Problem 85 - Chemistry An Atoms First Approach

Question

As \(\mathrm{O}_{2}(l)\) is cooled at 1 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 \(\mathrm{I} \rightarrow\) II phase transition is \(-743.1 \mathrm{J} / \mathrm{mol}\), and \(\Delta S\) for the same transition is \(-17.0 \mathrm{J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Step-by-Step Solution

Answer

The temperature at which solids I and II of \(\mathrm{O}_{2}(l)\) are in equilibrium is 43.7 K. This means that at this temperature, both solid phases can coexist without further change.

Step 1: Write the equation for Gibbs free energy change for the phase transition

We know that \(\Delta G = \Delta H - T\Delta S\), where \(\Delta G\) is the change in Gibbs free energy, \(\Delta H\) is the change in enthalpy, T is the temperature in Kelvin, and \(\Delta S\) is the change in entropy.

Step 2: Set \(\Delta G\) to 0 and plug in given values

Since the solids are in equilibrium, we can set \(\Delta G = 0\). Plug in the given values for \(\Delta H\) and \(\Delta S\): 0 = -743.1 \(\mathrm{J/mol}\) - T(-17.0 \(\mathrm{J/K\cdot mol}\))

Step 3: Solve for the equilibrium temperature T

Rearrange the equation to solve for T: T(-17.0 \(\mathrm{J/K\cdot mol}\)) = -743.1 \(\mathrm{J/mol}\) Now, divide both sides by -17.0 \(\mathrm{J/K\cdot mol}\): T = \(\frac{-743.1 \mathrm{J/mol}}{-17.0 \mathrm{J/K\cdot mol}}\) T ≈ 43.7 K

Step 4: State the answer and the meaning in context

The temperature at which solids I and II of \(\mathrm{O}_{2}(l)\) are in equilibrium is 43.7 K. This means that at this temperature, both solid phases can coexist without further change.

Key Concepts

Phase Transition

A phase transition occurs when a substance changes from one state of matter to another, such as a liquid turning into a solid or vice versa. During a phase transition, the physical properties of a substance change, but its chemical composition remains the same. In the context of our exercise, oxygen undergoes a phase transition from liquid to solid as it cools and then transitions between two different solid states.

Key points to understand about phase transitions are:

• They involve energy changes, which are either absorbed or released.
• The temperature remains constant during the transition.
• They depend on pressure, temperature, and other environmental conditions.

The phase transition from solid I to solid II in oxygen is particularly interesting because it involves a rearrangement of atoms resulting in a distinct crystal structure.

Enthalpy Change

The enthalpy change ((\(\Delta H\)),) associated with a phase transition, is the amount of heat absorbed or released when a substance undergoes a phase change at constant pressure. In this context, we calculate it for the transition between two solid forms of oxygen.

For the transition from solid I to solid II, (\(\Delta H = -743.1 \text{ J/mol}\)), indicating that the process releases heat, thus it is exothermic.

• Negative (\(\Delta H\)) values mean the system releases energy to the surroundings.
• Positive (\(\Delta H\)) values indicate energy absorption.

Tracking the enthalpy changes during such transitions provides insights into the energy dynamics of the reaction.

Entropy Change

Entropy ((\(S\))) is a measure of the disorder or randomness in a system. The change in entropy ((\(\Delta S\))) during a phase transition helps us understand how the molecular order changes. For the transition between two solid forms of oxygen, (\(\Delta S = -17.0 \text{ J/K . mol}\)) reflects a decrease in entropy, indicating an increase in order as the molecules move to a more structured form.

• A negative (\(\Delta S\)) suggests that the transition favors a more orderly arrangement.
• Positive (\(\Delta S\)) implies increasing disorder or randomness.

Understanding entropy changes is crucial as they are an integral part of the Gibbs free energy equation and influence the spontaneity of processes.

Equilibrium Temperature

Equilibrium temperature is the point at which the phase transition can occur without net change in the amount of phases. At this temperature, both phases coexist and balance each other in energetic terms.

In terms of Gibbs free energy ((\(\Delta G\))), equilibrium is reached when the free energy difference between the phases is zero, i.e., (\(\Delta G = 0\)). Here is how to calculate it:

• Use the equation (\(\Delta G = \Delta H - T\Delta S\)).
• Set (\(\Delta G = 0\)), and plug in values to find (\(T\)).
• For oxygen's solid phase transition, the equilibrium temperature is calculated to be 43.7 K.

This temperature is where solids I and II are in equilibrium, meaning both solid states can exist together without spontaneous transformation into one another.