Problem 15 - Chemistry: The Central Science in SI Units, Global Edition

Question

Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (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?

Step-by-Step Solution

Answer

The melting of ice to liquid water at a pressure of \(101.3\,\mathrm{kPa}\) is an endothermic process since it requires energy input (\(\Delta H > 0\)). The process is spontaneous in the temperature range where \(\Delta G < 0\) (higher temperatures), meaning \(T\Delta S > \Delta H\), and nonspontaneous in the temperature range where \(\Delta G > 0\) (lower temperatures), meaning \(T\Delta S < \Delta H\). The two phases are in equilibrium at a temperature of \(273.15\,\mathrm{K}\).

Step 1: (a) Determine if the process is endothermic or exothermic

To determine whether the process is endothermic or exothermic, we need to look at the change in enthalpy, denoted as \(\Delta H\). For melting of ice to water, the process requires energy input (usually as heat) to break down the hydrogen bonds between the water molecules in the solid state and create a liquid state. In an endothermic process, energy is absorbed, so \(\Delta H > 0\). Therefore, the melting of ice to liquid water at a pressure of \(101.3\,\mathrm{kPa}\) is an endothermic process.

Step 2: (b) Find the temperature range for the spontaneous process

To determine the temperature range for which the melting of ice is a spontaneous process, we need to examine the change in Gibbs free energy, or \(\Delta G\). The Gibbs free energy change can be calculated as follows: \(\Delta G = \Delta H - T\Delta S\) Where \(\Delta S\) is the change in entropy, and \(T\) is the temperature. A process is spontaneous if \(\Delta G < 0\). Since the process is endothermic from part (a), we know that \(\Delta H > 0\). Also, for the melting of ice, the entropy change is positive, as the solid ice transitions into a more disordered liquid water state, hence \(\Delta S > 0\). Therefore, if the product \(T\Delta S\) is larger than \(\Delta H\), then \(\Delta G\) will be negative, and the process will be spontaneous. This occurs at higher temperatures. So, the process is spontaneous in the temperature range where \(\Delta G < 0\), which means \(T\Delta S > \Delta H\) for higher temperatures.

Step 3: (c) Find the temperature range for the nonspontaneous process

Now, we need to find the temperature range where the melting process is nonspontaneous. A process is considered nonspontaneous when \(\Delta G > 0\). Since the process is endothermic and the melting of ice increases the entropy, \(\Delta H > 0\) and \(\Delta S >0\). If the product \(T\Delta S\) is less than \(\Delta H\), then \(\Delta G\) will be positive, and the process will be nonspontaneous. This occurs at lower temperatures. Thus, the process is nonspontaneous in the temperature range where \(\Delta G > 0\), which means \(T\Delta S < \Delta H\) for lower temperatures.

Step 4: (d) Determine the temperature at which the two phases are in equilibrium

At equilibrium, the Gibbs free energy change is zero, or \(\Delta G=0\). We can use the equation: \(\Delta G = \Delta H - T\Delta S\) Substituting \(\Delta G = 0\), we get: \(T\Delta S = \Delta H\) Since the Standard enthalpy of fusion for ice at \(1\,\mathrm{atm}\) is \(\Delta H = 6.01\,\mathrm{kJ/mol}\) and the Standard entropy of fusion for ice at \(1\,\mathrm{atm}\) is \(\Delta S = 21.97\,\mathrm{J/mol\cdot K}\), we can calculate the equilibrium temperature: \(T = \frac{\Delta H}{\Delta S} = \frac{6.01\times10^{3}\,\mathrm{J/mol}}{21.97\,\mathrm{J/mol\cdot K}} = 273.15\,\mathrm{K}\) So the equilibrium temperature between ice and liquid water at a pressure of \(101.3\,\mathrm{kPa}\) is \(273.15\,\mathrm{K}\).

Key Concepts

Endothermic processes

When we talk about endothermic processes, these are reactions or changes that absorb energy from their surroundings. A common example of an endothermic process is the melting of ice to form liquid water. To transition from a solid to a liquid, energy must be added to the system in the form of heat.

• This absorbed energy works to break the hydrogen bonds holding the water molecules together in the solid ice.
• As the bonds break, the structure becomes less rigid, leading to the liquid form.
• Because energy is absorbed during this transformation, the enthalpy change, \( \Delta H \), is positive.

This positive change in enthalpy signifies that the process requires an input of energy, a hallmark of endothermic reactions. Understanding this concept is crucial for analyzing many thermodynamic reactions and predicting spontaneous processes.

Gibbs free energy

Gibbs free energy is a concept used to predict whether a process or reaction will occur spontaneously under constant pressure and temperature. It combines enthalpy and entropy into a single value.

The Formula

The Gibbs free energy change is calculated using the formula:\[\Delta G = \Delta H - T\Delta S\]

• \( \Delta G \) is the Gibbs free energy change.
• \( \Delta H \) refers to the change in enthalpy.
• \( T \) represents temperature in Kelvin.
• \( \Delta S \) is the change in entropy.

A process is considered spontaneous if \( \Delta G < 0 \). For melting ice, the entropy change \( \Delta S \) is positive, as the system moves towards a more disordered state (from solid to liquid). This means that at higher temperatures, the term \( T\Delta S \) can outweigh \( \Delta H \), making \( \Delta G \) negative, thus ensuring spontaneity.
Understanding Gibbs free energy helps us determine under which conditions a thermodynamic process such as phase changes will naturally occur.

Phase equilibrium

Phase equilibrium denotes the condition where different phases of a substance coexist at equilibrium without any net change over time. For water, this involves the equilibrium between solid (ice) and liquid phases at a specific temperature and pressure.

Understanding Equilibrium

At equilibrium, the Gibbs free energy change \( \Delta G \) is zero. This means that the opposing transitions (from ice to water, and water to ice) occur at the same rate, maintaining balance.

• Using the formula \( \Delta G = \Delta H - T\Delta S \).
• At equilibrium, we set \( \Delta G = 0 \).
• Solving gives \( T = \frac{\Delta H}{\Delta S} \).

For water at standard atmospheric pressure, this equilibrium occurs at 273.15 K (or 0°C), which is the melting/freezing point. At this point, any heat added won't change the temperature until all ice melts.
Comprehending phase equilibrium is essential to understand how and why materials change from one state to another under steady conditions.