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Chapter 17: Problem 7

Predict the sign of \(\Delta S\) for each of the following and explain. a. the evaporation of alcohol b. the freezing of water c. compressing an ideal gas at constant temperature d. dissolving \(\mathrm{NaCl}\) in water

Short Answer

Expert verified
a. For the evaporation of alcohol, 螖S > 0 because the process increases disorder by turning the liquid into a gas. b. For the freezing of water, 螖S < 0 as the process increases order by turning the liquid into a solid. c. For compressing an ideal gas at constant temperature, 螖S < 0 because the process increases order by packing molecules closer together. d. For dissolving NaCl in water, 螖S > 0 as the process increases disorder by separating the ions and increasing the complexity of the solution.

Step by step solution

01

a. Evaporation of alcohol

When alcohol evaporates, it turns from a liquid state to a gaseous state. In the gaseous state, the alcohol molecules have more freedom to move around and are more disordered than in the liquid state. Thus, the entropy of the system increases during the evaporation process. We can predict that 螖S > 0 for the evaporation of alcohol.
02

b. Freezing of water

When water freezes, it turns from a liquid state to a solid state. In the solid state, the water molecules have less freedom to move around and are more ordered than in the liquid state. Thus, the entropy of the system decreases during the freezing process. We can predict that 螖S < 0 for the freezing of water.
03

c. Compressing an ideal gas at constant temperature

When an ideal gas is compressed at constant temperature, the volume of the gas decreases while the number of molecules remains the same. This results in the molecules being packed closer together, which makes the system more ordered. Hence, the entropy of the system decreases during the compression process. We can predict that 螖S < 0 for compressing an ideal gas at constant temperature.
04

d. Dissolving NaCl in water

When NaCl dissolves in water, the Na+ and Cl- ions become separated and surrounded by water molecules, introducing more possibilities for the arrangement of ions and water molecules in the system. This leads to an increase in disorder and complexity in the solution. Consequently, the entropy of the system increases during the dissolution process. We can predict that 螖S > 0 for dissolving NaCl in water.

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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?

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. \(\mathrm{NaCl}(s) \longrightarrow \mathrm{NaCl}(l)\)

Consider the following system at equilibrium at \(25^{\circ} \mathrm{C}\) : $$\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) \quad \Delta G^{\circ}=-92.50 \mathrm{~kJ}$$ What will happen to the ratio of partial pressure of \(\mathrm{PCl}_{5}\) to partial pressure of \(\mathrm{PCl}_{3}\) if the temperature is raised? Explain completely.

Monochloroethane \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\right)\) can be produced by the direct reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas or by the reaction of ethylene gas \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) with hydrogen chloride gas. The second reaction gives almost a \(100 \%\) yield of pure \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) at a rapid rate without catalysis. The first method requires light as an energy source or the reaction would not occur. Yet \(\Delta G^{\circ}\) for the first reaction is considerably more negative than \(\Delta G^{\circ}\) for the second reaction. Explain how this can be so.

Carbon monoxide is toxic because it bonds much more strongly to the iron in hemoglobin (Hgb) than does \(\mathrm{O}_{2}\). Consider the following reactions and approximate standard free energy changes: $$\begin{array}{clr}\mathrm{Hgb}+\mathrm{O}_{2} \longrightarrow \mathrm{HgbO}_{2} & \Delta G^{\circ}=-70 \mathrm{~kJ} \\ \mathrm{Hgb}+\mathrm{CO} \longrightarrow \mathrm{HgbCO} & \Delta G^{\circ}=-80 \mathrm{~kJ} \end{array}$$ Using these data, estimate the equilibrium constant value at \(25^{\circ} \mathrm{C}\) for the following reaction: $$\mathrm{HgbO}_{2}+\mathrm{CO} \rightleftharpoons \mathrm{HgbCO}+\mathrm{O}_{2}$$
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