Question

Consider the vaporization of liquid water to steam at a pressure of \(1 \mathrm{~atm}\). (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 water at 1 atm is an endothermic process. The process is spontaneous at temperatures above 373 K and non-spontaneous below 373 K. The two phases (liquid and vapor) are in equilibrium at 373 K.

Step-by-step solution

(a) Determine if the process is endothermic or exothermic

Water vaporization is the process of converting liquid water into steam. During this process, we need to supply enough energy (usually in the form of heat) in order to overcome the intermolecular forces in liquid water that hold the molecules together. As a result, the system absorbs energy from its surroundings, which indicates that the process is endothermic. So, the vaporization of water is an endothermic process.

(b) Determine the temperature range for spontaneous vaporization

For a process to be spontaneous, the change in Gibbs free energy (∆G) should be negative. The Gibbs free energy is given by the equation: ∆G = ∆H - T∆S where ∆H is the change in enthalpy (positive for an endothermic process), T is the temperature in Kelvin, and ∆S is the change in entropy. Because vaporization is an endothermic process (∆H > 0) and generally, the entropy of gases is greater than that of liquids, so ∆S > 0, we need to find the range of temperatures (T) where ∆G < 0. Since we are given that the process is spontaneous, T∆S > ∆H, which means that: T > ∆H / ∆S We need to know the values of ∆H and ∆S to find the specific temperature range. For water vaporization at 1 atm, we have: ∆H = 40.7 kJ/mol ∆S = 0.109 J/(mol·K) So, the process is spontaneous when T > 40,700 J/mol / (0.109 J/(mol·K)) or T > 373 K. Thus, the process of water vaporization is spontaneous at temperatures above 373 K.

(c) Determine the temperature range for non-spontaneous vaporization

We have already found that water vaporization is spontaneous when T > 373 K. This means that for T < 373 K, the vaporization process will be non-spontaneous. Hence, water vaporization is non-spontaneous at temperatures below 373 K.

(d) Determine the temperature where two phases are in equilibrium

When two phases of water (liquid and vapor) are in equilibrium, the Gibbs free energy ∆G = 0, meaning the vaporization process is reversible. Using the Gibbs free energy equation: 0 = ∆H - T∆S Rearranging for T, we get: T = ∆H / ∆S Substituting the values of ∆H and ∆S from part (b): T = 40,700 J/mol / (0.109 J/(mol·K)) T = 373 K Thus, the two phases of water are in equilibrium at 373 K.

Key concepts

Endothermic Process

When water vaporizes from liquid to steam, it requires energy to do so. This energy is absorbed from the surroundings, typically in the form of heat. As the molecules in liquid water are bound together by intermolecular forces, energy must be provided to overcome these forces for vaporization to occur. This process of taking in energy from the surroundings is known as an endothermic process.
In general, for any endothermic process, energy absorption is key. You can think of it as the system essentially "fueling up" to make the change happen. In terms of thermodynamics, an endothermic reaction increases the internal energy of the system. Water vaporization serves as an excellent example, highlighting how the phase change requires the input of energy.

Gibbs Free Energy

Gibbs free energy, often just called "Gibbs energy" or \( \Delta G \), is crucial for understanding whether a process will occur spontaneously. It takes into account two important thermodynamic quantities: enthalpy \( \Delta H \) and entropy \( \Delta S \). The Gibbs free energy equation is expressed as:

• \( \Delta G = \Delta H - T\Delta S \)

For a process to be spontaneous, such as the vaporization of water at the right temperature, \( \Delta G \) must be negative. This means that the term \( T\Delta S \) (temperature times the change in entropy) should surpass \( \Delta H \), which is positive for an endothermic process.
Spontaneity depends on temperature, so in the case of water vaporizing, it becomes spontaneous at temperatures above 373 K (Kelvin). This outcome is predicted by setting \( \Delta G < 0 \) in the equation, thereby determining the temperature condition.

Phase Equilibrium

Phase equilibrium occurs when two phases, like liquid and vapor, coexist in balance with no net change over time. For the system to be in equilibrium, the Gibbs free energy change, \( \Delta G \), must be zero. This indicates that neither phase is favored, and there is no spontaneous transition between phases.
In the context of water, at 373 K, liquid water and steam reach equilibrium under 1 atm pressure. At this temperature, the system is delicately balanced, and vaporization proceeds at exactly the rate at which condensation occurs. This demonstrates the reversible nature of the phase equilibrium point, maintaining steady states of both phases.

Vaporization

Vaporization is the process where a liquid is transformed into a gas. This requires energy to break intermolecular bonds in the liquid, turning into kinetic energy in the gas phase. Water vaporization, well-known as boiling, happens when water reaches its boiling point, around 373 K at 1 atm pressure. At this stage, water molecules have enough energy to phase into vapor.
The energy needed for vaporization manifests as the enthalpy of vaporization or \( \Delta H \). It highlights how energy-intensive the process is, thereby reinforcing its classification as endothermic. The shift in entropy \( \Delta S \) from the liquid to gas phase is significant, reflecting increased randomness and spreading of molecules in the vapor phase compared to the more organized liquid state.