Problem 24 - Chemistry The Central Science

Question

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C},\) which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? Explain.

Step-by-Step Solution

Answer

The entropy change of vaporizing water at 25°C (0.1369 kJ/mol K) is greater than the entropy change of vaporizing water at 100°C (0.1093 kJ/mol K). This result is derived from the formula \(\Delta S = \frac{\Delta H}{T}\), using a constant enthalpy of vaporization for water and the given temperatures in Kelvin. Since the initial and final states of the system remain the same, the entropy change does not depend on whether the process is carried out reversibly or not.

Step 1: Enthalpy of Vaporization

First, let's recall the relationship between enthalpy change (∆H) and entropy change (∆S) for a phase transition at constant temperature, T: \[\Delta G = \Delta H - T \Delta S\] Where ∆G is the Gibbs free energy change. For a phase transition, such as vaporization, ∆G = 0. Therefore, we can rearrange the equation to solve for ∆S: \[\Delta S = \frac{\Delta H}{T}\] Since we are given the assumption that the enthalpy of vaporization does not change significantly between the two temperatures, we can use a constant value for ∆H. The enthalpy of vaporization for water at 100°C is approximately 40.79 kJ/mol. We can then convert the given temperatures to Kelvin by adding 273.15: T1 = 298.15 K and T2 = 373.15 K.

Step 2: Calculate Entropy Change for Both Processes

Now, we'll use the enthalpy of vaporization and the temperatures to calculate the entropy change for each process: Process 1: Vaporizing water at 25°C (298.15 K) \[\Delta S_1 = \frac{\Delta H}{T_1} = \frac{40.79 \, \text{kJ/mol}}{298.15 \, \text{K}} = 0.1369 \, \text{kJ/mol K}\] Process 2: Vaporizing water at 100°C (373.15 K) \[\Delta S_2 = \frac{\Delta H}{T_2} = \frac{40.79 \, \text{kJ/mol}}{373.15 \, \text{K}} = 0.1093 \, \text{kJ/mol K}\]

Step 3: Compare Entropy Changes

Now that we have calculated the entropy change for both processes, we can directly compare them: \[\Delta S_1 = 0.1369 \, \text{kJ/mol K} > \Delta S_2 = 0.1093 \, \text{kJ/mol K}\] The entropy change of vaporizing water at 25°C is greater than the entropy change of vaporizing water at 100°C.

Step 4: Discuss Reversibility and Entropy Change

Entropy change depends on the heat transfer between the system and its surroundings. In a reversible process, the heat exchange is carried out infinitely slowly, allowing for the system to always be in equilibrium with the surroundings. In an irreversible process, the heat exchange occurs at a finite rate, and the system may not always be in equilibrium. However, in both processes (reversible and irreversible), the initial and final states of the system are the same, which means the values of the initial and final entropies do not depend on the path (reversible or irreversible) taken between them. Thus, the entropy change in either process does not depend on whether we carry out the process reversibly or not.

Key Concepts

Enthalpy of Vaporization

The enthalpy of vaporization is the amount of energy required to change a substance from a liquid to a gas at its boiling point. It's an integral part of the vaporization process. This energy is needed to overcome the intermolecular forces keeping the water molecules together as a liquid. For water, the enthalpy of vaporization is relatively high at 40.79 kJ/mol. This is because of the strong hydrogen bonds present between water molecules.

The presence of strong hydrogen bonds means that a substantial amount of energy is necessary to vaporize water. Given that the problem states that the enthalpy of vaporization doesn’t change much between 25°C and 100°C, we can simplify calculations by assuming a constant value. Remember, this constant value is used to calculate changes in entropy, which is critical for understanding the difference in entropy change at various temperatures.

Vaporization Process

Vaporization is a phase transition in which a liquid is transformed into a gas. This process can occur in two ways: boiling or evaporation. During vaporization, molecules within the liquid gain sufficient energy to escape as gas into the atmosphere. In our exercise, we focus on the vaporization of water at two different temperatures: 25°C and 100°C.

Despite the same enthalpy change, the vaporization process involves different changes in entropy based on temperature. Entropy, a measure of disorder or randomness, increases as a liquid turns into a gas. Obtaining the entropy change during vaporization is essential. We calculate it using the formula \( \Delta S = \frac{\Delta H}{T} \). The results show that vaporizing water at 25°C yields a larger entropy change than at 100°C. This is because at lower temperatures, the molecules are less energetic, providing a stark contrast when they transition to a gaseous state.

Reversible and Irreversible Processes

In thermodynamics, any process can occur reversibly or irreversibly. A reversible process is an ideal concept where the system changes states in infinite slow steps, maintaining equilibrium with its surroundings. Irreversible processes occur more quickly, with the system not in constant equilibrium.

Despite this difference in path, entropy hinges only on the initial and final states of the system. This means that whether a vaporization process occurs reversibly or irreversibly doesn't affect the overall entropy change. For our example with water, both vaporization at 25°C and at 100°C result in the same final and initial states, meaning the entropy changes remain unaffected by process reversibility. This principle underscores the path-independence property of entropy in thermodynamics.