Chloroform has \(\Delta H_{\text {vaporization }}=29.2 \mathrm{~kJ} / \mathrm{mol}\) and boils at \(61.2^{\circ} \mathrm{C} .\) What is the value of \(\Delta S_{\text {vaporization }}\) for chloroform? (a) \(87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (b) \(477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (c) \(-87.3 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\) (d) \(-477.1 \mathrm{~J} / \mathrm{mol}-\mathrm{K}\)

Thermochemistry is a fundamental branch of physical chemistry focused on the study of heat changes that accompany chemical reactions and phase transitions. To understand thermochemistry, we need to delve into the concept of enthalpy (\( \text{H} \) ), which is the total heat content of a system. It's particularly important when analyzing reactions or transitions where heat is either absorbed or released.

For instance, when a substance like chloroform vaporizes, it requires energy in the form of heat. This energy change associated with the phase change from liquid to vapor is quantified by the enthalpy of vaporization (\( \text{\text{ΔH}}_{\text{vaporization}} \)). In the exercise given, chloroform has an enthalpy of vaporization of 29.2 kJ/mol. Intuitively, this number signifies the amount of heat energy needed to convert one mole of liquid chloroform to chloroform vapor at its boiling point under constant pressure.

In terms of thermochemical calculations, this energy change is often related to properties such as temperature and entropy. The latter, which we will explore further in the context of chemical thermodynamics, serves as a key link in evaluating how energy disperses within a physical system during a phase transition.

Physical Chemistry is an integral part of the curriculum for the Joint Entrance Examination (JEE), which is a highly competitive test for engineering colleges in India. To excel in this section, students must have a grasp of concepts like thermochemistry and chemical thermodynamics. These topics are crucial for understanding the physical behaviors of chemical substances and predicting the outcomes of chemical reactions.

The JEE typically includes problems requiring calculations of thermodynamic properties such as enthalpy, entropy, and free energy. For example, calculating the entropy of vaporization like in the given problem is a typical type of question that could appear on the exam. It tests a student’s ability to apply the formula for entropy (\( \frac{\text{\text{ΔH}}_{\text{vaporization}}}{T} \) ) to find out how the energy is dispersed during the process of vaporization.

By practicing such problems, students enhance their problem-solving skills and their understanding of the physical properties of matter, which are both critical for achieving a high score on the JEE in the physical chemistry section.

Chemical Thermodynamics is the application of thermodynamic principles to chemical processes and reactions. Entropy (\( S \) ) is a central concept here, representing the degree of randomness or disorder within a system. When substances undergo phase transitions, such as vaporization, the entropy typically increases since gases have a greater degree of disorder compared to liquids.

In our exercise on chloroform, the calculation of the entropy of vaporization is a textbook example of chemical thermodynamics in action. By using the provided enthalpy of vaporization and the boiling point temperature, we're able to determine how much the disorder increases when chloroform transitions from a liquid to a gas at its boiling point.

The relationship between enthalpy (\( \text{\text{ΔH}} \) ) and entropy (\( \text{\text{ΔS}} \) ) is pivotal when analyzing thermodynamic processes. The formula \( \text{\text{ΔS}} = \frac{\text{\text{ΔH}}}{T} \) combines these properties and reflects how the energy spread behind chloroform's vaporization is balanced by the temperature in Kelvin. This calculation provides insights into the fundamental relationship between energy and the degree of disorder, two essential aspects of chemical thermodynamics.