Question

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?

Short answer

The vaporization of liquid water to steam at 1 atm is an endothermic process, as energy is absorbed to overcome the attractive forces between water molecules. The process is spontaneous at higher temperatures where the increase in entropy outweighs the enthalpy change, and non-spontaneous at lower temperatures where the increase in entropy cannot outweigh the enthalpy change. The two phases are in equilibrium at a temperature of approximately 336.4 K (63.4 °C) at 1 atm of pressure.

Step-by-step solution

Determine if the process is endothermic or exothermic

During the vaporization of liquid water to steam, energy must be provided to overcome the attractive forces between water molecules. This means that heat is absorbed by the water molecules during this process. Therefore, this process is endothermic. When a process is endothermic, the enthalpy change (∆H) is positive.

Determine the temperature range for spontaneity

A process is considered spontaneous if the Gibbs free energy change (∆G) is negative. The Gibbs free energy is given by the equation: \[ ∆G = ∆H - T∆S \] Here, ∆H is the enthalpy change (positive for endothermic), T is the temperature in Kelvin, and ∆S is entropy change (positive as the system becomes more disordered in the process of vaporization). For a spontaneous process, ∆G must be negative. We know that the process is endothermic, so ∆H is positive. The process also results in an increase in entropy (∆S is positive). In order to make ∆G negative, T∆S must be larger than ∆H. \[ -∆G = T∆S - ∆H > 0 \] The temperature range for a spontaneous process is when T∆S > ∆H. Essentially, the process becomes spontaneous at higher temperatures where the increase in entropy can outweigh the enthalpy change.

Determine the temperature range for non-spontaneity

A process is considered non-spontaneous if the Gibbs free energy change (∆G) is positive. In this case, the equation becomes: \[ ∆G = ∆H - T∆S > 0\] The temperature range for a non-spontaneous process is when ∆H > T∆S. Essentially, the process is non-spontaneous at lower temperatures, where the increase in entropy cannot outweigh the enthalpy change.

Determine the temperature of equilibrium between the two phases

At equilibrium, the Gibbs free energy change (∆G) is equal to zero. This means that the process is neither spontaneous nor non-spontaneous. The equation becomes: \[ ∆G = ∆H - T∆S = 0 \] To find the equilibrium temperature, we can rearrange the equation and solve for T: \[ T = \frac{∆H}{∆S} \] For the vaporization of liquid water to steam at 1 atm, the enthalpy change is ∆H = 40.7 kJ/mol, and the entropy change is ∆S = 121.0 J/(mol·K). Using these values, we can calculate the equilibrium temperature: \[ T = \frac{40.7 × 10^3 J/mol}{121.0 J/(mol·K)} \approx 336.4 K \] Therefore, the temperature at which the two phases are in equilibrium is approximately 336.4 K (63.4 °C) at 1 atm of pressure.