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Chapter 19: Problem 38

Propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\right)\) melts at \(-126.5^{\circ} \mathrm{C}\) and boils at \(97.4^{\circ} \mathrm{C}\). Draw a qualitative sketch of how the entropy changes as propanol vapor at \(150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short Answer

Expert verified
A qualitative sketch of entropy vs. temperature for the cooling of propanol vapor can be divided into five steps: 1. Cooling of propanol vapor from \(150^{\circ} \mathrm{C}\) to \(97.4^{\circ} \mathrm{C}\), where the entropy decreases with a negative slope. 2. Phase change - condensation at \(97.4^{\circ} \mathrm{C}\), where entropy abruptly decreases. 3. Cooling of propanol liquid from \(97.4^{\circ} \mathrm{C}\) to \(-126.5^{\circ} \mathrm{C}\), where the entropy decreases with a less negative slope than the vapor phase. 4. Phase change - freezing at \(-126.5^{\circ} \mathrm{C}\), where entropy abruptly decreases again. 5. Cooling of propanol solid from \(-126.5^{\circ} \mathrm{C}\) to \(-150^{\circ} \mathrm{C}\), where the entropy decreases with a smaller slope than the liquid phase.

Step by step solution

01

Cooling of Propanol Vapor

As the propanol vapor cools down from \(150^{\circ} \mathrm{C}\) to its boiling point, \(97.4^{\circ} \mathrm{C}\), the particles lose energy and become more ordered. As a result, the entropy of propanol vapor will decrease with a negative slope.
02

Phase Change - Condensation

Propanol vapor will undergo a phase change and become a liquid at its boiling point, \(97.4^{\circ} \mathrm{C}\). During this phase change, the propanol particles become even more ordered due to stronger intermolecular forces. Entropy will abruptly decrease at this point.
03

Cooling of Propanol Liquid

As the liquid propanol cools down from the boiling point to its melting point, \(-126.5^{\circ} \mathrm{C}\), the particles continue to lose energy. As a result, the entropy will continue to decrease, but with a less negative slope than the vapor phase.
04

Phase Change - Freezing

Propanol will undergo another phase change and become a solid at its melting point, \(-126.5^{\circ} \mathrm{C}\). During this change, the particles will become highly ordered and fixed in their positions within the solid crystal lattice. The entropy will again abruptly decrease at this point.
05

Cooling of Propanol Solid

Finally, as the solid propanol cools down from its melting point, \(-126.5^{\circ} \mathrm{C}\), to \(-150^{\circ} \mathrm{C}\), the particles continue to lose energy and slight movement. Therefore, the entropy will continue to decrease but with a smaller slope than the liquid phase. Based on these entropy changes, a qualitative sketch of entropy vs. temperature can be drawn with decreasing slopes during cooling phases, abrupt drops during phase changes, and different slopes for each type of phase (vapor, liquid, and solid).

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Most popular questions from this chapter

For each of the following pairs, indicate which substance possesses the larger standard entropy: (a) \(1 \mathrm{~mol}\) of \(\mathrm{P}_{4}(\mathrm{~g})\) at \(300{ }^{\circ} \mathrm{C}, 0.01 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{As}_{4}(\mathrm{~g})\) at \(300{ }^{\circ} \mathrm{C}, 0.01 \mathrm{~atm}\); (b) \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm}\), or \(1 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) at \(100^{\circ} \mathrm{C}, 1 \mathrm{~atm} ;\) (c) \(0.5 \mathrm{~mol}\) of \(\mathrm{N}_{2}(g)\) at \(298 \mathrm{~K}, 20\) - \(\mathrm{L}\) volume, or \(0.5 \mathrm{~mol} \mathrm{CH}_{4}(g)\) at \(298 \mathrm{~K}, 20-\mathrm{L}\) volume; (d) \(100 \mathrm{~g}\) \(\mathrm{Na}_{2} \mathrm{SO}_{4}(s)\) at \(30^{\circ} \mathrm{C}\) or \(100 \mathrm{~g} \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)\) at \(30^{\circ} \mathrm{C} .\)

For a particular reaction, \(\Delta H=-32 \mathrm{~kJ}\) and \(\Delta S=\) \(-98 \mathrm{~J} / \mathrm{K}\). Assume that \(\Delta H\) and \(\Delta S\) do not vary with temperature. (a) At what temperature will the reaction have \(\Delta G=0 ?\) (b) If \(T\) is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

(a) For a process that occurs at constant temperature, express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant \(T\) and \(P\), the value of \(\Delta G\) is positive. What can you conclude? (c) What is the relationship between \(\Delta G\) for a process and the rate at which it occurs?

A system goes from state 1 to state 2 and back to state 1 . (a) What is the relationship between the value of \(\Delta E\) for going from state 1 to state 2 to that for going from state 2 back to state 1 ? (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\) (c) Suppose the changes in state are reversible processes. Can you conclude any thing about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state \(1 ?\)

Ammonium nitrate dissolves spontaneously and endothermally in water at room temperature. What can you deduce about the sign of \(\Delta S\) for this solution process?
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