As \(\mathrm{O}_{2}(l)\) 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 \(-17.0 \mathrm{~J} / \mathrm{K} \cdot\) mol. At what temperature are solids I and II in equilibrium?

Short Answer

Expert verified
Solid phases I and II of O2 are in equilibrium at 43.7 K. This is found by setting the Gibbs free energy equation (\(ΔG = ΔH - TΔS\)) equal to 0 and solving for temperature T. Given \(ΔH = -743.1\ J/mol\) and \(ΔS = -17.0\ J/K \cdot mol\), equilibrium is achieved when \(T = \frac{743.1\ J/mol}{17.0\ J/K \cdot mol} = 43.7\ K\).

Step by step solution

01

Understand Gibbs free energy

In a spontaneous reaction, the Gibbs free energy change (ΔG) is negative. At equilibrium, ΔG = 0, this means that there is no net change in the system, and both phases coexist. So we have to find the temperature at which ΔG = 0 for the I → II phase transition.
02

Apply Gibbs free energy formula

Using the Gibbs free energy equation: \(ΔG = ΔH - TΔS\) Since we're looking for the temperature at equilibrium: \(0 = ΔH - TΔS\) We have \(ΔH = -743.1\ J/mol\) and \(ΔS = -17.0\ J/K \cdot mol\). Insert these values into the equation and solve for T (the temperature).
03

Solve for T

Now, we need to solve the equation for the temperature (T): \(0 = -743.1\ J/mol - T \times (-17.0\ J/K \cdot mol)\) Rearrange this equation to find T: \(T = \frac{743.1\ J/mol}{17.0\ J/K \cdot mol}\) Divide the terms to get the temperature: \(T = 43.7\ K\)
04

Conclusion

Solid phases I and II of O2 are in equilibrium at 43.7 K. At this temperature, both phases coexist due to the balance between enthalpy and entropy changes in the phase transition.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs free energy, symbolized as \textbf{G}, is a thermodynamic quantity that can predict the direction of a chemical reaction under constant temperature and pressure conditions. It is defined by the equation \( G = H - TS \) where \( H \) is the enthalpy, \( T \) is the temperature in Kelvin, and \( S \) is the entropy. A negative change in Gibbs free energy (\(\Delta G < 0\)) signals a spontaneous reaction, while a positive change (\(\Delta G > 0\)) implies non-spontaneity. At equilibrium, \(\Delta G\) is zero, indicating no net change in a system's free energy and thus, no further spontaneous reactions will occur.

Understanding the role of Gibbs free energy is crucial when studying phase transitions, as it helps identify the temperature and pressure at which two phases, such as solids I and II of \(\mathrm{O}_{2}\), coexist without the net transfer of matter from one phase to the other.
Enthalpy Change (\(\Delta H\))
Enthalpy change, expressed as \(\Delta H\), is a measure of heat exchange in a system at constant pressure. It is one of the key factors affecting the Gibbs free energy of the system. An endothermic reaction, where the system absorbs heat, results in a positive \(\Delta H\), while an exothermic reaction, where the system releases heat, leads to a negative \(\Delta H\).

In the context of the I → II phase transition of \(\mathrm{O}_{2}\), the negative enthalpy change (\(\Delta H = -743.1\ J/mol\)) signifies that the transition from solid phase I to solid phase II releases heat to the surroundings, making it an exothermic process. It's important to note that enthalpy change alone does not determine spontaneity, but when combined with entropy change, it greatly influences the Gibbs free energy of a reaction.
Entropy Change (\(\Delta S\))
Entropy change, denoted by \(\Delta S\), quantifies the disorder or randomness in a system. A positive \(\Delta S\) indicates that the system has become more disordered, and a negative value implies a decrease in disorder. Entropy plays a fundamental role in determining the spontaneity of a process according to the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time.

For the phase transition from solid I to solid II of \(\mathrm{O}_{2}\), the negative entropy change (\(\Delta S = -17.0\ J/K \cdot mol\)) tells us that the system becomes more ordered as the transition takes place. This change in entropy is a critical factor in finding the equilibrium temperature where the Gibbs free energy change for the transition is zero.
Spontaneous Reaction
A spontaneous reaction is a process that occurs naturally without external intervention. Such reactions are characterized by the release of free energy, resulting in a negative Gibbs free energy change (\(\Delta G < 0\)). However, not all spontaneous reactions happen quickly; some may occur over an extended period, such as the rusting of iron.

It's essential to recognize that spontaneity is determined by both the enthalpy and entropy changes in a system. While a highly exothermic reaction (large negative \(\Delta H\)) tends to be spontaneous, this is not a hard-and-fast rule. The entropy change (\(\Delta S\)) and the temperature of the system also impact spontaneity, as evidenced by the Gibbs free energy equation.
Equilibrium in Chemical Systems
Equilibrium in a chemical system is the state where the concentrations of all reactants and products remain constant over time. This occurs when the rates of the forward and reverse reactions are equal. For phase transitions, equilibrium signifies that the amount of each phase remains unchanged because the rate at which one phase turns into another is the same as the rate of the reverse transition.

In the case of solid phases I and II of \(\mathrm{O}_{2}\), equilibrium is reached at a specific temperature where the Gibbs free energy of the phase transition is zero. This temperature, called the equilibrium temperature, marks the point where both solid phases can coexist stably. The concept of equilibrium is vital for understanding and predicting the conditions necessary for the coexistence of different phases in a material.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{~g} / \mathrm{mol}\) ) was dissolved in \(150 . \mathrm{mL}\) of a solvent \(\left(\right.\) density \(\left.=0.879 \mathrm{~g} / \mathrm{cm}^{3}\right)\). The elevated boiling point of the solution was \(355.4 \mathrm{~K}\). What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is \(33.90 \mathrm{~kJ} / \mathrm{mol}\), the entropy of vaporization is \(95.95 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\), and the boiling-point elevation constant is \(2.5 \mathrm{~K} \cdot \mathrm{kg} / \mathrm{mol}\).

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(\mathrm{I}_{2}(g) \longrightarrow \mathrm{I}_{2}(s)\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free