Question

(a) Using the heat of vaporization in Appendix \(\mathrm{B}\), calculate the entropy change for the vaporization of water at \(25^{\circ} \mathrm{C}\) and at \(100^{\circ} \mathrm{C}\). (b) From your knowledge of microstates and the structure of liquid water, explain the difference in these two values.

Short answer

The entropy change for the vaporization of water at \(25^{\circ} \mathrm{C}\) (298.15K) is 136.5 J/mol⋅K, while at \(100^{\circ} \mathrm{C}\) (373.15K) it is 109.0 J/mol⋅K. The higher entropy change at \(25^{\circ} \mathrm{C}\) can be attributed to a stronger network of hydrogen bonds restricting the movement of water molecules, and thus requiring more energy to overcome these restrictions. At \(100^{\circ} \mathrm{C}\), the hydrogen bond network is weaker, and lower energy is needed to vaporize water, resulting in a smaller entropy change.

Step-by-step solution

Convert temperatures to Kelvin and find the heat of vaporization

First, we need to convert the given temperatures to Kelvin by adding 273.15 to each value: \(T_1 = 25^{\circ} \mathrm{C} + 273.15 = 298.15 \mathrm{K}\) \(T_2 = 100^{\circ} \mathrm{C} + 273.15 = 373.15 \mathrm{K}\) Next, find the heat of vaporization of water from Appendix B. The heat of vaporization for water is: ΔH = \(40.7 \times 10^3 \: \mathrm{J/mol}\)

Calculate the entropy change for each temperature

Now, we can use the formula mentioned earlier to find the entropy change for each temperature. For \(25^{\circ} \mathrm{C}\) (298.15K): \[\Delta S_1 = \frac{\Delta H}{T_1} = \frac{40.7 \times 10^3 \: \mathrm{J/mol}}{298.15 \: \mathrm{K}} = 136.5 \: \mathrm{J/mol \cdot K}\] For \(100^{\circ} \mathrm{C}\) (373.15K): \[\Delta S_2 = \frac{\Delta H}{T_2} = \frac{40.7 \times 10^3 \: \mathrm{J/mol}}{373.15 \: \mathrm{K}} = 109.0 \: \mathrm{J/mol \cdot K}\]

Analyze the difference between the two calculated entropy values

Now let's analyze the difference between the entropy changes at the two temperatures. We can see that the entropy change at \(25^{\circ} \mathrm{C}\) is higher than that at \(100^{\circ} \mathrm{C}\). To understand the difference, let's consider the microstates and the structure of liquid water. In liquid water, hydrogen bonds form a network that holds water molecules together. This network restricts the movement of water molecules and reduces the number of accessible microstates. At \(25^{\circ} \mathrm{C}\), the heat needed to break these hydrogen bonds is higher, so more energy is required to overcome these restrictions and increase the number of allowable microstates. This higher energy requirement results in a higher entropy change. When the temperature increases to \(100^{\circ} \mathrm{C}\), the hydrogen bond network is weaker, and more molecules have already gained enough kinetic energy to break some hydrogen bonds. Consequently, the energy required to vaporize water at this temperature is lower, leading to a lower entropy change.

Key concepts

Heat of Vaporization

The heat of vaporization is the amount of energy needed to turn a liquid into a gas at its boiling point, without changing the temperature. It's an important concept because it tells us how much energy is required for a phase transition from liquid to vapor. For water, the heat of vaporization is about 40.7 kJ/mol. This means each mole of water requires 40,700 joules of energy to vaporize. This value is crucial in understanding the energy dynamics of phase changes and is often related to the change in entropy, which measures the disorder or randomness in a system when vaporization occurs.

Microstates

Microstates represent the different possible configurations that a system can have at a given energy level. More microstates mean more ways to arrange the energy, and this corresponds to greater entropy, or disorder, within the system. In the context of water vaporization:

• Liquid water has fewer accessible microstates due to hydrogen bonds that constrain the water molecules.
• When water vaporizes, these hydrogen bonds are broken, and the molecules can move more freely, greatly increasing the number of possible microstates.

This increase in microstates results in a higher entropy when water changes from liquid to vapor.

Hydrogen Bonds

Hydrogen bonds are special types of dipole-dipole attractions occurring in molecules where hydrogen is bonded to a highly electronegative atom like oxygen. In water, hydrogen bonds are pivotal. They form a network that holds water molecules together tightly in the liquid state. Here's how they work:

• These bonds are relatively strong intermolecular forces, which is why water has a high boiling point compared to other molecules of similar size.
• During vaporization at lower temperatures, significant energy is needed to break these bonds, leading to a substantial increase in entropy.

The ability to break and reform these bonds gives water its unique properties, such as high specific heat and surface tension.

Temperature Conversion

Temperature conversion is essential when performing calculations in thermodynamics, as it allows us to work with different temperature scales. Typically, temperatures are given in Celsius, but for scientific calculations, especially those involving gases, Kelvin is used because it starts from absolute zero:

• To convert Celsius to Kelvin, add 273.15 to the Celsius temperature.
• For example, converting 25°C involves adding 273.15, resulting in 298.15 K.

Using Kelvin ensures consistency and accuracy in calculations related to physical chemistry, such as evaluating entropy changes.