Predict which member of each of the following pairs has the greater standard entropy at \(25^{\circ} \mathrm{C} :(\mathbf{a}) \mathrm{C}_{6} \mathrm{H}_{6}(l)\) or \(\mathrm{C}_{6} \mathrm{H}_{6}(g)\) (b) \(\mathrm{CO}(g)\) or \(\mathrm{CO}_{2}(g),(\mathbf{c}) 1 \mathrm{mol} \mathrm{N}_{2} \mathrm{O}_{4}(g)\) or 2 \(\mathrm{mol} \mathrm{NO}_{2}(\mathrm{g})\) (d) \(\mathrm{HCl}(g)\) or \(\mathrm{HCl}(a q) .\) Use Appendix \(\mathrm{C}\) to find the standard entropy of each substance.

Short Answer

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Pair A: C6H6(g) has the greater standard entropy than C6H6(l) at 25°C. Pair B: CO2(g) has the greater standard entropy than CO(g) at 25°C. Pair C: 2 moles of NO2(g) have the greater standard entropy than 1 mol N2O4(g) at 25°C. Pair D: HCl(aq) has the greater standard entropy than HCl(g) at 25°C.

Step by step solution

01

Pair A: C6H6(l) vs C6H6(g)

In general, the entropy of a substance increases when it changes from a solid to liquid, and from a liquid to gas. This is due to the increased disorder associated with the increased freedom of motion of the particles. Thus, C6H6(g) will have a greater standard entropy than C6H6(l) at 25°C.
02

Pair B: CO(g) vs CO2(g)

Both substances are in the gaseous state. To compare their standard entropy, we need to consider the molecular complexity. CO2 is more complex than CO because it has more atoms in the molecule. Therefore, CO2(g) will have greater standard entropy than CO(g) at 25°C.
03

Pair C: 1 mol N2O4(g) vs 2 moles of NO2(g)

When comparing a reaction with one mole of a reactant to one with double the number of moles of the same reactant, we need to consider the increased number of particles in the system. An increase in the number of particles in the system contributes to increased entropy. Here, we have two times the moles of NO2(g) compared to one mole of N2O4(g), which implies that the system with 2 moles of NO2(g) will have greater standard entropy at 25°C.
04

Pair D: HCl(g) vs HCl(aq)

The difference between HCl(g) and HCl(aq) is that the former state is gaseous, while the latter is an aqueous solution. Generally, the entropy of a substance increases when it changes from a gaseous state to a dissolved state in a solvent (such as water). This is due to the increased disorder associated with the interaction of the solute particles with the solvent particles. Thus, HCl(aq) will have a greater standard entropy than HCl(g) at 25°C. Note: Appendix C refers to the thermodynamic tables provided with the exercise. These tables include values of the standard entropy (S°) for various substances. You can use these values to confirm the results obtained above.

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Key Concepts

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

Entropy and Molecular Complexity
Entropy, often denoted by the symbol S, is a fundamental concept in thermodynamics that measures the degree of disorder or randomness in a system. When we talk about molecular complexity, we refer to the number of atoms and the complexity of the interactions between them within a molecule. Generally, the more complex a molecule is, the higher its entropy.

For instance, comparing carbon monoxide (CO) with carbon dioxide (CO2), CO2 has a greater number of atoms which leads to more possible ways the energy can be distributed throughout the molecule. As a result, CO2 has a higher standard entropy than CO. Molecular complexity significantly influences a system's entropy, as a larger, more complex molecule will have more vibrational modes, allowing for more microstates that a system can exist in.
Entropy of States of Matter
The concept of entropy of states of matter is crucial to understanding the behavior of substances under different conditions. Entropy varies with the physical state of a material: solids, liquids, and gases each have distinct entropy values.

In general, gases have the highest standard entropy because their molecules are far apart and move with considerable freedom. Liquids follow since they have some structure but less than solids. Finally, solids possess the lowest entropy due to their relatively fixed and orderly arrangement of atoms. To illustrate, benzene (C6H6) in gaseous form has greater standard entropy than in its liquid form, because the gas molecules can spread out and move independently, creating a state of higher disorder.
Thermodynamics
The branch of physics called thermodynamics deals with heat and temperature and their relation to energy and work. It lays out the fundamental principles governing the transfer and transformation of energy in systems. Its laws help us understand phenomena such as why heat flows from hot to cold objects, how engines produce work, and why certain reactions are spontaneous.

At the heart of thermodynamics is the concept of entropy, which can help explain physical processes and predict the direction of spontaneous change. For example, the second law of thermodynamics states that the total entropy of an isolated system can never decrease over time. This explains why substances spontaneously transition from solid to liquid to gas at increasing temperatures, since these changes typically lead to an increase in entropy.
Molecular Disorder
The concept of molecular disorder is intertwined with entropy. It refers to the degree to which molecules within a system are organized. An ordered state, such as a crystal lattice of a solid, has low entropy because the molecules are arranged in a predictable pattern. Conversely, in a disordered state, like a gas, molecules are distributed randomly and move freely, contributing to high entropy.

Molecular disorder becomes particularly apparent when comparing different states of matter or the same substance under different conditions. Consider hydrochloric acid (HCl) in gas versus aqueous form. When HCl is dissolved in water, it interacts with water molecules, dramatically increasing the disorder of the system compared to its gaseous state, and hence also increasing its standard entropy.

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Most popular questions from this chapter

Indicate whether each statement is true or false. (a) \(\Delta S\) is a state function. ( b) If a system undergoes a reversible change, the entropy of the universe increases.(c) If a system undergoes a reversible process, the change in entropy of the system is exactly matched by an equal and opposite change in the entropy of the surroundings. (d) If a system undergoes a reversible process, the entropy change of the system must be zero.

The oxidation of glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) in body tissue produces \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) . In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) and \(\mathrm{CO}_{2} .\) (a) Using data given in Appendix \(\mathrm{C},\) compare the equilibrium constants for the following reactions: $$ \begin{array}{c}{\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \rightleftharpoons 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)} \\ {\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \rightleftharpoons 2 \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+2 \mathrm{CO}_{2}(g)}\end{array} $$ (b) Compare the maximum work that can be obtained from these processes under standard conditions.

Acetylene gas, \(\mathrm{C}_{2} \mathrm{H}_{2}(g),\) is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) (b) How much heat is produced in burning 1 \(\mathrm{mol}\) of \(\mathrm{C}_{2} \mathrm{H}_{2}\) under standard conditions if both reactants and products are brought to 298 \(\mathrm{K?}\) (c) What is the maximum amount of useful work that can be accomplished under standard conditions by this reaction?

(a) What sign for \(\Delta S\) do you expect when the pressure on 0.600 mol of an ideal gas at 350 \(\mathrm{K}\) is increased isothermally from an initial pressure of 0.750 atm? (b) If the final pressure on the gas is 1.20 atm, calculate the entropy change for the process. (c) Do you need to specify the temperature to calculate the entropy change?

Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbonylation of acetic acid proceeds according to: $$ \mathrm{CH}_{3} \mathrm{COOH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)+\mathrm{CO}(g) $$ By using data from Appendix \(\mathrm{C}\) , calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not vary with temperature.

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