Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two crystalline forms of white phosphorus are known. Both forms contain \(\mathrm{P}_{4}\) molecules, but the molecules are packed together in different ways. The \(\alpha\) form is always obtained when the liquid freezes. However, below \(-76.9^{\circ} \mathrm{C}\), the \(\alpha\) form spontaneously converts to the \(\beta\) form: $$\mathrm{P}_{4}(s, \alpha) \longrightarrow \mathrm{P}_{4}(s, \beta)$$ a. Predict the signs of \(\Delta H\) and \(\Delta S\) for this process. b. Predict which form of phosphorus has the more ordered crystalline structure (has the smaller positional probability).

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.

Consider a weak acid, HX. If a \(0.10 M\) solution of HX has a pH of \(5.83\) at \(25^{\circ} \mathrm{C}\), what is \(\Delta G^{\circ}\) for the acid's dissociation reaction at \(25^{\circ} \mathrm{C}\) ?

Consider the following reaction: \(\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) \quad K_{298}=0.090\) For \(\mathrm{Cl}_{2} \mathrm{O}(g)\), \(\Delta G_{\mathrm{f}}^{\circ}=97.9 \mathrm{~kJ} / \mathrm{mol}\) \(\Delta H_{\mathrm{f}}^{\circ}=80.3 \mathrm{~kJ} / \mathrm{mol}\) \(S^{\circ}=266.1 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\) a. Calculate \(\Delta G^{\circ}\) for the reaction using the equation \(\Delta G^{\circ}=\) \(-R T \ln (K)\) b. Use bond energy values (Table 8.4) to estimate \(\Delta H^{\circ}\) for the reaction. c. Use the results from parts a and \(b\) to estimate \(\Delta S^{\circ}\) for the reaction. d. Estimate \(\Delta H_{\mathrm{f}}^{\circ}\) and \(S^{\circ}\) for \(\mathrm{HOCl}(g)\). e. Estimate the value of \(K\) at \(500 . \mathrm{K}\). f. Calculate \(\Delta G\) at \(25^{\circ} \mathrm{C}\) when \(P_{\mathrm{H}_{2} \mathrm{O}}=18\) torr, \(P_{\mathrm{Cl}_{2} \mathrm{O}}=2.0\) torr, and \(P_{\mathrm{HOCl}}=0.10\) torr.

Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free