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Osmium dioxide occurs either as a black powder or as brown crystals. The density of the black powder form is \(7.7 \mathrm{~g} \cdot \mathrm{cm}^{-3}\) and the density of the brown crystalline form is \(11.4 \mathrm{~g} \cdot \mathrm{cm}^{-3}\). Which is the more stable form at high pressure?

Short Answer

Expert verified
The brown crystalline form is more stable at high pressure.

Step by step solution

01

Understand the Concept

The stability of a form of a substance under high pressure depends on its density. Higher density materials are generally more stable at high pressures. This is because higher density indicates that the atoms or molecules are packed more closely together, making it harder for the pressure to cause changes to the structure.
02

Compare the Densities

Compare the densities of the two forms of osmium dioxide given in the problem. The density of the black powder form is \(7.7 \text{ g} \cdot \text{cm}^{-3}\) and the density of the brown crystalline form is \(11.4 \text{ g} \cdot \text{cm}^{-3}\).
03

Determine Stability Based on Density

Since the brown crystalline form has a higher density of \(11.4 \text{ g} \cdot \text{cm}^{-3}\), compared to \(7.7 \text{ g} \cdot \text{cm}^{-3}\) for the black powder form, the brown crystalline form will be more stable under high pressure.

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Key Concepts

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

Density and Its Impact on Stability
Density is a measure of how much mass is contained in a given volume. In simple terms, it can be described by the formula \( \text{Density} = \frac{\text{Mass}}{\text{Volume}} \). When considering phase stability, particularly under varying conditions like high pressure, the density of a material plays a crucial role.

A higher-density material means that its particles—atoms or molecules—are packed more closely together. This tight packing helps the material resist changes when subjected to external forces such as pressure. In the case of osmium dioxide, the brown crystalline form has a density of \( 11.4 \text{ g/cm}^3 \), whereas the black powder form has a density of \( 7.7 \text{ g/cm}^3 \). Thus, the brown crystalline form is more compact.

Why does this matter for stability? High density implies that the material can endure higher external pressures without changing its structure. When comparing two forms of a substance, the form with greater density often proves to be the more stable one under high pressure conditions.
High Pressure and Material Behavior
High pressure is an external force that can significantly affect the stability of materials. It can compact the structure of materials, altering their properties and sometimes even forcing them into a new phase with different attributes.
  • At high pressures, materials with a dense structure are better able to maintain stability because their closely packed particles resist further compression.
  • Conversely, less dense materials might experience more dramatic structural changes under high pressure, sometimes leading to phases that are less stable or entirely different.
For osmium dioxide, the brown crystalline form is more stable than the black powder form under high-pressure conditions. This is due to its higher initial density, which makes it naturally more resistant to structural changes when pressure is applied. Understanding how materials react to pressure helps in various fields, from geology, where pressures affect rock formations deep within Earth, to materials science, where high-pressure conditions are used to synthesize new compounds.
Crystalline Structure and Its Relevance
Crystalline structures are arrangements of atoms or molecules in a highly ordered manner, extending in a repeating pattern through space. This orderliness is central to understanding the stability and physical properties of materials.

The crystalline form of osmium dioxide is not just more dense; it's more ordered than the powder form. This high degree of order means that when external pressures are applied, the forces are distributed uniformly, thus maintaining structural integrity much better.
  • Crystals form when particles assemble in a consistent pattern. This pattern helps in maintaining solid strength and resistance to deformation.
  • In contrast, amorphous structures—like powders—lack this regular pattern, leading to less overall strength when subjected to stress.
Therefore, in high-pressure environments, crystalline forms like the brown osmium dioxide maintain their stability better than their amorphous counterparts, emphasizing the role of crystalline structure in the broader context of phase stability.

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Most popular questions from this chapter

Consider the chemical equilibrium described by the equation $$ 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \leftrightharpoons 2 \mathrm{NOBr}(g) $$ Use Le Châtelier's principle to predict the effect on the equilibrium concentration of \(\operatorname{NOBr}(g)\) and of \(\mathrm{NO}(g)\) resulting from (a) an increase in the concentration of \(\mathrm{Br}_{2}(g)\) (b) a twofold decrease in the reaction volume

Sodium hydrogen carbonate, commonly called sodium bicarbonate, is used in baking soda and in fire extinguishers as a source of \(\mathrm{CO}_{2}(g) .\) It decomposes according to the equation $$ 2 \mathrm{NaHCO}_{3}(s) \leftrightharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Given that \(K_{\mathrm{p}}=0.26 \mathrm{bar}^{2}\) at \(125^{\circ} \mathrm{C}\), calculate the partial pressures of \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) in units of bars at equilibrium when \(\mathrm{NaHCO}_{3}(s)\) is heated to \(125^{\circ} \mathrm{C}\) in a closed vessel.

Prior to learning about equilibrium states, we solved stoichiometric problems using the concept of "limiting reactants." Under what conditions does the method of limiting reactants apply?

Consider the chemical equilibrium described by the equation $$ \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{4}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=-74.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ Predict the way in which the equilibrium will shift in response to each of the following changes in conditions (if the equilibrium is unaffected by the change, then write no change): (a) decrease in temperature (b) decrease in reaction volume (c) decrease in \(\bar{P}_{\mathrm{H}_{2}}\) (d) increase in \(\bar{P}_{\mathrm{CH}_{4}}\) (e) addition of \(\mathrm{C}(s)\)

For the exothermic dissolution reaction of hydrofluoric acid described by $$ \mathrm{HF}(a q) \leftrightharpoons \mathrm{H}^{*}(a q)+\mathrm{F}^{-}(a q) $$ in which direction will the equilibrium shiftin response to the following changes? (a) decrease in temperature (b) addition of \(\mathrm{H}_{2} \mathrm{O}(l)\) (c) addition of \(\mathrm{KF}(s)\) (d) addition of \(\mathrm{NaOH}(s)\) (Hint: What will this strong base react with in solution?)

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