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Chapter 17: Problem 17

The third law of thermodynamics states that the entropy of a perfect crystal at \(0 \mathrm{~K}\) is zero. In Appendix \(4, \mathrm{~F}^{-}(a q), \mathrm{OH}^{-}(a q)\), and \(\mathrm{S}^{2-}(a q)\) all have negative standard entropy values. How can \(S^{\circ}\) values be less than zero?

Short Answer

Expert verified
In conclusion, the negative standard entropy values of F^-(aq), OH^-(aq), and S^2-(aq) are due to the increased order or decreased randomness in the system when they dissolve in water and form hydration shells. These values are in reference to their specific reference states, not absolute zero. Therefore, the third law of thermodynamics, stating the entropy of a perfect crystal at 0K is zero, does not apply to these ions in solution.

Step by step solution

01

Understanding Entropy and Third Law of Thermodynamics

Entropy is a measure of disorder or randomness in a system. The third law of thermodynamics states that, at absolute zero (0K), a perfect crystal will have zero entropy, meaning it has a perfectly ordered state.
02

Understanding Standard Entropy

Standard entropy (S°) is the entropy of a substance at a standard state, typically 1 atm pressure and 298 K (25°C). A positive entropy value indicates an increase in the disorder of a substance, while a negative value indicates a decrease in the disorder compared to the reference state.
03

Investigating Negative Standard Entropy Values

The negative standard entropy values of F^-(aq), OH^-(aq), and S^2-(aq) indicate that their formation from their respective reference states increases the order or decreases the randomness in the system. It's important to understand that the negative values are not in reference to absolute zero, but rather to their specific reference states.
04

Dissolving as an Ordered Process

When ions like F^-(aq), OH^-(aq), and S^2-(aq) dissolve in water, they interact with water molecules and form ordered structures around them (hydration shells). This ordering of the interactions between the ions and water molecules leads to a decrease in entropy, resulting in a negative entropy for the ions.
05

Entropy at 0K should not be Considered for Ions in Solution

The third law of thermodynamics and its relation to the entropy of a perfect crystal at 0K does not apply for ions in solutions. Ions in solutions exist at conditions far above 0K, so we cannot compare their entropy values to that of a perfect crystal at absolute zero. In conclusion, F^-(aq), OH^-(aq), and S^2-(aq) have negative standard entropy values because their formation increases the order or decreases the randomness in the system, not because their entropy is less than that of a perfect crystal at 0K. The third law of thermodynamics does not apply to these ions, as their entropy values are not compared to the absolute zero state.

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Most popular questions from this chapter

Hydrogen cyanide is produced industrially by the following exothermic reaction: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \stackrel{\mathrm{low}^{\circ} \mathrm{c}}{\mathrm{Pt} \cdot \mathrm{Rh}} 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ Is the high temperature needed for thermodynamic or kinetic reasons?

Consider the reaction $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ where \(\Delta H^{\circ}=-103.8 \mathrm{~kJ} / \mathrm{mol} .\) In a particular experiment, equal moles of \(\mathrm{H}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) and \(\mathrm{Br}_{2}(\mathrm{~g})\) at \(1.00 \mathrm{~atm}\) were mixed in a \(1.00\) -L flask at \(25^{\circ} \mathrm{C}\) and allowed to reach equilibrium. Then the molecules of \(\mathrm{H}_{2}\) at equilibrium were counted using a very sensitive technique, and \(1.10 \times 10^{13}\) molecules were found. For this reaction, calculate the values of \(K, \Delta G^{\circ}\), and \(\Delta S^{\circ}\).

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.

Predict the sign of \(\Delta S_{\text {surr }}\) for the following processes. a. \(\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)\) b. \(I_{2}(g) \longrightarrow I_{2}(s)\)

Hydrogen sulfide can be removed from natural gas by the reaction $$2 \mathrm{H}_{2} \mathrm{~S}(g)+\mathrm{SO}_{2}(g) \rightleftharpoons 3 \mathrm{~S}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$$ Calculate \(\Delta G^{\circ}\) and \(K\) (at \(298 \mathrm{~K}\) ) for this reaction. Would this reaction be favored at a high or low temperature?
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