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Chapter 17: Problem 2

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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Most popular questions from this chapter

Using the following data, calculate the value of \(K_{\mathrm{sp}}\) for \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\), one of the least soluble of the common nitrate salts. $$ \begin{array}{lc} \text { Species } & \Delta G_{\mathrm{f}}^{\circ} \\ \mathrm{Ba}^{2+}(a q) & -561 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{NO}_{3}^{-}(a q) & -109 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}(s) & -797 \mathrm{~kJ} / \mathrm{mol} \\ \hline \end{array} $$

Consider the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g) $$ For each of the following mixtures of reactants and products at \(25^{\circ} \mathrm{C}\), predict the direction in which the reaction will shift to reach equilibrium. a. \(P_{\mathrm{NO}_{2}}=P_{\mathrm{N}_{2} \mathrm{O}_{4}}=1.0 \mathrm{~atm}\) b. \(P_{\mathrm{NO}_{2}}=0.21 \mathrm{~atm}, P_{\mathrm{N}_{2} \mathrm{O}_{4}}=0.50 \mathrm{~atm}\) c. \(P_{\mathrm{NO},}=0.29 \mathrm{~atm}, P_{\mathrm{N}, \mathrm{O}_{4}}=1.6 \mathrm{~atm}\)

For the process \(\mathrm{A}(l) \longrightarrow \mathrm{A}(\mathrm{g})\), which direction is favored by changes in energy probability? Positional probability? Explain your answers. If you wanted to favor the process as written, would you raise or lower the temperature of the system? Explain.

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

Human DNA contains almost twice as much information as is needed to code for all the substances produced in the body. Likewise, the digital data sent from Voyager II contained one redundant bit out of every two bits of information. The Hubble space telescope transmits three redundant bits for every bit of information. How is entropy related to the transmission of information? What do you think is accomplished by having so many redundant bits of information in both DNA and the space probes?
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