Question

At \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}, \Delta H^{\circ}=40.6 \mathrm{~kJ} / \mathrm{mol}\) for the vaporiza- tion of water. Estimate \(\Delta G^{\circ}\) for the vaporization of water at \(90 .{ }^{\circ} \mathrm{C}\) and \(110 .{ }^{\circ} \mathrm{C}\). Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) at \(100 .{ }^{\circ} \mathrm{C}\) and \(1.00 \mathrm{~atm}\) do not depend on temperature.

Short answer

At 90°C (363 K), the estimated ΔG for the vaporization of water is approximately \(0.407\text{ kJ/mol}\), and at 110°C (383 K), the estimated ΔG is approximately \(-0.587\text{ kJ/mol}\).

Step-by-step solution

Write down the given information

We are given the following information: \[\Delta H^{\circ} = 40.6 \text{ kJ/mol}\] Temperature: \(T_1 = 100 °C = 373 K\) - Estimate \(\Delta G^{\circ}\) at: - \(T_2 = 90 °C = 363 K\) - \(T_3 = 110 °C = 383 K\) Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Use the relationship between ΔH, ΔG, and ΔS

We know the relationship between Gibbs free energy change (\(\Delta G^{\circ}\)), enthalpy change (\(\Delta H^{\circ}\)), and entropy change (\(\Delta S^{\circ}\)) is given by: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] First, we'll find the \(\Delta S^{\circ}\) at 100°C and 1.00 atm.

Rearrange the equation and find ΔS

Rearrange the equation to solve for \(\Delta S^{\circ}\): \[\Delta S^{\circ} = \frac{\Delta H^{\circ} - \Delta G^{\circ}}{T}\] At 100°C and 1.00 atm, we know water is at its boiling point, so its \(\Delta G^{\circ}\) should be 0. Therefore, we can find the entropy change at 100°C: \[\Delta S^{\circ} = \frac{40.6 \text{ kJ/mol}}{373 \text{ K}} = 0.1087\text{ kJ/(mol K)}\]

Estimate ΔG at the target temperatures

Now that we have the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\), we can estimate \(\Delta G^{\circ}\) at the target temperatures, 363 K (90°C) and 383 K (110°C), using the equation from Step 2: 1. For \(T_2 = 363 K\) (90°C): \[\Delta G^{\circ}_{90} = \Delta H^{\circ} - T_2\Delta S^{\circ} = 40.6 \text{ kJ/mol} - (363 \text{ K})(0.1087\text{ kJ/(mol K)}) \approx 0.407\text{ kJ/mol}\] 2. For \(T_3 = 383 K\) (110°C): \[\Delta G^{\circ}_{110} = \Delta H^{\circ} - T_3\Delta S^{\circ} = 40.6 \text{ kJ/mol} - (383 \text{ K})(0.1087\text{ kJ/(mol K)}) \approx -0.587\text{ kJ/mol}\]

Write down the final answer

So, the estimated \(\Delta G^{\circ}\) for the vaporization of water at the target temperatures are: - At \(90 °C (363 K)\): \(\Delta G^{\circ} \approx 0.407\text{ kJ/mol}\) - At \(110 °C (383 K)\): \(\Delta G^{\circ} \approx -0.587\text{ kJ/mol}\)

Key concepts

Enthalpy Change

Enthalpy change, represented by \(\Delta H^\circ\), is a measure of the total energy change in a system during a chemical reaction under constant pressure. It reflects the heat absorbed or released when substances react. In the context of the vaporization of water, enthalpy change indicates the amount of energy required to convert liquid water into water vapor.

When water vaporizes at its boiling point, the enthalpy change is positive, as it is an endothermic process requiring the input of heat. The given value of \(\Delta H^\circ = 40.6 \text{ kJ/mol}\) is a standard molar enthalpy of vaporization, implying that 40.6 kJ of energy is needed to vaporize one mole of water under standard conditions (100°C and 1 atm).

Entropy Change

Entropy change, denoted as \(\Delta S^\circ\), captures the change in a system's disorder or randomness during a chemical process. It is a fundamental concept in the second law of thermodynamics, which states that for any spontaneous process, the entropy of the universe must increase.

In the exercise, once \(\Delta H^\circ\) is known, the entropy change for the vaporization of water can be calculated at the boiling point, where \(\Delta G^\circ = 0\). Using the formula \(\Delta S^\circ = \frac{\Delta H^\circ - \Delta G^\circ}{T}\), we find that the entropy of water increases upon vaporization. This increase corresponds to the greater disorder as water molecules move from a more structured liquid state to a less structured gaseous state.

Chemical Thermodynamics

Chemical thermodynamics is the branch of physical chemistry concerned with the relationship between heat, work, and chemical reactions or physical transformations within a system. It combines the concepts of enthalpy and entropy to understand the energetics and spontaneity of chemical processes through Gibbs free energy (\(\Delta G^\circ\)).

The Gibbs free energy equation, \(\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ\), allows us to predict whether a reaction will proceed spontaneously. A negative \(\Delta G^\circ\) indicates a spontaneous process, while a positive value suggests non-spontaneity. In the provided exercise, the calculated \(\Delta G^\circ\) values at different temperatures help estimate the spontaneity of water vaporization at those temperatures.

Vaporization of Water

The vaporization of water is a phase change where liquid water is transformed into vapor. This process occurs when the kinetic energy of water molecules is sufficient to overcome the intermolecular forces holding them in the liquid state.

The enthalpy of vaporization is substantial because of the strong hydrogen bonds between water molecules that must be broken. As temperature increases, so does the molecular kinetic energy, facilitating the transition to the gaseous phase. This is reflected in the calculated Gibbs free energy values at 90°C and 110°C, demonstrating that as temperature raises past the boiling point, water vaporization becomes more spontaneous.