Question

As \(\mathrm{O}_{2}(l)\) is cooled at \(1 \mathrm{atm},\) it freezes at 54.5 \(\mathrm{K}\) to form solid I. At a lower temperature, solid I rearranges to solid II, which has a different crystal structure. Thermal measurements show that \(\Delta H\) for the \(I \rightarrow\) II phase transition is \(-743.1 \mathrm{J} / \mathrm{mol}\) , and \(\Delta S\) for the same transition is \(-17.0 \mathrm{J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Short answer

The temperature at which solids I and II are in equilibrium can be found using the Clapeyron equation, \( \Delta G = \Delta H - T \Delta S = 0 \). We are given ∆H = -743.1 J/mol, and ∆S = -17.0 J/(K⋅mol). Rearrange the equation to get T: \( T = \frac{\Delta H}{\Delta S} \). Substitute the given values and solve for T: \( T = \frac{-743.1 \, \mathrm{J/mol}}{-17.0 \, \mathrm{J/(K \cdot mol)}} = 43.7 \, \mathrm{K} \). The equilibrium temperature for solids I and II is 43.7 K.

Step-by-step solution

Identify the Clapeyron Equation

In this case, since we're dealing with phase transitions, the equilibrium condition in terms of temperature can be found using the Clapeyron equation, which is: \( \Delta G = \Delta H - T \Delta S = 0 \) where, ∆G is the Gibbs free energy change, ∆H is the enthalpy change, T is the temperature, and ∆S is the entropy change. Since equilibrium implies that ∆G = 0, we will rearrange the above equation to solve for T.

Rearrange the Clapeyron Equation to Solve for T

Rearrange the Clapeyron equation to get T: \( T = \frac{\Delta H}{\Delta S} \) We are given ∆H = -743.1 J/mol, and ∆S = -17.0 J/(K⋅mol). Now we can substitute these values into the equation and solve for T.

Substitute the Given Values and Solve for T

Now substitute the given ∆H and ∆S values into the equation: \( T = \frac{-743.1 \, \mathrm{J/mol}}{-17.0 \, \mathrm{J/(K \cdot mol)}} \) Solve for T: \( T = 43.7 \, \mathrm{K} \)

State the Equilibrium Temperature for Solids I and II

The temperature at which solids I and II are in equilibrium is found to be 43.7 K.

Key concepts

Phase Transition

A phase transition occurs when a substance undergoes a change from one state of matter to another. This could involve transitions such as solid to liquid (melting), liquid to gas (vaporization), or solid to solid, as with our example of oxygen transitioning between two different crystal structures. These transitions are characterized by changes in thermodynamic properties such as enthalpy (ΔH) and entropy (ΔS).

In the context of thermodynamics, during a phase transition, the substance absorbs or releases energy, often in the form of heat, while its temperature remains constant. It's important to note that these phase transitions occur at specific conditions of pressure and temperature, known as phase equilibrium.

• Melting occurs when a solid turns into a liquid.
• Freezing happens when a liquid becomes a solid.
• The transition from one solid phase to another, as in the oxygen example, involves a rearrangement of the crystal structure.

Understanding phase transitions helps in predicting the behavior of substances under different physical conditions. This knowledge plays a critical role in many industries, from metallurgy to the synthesis of new materials.

Gibbs Free Energy

Gibbs Free Energy, often denoted as G, is a thermodynamic potential that measures the maximum reversible work a thermodynamic system can perform at constant temperature and pressure. It is a way to predict the direction of phase transitions and chemical reactions.

The concept of Gibbs Free Energy is crucial when dealing with phase transitions, as it allows us to determine the conditions under which a phase change will occur by its equilibrium condition, which is when the change in Gibbs Free Energy (\( \Delta G \)) is zero. This directly relates to the Clapeyron equation, simplifying the calculation of equilibrium conditions.

• If \( \Delta G \) is negative, the process or reaction is spontaneous.
• If \( \Delta G \) is positive, the process is non-spontaneous.
• When \( \Delta G = 0 \), the system is in equilibrium. For phase transitions, this means that the two phases are balanced.

Understanding Gibbs Free Energy helps one predict not just equilibrium conditions but also spontaneity and feasibility of reactions under given circumstances.

Clapeyron Equation

The Clapeyron Equation is a way to describe the relationship between temperature, pressure, and volume change during a phase transition. It is particularly useful in phase equilibrium scenarios, like in our example of oxygen's solid-state transformation.

For phase transitions like solid to solid (as a change in crystal structure), the simplified form of the Clapeyron equation is used in terms of Gibbs Free Energy:\[ \Delta G = \Delta H - T \Delta S = 0 \]

This equation allows us to solve for the temperature at which two phases are in equilibrium, utilizing the relationship that the change in enthalpy (\( \Delta H \)) and entropy (\( \Delta S \)) determine the temperature.

• Rearranging gives us \( T = \frac{\Delta H}{\Delta S} \).
• This relationship is crucial for finding equilibrium points during phase changes.
• The Clapeyron Equation simplifies the understanding of phase transitions by directly connecting thermodynamic properties like enthalpy and entropy with equilibrium temperature.

Therefore, mastering the Clapeyron Equation is essential for those studying chemical thermodynamics and the behavior of substances under varying conditions.