Consider the vaporization of liquid water to steam at a pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

To answer this question, we need to think about the process of vaporization. When liquid water turns into steam, it needs to absorb energy in the form of heat to overcome the intermolecular forces holding the water molecules together in the liquid phase. As the water absorbs heat, the process is endothermic.

To determine if the process is spontaneous, we will consider the sign of Gibbs free energy change (ΔG). If ΔG is negative, the process is spontaneous. The equation for Gibbs free energy change is: ΔG = ΔH - TΔS, where ΔH is the enthalpy change, T is the temperature in Kelvin, and ΔS is the entropy change. In the case of vaporization, we know that ΔH > 0 (since it's endothermic) and ΔS > 0 (since gases have higher entropy than liquids). So, for the process to be spontaneous (ΔG < 0), we need: 0 > ΔH - TΔS, which implies: T > ΔH/ΔS. Thus, the process is spontaneous for temperatures higher than ΔH/ΔS.

For nonspontaneous processes, ΔG > 0. Using the same Gibbs free energy change equation (ΔG = ΔH - TΔS), we can write the condition for nonspontaneity as: 0 < ΔH - TΔS, which implies: T < ΔH/ΔS. Thus, the process is nonspontaneous for temperatures lower than ΔH/ΔS.

At equilibrium, the Gibbs free energy change (ΔG) is zero; both phases coexist, and the rate of vaporization equals the rate of condensation. Using the Gibbs free energy change equation (ΔG = ΔH - TΔS), we can write the condition for equilibrium as: 0 = ΔH - TΔS, which implies: T = ΔH/ΔS. The Clapeyron equation is often used to find the equilibrium temperature, but it requires additional information such as the specific enthalpy and entropy changes for the process or the saturation vapor pressure at different temperatures. In the case of water, at 1 atm pressure, the equilibrium temperature between liquid water and steam is 373.15 K (100°C), which is the normal boiling point of water. In summary: a) The vaporization of liquid water to steam is an endothermic process. b) It is spontaneous for temperatures higher than ΔH/ΔS. c) It is nonspontaneous for temperatures lower than ΔH/ΔS. d) The two phases are in equilibrium at 100°C (373.15 K) at 1 atm pressure.