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The third law of thermodynamics states that the entropy of a perfect crystal at 0 \(\mathrm{K}\) is zero. In Appendix \(4, \mathrm{F}^{-}(a q), \mathrm{OH}^{-}(a q)\) and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Short Answer

Expert verified
In conclusion, the negative entropy values of F⁻(aq), OH⁻(aq), and S²⁻(aq) are due to their highly ordered structures formed by electrostatic interactions with water molecules. These compounds do not meet the conditions of the Third Law of Thermodynamics (perfect crystal, 0 K), allowing their entropy values to be negative representing high order and low randomness at specific temperature and pressure.

Step by step solution

01

Understand what entropy represents

Entropy (S) is a measure of the disorder or randomness of a system. The higher the entropy value, the more disordered the system is. When a substance has a negative entropy value, it means that the substance is highly ordered at that specific temperature and pressure.
02

The Third Law of Thermodynamics and Perfect Crystals

The Third Law of Thermodynamics states that the entropy of a perfect crystal at absolute zero (0 K) is zero. This law is applied to perfect crystals because these systems have only one possible configuration: all the atoms within the crystal lattice are ordered and symmetrically arranged. As a result, there is no randomness or disorder in the system at absolute zero, meaning the entropy is zero.
03

Investigate the compounds with negative entropy

The compounds mentioned, F⁻(aq), OH⁻(aq), and S²⁻(aq), have negative standard entropy values at 298 K. These compounds are ions in aqueous solution, and they have very high order due to their strong electrostatic interactions with surrounding water molecules. These interactions can create a highly ordered structure and reduce the overall randomness of the system.
04

Relate the Third Law of Thermodynamics to Compounds with Negative Entropy

The third law of thermodynamics applies to perfect crystals at absolute zero. The compounds mentioned are not perfect crystals and are not at a temperature of 0 K. The concept of perfect crystals is an ideal approximation that doesn't necessarily apply to all substances. For the compounds mentioned, the highly ordered structures formed due to their electrostatic interactions with water molecules results in negative entropy values. The Third Law of Thermodynamics does not apply directly to these conditions, which allows for these compounds to have negative entropy values. In conclusion, the compounds F⁻(aq), OH⁻(aq), and S²⁻(aq) can have negative standard entropy values because their conditions do not match the requirements of the Third Law of Thermodynamics. Their highly ordered structures formed due to interaction with water molecules lead to negative entropy measures, indicating that the system is highly ordered and less random at the specific temperature and pressure.

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Key Concepts

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

Entropy
Entropy is a fundamental concept in thermodynamics, describing the degree of disorder or randomness in a system. Think of it as the measure of how chaotic a system is. When a system has high entropy, it is more disordered, with molecules moving randomly.
On the flip side, low entropy means the system is more ordered, with molecules in more predictable positions. This order versus disorder is crucial in understanding how energy is distributed within a system. It often increases with temperature, since heat can cause molecules to move more vigorously.
The concept of entropy is central to the second law of thermodynamics, which states that the total entropy of an isolated system can never decrease over time. However, there are circumstances where entropy might appear to be negative. In the specific case of certain ions in aqueous solutions, their interactions with surrounding water molecules can create highly ordered structures. These result in a negative entropy because the system is less random than it would otherwise be.
Perfect Crystal
A perfect crystal is a theoretical construct used in the third law of thermodynamics. It is an idealized state where every atom or molecule is in its place in a perfectly aligned lattice without any defects. At absolute zero temperature (0 K), a perfect crystal has exactly one possible configuration because all its atoms are arranged in a completely ordered fashion.
The third law of thermodynamics asserts that the entropy of a perfect crystal at absolute zero is zero. This is because there is no randomness in such a perfectly ordered system. The atoms have no energy to move and therefore remain perfectly aligned.
This concept helps scientists understand the baseline or minimal entropy a system can have, providing a reference point. In practical terms, achieving a perfect crystal at absolute zero is impossible, but the concept still aids in predicting and understanding real-world behaviors of materials and substances as they approach low temperatures.
Negative Entropy
Negative entropy might sound puzzling, as it suggests an ordered state even lower than the perceived baseline of zero entropy. In reality, negative entropy often refers to systems where the order is exceptionally high.
This can occur in certain ions dissolved in water, like F⁻, OH⁻, and S²⁻, which show negative standard entropy values. These ions strongly interact with water molecules surrounding them, leading to a structured and highly ordered system. This high order reduces the randomness to the point where entropy takes on negative values.
While the third law of thermodynamics does not directly apply to non-crystalline substances or to systems not at 0 K, negative entropy illustrates how specific interactions at higher temperatures can lead to a notable level of order. Therefore, negative entropy in aqueous systems represents an unusual but fascinating alignment, contrary to the typical association of increasing entropy with increased disorder.

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Most popular questions from this chapter

Liquid water at \(25^{\circ} \mathrm{C}\) is introduced into an evacuated, insulated vessel. Identify the signs of the following thermodynamic functions for the process that occurs: \(\Delta H, \Delta S, \Delta T_{\text { water }} \Delta S_{\text { surr }}\) \(\Delta S_{\text { univ }}\)

When the environment is contaminated by a toxic or potentially toxic substance (for example, from a chemical spill or the use of insecticides), the substance tends to disperse. How is this consistent with the second law of thermodynamics? In terms of the second law, which requires the least work: cleaning the environment after it has been contaminated or trying to prevent the contamination before it occurs? Explain.

Which of the following reactions (or processes) are expected to have a negative value for \(\Delta S^{\circ} ?\) a. \(\mathrm{SiF}_{6}(a q)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{HF}(g)+\mathrm{SiF}_{4}(g)\) b. \(4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)\) c. \(\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow \mathrm{COCl}_{2}(g)\) d. \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)\) e. \(\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)\)

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving NaCl in water

The enthalpy of vaporization of ethanol is 38.7 kJ/mol at its boiling point \(\left(78^{\circ} \mathrm{C}\right) .\) Determine \(\Delta S_{\mathrm{sys}}, \Delta S_{\mathrm{surr}},\) and \(\Delta S_{\mathrm{univ}}\) when 1.00 mole of ethanol is vaporized at \(78^{\circ} \mathrm{C}\) and 1.00 atm.

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