Question

Consider the vaporization of water: ${\mathrm{H}}_2\mathrm{O}(1)⟶$ ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ at ${100}^{\circ }\mathrm{C},$ with ${\mathrm{H}}_2\mathrm{O}(1)$ in its standard state, but with the partial pressure of ${\mathrm{H}}_2\mathrm{O}(\mathrm{g})$ at $2.0\mathrm{atm}$ Which of the following statements about this vaporization at ${100}^{\circ }\mathrm{C}$ are true? (a) $\mathrm{\Delta }{G}^{\circ }=0,$ (b) $\mathrm{\Delta }G=0$ (c) $\mathrm{\Delta }{G}^{\circ }>0,$ (d) $\mathrm{\Delta }G>0?$ Explain.

Short answer

Statements A: $\mathrm{\Delta }{G}^{\circ }=0$ and D: $\mathrm{\Delta }G>0$ are correct.

Step-by-step solution

Understand the Statements

Four statements are provided regarding $\mathrm{\Delta }G$ and $\mathrm{\Delta }{G}^{\circ }$ for the vaporization of water. Analyze each one in terms of the standard and actual conditions of the system.

Evaluate Statement A

Statement A states that $\mathrm{\Delta }{G}^{\circ }=0$. Under standard conditions, the pressure of water vapor should be 1 atm. But in this case, we are considering vaporization at 100°C, where liquid water and water vapor are at equilibrium. Hence, $\mathrm{\Delta }{G}^{\circ }=0$ is correct under these conditions.

Evaluate Statement B

Statement B states that $\mathrm{\Delta }G=0$. This statement would be true if the actual conditions were the standard conditions. However, the conditions for this process are not standard (pressure is 2.0 atm instead of 1 atm). Therefore, $\mathrm{\Delta }G$ is not equal to zero in this case.

Evaluate Statement C

Statement C states that $\mathrm{\Delta }{G}^{\circ }>0$. This statement is inconsistent with Step 2, where we found that $\mathrm{\Delta }{G}^{\circ }=0$ for the vaporization of water at 100°C under standard conditions. Therefore, statement C is incorrect.

Evaluate Statement D

Statement D asserts that $\mathrm{\Delta }G>0$. When the pressure of water vapor is higher than the equilibrium pressure (2.0 atm vs 1 atm), it implies that the system will respond by shifting the reaction direction to lower the pressure, that is, towards the liquid state. Thus, $\mathrm{\Delta }G>0$, so statement D is correct.

Key concepts

Chemical Thermodynamics

Chemical thermodynamics is the study of how energy changes within chemical reactions and changes of states. It provides a framework to understand how thermal energy is converted to and from other forms of energy and how it affects matter. In the context of the vaporization of water, thermodynamics helps us to analyze how heat energy is absorbed to convert liquid water into water vapor.

Role of Heat and Energy Transfer

During phase transitions such as vaporization, heat is absorbed by the water molecules, increasing their kinetic energy until they can overcome intermolecular forces and escape as gas. This process is endothermic, meaning it absorbs heat from the surroundings, and, in this case, occurs at 100°C, which is the boiling point of water at 1 atm pressure.

Understanding Vaporization with Thermodynamics

Utilizing thermodynamics, we can determine if the process is spontaneous by analyzing Gibbs free energy changes (which we will discuss in detail in the next section) and evaluate how conditions such as temperature and pressure affect the phase transition.

Gibbs Free Energy

Gibbs free energy (G) is a thermodynamic quantity used to predict whether a process will occur spontaneously at constant temperature and pressure. It is defined by the equation:
$$G=H-TS$$
where H is the enthalpy, T is the temperature in Kelvin, and S is the entropy of the system.

Gibbs Free Energy in Vaporization

The change in Gibbs free energy ($\mathrm{\Delta }G$) during vaporization gives us valuable information about the spontaneity of the process. When $\mathrm{\Delta }G<0$, the process is spontaneous. Conversely, if $\mathrm{\Delta }G>0$, additional energy is required to drive the process. At equilibrium, $\mathrm{\Delta }G=0$, indicating no net change in the system. Therefore, when considering the vaporization of water at its boiling point under standard conditions (1 atm), $\mathrm{\Delta }{G}^{\circ }$ is zero because the system is at equilibrium. However, if the water vapor is at 2 atm, as described in the exercise, the condition is no longer at equilibrium, and $\mathrm{\Delta }G>0$, which is consistent with the provided solution that the system is non-spontaneous under these conditions.

Equilibrium

Equilibrium in chemical reactions is the state where the rates of the forward and reverse reactions are equal, resulting in no overall change in concentration of reactants and products. It is a dynamic balance, not a static one, because molecules are continuously reacting while maintaining a constant state.

Phase Equilibrium of Water

In the case of vaporization of water at 100°C, the liquid and vapor phases are in equilibrium when the system is closed and at a pressure of 1 atm. This means that the rate at which water molecules evaporate from the liquid phase is equal to the rate at which they condense back into the liquid.

Shifting Equilibrium

Altering conditions like pressure can shift the equilibrium. In our example, increasing the pressure of water vapor to 2 atm pushes the system away from equilibrium because the increased pressure favors the condensation process, in accordance with Le Chatelier's Principle. Understanding equilibrium is essential to determining the direction in which a reaction will proceed when subjected to changes in conditions.

Partial Pressure

Partial pressure is the pressure exerted by a single gas component in a mixture of gases. It is proportional to the mole fraction of the gas in the mixture and is a measure of its contribution to the total pressure.

Partial Pressure in Vaporization

In the context of water vapor, its partial pressure is critical in determining the direction of the vapor-liquid equilibrium. Under standard conditions for vaporization of water (1 atm total pressure at 100°C), the partial pressure of water vapor is also 1 atm. However, if the partial pressure is above this value, as in the example question where it's 2 atm, it indicates an over-saturation of the vapor phase which leads to a shift back towards the liquid.

Implications of Altered Partial Pressure

By setting a partial pressure higher than the equilibrium pressure, the balance is disturbed, favoring the reverse process, which in this situation, is condensation. This is central to the solution of the problem, showing why $\mathrm{\Delta }G$ is greater than zero; the process of vaporization requires work to be done against the prevailing conditions.