The enthalpy of vaporization of ethanol is \(38.7 \mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\text {surr }}\) and \(\Delta S_{\text {univ }}\) when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.

The enthalpy of vaporization (ΔH_{vap}) is a crucial concept when considering the phase transition of a substance from a liquid to a gas. This term refers to the amount of energy needed to vaporize one mole of a liquid at constant pressure. It is measured in kilojoules per mole (kJ/mol).

In the context of the exercise, for ethanol, the enthalpy of vaporization is given as 38.7 kJ/mol at its boiling point. Understanding this concept is essential because it reflects the strength of intermolecular forces within the liquid; more energy is required to vaporize substances with stronger bonds between molecules. The calculation of the system's entropy change (ΔS_{sys}) during vaporization uses this value along with the substance's boiling point converted to Kelvin.

When a substance vaporizes, the system absorbs heat, which means that the enthalpy of vaporization is a positive value since the process is endothermic. Intuitively, a high enthalpy of vaporization indicates that the substance requires a significant amount of energy to undergo phase change from liquid to gas, which can infer a high degree of order or strong molecular interactions in the liquid phase.

Temperature conversion to Kelvin (K) is a fundamental step in thermodynamics to ensure that calculations are accurate. Kelvin is the SI base unit of temperature and is used in scientific equations to measure absolute temperature. To convert Celsius (C) to Kelvin, one simply adds 273.15 to the Celsius temperature.

In the provided exercise, the conversion was necessary to calculate the entropy change. Students frequently make the mistake of either forgetting to convert Celsius to Kelvin or using an incorrect conversion factor. It is imperative to remember that the Kelvin scale starts at absolute zero, which is theoretically the coldest possible temperature where particles have minimal thermal motion. Therefore, 0 K (absolute zero) is equivalent to -273.15°C. Using Kelvin allows for the use of true ratios and differences in thermodynamic equations — vital when calculating the entropy change during vaporization, where being even a small amount off in temperature can result in a considerable error in the calculated entropy.

The concept of entropy of the universe (ΔS_{univ}) is central to the second law of thermodynamics, which states that in any spontaneous process, the total entropy of the universe must increase. Entropy measures the disorder or randomness of a system; an increase in entropy signifies a system's progression towards more disorder.

In thermodynamic processes such as vaporization, we look at the entropy change in the system (ΔS_{sys}) and the surroundings (ΔS_{surr}) to determine the overall entropy change of the universe. The exercise demonstrates that the entropy gained by the system is exactly counterbalanced by the entropy lost by the surroundings, leading to no net change in the universe's entropy. However, this result is peculiar for processes occurring at the boiling point under equilibrium conditions, since it is assumed that the process is reversible and does not consider external irreversibilities.

Under typical conditions, when 1 mole of a liquid vaporizes, the disorder of the system increases (positive ΔS_{sys}) because the molecules in the vapor phase are more dispersed than in the liquid phase. Simultaneously, the surroundings generally decrease in entropy (negative ΔS_{surr}) because energy is removed from the surroundings to accomplish this phase change. Nevertheless, the second law assures us that the sum of these changes (ΔS_{univ}) will be positive for any real, irreversible process.