Question

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

Short answer

The temperature at which solids I and II of O₂ are in equilibrium is \(43.7 \mathrm{~K}\), which is found using the Gibbs free energy change formula and the given enthalpy and entropy change values for the phase transition.

Step-by-step solution

Write the Gibbs free energy change formula

Gibbs free energy change (ΔG) is related to enthalpy change (ΔH) and entropy change (ΔS) by the following equation: ΔG = ΔH - TΔS At equilibrium, the Gibbs free energy change (ΔG) for the phase transition is zero, so we have 0 = ΔH - TΔS

Rearrange the formula to solve for temperature

We need to find the temperature (T) at which the two solids are in equilibrium, so we rearrange the equation to solve for T: T = ΔH / ΔS

Plug in the given values for ΔH and ΔS

We are given that the enthalpy change (ΔH) for the phase transition is -743.1 J/mol and the entropy change (ΔS) is -17.0 J/(K⋅mol). Substituting these values into the formula, we get: T = (-743.1 J/mol) / (-17.0 J/(K⋅mol))

Calculate the temperature

Now, calculate the temperature where the two phases are in equilibrium: T = 43.7 K Therefore, at a temperature of \(43.7 \mathrm{~K}\), solids I and II of O₂ are in equilibrium.

Key concepts

Phase Transition

A phase transition occurs when a substance changes its state, such as from solid to liquid or liquid to gas. In the exercise above, the phase transition involves the rearrangement of the crystal structure between two solid phases of \(\mathrm{O}_{2}\), known as Solid I and Solid II. These transitions can happen when energy is added or removed from the system. The transition from Solid I to Solid II is unique because both are solids, yet they differ in their atomic arrangements or structure.

Key points to remember about phase transitions in this context:

• They are characterized by changes in enthalpy and entropy.
• The phase transition does not involve a change in the physical state (e.g., from solid to liquid), but a change in structure.
• Gibbs free energy change (\(\Delta G\)) is zero at equilibrium during a phase transition.

Enthalpy Change

Enthalpy change (\(\Delta H\)) measures the heat absorbed or released during a phase transition at constant pressure. For the transition from Solid I to Solid II in the exercise, \(\Delta H\) is given as \(-743.1 \mathrm{~J/mole}\). This negative value indicates an exothermic process where heat is released as Solid I rearranges into Solid II.

In general, enthalpy change provides us with insight into the amount of energy needed for the transition. In this case:

• Negative \(\Delta H\): Energy is released, implying that the new phase (Solid II) is lower in energy compared to Solid I.
• This release of energy often results in a more stable structural arrangement.

Entropy Change

Entropy change (\(\Delta S\)) reflects the degree of disorder or randomness in a system. For the given transition, \(\Delta S\) is \(-17.0 \mathrm{~J/(K \cdot mole)}\), which suggests that as Solid I transitions to Solid II, the system's disorder decreases. In simple terms, the molecules in Solid II are more orderly arranged than in Solid I.

Understanding entropy change can help us appreciate:

• Negative \(\Delta S\) indicates a decrease in randomness or disorder of the system.
• A lower entropy in Solid II implies a more ordered and possibly tighter crystal lattice.
• Changes in entropy tell us about the feasibility and spontaneity of the phase transition when considered alongside enthalpy changes.

Equilibrium Temperature

The equilibrium temperature is the point at which two phases coexist in balance. For the transition between Solids I and II, this temperature is where their Gibbs free energy is equivalent, reflected by \(\Delta G = 0\). At this point, no net energy is exchanged, meaning the transition can readily occur in either direction.

As detailed in the solution:

• The equilibrium temperature is calculated using the equation \([T = \Delta H / \Delta S]\).
• For the transition described, it's \([43.7 \mathrm{~K}]\), indicating the temperature at which both solid structures are equally stable.
• This concept helps in understanding at what conditions, especially temperatures, a system favors a particular structural arrangement.