Question

The enthalpy of vaporization of chloroform \(\left(\mathrm{CHCl}_{3}\right)\) is \(31.4\) \(\mathrm{kJ} / \mathrm{mol}\) at its boiling point \(\left(61.7^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{sur}}\), and \(\Delta S_{\text {univ }}\) when \(1.00 \mathrm{~mol}\) chloroform is vaporized at \(61.7^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm} .\)

Short answer

The change in entropy for the system (ΔS_sys) when 1.00 mol chloroform vaporizes at 61.7°C and 1.00 atm is approximately 93.9 J / (mol K). The change in entropy for the surroundings (ΔS_sur) is approximately -93.9 J / (mol K). The change in entropy for the universe (ΔS_univ) is 0 J / (mol K), which indicates that the process is at equilibrium.

Step-by-step solution

Calculate ΔS_sys (Change in Entropy of the System)

: To calculate the change in entropy for the system (ΔS_sys), we can use the formula: ΔS_sys = q_sys / T Where q_sys is the heat absorbed by the system at constant pressure, and T is the temperature in Kelvin. We have the enthalpy of vaporization (ΔH_vap) given as 31.4 kJ/mol. The heat absorbed (q_sys) is equal to the enthalpy of vaporization multiplied by the number of moles: q_sys = ΔH_vap × n Converting the boiling point from Celsius to Kelvin: T = 61.7°C + 273.15 = 334.85 K Now, plug in the given values and calculate ΔS_sys: ΔS_sys = (31.4 kJ/mol × 1.00 mol) / 334.85 K

Convert ΔS_sys units

: To obtain the result in J / (mol K), we need to convert kJ to J: ΔS_sys = (31.4 × 10^3 J/mol) / 334.85 K ΔS_sys ≈ 93.9 J / (mol K)

Calculate ΔS_sur (Change in Entropy of the Surroundings)

: To calculate the change in entropy for the surroundings (ΔS_sur), we can use the formula: ΔS_sur = -q_sur / T Here, q_sur (heat gained by the surroundings) is equal to -q_sys (heat absorbed by the system) because the energy is conserved. Therefore, the formula becomes: ΔS_sur = -(-31.4 kJ/mol × 1.00 mol) / 334.85 K

Convert ΔS_sur units

: As before, we convert kJ to J for ΔS_sur: ΔS_sur = (31.4 × 10^3 J/mol) / 334.85 K ΔS_sur ≈ -93.9 J / (mol K)

Calculate ΔS_univ (Change in Entropy of the Universe)

: To calculate the change in entropy for the universe (ΔS_univ), we can sum the changes in entropy for the system and surroundings: ΔS_univ = ΔS_sys + ΔS_sur ΔS_univ = 93.9 J / (mol K) - 93.9 J / (mol K) ΔS_univ = 0 J / (mol K) The change in entropy for the universe is 0 J / (mol K) when 1.00 mol chloroform is vaporized at 61.7°C and 1.00 atm, which indicates that the process is at equilibrium.

Key concepts

Enthalpy of Vaporization

The enthalpy of vaporization is a vital concept in thermodynamics. It refers to the amount of heat required to convert a given amount of a liquid into a vapor at a constant temperature and pressure. In simpler terms, it is the energy needed to break the intermolecular forces that hold the liquid together. For chloroform (CHCl extsubscript{3}), the enthalpy of vaporization is 31.4 kJ/mol at its boiling point of 61.7°C.
This figure tells us how much energy is necessary to change chloroform from a liquid to a gaseous state when it is at 61.7°C and 1 atmosphere of pressure. Knowing this value is essential for calculating other thermodynamic properties, such as the change in entropy of the system and surroundings during vaporization.

Entropy Change

Entropy is a measure of disorder or randomness in a system, and it increases when a system becomes more disordered. The change in entropy ( abla S) of a system can be calculated when a substance undergoes a phase change, such as vaporization. During vaporization, the entropy increases because the molecules transfer from a more ordered liquid state to a less ordered gaseous state.
For chloroform vaporization at 61.7°C, the formula abla S = q/T is used, where q is the heat exchanged and T is the temperature in Kelvin. Chloroform's heat of vaporization ( abla H_{vap}) is 31.4 kJ/mol, converted to 31,400 J/mol. At its boiling point, 61.7°C, the temperature in Kelvin is 334.85 K. Therefore, the entropy change of the system ( abla S_{sys}) is calculated to be approximately 93.9 J/(mol·K), indicating a significant increase in disorder as chloroform vaporizes.

Equilibrium Process

An equilibrium process in thermodynamics is a state where the forward and reverse reactions occur at equal rates, resulting in no net change in the system's properties. In this exercise, since abla S_{univ} equals 0, it implies the process of vaporizing chloroform is at equilibrium.
This means the entropy increase of the system is exactly balanced by the entropy decrease of the surroundings. Such processes are reversible and occur very slowly, ensuring the system is always close to equilibrium, preventing net changes in the energy of the universe. Understanding this concept helps in predicting the behavior of substance transformations under different conditions and finding optimal conditions for specific processes.

Entropy of Surroundings

When discussing the entropy of surroundings ( abla S_{sur}), it is important to know that it represents the entropy change experienced by everything outside the system during a thermodynamic process. It is calculated similarly to the system's entropy but accounts for the inverse heat energy transfer. Thus, abla S_{sur} is negative when the system absorbs heat.
In this context, the chloroform absorbs heat to vaporize, matching the amount of heat given off by the surroundings. Calculated as abla S_{sur} = -q_{sys}/T, it results in the same magnitude but opposite at around -93.9 J/(mol·K). This change reflects how the surroundings counterbalance the system’s increased disorder, providing an essential equilibrium state, crucial for both scientific understanding and industrial applications.