Question

At \(25^{\circ} \mathrm{C}, \Delta G^{\circ}\) for the process: $$ \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) $$ is \(8.6 \mathrm{~kJ} / \mathrm{mol} .\) Calculate the vapor pressure of water at this temperature.

Short answer

The vapor pressure of water at 25°C is approximately 0.031 atm.

Step-by-step solution

Understanding the Problem

We need to find the vapor pressure of water at \( 25^{\circ} \mathrm{C} \) using the given \( \Delta G^{\circ} = 8.6 \, \mathrm{kJ/mol} \) for the transition from liquid to gas.

Recall the Relationship between Free Energy and Equilibrium

The relation between \( \Delta G^{\circ} \) and the equilibrium constant \( K \) is given by \( \Delta G^{\circ} = -RT \ln K \), where \( R \) is the gas constant and \( T \) is the temperature in Kelvin.

Convert Temperature to Kelvin

Since \( T = 25^{\circ} \mathrm{C} \), convert this to Kelvin: \( T = 25 + 273.15 = 298.15 \, \mathrm{K} \).

Solve for Equilibrium Constant

Rearrange the formula to find \( K \): \( K = e^{-\Delta G^{\circ} /(RT)} \). Use \( R = 8.314 \, \mathrm{J/(mol \cdot K)} \), and \( \Delta G^{\circ} = 8600 \, \mathrm{J/mol} \).

Calculate \( K \)

Substitute the values: \[ K = e^{-\frac{8600}{8.314 \times 298.15}} \]. Evaluate this expression to find \( K \).

Interpret \( K \) as Vapor Pressure

For the phase transition \( \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g) \), \( K \) is the vapor pressure in \( \mathrm{atm} \), since the equilibrium is between the liquid and gas phases.

Final Calculation

Evaluate \( K = e^{-\frac{8600}{8.314 \times 298.15}} \) to obtain the vapor pressure in \( \mathrm{atm} \).

Key concepts

Gibbs Free Energy

Gibbs Free Energy, denoted as \( \Delta G \), is an essential concept in thermodynamics that helps predict whether a process will occur spontaneously at constant temperature and pressure. For a phase transition, like water shifting from liquid to gas, the change in Gibbs free energy \( \Delta G^\circ \) indicates how favorable the process is. If \( \Delta G^\circ \) is negative, the process occurs spontaneously. In the case where \( \Delta G^\circ \) is positive, like in the given problem at \( 25^\circ \mathrm{C} \) for water, this means the vaporization process is not spontaneous under standard conditions.

• Free energy change combines both enthalpy \( \Delta H \) and entropy \( \Delta S \) changes of a system: \( \Delta G = \Delta H - T\Delta S \).
• The standard Gibbs free energy change \( \Delta G^\circ \) is calculated under standard conditions of 1 atm and 298K for temperature.

Understanding these fundamentals of Gibbs free energy helps explain why certain processes need external energy, such as heating, to proceed.

Equilibrium Constant

The equilibrium constant, denoted as \( K \), is a value that quantifies the ratio of concentrations of products to reactants at equilibrium for a reversible reaction. In vapor pressure calculations, \( K \) is derived from the relationship with Gibbs free energy: \( \Delta G^\circ = -RT \ln K \). This equation connects the spontaneity of a reaction (or phase change) to the position of equilibrium.

• \( R \) in the equation is the universal gas constant, \( 8.314 \, \mathrm{J/(mol \cdot K)} \).
• \( T \) represents the temperature in Kelvin, hence the need for temperature conversion if given in Celsius.

For the process of water vaporizing, converting \( \Delta G^\circ \) to \( K \) helps determine the vapor pressure of water in atmospheric units. Since \( K \) represents the vapor pressure for this liquid to gas transition, calculating it provides insight into the system's behavior at equilibrium.

Phase Transition

Phase transitions occur when a substance changes from one state of matter to another, such as solid to liquid or liquid to gas. In our scenario, we are interested in the transition of water from liquid (\( \mathrm{H}_2\mathrm{O}(l) \)) to gas (\( \mathrm{H}_2\mathrm{O}(g) \)). This particular phase transition encompasses vaporization, where molecules escape from a liquid and enter into a gaseous state.

• This transformation is significant because it involves a change in energy and entropy.
• The boiling point of water at 1 atm is 100°C, but below this temperature, the transition isn't spontaneous without energy input, as indicated by the positive \( \Delta G^\circ \).

Understanding phase transitions helps explain why certain substances require specific conditions to change states and is pivotal for predicting natural phenomena and industrial processes.

Temperature Conversion

In scientific calculations, especially those involving thermodynamics, converting temperatures to Kelvin from Celsius is vital. Calculations with formulas like \( \Delta G^\circ = -RT \ln K \) necessitate temperature to be in Kelvin for consistency and accuracy. The Kelvin scale starts at absolute zero, making it crucial for equations involving energy.

• To convert Celsius to Kelvin, simply add \( 273.15 \) to the Celsius temperature.
• For our exercise, converting \( 25^\circ \mathrm{C} \) to Kelvin gives us \( T = 298.15 \, \mathrm{K} \).

Being comfortable with temperature conversions ensures precision in calculations, which is fundamental to successful problem-solving in chemistry and physics questions.