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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 NaCl in water

Short Answer

Expert verified
a. Evaporation of alcohol: 螖S is positive because the molecules become more spaced out and disordered. b. Freezing of water: 螖S is negative because the molecules become more ordered in the solid phase. c. Compressing an ideal gas at constant temperature: 螖S is negative because the molecules become more constrained and ordered. d. Dissolving NaCl in water: 螖S is positive because the ions increase the disorder in the solution.

Step by step solution

01

a. Evaporation of alcohol

The evaporation of alcohol is a process in which alcohol molecules go from the liquid phase to the gaseous phase. In the gaseous phase, the molecules are more spaced out and move freely as compared to their more ordered and compact state in the liquid phase, leading to increased disorder. Therefore, the entropy increases during evaporation. The sign of 螖S for the evaporation of alcohol is positive.
02

b. Freezing of water

The freezing of water is a process in which water molecules in the liquid phase get converted into the solid phase (ice). In the solid phase, the molecules have a more regular and ordered arrangement as compared to the liquid phase. This leads to decreased disorder, which means that during the process of freezing, the entropy decreases. The sign of 螖S for the freezing of water is negative.
03

c. Compressing an ideal gas at constant temperature

When an ideal gas is compressed at constant temperature, the volume of the gas decreases, causing the molecules to occupy a smaller space. As the molecules get closer together and their arrangement becomes more constrained, the disorder of the system decreases. Thus, the entropy decreases during compression at constant temperature. The sign of 螖S for compressing an ideal gas at constant temperature is negative.
04

d. Dissolving NaCl in water

When NaCl dissolves in water, the individual ions (Na+ and Cl-) are separated from the solid structure and become surrounded by water molecules, forming a homogeneous solution. The increase in the number of particles in the solution increases the disorder of the system, leading to an increase in entropy. The sign of 螖S for dissolving NaCl in water is positive.

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Most popular questions from this chapter

Describe how the following changes affect the positional probability of a substance. a. increase in volume of a gas at constant T b. increase in temperature of a gas at constant V c. increase in pressure of a gas at constant T

Gas \(A_{2}\) reacts with gas \(B_{2}\) to form gas AB at a constant temperature. The bond energy of \(A B\) is much greater than that of either reactant. What can be said about the sign of \(\Delta H ? \Delta S_{\mathrm{surr}} ?\) \(\Delta S\)? Explain how potential energy changes for this process. Explain how random kinetic energy changes during the process.

The equilibrium constant K for the reaction $$2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g)$$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus 1\(/ T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{K}\) and a y-intercept of \(-14.51 .\) Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise \(85 .\)

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?

Consider the reactions $$\mathrm{Ni}^{2+}(a q)+6 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ni}\left(\mathrm{NH}_{3}\right)_{6}^{2+}(a q)$$ $$\mathrm{Ni}^{2+}(a q)+3 \mathrm{en}(a q) \longrightarrow \mathrm{Ni}(\mathrm{en})_{3}^{2+}(a q)$$ where $$\mathrm{en}=\mathrm{H}_{2} \mathrm{N}-\mathrm{CH}_{2}-\mathrm{CH}_{2}-\mathrm{NH}_{2}$$ The \(\Delta H\) values for the two reactions are quite similar, yet \(\mathrm{K}_{\text { reaction } 2}>K_{\text { reaction }}\) . Explain.
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