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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: ΔS is positive because the molecules become more spaced out and disordered. b. Freezing of water: ΔS is negative because the molecules become more ordered in the solid phase. c. Compressing an ideal gas at constant temperature: ΔS is negative because the molecules become more constrained and ordered. d. Dissolving NaCl in water: ΔS 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 ΔS 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 ΔS 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 ΔS 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 ΔS for dissolving NaCl in water is positive.

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Most popular questions from this chapter

For rubidium \(\Delta H_{\mathrm{vap}}^{\circ}=69.0 \mathrm{kJ} / \mathrm{mol}\) \(686^{\circ} \mathrm{C},\) its boiling point. Calculate $\Delta S^{\circ}, q, w,\( and \)\Delta E$ for the vaporization of 1.00 mole of rubidium at \(686^{\circ} \mathrm{C}\) and 1.00 atm pressure.

Some nonelectrolyte solute (molar mass \(=142 \mathrm{g} / \mathrm{mol} )\) was dissolved in \(150 . \mathrm{mL}\) of a solvent (density $=0.879 \mathrm{g} / \mathrm{cm}^{3} )$ The elevated boiling point of the solution was 355.4 \(\mathrm{K}\) . What mass of solute was dissolved in the solvent? For the solvent, the enthalpy of vaporization is 33.90 \(\mathrm{kJ} / \mathrm{mol}\) , the entropy of vaporization is 95.95 \(\mathrm{J} / \mathrm{K} \cdot\) mol, and the boiling- point elevation constant is 2.5 \(\mathrm{K} \cdot \mathrm{kg} / \mathrm{mol}\) .

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?

The following reaction occurs in pure water: $$\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ which is often abbreviated as $$\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q)$$ For this reaction, \(\Delta G^{\circ}=79.9 \mathrm{kJ} / \mathrm{mol}\) at \(25^{\circ} \mathrm{C}\) . Calculate the value of \(\Delta G\) for this reaction at \(25^{\circ} \mathrm{C}\) when \(\left[\mathrm{OH}^{-}\right]=0.15 M\) and \(\left[\mathrm{H}^{+}\right]=0.71 M .\)

Predict the sign of \(\Delta S^{\circ}\) and then calculate \(\Delta S^{\circ}\) for each of the following reactions. a. $2 \mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}_{\text { rhombic}}(s)+2 \mathrm{H}_{2} \mathrm{O}(g)$ b. $2 \mathrm{SO}_{3}(g) \longrightarrow 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$ c. $\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$
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