance from a liquid into a gas at constant pressure. This is crucial to understanding how substances like ethanol transform between phases. The molar heat of vaporization is typically measured in kilojoules per mole (kJ/mol), indicating the energy required to overcome intermolecular forces during vaporization. For ethanol, the given molar heat of vaporization is 39.3 kJ/mol. This value tells us how much energy is necessary to convert one mole of liquid ethanol into its gaseous state under normal atmospheric conditions. Understanding this concept is a stepping stone to calculating entropy changes, which are important for predicting the spontaneity of phase changes. Additionally, it plays a vital role in various industrial processes, such as distillation, where separation of substances is achieved by exploiting their differing vaporization energies.

Boiling Point

The boiling point of a substance is the temperature at which its vapor pressure equals the external pressure surrounding the liquid. This is the point at which the liquid turns into vapor. For ethanol, this occurs at 78.3°C. In thermodynamics, the boiling point is a critical factor because it provides the temperature at which phase changes entail neither the absorption nor release of energy. Instead, all energy applied contributes to the phase change itself. When considering thermodynamic equations, we need to use the Kelvin scale for temperatures to ensure consistency and accuracy. For ethanol, the boiling point converts to 351.45 K. This conversion is pivotal because many thermodynamic calculations, such as entropy change, require temperatures expressed in Kelvin to maintain correct units and values. The boiling point thus serves as a gateway to understanding and computing crucial properties like entropy.

Thermodynamic Equations

Thermodynamic equations are mathematical expressions that describe the relationships between different thermodynamic quantities, such as pressure, volume, temperature, and energy. These equations are essential tools for predicting the behavior of substances during phase changes and other processes.One key thermodynamic equation is the entropy change formula: \[ \Delta S = \frac{\Delta H_{vap}}{T}\]where \(\Delta S\) is the change in entropy, \(\Delta H_{vap}\) is the molar heat of vaporization, and \(T\) is the temperature in Kelvin. This equation relates the change in energy associated with the vaporization process to the entropy change. To accurately apply this equation, all input quantities must be converted to consistent units, such as temperature in Kelvin and enthalpy in joules, ensuring precise calculations. Understanding and applying these equations enables scientists and engineers to control and predict phase changes in various applications, such as chemical manufacturing and climatology.

Entropy Change Calculation

Calculating the change in entropy, \(\Delta S\), during a phase change such as vaporization involves using the relationship between enthalpy and temperature. Entropy, a measure of disorder or randomness in a system, is an essential concept in thermodynamics, reflecting how systems evolve over time.In the case of ethanol, the given molar heat of vaporization, \(39.3 \text{ kJ/mol}\), and the boiling point in Kelvin, \(351.45 \text{ K}\), are used to find \(\Delta S\). Using the equation:\[\Delta S = \frac{\Delta H_{vap}}{T} = \frac{39.3 \text{ kJ/mol}}{351.45 \text{ K}}\]This computation gives an entropy change of approximately \(0.1119 \text{ kJ/mol} \cdot \text{K}\) for one mole. For 0.50 moles, the entropy change is calculated by multiplying by the number of moles:\[\Delta S = 0.1119 \text{ kJ/mol} \cdot \text{K} \times 0.50 = 0.05595 \text{ kJ/K}\]Converting to joules provides the final answer: \(55.95 \text{ J/K}\). This final conversion ensures the results are in units commonly used for entropy, making them more accessible for comparison and analysis.