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If a mixture consisting of many small crystals in contact with a saturated solution is studied for a period of time, the smallest of the crystals are observed to dissolve while the larger ones grow even larger. Explain this phenomenon in terms of a dynamic equilibrium.

Short Answer

Expert verified
In a dynamic equilibrium, smaller crystals dissolve more easily due to their higher surface energy, while larger crystals grow as the solute from the smaller crystals deposits onto them.

Step by step solution

01

Understanding Dynamic Equilibrium

Recognize that a dynamic equilibrium exists when the rate of the forward reaction (crystals dissolving) is equal to the rate of the backward reaction (crystals forming). In a saturated solution, this means that, over time, there is a constant exchange of the substance between the solid and dissolved states.
02

Explaining Dissolution of Smaller Crystals

Acknowledge that smaller crystals have a higher surface area to volume ratio compared to larger ones. This results in higher surface energy, making them more soluble. At dynamic equilibrium, these smaller crystals tend to dissolve more readily to reduce their surface energy.
03

Describing the Growth of Larger Crystals

Understand that as the smaller crystals dissolve, additional solute particles become available in the solution. These particles are more likely to be deposited onto the larger crystals, which are less energetically costly to add to due to their lower surface area to volume ratio. As a result, the larger crystals tend to grow.

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Key Concepts

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

Saturated Solution
A saturated solution is a key concept in understanding various chemical processes, including dynamic equilibrium in crystal growth. It's a scenario where the solution contains the maximum concentration of solute that can be dissolved at a given temperature. Additional solute will not dissolve in a saturated solution under stable conditions; instead, it reaches a state of balance.

Factors like temperature and pressure can affect saturation levels—raising the temperature typically increases a solute's solubility before once again reaching a new saturation point. Think of it as a crowded room; once there's no more space, no additional people can enter unless the room itself expands.
Dissolution of Crystals
When crystals dissolve, they transition from the solid to the aqueous phase, which we call dissolution. It's essentially the breaking down of the crystal lattice structure and the dispersion of its particles throughout the solvent. This behavior is influenced by factors such as temperature, the nature of the solute and solvent, and the presence of other ions or compounds in the solution.

Within a saturated solution, crystals dissolve until they reach a point where the solid and dissolved phases are in equilibrium. This dissolution process is crucial for many biological and environmental functions, as well as in industrial applications where precise control over solubility is necessary.
Surface Area to Volume Ratio
The surface area to volume ratio is a vital concept to grasp when discussing dynamic equilibrium in crystal growth. Smaller crystals have a higher ratio compared to larger ones. Imagine breaking a large piece of chalk into smaller pieces—the total surface area increases relative to the volume.

This increased ratio means more particle exposure to the solvent, leading to enhanced dissolution rates. Conversely, larger crystals with a smaller surface area to volume ratio are less prone to dissolve, favoring crystal growth instead. This concept is also pivotal in various biological processes and industrial applications, particularly where the efficiency of reactions is dependent on the surface area available.
Solubility
Solubility is the property that determines how much of a solute can be dissolved in a solvent at a particular temperature and pressure. Understanding solubility is crucial for grasping the dynamics of crystal creation and dissolution. It's not a static figure; rather, it's influenced by environmental conditions.

The solubility of a substance is a key player in determining whether a solution becomes saturated and how dynamic equilibrium is established in a given system. When dealing with solutions, the solubility can predict how substances react under changing conditions, such as temperature fluctuations or the addition of other chemical species.
Crystal Growth
Crystal growth occurs when the particles dissolved in a solution begin to form a solid structure, under certain conditions. This process is dictated by the rate at which the solute particles leave the solution and attach themselves to the crystal surface. As hinted in the solution, larger crystals grow at the expense of smaller ones in a saturated solution due to their favorable surface energy situations.

In practical applications, the controlled growth of crystals is paramount in manufacturing semiconductors and pharmaceuticals. Knowing how to manipulate conditions to facilitate appropriate crystal growth is vital in these industries to ensure high-quality products.

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Most popular questions from this chapter

Suppose we set up a system in which water is poured into a vessel having a hole in the bottom. If the rate of water inflow is adjusted so that it matches the rate at which water drains through the hole, the amount of water in the vessel remains constant over time. Is this an equilibrium system? Explain.

How will the value of \(K_{\mathrm{P}}\) for the following reactions be affected by an increase in temperature? (a) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) $$ \Delta H^{\circ}=-18 \mathrm{~kJ} $$ (b) \(\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons 3 \mathrm{NO}(g)\) $$ \Delta H^{\circ}=+155.7 \mathrm{~kJ} $$ (c) \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{NOCl}(g)\) \(\Delta H^{\circ}=-77.07 \mathrm{~kJ}\)

The reaction \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) has a \(K_{\mathrm{c}}=\) \(2.0 \times 10^{9}\) at \(25^{\circ} \mathrm{C}\). If \(0.100 \mathrm{~mol}\) of \(\mathrm{H}_{2}\) and \(0.200 \mathrm{~mol}\) of \(\mathrm{Br}_{2}\) are placed in a \(10.0 \mathrm{~L}\) container, what will all the equilibrium concentrations be at \(25^{\circ} \mathrm{C}\) ? (Hint: Where does the position of equilibrium lie when \(K\) is very large?)

Calculate the molar concentration of water in (a) \(18.0 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}(l),\) (b) \(100.0 \mathrm{~mL}\) of \(\mathrm{H}_{2} \mathrm{O}(l),\) and \(\mathbf{( c )} 1.00 \mathrm{~L}\) of \(\mathrm{H}_{2} \mathrm{O}(l)\). Assume that the density of water is \(1.00 \mathrm{~g} / \mathrm{mL}\)

Write the equilibrium law corresponding to \(K_{c}\) for each of the following heterogeneous reactions. (a) \(\mathrm{CaCO}_{3}(s)+\mathrm{SO}_{2}(g) \rightleftharpoons \mathrm{CaSO}_{3}(s)+\mathrm{CO}_{2}(g)\) (b) \(\mathrm{AgCl}(s)+\mathrm{Br}^{-}(a q) \rightleftharpoons \mathrm{AgBr}(s)+\mathrm{Cl}^{-}(a q)\) (c) \(\mathrm{Cu}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{Cu}^{2+}(a q)+2 \mathrm{OH}^{-}(a q)\) (d) \(\mathrm{Mg}(\mathrm{OH})_{2}(s) \rightleftharpoons \mathrm{MgO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (e) \(3 \mathrm{CuO}(s)+2 \mathrm{NH}_{3}(g) \rightleftharpoons\) \(3 \mathrm{Cu}(s)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)\)

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