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Provide examples to support the claim that processes involving the breaking and forming of intermolecular attractions are reversible.

Short Answer

Expert verified
Processes involving intermolecular attractions, such as melting/freezing and dissolving/crystallization, are reversible because they can proceed both forward and backward.

Step by step solution

01

Understanding Intermolecular Attractions

Intermolecular attractions refer to forces that hold molecules together, such as hydrogen bonds, Van der Waals forces, and dipole-dipole interactions. These forces are typically weaker than covalent or ionic bonds within a molecule.
02

Introducing Reversible Processes

Reversible processes are those that can proceed in both directions — forward and backward — such that the system can return to its original state by reversing the process.
03

Example 1 - Melting and Freezing

Consider water ice melting into liquid water. Ice melts (breaking hydrogen bonds) when heated. If the water is then cooled, it refreezes, with hydrogen bonds reforming between water molecules, demonstrating reversibility.
04

Example 2 - Sublimation and Deposition

Dry ice (solid CO2) undergoes sublimation to gas as intermolecular forces are overcome. If the gaseous CO2 is cooled, it deposits back into solid form, restoring the intermolecular attractions.
05

Example 3 - Dissolving and Crystallization

Salt dissolving in water is an example where ionic bonds in salt are broken and ion-dipole attractions are formed. Upon evaporation of water, salt molecules recrystallize, re-establishing ionic bonds.

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Key Concepts

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

Reversible Processes
Reversible processes are fascinating because they illustrate the dynamic nature of molecular interactions. These processes can move forward and backward, allowing a system to return to its original state. In the microscopic world of molecules, this means that the attractions between them can be rearranged only to be returned to their initial configurations. For instance, when ice melts into water, hydrogen bonds break, and if frozen again, the molecules re-establish those hydrogen bonds, returning to solid ice. The ability to reverse these steps accurately makes such processes invaluable in both natural phenomena and industrial applications.
Hydrogen Bonds
Hydrogen bonds play a crucial role in many reversible processes. These are special types of dipole-dipole attractions that occur when hydrogen is covalently bonded to a more electronegative element, creating a partially positive charge. This positive charge is attracted to a negative charge nearby, often from oxygen, nitrogen, or fluorine, forming a hydrogen bond.
  • Hydrogen bonds are weaker than covalent bonds within molecules but significant enough to influence the physical state (solid, liquid, gas) of substances like water.
  • These bonds are responsible for water's high boiling point and its solid form being less dense than its liquid form, which is unique among molecules.
Understanding hydrogen bonding is essential for explaining how certain materials respond to temperature changes, as they are easily broken and reformed, leading to reversible processes such as melting and freezing.
Sublimation and Deposition
Although we don't often observe them, sublimation and deposition are quite common in nature. Sublimation refers to the transition of a substance from a solid directly to a gas, bypassing the liquid phase. Deposition is the reverse, where a gas turns directly into a solid. This occurs when molecules gain or lose energy quickly enough that they bypass the intermediate liquid state.
  • The process of sublimation requires energy input to overcome intermolecular forces binding molecules in the solid.
  • During deposition, the release of energy allows gases to form solid structures, often in a crystalline pattern, as seen with substances like frost.
These reversible processes are evident in materials such as dry ice (solid carbon dioxide), which sublimates at room temperature and deposits back into the solid form under cold conditions, showcasing the reversibility of intermolecular attractions.
Dissolving and Crystallization
Dissolving and crystallization clearly highlight the role of intermolecular forces in reversible processes. During dissolution, a solute like salt is mixed in a solvent like water. The ionic bonds in the salt are disrupted, and new, weaker ion-dipole attractions form between the salt ions and water molecules.
  • As water evaporates, these weaker interactions are undone, and the salt molecules come back together, recrystallizing into their solid form.
  • The ability to switch back and forth between these states emphasizes the reversible nature of breaking and forming molecular attractions.
This principle is used not only in culinary applications like making saltwater but also in manufacturing, where crystallization processes help develop pure substances from solutions.

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Most popular questions from this chapter

Why is it useful to know the value of the equilibrium constant \(K\) for a reversible process?

Imagine that you dissolve sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s),\) in water. a. Write a balanced equation for this reversible process. b. In which direction does the process proceed if you place a tiny amount of sugar in a large pot filled with water? c. In which direction does the process proceed if you allow the water to evaporate from a sugar solution?

When phosphorus pentachloride, \(\mathrm{PCl}_{5}(g),\) is placed in a sealed container, it breaks apart reversibly. h e concentrations of the starting substance and products are monitored. Use the data in the table to create a graph showing the concentrations of the starting substance and products over time. $$\mathrm{PCl}_{5}(g) \leftrightharpoons \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)$$ \(\begin{array}{|c|c|c|c|}\hline \text{ Time (s)} & \text{ \)\left[\mathbf{P} \mathrm{I}_{\mathrm{5}}\right](\mathrm{mol} / \mathrm{L})\(} & \text{ \)\left[\mathbf{P} \mathrm{I}_{\mathrm{3}}\right](\mathrm{mol} / \mathrm{L})\(} & \text{ \)\left[\mathrm{Cl}_{2}\right](\mathrm{mol} / \mathrm{L})\(} \\ \hline 0 & {0.00} & {1.00} & {1.00} \\ \hline 20 & {0.10} & {0.90} & {0.90} \\ 40 & {0.20} & {0.80} & {0.80} \\ \hline 60 & {0.25} & {0.75} & {0.75} \\ \hline 80 & {0.29} & {0.71} & {0.71} \\ \hline 100 & {0.29} & {0.71} & {0.71} \\\ \hline 120 & {0.29} & {0.71} & {0.71} \\ \hline\end{array}\) a. What are the equilibrium concentrations of the starting substance and the products? b. At what time did the mixture reach equilibrium? How do you know? c. At 20 seconds, which process is faster, the forward or the reverse process? At 100 seconds? d. Why is the amount of starting substance, \(\mathrm{PCl}_{5},\) not equal to the amounts of each of the products, \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2},\) at equilibrium? Support your answer with evidence. e. Why are the amounts of \(\mathrm{PCl}_{3}\) and \(\mathrm{Cl}_{2}\) always equal to one another?

Imagine that you breathe in oxygen and it binds to hemoglobin in your blood to form \(\left[\mathrm{O}_{2} : \text { hemoglobin }\right] .\) a. Write a balanced equation for the process. b. The \(\left[\mathrm{O}_{2} : \text { hemoglobin delivers oxygen to your cells. Describe this process. }\right.\) c. Provide evidence that this process is reversible.

Methane, \(\mathrm{CH}_{4}(g),\) burns in oxygen, \(\mathrm{O}_{2}(g),\) to produce carbon dioxide, \(\mathrm{CO}_{2}(g),\) and water, \(\mathrm{H}_{2} \mathrm{O}(l) .\) a. Write a balanced equation for this process. b. This process is described as irreversible. What does this mean?

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