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For each of the following pairs, predict which substance has the higher entropy per mole at a given temperature: (a) \(\mathrm{I}_{2}(s)\) or \(\mathrm{I}_{2}(g)\) (b) \(\mathrm{O}_{2}(g)\) at \(50.7 \mathrm{kPa}\) or \(\mathrm{O}_{2}\) at \(101.3 \mathrm{kPa}\) (c) 1 mol of \(\mathrm{N}_{2}\) in 22.4 Lor \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\). (d) \(\mathrm{CH}_{3} \mathrm{OH}(l)\) or \(\mathrm{CH}_{3} \mathrm{OH}(s)\)

Short Answer

Expert verified
(a) I2(g), (b) O2(g) at 50.7 kPa, (c) N2 in 44.8 L, (d) CH3OH(l) have higher entropy.

Step by step solution

01

Understanding the Entropy Concept

Entropy is a measure of the disorder or randomness in a system. In general, gases have higher entropy than liquids, which in turn have higher entropy than solids. Additionally, increasing volume or decreasing pressure of a gas increases its entropy.
02

Analyzing Part (a): I2(s) vs. I2(g)

Between iodine as a solid (I_{2}(s)) and iodine as a gas (I_{2}(g)), we expect the gaseous state to have higher entropy because gas particles are more randomly distributed than those in a solid.
03

Analyzing Part (b): O2(g) at Different Pressures

O_{2}(g) at lower pressure (50.7 kPa) would have higher entropy compared to O_{2}(g) at 101.3 kPa. Decreasing pressure allows gas particles to occupy a larger volume, leading to higher entropy.
04

Analyzing Part (c): N2 in Different Volumes

1 mol of N_{2} in 44.8 L will have higher entropy than in 22.4 L because increasing volume allows the gas molecules to spread out more, thus increasing disorderness and entropy.
05

Analyzing Part (d): CH3OH(l) vs. CH3OH(s)

Between methanol as a liquid (CH_{3}OH(l)) and methanol as a solid (CH_{3}OH(s)), the liquid state has higher entropy. Molecules in a liquid can move more freely than those in a solid crystal lattice.

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Key Concepts

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

Disorder
Entropy is often described as a measure of "disorder" within a system. This notion of disorder refers to how energy is distributed among the particles, and how freedom of movement varies across different states of matter. In a more ordered system, like a solid, particles are closely packed and have less freedom to move. In contrast, a gas represents a state of high disorder, since its particles move freely and are spread out over a larger volume.
This concept is key to understanding why a gas generally has a higher entropy than a solid or liquid.
  • Solids: Particles are in fixed positions, creating low disorder.
  • Liquids: Particles have more freedom to move, increasing disorder.
  • Gases: Particles are widely dispersed and have high freedom, leading to high disorder.
Thus, the more distributed and less confined the particles, the higher the entropy, or disorder, of the system.
Phase Transitions
Phase transitions such as melting, evaporation, or sublimation involve changes in entropy due to alterations in the arrangement of molecules. When a substance undergoes a phase transition from solid to liquid or from liquid to gas, its entropy typically increases.
During these transitions, the structured order of solids gives way to the more disordered, freely moving particles of liquid or gas states. This happens because it takes energy to overcome the forces holding the molecules in their structured form, allowing them to move more independently.
  • Melting: Solid to liquid, increases entropy.
  • Evaporation: Liquid to gas, significantly increases entropy.
  • Sublimation: Direct transition from solid to gas, dramatically increases entropy.
Each transition from a more ordered phase to a less ordered phase corresponds to an increase in entropy, demonstrating the natural tendency for systems to move towards greater disorder.
Gas Laws
The behavior of gases can be predicted using various gas laws, which explain how pressure, volume, and temperature interact in a gaseous state. These laws also help clarify how changes in these variables affect entropy.
  • Boyle's Law states that at constant temperature, as the volume of a gas increases, the pressure decreases. This increase in volume leads to higher entropy because gas particles can occupy more space and thus have increased randomness.
  • Charles's Law indicates that at constant pressure, increasing the temperature of a gas causes an increase in volume and, hence, entropy.
  • Avogadro’s Law explains that increasing the number of moles of a gas, while keeping pressure and temperature constant, results in higher volume and entropy.
In summary, any changes that allow gas particles to occupy a larger volume or move more freely contribute to increasing the system's entropy.
Pressure Effects
Pressure can significantly affect entropy, particularly in gaseous systems. When the pressure on a gas is decreased, its volume typically increases assuming constant temperature, leading to higher entropy. This is because the gas particles can spread out more, thereby increasing the disorder and the number of ways energy can be distributed among them.
Lower pressure means fewer collisions between gas molecules and more available space for movement.
  • At high pressure, gas particles are more confined, leading to lower entropy.
  • At low pressure, particles spread out, leading to higher entropy.
Therefore, by decreasing pressure and increasing volume, a gas system achieves greater entropy, embodying the naturally preferred state of higher disorder.

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Most popular questions from this chapter

The normal boiling point of \(n\) -octane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) is \(125^{\circ} \mathrm{C}\). (a) Is the condensation of gaseous \(n\) -octane to liquid \(n\) -octane an endothermic or exothermic process? (b) In what temperature range is the boiling of \(n\) -octane a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid \(n\) -octane and gaseous \(n\) -octane are in equilibrium? Explain.

A certain reaction has \(\Delta H^{\circ}=+20.0 \mathrm{~kJ}\) and \(\Delta S^{\circ}=\) \(+100.0 \mathrm{~J} / \mathrm{K} .\) (a) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the surroundings? (c) Calculate \(\Delta G^{\circ}\) for the reaction at \(298 \mathrm{~K} .(\mathbf{d})\) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Predict the sign of \(\Delta S_{\text {sys }}\) for each of the following processes: (a) Gaseous \(\mathrm{H}_{2}\) reacts with liquid palmitoleic acid \(\left(\mathrm{C}_{16} \mathrm{H}_{30} \mathrm{O}_{2},\right.\) unsaturated fatty acid) to form liquid palmitic acid \(\left(\mathrm{C}_{16} \mathrm{H}_{32} \mathrm{O}_{2}\right.\) saturated fatty acid). (b) Liquid palmitic acid solidifies at \(1^{\circ} \mathrm{C}\) to solid palmitic acid. (c) Silver chloride precipitates upon mixing \(\mathrm{AgNO}_{3}(a q)\) and \(\mathrm{NaCl}(a q)\). (d) Gaseous \(\mathrm{H}_{2}\) dissociates in an electric arc to form gaseous Hatoms (used in atomic hydrogen welding).

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel at room temperature: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H^{\circ}\) is positive at room temperature. (a) What is the sign of \(\Delta S^{\circ}\) at room temperature? (b) If the hydrated compound is placed in a large, closed vessel that already contains a large amount of water vapor, does \(\Delta S^{\circ}\) change for this reaction at room temperature?

From the values given for \(\Delta H^{\circ}\) and \(\Delta S^{\circ},\) calculate \(\Delta G^{\circ}\) for each of the following reactions at \(298 \mathrm{~K}\). If the reaction is not spontaneous under standard conditions at \(298 \mathrm{~K}\), at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} $$

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