/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 For each of the following pairs,... [FREE SOLUTION] | 91影视

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Chapter 19: Problem 39

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 molof \(\mathrm{N}_{2}\) in 22.4 Lor \(1 \mathrm{~mol}\) of \(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\). (d) \(\mathrm{CH}_{3} \mathrm{OH}(I)\) or \(\mathrm{CH}_{3} \mathrm{OH}(s)\)

Short Answer

Expert verified
In summary: (a) \(\mathrm{I}_{2}(g)\) has higher entropy than \(\mathrm{I}_{2}(s)\). (b) \(\mathrm{O}_{2}(g)\) at \(50.7 \mathrm{kPa}\) has higher entropy than \(\mathrm{O}_{2}(g)\) at \(101.3 \mathrm{kPa}\). (c) \(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\) has higher entropy than \(\mathrm{N}_{2}\) in \(22.4 \mathrm{~L}\). (d) \(\mathrm{CH}_{3}\mathrm{OH}(l)\) has higher entropy than \(\mathrm{CH}_{3}\mathrm{OH}(s)\).

Step by step solution

01

Pair (a) Analysis

For iodine, we have to compare the entropy of the solid phase and the gas phase. As mentioned, entropy is higher for gases than for solids.
02

Pair (a) Conclusion

In pair (a), Iodine gas (\(\mathrm{I}_{2}(g)\)) has the higher entropy per mole at a given temperature compared to solid iodine (\(\mathrm{I}_{2}(s)\)).
03

Pair (b) Analysis

In this case, we have to compare the entropy of oxygen gas at two different pressures. As pressure increases, entropy decreases.
04

Pair (b) Conclusion

In pair (b), oxygen gas (\(\mathrm{O}_{2}\)) at \(50.7 \mathrm{kPa}\) pressure has higher entropy per mole at a given temperature compared to oxygen gas at \(101.3 \mathrm{kPa}\).
05

Pair (c) Analysis

In this case, we have to compare the entropy of 1 mole of nitrogen gas in different volumes. As volume increases, entropy also increases.
06

Pair (c) Conclusion

In pair (c), 1 mole of nitrogen gas in 44.8 L (\(\mathrm{N}_{2}\) in \(44.8 \mathrm{~L}\)) has higher entropy per mole at a given temperature compared to 1 mole of nitrogen gas in 22.4 L (\(\mathrm{N}_{2}\) in 22.4 L).
07

Pair (d) Analysis

For methanol (\(\mathrm{CH}_{3}\mathrm{OH}\)), we have to compare the entropy of the liquid phase and the solid phase. As mentioned, entropy is higher for liquids than for solids.
08

Pair (d) Conclusion

In pair (d), liquid methanol (\(\mathrm{CH}_{3}\mathrm{OH}(l)\)) has higher entropy per mole at a given temperature compared to solid methanol (\(\mathrm{CH}_{3}\mathrm{OH}(s)\)).

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Key Concepts

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

Phase Transition
Phase transitions are changes in the state of matter. They involve transformations like solid to liquid, liquid to gas, or vice versa. These changes play a crucial role in altering the entropy of a substance.
Entropy, which measures the disorder of a system, often increases when there is a phase transition. Here's why:
  • In solids, particles are tightly packed, with minimal freedom of movement. This results in lower entropy.
  • Liquids allow more particle movement, leading to higher entropy compared to solids.
  • Gases have particles widely dispersed and moving freely, thus possessing the highest entropy among the phases.
Consider iodine transitioning from a solid to a gas. The dramatic increase in disorder as particles move from a fixed arrangement in a solid to freely moving in a gas leads to a higher entropy in the gaseous state.
Gas Laws
Gas laws describe the behavior of gases and relate different gas properties such as pressure, volume, and temperature. These laws are crucial to understanding gas entropy.
One primary law is the **Ideal Gas Law**, expressed as \( PV = nRT \), where \(P\) is pressure, \(V\) is volume, \(n\) is moles, \(R\) is the gas constant, and \(T\) is temperature.
In the context of entropy:
  • An increase in volume (\(V\)) at constant pressure allows gas particles more space to move, increasing entropy.
  • A decrease in pressure (\(P\)) could imply a volume increase or a drop in temperature, affecting entropy.
For nitrogen gas, increasing the volume results in greater entropy due to enhanced molecular movement.
Pressure and Volume
Pressure and volume are key factors in a gas's entropy. They are inversely related in the Ideal Gas Law, meaning a change in one affects the other.
**Understanding their relationship: **
  • When volume increases, gases expand, allowing molecules more room to move freely. This leads to an increase in entropy.
  • Conversely, when pressure rises, the volume decreases, limiting molecular movement and reducing entropy.
For example, oxygen gas at a lower pressure of 50.7 kPa will have higher entropy than at a higher pressure of 101.3 kPa, as molecules can disperse more freely in less restrictive environments.
Thermodynamics
Thermodynamics assesses the principles that govern energy and matter interactions, including entropy. It offers insight into how and why changes occur in physical systems.
**The four laws of thermodynamics** primarily guide these interactions:
  • The **zeroth law** establishes temperature as a uniform measure.
  • The **first law**, or conservation of energy, highlights that energy cannot be created or destroyed.
  • The **second law** states that entropy tends to increase, driving spontaneous processes.
  • The **third law** posits that absolute zero is unattainable as it requires zero entropy.
Entropy is a central concept. It predicts the direction of state changes, favoring states with higher disorder.
This understanding explains phenomena such as the persistence of methanol in a liquid state or the higher entropy of gases compared to solids.

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Most popular questions from this chapter

Consider the melting of ice (solid water) to liquid water at a pressure of \(101.3 \mathrm{kPa}\). (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

For a certain chemical reaction, \(\Delta H^{\circ}=-40.0 \mathrm{k} \mathrm{J}\) and \(\Delta S^{\circ}=-150.0 \mathrm{~J} / \mathrm{K}\). (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 \(298 \mathrm{~K}\). (d) Is the reaction spontaneous at \(298 \mathrm{~K}\) under standard conditions?

Carbon disulfide \(\left(C S_{2}\right)\) is a toxic, highly flammable substance. The following thermodynamic data are available for \(\mathrm{CS}_{2}(I)\) and \(\mathrm{CS}_{2}(g)\) at \(298 \mathrm{~K}\) \begin{tabular}{lcc} \hline & \(\Delta H_{i}(\mathrm{k} / \mathrm{mol})\) & \(\Delta G_{i}^{\prime}(\mathrm{kJ} / \mathrm{mol})\) \\ \hline\(C S_{2}(l)\) & 89.7 & 65.3 \\ \(C S_{2}(g)\) & 117.4 & 67.2 \\ \hline \end{tabular} (a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the \(\mathrm{C}-\mathrm{S}\) bonds? \((\mathbf{b})\) Use the VSEPR method to predict the structure of the \(\mathrm{CS}_{2}\) molecule. (c) Liquid \(\mathrm{CS}_{2}\) burns in \(\mathrm{O}_{2}\) with a blue flame, forming \(\mathrm{CO}_{2}(g)\) and \(\mathrm{SO}_{2}(g)\). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix \(C,\) calculate \(\Delta H^{\circ}\) and \(\Delta G^{\circ}\) for the reaction in part \((c) .\) Is the reaction exothermic? Is it spontaneous at \(298 \mathrm{~K} ?\) (e) Use the data in the table to calculate \(\Delta S^{\circ}\) at \(298 \mathrm{~K}\) for the vaporization of \(\mathrm{CS}_{2}(I) .\) Is the sign of \(\Delta S^{\circ}\) as you would expect for a vaporization? (f) Using data in the table and your answer to part (e), estimate the boiling point of \(\mathrm{CS}_{2}(l)\). Do you predict that the substance will be a liquid or a gas at \(298 \mathrm{~K}\) and \(101.3 \mathrm{kPa}\) ?

For a particular reaction, \(\Delta H=30.0 \mathrm{~kJ}\) and \(\Delta S=90.0 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(\mathrm{T}\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

(a) What is the difference between a state and a microstate of a system? (b) As a system goes from state A to state B, its entropy decreases. What can you say about the number of microstates corresponding to each state? (c) In a particular spontaneous process, the number of microstates available to the system decreases. What can you conclude about the sign of \(\Delta S\) surr?
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