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

Short Answer

Expert verified
For a liquid, we would expect \(\Delta S_{\text{evaporation}}\) to be larger than \(\Delta S_{\text {fusion}}\) because the increase in disorder is greater when transitioning from a liquid to a gas, compared to transitioning from a solid to a liquid. This is due to the gas phase having no defined structure and more freedom to move, leading to a higher degree of randomness.

Step by step solution

01

Understanding Entropy

Entropy (S) is a measure of the disorder or randomness in a system. The second law of thermodynamics states that any natural process will tend to increase the total entropy of the universe. In the context of phase transitions, it means that as a substance moves from a more organized phase to a less organized phase, its entropy increases.
02

Comparing Fusion and Evaporation

In the process of fusion, a solid becomes a liquid, whereas in the process of evaporation, a liquid transitions to a gas. Recall from step 1 that entropy increases when a substance moves from a more organized phase to a less organized phase. Let's assess the order in each phase: 1. Solid: Particles are closely packed and have a well-defined structure. 2. Liquid: Particles are more spaced apart than in a solid and have less defined structure. 3. Gas: Particles are very spread out and have no defined structure. Comparing the phases, we can see that going from a solid to a liquid (fusion) introduces some disorder, but going from a liquid to a gas (evaporation) introduces much more disorder.
03

Determining Which Entropy Change is Larger

Given the increase in disorder for each phase transition, we can now identify which entropy change is larger. Since evaporation introduces a larger increase in disorder compared to fusion, we can conclude that the entropy change for evaporation, \(\Delta S_{\text {evaporation}}\), is larger than the entropy change for fusion, \(\Delta S_{\text {fusion}}\).
04

Providing an Explanation

The larger entropy change for evaporation is due to the substantial increase in disorder when a substance transitions from a liquid (with some degree of structure) to a gas (with no structure). In the gas phase, particles are much more spread apart and have more freedom to move around, leading to a higher degree of randomness. On the other hand, while fusion also results in an increase in entropy, the increase is not as significant because the disorder between a solid and a liquid is not as substantial as between a liquid and a gas. In conclusion, \(\Delta S_{\text {evaporation}}\) is larger than \(\Delta S_{\text {fusion}}\) because the increase in disorder during evaporation is greater than that during fusion.

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Most popular questions from this chapter

Predict the sign of \(\Delta S^{\circ}\) for each of the following changes. a. \(\mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{NaCl}(s)\) b. \(\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{~g})\) c. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{Cl}^{-}(a q)\) d. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\)

List three different ways to calculate the standard free energy change, \(\Delta G^{\circ}\), for a reaction at \(25^{\circ} \mathrm{C}\). How is \(\Delta G^{\circ}\) estimated at temperatures other than \(25^{\circ} \mathrm{C}\) ? What assumptions are made?

Consider the following reaction: $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$Calculate \(\Delta G\) for this reaction under the following conditions (assume an uncertainty of \(\pm 1\) in all quantities): a. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=P_{\mathrm{H}_{2}}=200 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=50 \mathrm{~atm}\) b. \(T=298 \mathrm{~K}, P_{\mathrm{N}_{2}}=200 \mathrm{~atm}, P_{\mathrm{H}_{2}}=600 \mathrm{~atm}, P_{\mathrm{NH}_{3}}=200 \mathrm{~atm}\)

Given the following data: $$\begin{aligned}2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) & & \Delta G^{\circ}=-51 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{O}_{2}(g) & \Delta \mathrm{H}_{2} \mathrm{O}(l) & & \Delta G^{\circ}=-474 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ} &=-394 \mathrm{~kJ}\end{aligned}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(\mathrm{~g})+2 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(l) .\)

Entropy has been described as "time's arrow." Interpret this view of entropy.

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