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Which substance has the higher entropy? (a) dry ice (solid \(\mathrm{CO}_{2}\) ) at \(-78^{\circ} \mathrm{C}\) or \(\mathrm{CO}_{2}(\mathrm{g})\) at \(0^{\circ} \mathrm{C}\) (b) liquid water at \(25^{\circ} \mathrm{C}\) or liquid water at \(50^{\circ} \mathrm{C}\) (c) pure alumina, \(\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s}),\) or ruby (ruby is \(\mathrm{Al}_{2} \mathrm{O}_{3}\) in which some \(\mathrm{Al}^{3+}\) ions in the crystalline lattice are replaced with \(\mathrm{Cr}^{3+}\) ions) (d) one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 1 bar pressure or one mole of \(\mathrm{N}_{2}(\mathrm{g})\) at 10 bar pressure (both at \(298 \mathrm{K})\)

Short Answer

Expert verified
(a) \(\mathrm{CO}_2(\mathrm{g})\) at \(0^{\circ} \mathrm{C}\); (b) liquid water at \(50^{\circ} \mathrm{C}\); (c) ruby; (d) \(\mathrm{N}_2(\mathrm{g})\) at 1 bar.

Step by step solution

01

Understanding Entropy

Entropy is a measure of the disorder or randomness in a system. Therefore, the more disordered a system, the higher its entropy. Generally, entropy increases when a substance transitions from solid to liquid to gas because gas molecules are more disordered and dispersed.
02

Comparing States of Matter: Part (a)

For part (a), dry ice is solid \( ext{CO}_2\) at \(-78^{\circ} ext{C}\) and \( ext{CO}_2( ext{g})\) is the gaseous form. Since gases have more disorder than solids, \( ext{CO}_2( ext{g})\) at \(0^{\circ} ext{C}\) has higher entropy than dry ice.
03

Temperature Effect on Entropy: Part (b)

For part (b), we consider liquid water at two different temperatures: \(25^{\circ} ext{C}\) and \(50^{\circ} ext{C}\). As temperature increases, the molecular motion in the liquid increases, leading to higher entropy. Thus, liquid water at \(50^{\circ} ext{C}\) has higher entropy than at \(25^{\circ} ext{C}\).
04

Effect of Impurities: Part (c)

In part (c), we have pure alumina \( ext{Al}_2 ext{O}_3(s)\) compared to ruby, a form of \( ext{Al}_2 ext{O}_3\) with some \( ext{Al}^{3+}\) ions replaced by \( ext{Cr}^{3+}\). This substitution introduces more disorder in the crystalline structure, making ruby have higher entropy than pure alumina.
05

Effect of Pressure: Part (d)

For part (d), we have \( ext{N}_2( ext{g})\) at two pressures: 1 bar and 10 bar, both at 298 K. Lower pressure leads to gases occupying a larger volume and therefore more disorder. Consequently, \( ext{N}_2( ext{g})\) at 1 bar has higher entropy than at 10 bar.

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Key Concepts

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

States of Matter
The concept of entropy is heavily influenced by the state of matter of a substance. Matter can exist in three primary states: solid, liquid, and gas. Each state has distinct characteristics that affect entropy. Solid states possess tightly packed molecules with minimal movement, resulting in lower entropy as the structure is organized and predictable.
In contrast, liquid states have molecules that can move around more freely, which allows for increased randomness and thus higher entropy than solids. Lastly, gases have highly dispersed molecules that move rapidly in all directions. This movement and spread create the most disorder, which means gases exhibit the highest entropy.
  • Solids: Ordered structure, low entropy.
  • Liquids: More movement and disorder, moderate entropy.
  • Gases: High dispersion and randomness, high entropy.
Understanding these differences helps explain why gas substances, like \( ext{CO}_2( ext{g})\), at 0°C, have more entropy compared to their solid form at lower temperatures, such as dry ice at -78°C.
Temperature Effect on Entropy
Temperature plays a significant role in determining the entropy of a substance. When temperature increases, the kinetic energy of molecules also increases, causing them to move more vigorously. This heightened molecular motion contributes to greater disorder, or higher entropy, within a system.
For example, if we compare liquid water at temperatures of 25°C and 50°C, the water at the higher temperature will have molecules moving more excitedly and randomly. This increase in randomness equates to an increase in entropy.
  • Higher temperatures: More molecular motion, increased entropy.
  • Lower temperatures: Less molecular motion, decreased entropy.
This principle explains why raising the temperature of any material tends to increase its entropy, aligning with our observation in the exercise that water at 50°C has more entropy than at 25°C.
Crystalline Structure
A crystalline structure refers to a well-organized arrangement of atoms or ions within a solid. This orderly setup can have significant implications for entropy. Pure crystalline substances, such as pure alumina \( ext{Al}_2 ext{O}_3 \), have lower entropy because of their highly repetitive and predictable arrangements.
When impurities are introduced into a crystalline structure, as in the case of ruby, some \( ext{Al}^{3+} \) ions are replaced with \( ext{Cr}^{3+} \) ions. This substitution disrupts the orderly pattern, introducing imperfections and irregularities.
  • Pure crystalline substances: Low disorder, low entropy.
  • Impure crystalline substances: Increased disorder, higher entropy.
The presence of these impurities increases the system's disorder, thereby increasing its entropy. This helps us understand why ruby, with its modified lattice, has higher entropy than pure alumina.
Pressure Effect on Entropy
Pressure is another crucial factor when considering entropy, particularly in gases. Entropy varies inversely with pressure. This means that when pressure increases, the molecules of a gas are forced closer together, creating an environment of decreased volume and disorder.
Conversely, at lower pressures, gas molecules can spread out more, occupying a larger volume, which results in increased disorder and, therefore, higher entropy. For example, one mole of \( ext{N}_2( ext{g}) \) at 1 bar pressure will have higher entropy compared to the same amount at 10 bar pressure.
  • High pressure: Less molecular spread, decreased entropy.
  • Low pressure: More molecular spread, increased entropy.
This concept illustrates how changes in pressure can markedly affect the entropy of gaseous systems, as highlighted by the difference in entropy between \( ext{N}_2( ext{g}) \) at varying pressures.

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Most popular questions from this chapter

Indicate which of the following processes are reversible. (a) Nitrogen gas expands into a vacuum. (b) Dry ice, \(\mathrm{CO}_{2}(\mathrm{s}),\) sublimes at \(25^{\circ} \mathrm{C}\) and 1.0 atm. (c) Energy as heat is added to a mixture of ice and water at \(0^{\circ} \mathrm{C},\) causing some of the ice to melt. (d) Methanol and ethanol mix forming a homogeneous solution.

Calculate \(\Delta_{f} G^{\circ}\) for \(\mathrm{HI}(\mathrm{g})\) at \(350^{\circ} \mathrm{C},\) given the following equilibrium partial pressures: \(P\left(\mathrm{H}_{2}\right)=\) 0.132 bar, \(P\left(\mathrm{I}_{2}\right)=0.295\) bar, and \(P(\mathrm{HI})=1.61\) bar. At \(350^{\circ} \mathrm{C}\) and \(1 \mathrm{bar}, \mathrm{I}_{2}\) is a gas. $$1 / 2 \mathrm{H}_{2}(\mathrm{g})+1 / 2 \mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{HI}(\mathrm{g})$$

Titanium(IV) oxide is converted to titanium carbide with carbon at a high temperature. $$\mathrm{TiO}_{2}(\mathrm{s})+3 \mathrm{C}(\mathrm{s}) \rightarrow 2 \mathrm{CO}(\mathrm{g})+\mathrm{TiC}(\mathrm{s})$$ $$\begin{array}{lc} & \text { Free Energies of Formation at } \\ \text { Compound } & 727^{\circ} \mathrm{C}, \mathrm{kJ} / \mathrm{mol} \\ \hline \mathrm{TiO}_{2}(\mathrm{s}) & -757.8 \\ \mathrm{TiC}(\mathrm{s}) & -162.6 \\\ \mathrm{CO}(\mathrm{g}) & -200.2 \end{array}$$ (a) Calculate \(\Delta_{\mathrm{r}} G^{\circ}\) and \(K\) at \(727^{\circ} \mathrm{C}.\) (b) Is the reaction product-favored at equilibrium at this temperature? (c) How can the reactant or product concentrations be adjusted for the reaction to proceed at \(727^{\circ} \mathrm{C} ?\)

Calculate the change in entropy for a system in going from a condition of 5 accessible microstates to 30 accessible microstates.

The standard free energy change, \(\Delta_{\mathrm{r}} G^{\circ},\) for the formation of \(\mathrm{O}_{3}(\mathrm{g})\) from \(\mathrm{O}_{2}(\mathrm{g})\) is \(+163.2 \mathrm{kJ} / \mathrm{mol}\) -ren at \(25^{\circ} \mathrm{C} .\) Calculate \(K_{\mathrm{p}}\) at this temperature for the equilibrium $$3 \mathrm{O}_{2}(\mathrm{g}) \rightleftarrows 2 \mathrm{O}_{3}(\mathrm{g})$$ Comment on the sign of \(\Delta_{r} G^{\circ}\) and the magnitude of \(K_{\mathrm{p}}.\)

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