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For a liquid, which would you expect to be larger, \(\Delta S_{\text {fusion }}\) or \(\Delta S_{\text {evaporation }}\) ? Why?

Short Answer

Expert verified
In summary, for a liquid, the change in entropy for the evaporation process, \(\Delta S_{\text{evaporation}}\), is larger than the change in entropy for the fusion process, \(\Delta S_{\text{fusion}}\). This is because the transition from a liquid to a gaseous state involves a more significant increase in the molecule's freedom and randomness compared to the transition from a solid to a liquid state.

Step by step solution

01

Understand Entropy Change

The change in entropy, represented by \(\Delta S\), represents the degree of disorder or randomness in a system. An increase in entropy indicates an increase in the disorder of a system, while a decrease indicates a more ordered state.
02

Fusion vs Evaporation

The fusion is the process of turning from solid to liquid, whereas evaporation is the process of turning from liquid to gas. Generally, gases have more disorder and randomness as their molecules move much more freely and occupy larger volumes than liquids or solids. Thus, we can expect the entropy change for the two processes to be different.
03

Compare Entropy Change for Fusion and Evaporation

Comparing fusion and evaporation, we can realize that turning a substance from a liquid to a gas (evaporation) involves a significant increase in the degree of randomness and disorder as molecules have much more freedom to move and spread out in a gaseous state than they do in a liquid state. On the other hand, going from a solid to a liquid (fusion) also increases the degree of randomness and disorder but to a lesser extent as the molecules in the liquid state are still relatively close to each other.
04

Determine Which Entropy Change is Larger

As noted, the process of evaporation results in a greater increase in randomness and disorder in comparison to the process of fusion. Consequently, we can conclude that: \[\Delta S_{\text{evaporation}} > \Delta S_{\text{fusion}}\]
05

Conclusion

In sum, for a liquid, the change in entropy for the evaporation process is larger than the change in entropy for the fusion process. This is because the transition from a liquid to a gaseous state involves a more significant increase in the molecule's freedom and randomness compared to the transition from a solid to a liquid state.

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Key Concepts

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

Phase Transition
Phase transitions involve changes in the state of a substance, such as going from solid to liquid (fusion) or from liquid to gas (evaporation). These transitions occur because substances gain or lose energy, leading to a rearrangement of their molecular structure.
  • During the fusion process, a solid absorbs heat, causing its particles to move more freely, turning into a liquid.
  • In evaporation, a liquid absorbs energy to become a gas, with molecules gaining even more freedom and spreading out significantly.
Phase transitions play a crucial role in understanding entropy changes because the degree of molecular movement differs in each state. The larger the energy required or the greater the freedom of movement, the more significant the phase change, affecting the entropy.
Entropy Change
Entropy is a measure of a system's disorder or randomness, with changes in entropy reflecting how molecular arrangements shift during phase transitions. In any given change:
  • An increase in entropy indicates more randomness; molecules move more freely.
  • A decrease in entropy means a more ordered state, as molecular movement is more restricted.
For example, when ice melts into water (fusion), molecules become more disordered than in their rigid ice state, meaning entropy increases. However, moving from water to vapor (evaporation) signifies an even more substantial entropy increase because gas molecules spread out and possess greater freedom. Understanding entropy is crucial in predicting and explaining the energetics behind different phase transitions.
Fusion vs Evaporation
Comparing fusion and evaporation helps to better understand how phase transitions affect entropy. Consider the molecular behavior:
  • Fusion (solid to liquid) involves molecules gaining enough energy to break free from their fixed positions but remaining relatively close.
  • Evaporation (liquid to gas) provides molecules with the energy to break loose entirely and move independently, dispersing throughout a container.
Since evaporation allows molecules much more freedom compared to fusion, the change in entropy is more pronounced. Thus, for a given substance, the change in entropy during evaporation e ( Δ S _ e v a p o r a t i o n ) is greater than that during fusion e ( Δ S _ f u s i o n ). This concept underscores the idea that gases represent a state of higher entropy than solids or liquids.

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Most popular questions from this chapter

Gas \(\mathrm{A}_{2}\) reacts with gas \(\mathrm{B}_{2}\) to form gas \(\mathrm{AB}\) at a constant temperature. The bond energy of \(\mathrm{AB}\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\text {surr }}\) ? \(\Delta S\) ? Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

Which of the following involve an increase in the entropy of the system? a. melting of a solid b. sublimation c. freezing d. mixing e. separation f. boiling

For the reaction $$ \mathrm{SF}_{4}(g)+\mathrm{F}_{2}(g) \longrightarrow \mathrm{SF}_{6}(g) $$ the value of \(\Delta G^{\circ}\) is \(-374 \mathrm{~kJ}\). Use this value and data from Appendix 4 to calculate the value of \(\Delta G_{\mathrm{f}}^{\circ}\) for \(\mathrm{SF}_{4}(g)\).

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.

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

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