Question

Chloroform has \(\Delta H_{\text {vaporimation }}=29.2 \mathrm{~kJ} / \mathrm{mol}\) and boils at \(61.2{ }^{\circ} \mathrm{C}\). What is the value of \(\Delta S_{\text {vaporization }}\) for chloroform?

Short answer

\(\Delta S_{\text{vaporization}} \approx 87.3 \, \mathrm{J/mol}\cdot\mathrm{K}\).

Step-by-step solution

Understand the Relationship Between ΔH and ΔS

The process of vaporization can be described using the equation for Gibbs free energy change: \[\Delta G = \Delta H - T \Delta S\]At the boiling point, equilibrium is established, and \(\Delta G = 0\). Therefore, this equation simplifies to: \[0 = \Delta H - T \Delta S\]which can be rearranged to find \(\Delta S\):\[\Delta S = \frac{\Delta H}{T}\].

Convert Temperature to Kelvin

The temperature given in the problem is \(61.2^{\circ} \mathrm{C}\). To use it in our equations correctly, we must convert it to Kelvin. The formula for converting Celsius to Kelvin is:\[T(K) = T(\degree C) + 273.15\]So, \(T = 61.2 + 273.15 = 334.35 \,\mathrm{K}\).

Convert ΔH from kJ/mol to J/mol

Since entropy \(\Delta S\) is often given in \(\mathrm{J/mol}\cdot\mathrm{K}\), we should convert \(\Delta H\) from kJ/mol to J/mol by multiplying by 1000:\[\Delta H = 29.2 \, \mathrm{kJ/mol} \times 1000 \, \mathrm{J/kJ} = 29200 \, \mathrm{J/mol}\].

Calculate ΔS for Vaporization

With both \(\Delta H\) and \(T\) in the correct units, substitute the values into the equation from Step 1:\[\Delta S = \frac{29200 \, \mathrm{J/mol}}{334.35 \, \mathrm{K}}\]Perform the division to find:\[\Delta S \approx 87.3 \, \mathrm{J/mol}\cdot\mathrm{K}\].

Key concepts

Vaporization

Vaporization is a phase transition where a substance changes from a liquid to a gas. This process requires energy to overcome the intermolecular forces that hold the liquid together. As the liquid is heated, its molecules gain kinetic energy and, at a certain temperature, they have enough energy to transition into the gas phase.
This temperature is known as the boiling point. For chloroform, the boiling point is 61.2°C. At this temperature, vaporization occurs at a constant rate, and both liquid and vapor are in equilibrium. It's important to note that during this phase change, the temperature of the substance remains constant while it absorbs energy. This energy is termed "enthalpy of vaporization." Understanding vaporization helps to grasp other thermodynamic concepts like enthalpy and entropy.

Enthalpy

Enthalpy, represented by \( \Delta H \), is a measure of the total energy in a thermodynamic system. It includes both internal energy and the energy required to make space for the substance by displacing its environment (or pressure-volume work). In the context of phase changes like vaporization, enthalpy refers to the energy change as the substance transitions between phases.
During the vaporization of chloroform, its enthalpy of vaporization (\( \Delta H_{\text{vaporization}} \)) is 29.2 kJ/mol. This is the energy required to vaporize one mole of chloroform at its boiling point. The process involves breaking the liquid's intermolecular bonds, hence this energy is positive, indicating that the system absorbs energy from its surroundings.

Entropy

Entropy, denoted by \( \Delta S \), is a measure of the amount of disorder or randomness in a system. In thermodynamics, a system's entropy increases when it becomes more disordered. For example, the transition from liquid to gas represents an increase in entropy because gas molecules are more free to move than those in a liquid state.
When calculating entropy changes during vaporization, it is important to consider the relationship outlined by \( \Delta G = \Delta H - T \Delta S\) where \( \Delta G \) is the change in Gibbs free energy. At equilibrium, such as a substance at its boiling point, \( \Delta G\) is zero, simplifying the calculation to \( \Delta S = \frac{\Delta H}{T}\). For chloroform, using the enthalpy and temperature (converted to Kelvin), we determined \( \Delta S\) to be approximately 87.3 J/mol•K, signifying an increase in disorder as chloroform vaporizes.

Gibbs Free Energy

Gibbs free energy, denoted as \( \Delta G\), is a valuable concept in predicting the spontaneity of a thermodynamic process. It's defined as the energy associated with a chemical reaction that can do work at constant temperature and pressure. The equation \( \Delta G = \Delta H - T \Delta S\) helps understand the balance between enthalpy and entropy changes.
In vaporization, at a substance's boiling point, \( \Delta G = 0\), indicating that the system is in equilibrium between liquid and vapor. If \( \Delta G\) is negative, the process is spontaneous, while a positive \( \Delta G\) suggests non-spontaneity. The zero value at equilibrium means the energy needed to vaporize (expressed in \( \Delta H\)) is exactly balanced by the entropy's effect times the temperature (represented by \( T \Delta S\)). Understanding \( \Delta G\) provides insights into how energy is utilized effectively in phase transitions.