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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
The negative standard entropy values for F^-, OH^-, and S^2- ions do not contradict the third law of thermodynamics because these values are relative quantities compared to a reference substance at a pressure of 1 bar and temperature of 298 K. The negative values indicate that these ions are more ordered compared to the reference substances, in their aqueous solution, due to ion-dipole interactions with water molecules. These values should not be misinterpreted as implying negative entropy at absolute zero.

Step by step solution

01

Understand Entropy

Entropy (S) is a measure of the disorder or randomness of a system. According to the third law of thermodynamics, the entropy of a perfect crystal at 0 K is assumed to be zero. This is because, in a perfect crystal, all particles are in a well-ordered state at absolute zero (0 K).
02

Comparing Entropy Values to Absolute Zero

The negative values of the standard entropy (S°) of F^-, OH^-, and S^2- indicate that their entropy values are less than that of a perfect crystal at 0 K. This might seem counterintuitive, given the third law of thermodynamics. However, it is important to note that these negative entropy values are obtained through comparisons and measurements relative to other substances, not by direct measurements of the entropy at absolute zero.
03

Understand Standard Entropy

Standard entropy (S°) refers to the entropy of a substance at a reference state, typically at a pressure of 1 bar and temperature of 298 K. The standard entropy values, like any other thermodynamic properties, are based on a reference standard. Just like how the standard enthalpy of formation for an element in its standard state is defined to be zero, the standard entropy values are relative quantities based on the reference.
04

Comparing to a Reference Substance

The negative standard entropy values for F^-, OH^-, and S^2- ions essentially indicate that these ions are more ordered than the reference substances to which they are compared. It should be noted that these negative values are for the entropy of the solvated ions in their aqueous solution, which means that we are comparing the entropy of these ions in water to their reference state. In that respect, the negative entropy values make sense as adding ions in an aqueous solution creates a more ordered environment around the ions due to the formation of ion-dipole interactions with water molecules. In conclusion, the negative values of standard entropy for F^-, OH^-, and S^2- ions do not contradict the third law of thermodynamics. These values signify that the disorder or randomness is less than that of the reference substances with which they are compared, and should not be misinterpreted as implying negative entropy at absolute zero.

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Key Concepts

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

Entropy
Entropy, symbolized as \( S \), is a fundamental concept in thermodynamics. It quantifies the degree of disorder or randomness in a system. Imagine a room: if it's tidy, things are in order, reflecting low entropy. However, a messy room, with things scattered everywhere, represents higher entropy. Entropy gives us a lens to view energy dispersal in a system, and how spontaneous changes might occur. In nature, systems tend to progress toward higher entropy or disorder.
  • The more the system's particles are spread out, the higher the entropy.
  • Systems at high temperatures typically have higher entropy than those at lower temperatures, due to increased molecular motion.
The third law of thermodynamics tells us about the behavior of entropy at extremely low temperatures, paving our understanding of perfectly ordered systems.
Standard Entropy
Standard entropy, denoted as \( S^\circ \), is the absolute entropy of a substance measured under standard conditions: pressure of 1 bar and a temperature of 298 K. This measurement provides a baseline to compare the entropy of different substances. Unlike absolute values, standard entropy values are relative to certain standard conditions.
  • Think of standard entropy like a scale we use to compare substances under common conditions.
  • Helps chemists understand how much disorder exists in a system compared to others.
The values can sometimes be less than zero, which can seem surprising. However, negative \( S^\circ \) values do not indicate a system has negative entropy but rather provide an insight into the comparability of disorderliness relative to reference states.
Perfect Crystal
A perfect crystal represents an idealized solid where atoms or molecules are in a flawless, orderly pattern. According to the third law of thermodynamics, a perfect crystal at absolute zero has entropy equal to zero. This is because all its particles are at their lowest possible energy state, forming a perfectly structured configuration with no randomness.
  • Such a state represents the ultimate order: a property unique to perfect crystals at 0 Kelvin.
  • In practice, achieving a perfect crystal is extremely challenging, and real substances never quite reach this state but are instead approximations.
Hence, discussing perfect crystals helps in understanding the extremes of the thermodynamic spectrum where entropy reaches its minimal limit.
Absolute Zero
Absolute zero, labeled as 0 Kelvin or \(-273.15\) Celsius, defines the lowest theoretical temperature. At this point, a system's thermal motion ceases, theoretically stopping all atomic movement. It provides a crucial reference point, especially in the third law of thermodynamics.
  • The third law asserts that as a system approaches absolute zero, its entropy approaches a constant minimum.
  • Absolute zero is more of a theoretical limit; in practice, it is impossible to reach completely.
This concept reinforces the understanding of how temperature and disorder are intertwined, demonstrating the broader implications of thermodynamic principles in both theoretical and practical applications.

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Most popular questions from this chapter

Consider the reactions $$ \begin{aligned} \mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) & \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q) \\ \mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) & \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q) \end{aligned} $$ where $$ \text { en }=\mathrm{H}_{2} \mathrm{~N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2} $$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text {reaction } 2}>K_{\text {reaction } 1} .\) Explain.

The following reaction occurs in pure water: $$ \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ which is often abbreviated as $$ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) $$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{~kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\). Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 \mathrm{M}\) and \(\left[\mathrm{H}^{+}\right]=0.71 M\).

Which of the following processes are spontaneous? a. Salt dissolves in \(\mathrm{H}_{2} \mathrm{O}\). b. A clear solution becomes a uniform color after a few drops of dye are added. c. Iron rusts. d. You clean your bedroom.

Consider the following reaction at \(800 . \mathrm{K}\) : $$ \mathrm{N}_{2}(g)+3 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{NF}_{3}(g) $$ An equilibrium mixture contains the following partial pressures: \(P_{\mathrm{N}_{2}}=0.021 \mathrm{~atm}, P_{\mathrm{F}_{2}}=0.063 \mathrm{~atm}, P_{\mathrm{NF}_{3}}=0.48 \mathrm{~atm}\). Calculate \(\Delta G^{\circ}\) for the reaction at \(800 . \mathrm{K}\).

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ} $$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

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