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Does the entropy of the system increase, decrease, or stay the same when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

Short Answer

Expert verified
The entropy of the system (a) increases when a solid melts, (b) decreases when a gas liquefies, and (c) increases when a solid sublimes, due to the changes in disorder and freedom of movement of the particles in each process.

Step by step solution

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a) Solid melting

When a solid melts, the particles in the solid gain more energy, which allows them to move more freely and take on more possible configurations. This increased freedom of motion leads to an increase in the disorder of the system. Therefore, the entropy of the system increases when a solid melts.
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b) Gas liquefying

When a gas liquefies, the particles in the gas lose some energy, which causes them to become more closely bound together and less free to move. This reduction in freedom of movement leads to a decrease in the disorder of the system. As a consequence, the entropy of the system decreases when a gas liquefies.
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c) Solid subliming

Sublimation is the process of a solid going directly to the gas phase without passing through the liquid phase. When a solid sublimes, the particles in the solid gain a significant amount of energy, causing them to become less ordered and more random in their position and movements. The gas phase has a higher level of disorder compared to the solid state. Therefore, the entropy of the system increases when a solid sublimes.

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Key Concepts

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

Solid Melting Entropy
Understanding the entropy change involved when a solid melts is fundamental to grasping the nature of phase transitions. Entropy, a measure of disorder or randomness within a system, typically increases during a melting process. This is because in the solid state, molecules are arranged in a structured lattice with limited freedom of movement. As heat is applied, the added energy overcomes the forces holding the lattice together, allowing the molecules to move about more freely in the liquid state.

As the solid transitions to a liquid, the number of possible configurations for the molecules increases. This increase in microstates translates directly into an increase in entropy. A common example of this process is the melting of ice to water. In essence, the entropy increase reflects the system's shift from a more ordered to a less ordered state, a key point for students to remember when considering phase changes and entropy.
Gas Liquefying Entropy
In contrast to the melting of a solid, the liquefaction of a gas represents a decrease in entropy. In the gaseous phase, particles have the greatest freedom of movement and the largest volume to occupy, resulting in a high degree of disorder and a high number of microstates. When a gas condenses into a liquid, the particles lose energy and become restricted in motion due to intermolecular forces drawing them closer together.

During this transition from gas to liquid, the entropy decrease is due to a reduction in the system's disorder and the number of ways particles can be arranged. A practical illustration is the condensation of steam into liquid water. By recognizing this decrease in entropy, students can better understand how energy and particle dynamics influence phase changes, linking entropy to the observable phenomena around them.
Solid Subliming Entropy
The phase transition known as sublimation, where a solid turns directly into a gas, is intriguing because it skips the liquid phase altogether. A prime example of sublimation that can be readily observed is dry ice (solid carbon dioxide) turning into carbon dioxide gas. During sublimation, the solid gains enough energy for the particles to overcome not just the intermolecular forces that bind them in a solid state, but also the ones that would keep them as a liquid.

This leap to the gaseous phase means particles can spread out and move freely and independently, which represents a significant increase in the number of possible microstates. As a result, sublimation leads to a large increase in entropy, signifying a substantial rise in disorder within the system. This concept helps students visualize how solids can transform directly into gases and the dramatic entropy changes associated with such phase transitions.

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Most popular questions from this chapter

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.

For a particular reaction, \(\Delta H=-32 \mathrm{kJ}\) and \(\Delta S=-98 \mathrm{J} / \mathrm{K}\) . 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 \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

Consider the following three reactions: $$ \begin{array}{l}{\text { (i) } \operatorname{Ti}(s)+2 \mathrm{Cl}_{2}(g) \longrightarrow \operatorname{TiCl}_{4}(g)} \\ {\text { (ii) } \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{CCl}_{4}(g)+6 \mathrm{HCl}(g)} \\ {\text { (iii) } \mathrm{BaO}(s)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{BaCO}_{3}(s)}\end{array} $$ (a) For each of the reactions, use data in Appendix \(C\) to calculate \(\Delta H^{\circ}, \Delta G^{\circ}, K,\) and \(\Delta S^{\circ}\) at \(25^{\circ} \mathrm{C}\) . (b) Which of these reactions are spontaneous under standard conditions at \(25^{\circ} \mathrm{C} ?(\mathbf{c})\) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature.

The value of \(K_{a}\) for nitrous acid \(\left(\mathrm{HNO}_{2}\right)\) at \(25^{\circ} \mathrm{C}\) is given in Appendix D. (a) Write the chemical equation for the equilibrium that corresponds to \(K_{a \cdot}\) (b) By using the value of \(K_{a},\) calculate \(\Delta G^{\circ}\) for the dissociation of nitrous acid in aqueous solution. (c) What is the value of \(\Delta G\) at equilibrium? (d) What is the value of \(\Delta G\) when \(\left[\mathrm{H}^{+}\right]=5.0 \times 10^{-2} M\) \(\left[\mathrm{NO}_{2}^{-}\right]=6.0 \times 10^{-4} M,\) and \(\left[\mathrm{HNO}_{2}\right]=0.20 \mathrm{M} ?\)

(a) In a chemical reaction, two gases combine to form a solid. What do you expect for the sign of \(\Delta S ?\) (b) How does the entropy of the system change in the processes described in Exercise 19.12\(?\)

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