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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 atm is cooled to solid propanol at \(-150^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Short Answer

Expert verified
The qualitative sketch of how the entropy changes as propanol goes from vapor at \(150^{\circ}\mathrm{C}\) to solid at \(-150^{\circ}\mathrm{C}\) will show a general continuous decreasing trend with steeper slopes during the points of phase transition, which are at \(97.4^{\circ}\mathrm{C}\) and \(-126.5^{\circ}\mathrm{C}\). The sketch will have the following features: 1) Initial decrease of entropy as vapor cools down, 2) A more dramatic decrease in entropy at the vapor to liquid phase transition, 3) Decrease of entropy as liquid cools down, 4) Another dramatic decrease in entropy at the liquid to solid phase transition, and 5) Decrease of entropy as solid cools down to its final temperature.

Step by step solution

01

Identify phase transitions

From the given information, we know that propanol melts at \(-126.5^{\circ}\mathrm{C}\) and boils at \(97.4^{\circ}\mathrm{C}\). Therefore, the phase transitions are as follows: 1. Cooling from vapor to liquid occurs at \(97.4{ }^{\circ} \mathrm{C}\). 2. Cooling from liquid to solid occurs at \(-126.5^{\circ} \mathrm{C}\).
02

Cooling from vapor to liquid

When cooling the propanol vapor from \(150^{\circ} \mathrm{C}\) to \(97.4^{\circ} \mathrm{C}\), the entropy decreases as the molecules lose energy and come closer together. As the temperature reaches the boiling point, the entropy of the system will decrease more dramatically due to the phase transition from vapor to liquid.
03

Cooling from liquid to solid

After the phase transition from vapor to liquid, the entropy continues to decrease as the liquid cools down to \(-126.5^{\circ}\mathrm{C}\). At the melting point, the liquid changes into a solid. Just like in the previous case, the entropy of the system decreases more dramatically due to the phase transition.
04

Cooling the solid

Once the propanol has changed into a solid, the entropy continues to decrease as it cools down to \(-150^{\circ}\mathrm{C}\). In the solid phase, the entropy decreases at a slower rate compared to the phase transitions.
05

Drawing the qualitative sketch

Combine the information from the previous steps to draw a qualitative sketch of how the entropy changes during the cooling process of propanol from \(150^{\circ}\mathrm{C}\) vapor to \(-150^{\circ}\mathrm{C}\) solid. The sketch should show the following major features: 1. Initial decrease of entropy as vapor cools down. 2. A more dramatic decrease in entropy at the vapor to liquid phase transition. 3. Decrease of entropy as liquid cools down. 4. Another dramatic decrease in entropy at the liquid to solid phase transition. 5. Decrease of entropy as solid cools down to its final temperature. The entropy will have a general continuous decreasing trend with steeper slopes during the points of phase transition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Phase transitions
A phase transition refers to the change of a substance from one state of matter to another, like from vapor to liquid or liquid to solid.
Phase transitions are significant because they involve substantial energy exchanges, known as latent heat, despite the temperature remaining constant during the transition point itself.
In the case of propanol, we observe two important phase transitions when cooling it from vapor at 150°C to solid at -150°C:
  • Vapor to liquid at its boiling point, 97.4°C.
  • Liquid to solid at its melting point, -126.5°C.
The entropy, which measures molecular disorder, decreases notably at these transitions as molecules become more organized.
Propanol cooling
When cooling propanol, the process involves a series of temperature reductions and phase changes. Starting with propanol in its vapor phase at 150°C, it transitions through several temperature ranges until it becomes solid at -150°C.
During cooling:
  • The molecular movement decreases.
  • Energy is removed from the system, resulting in a lower entropy.
  • The once loosely held vapor molecules in vapor become more orderly in the subsequent phases.
This understanding of cooling and phase changes gives insight into energy conservation and molecular interactions in substances like propanol.
Entropy sketch
Entropy sketches are useful visuals for understanding how entropy changes across different temperature ranges and phases. In our case:
  • Starting as vapor, propanol begins with a high entropy due to dispersed molecular movement.
  • As temperature lowers towards the boiling point of 97.4°C, the entropy decreases, and we observe an abrupt decline when phase transition occurs.
  • Between 97.4°C and -126.5°C, in liquid form, the entropy reduces gradually until reaching the melting point.
  • An additional sharp decline in entropy happens as propanol turns into solid.
The sketch would show a continuous but uneven downward slope, with the sharpest drops at the phase transition temperatures.
Vapor to liquid
The vapor to liquid transition occurs at the boiling point of 97.4°C in propanol. During this change:
  • Vapor molecules lose significant energy.
  • They condense into a liquid, thus becoming more densely packed.
  • Entropy sharply decreases because the molecules become more ordered compared to the disorganized vapor phase.
This phase change is marked by latent heat release, where energy disperses without temperature change, altering molecule organization dramatically.
Liquid to solid
The transition from liquid to solid in propanol occurs at a temperature of -126.5°C. As this phase change happens:
  • Liquid molecules start arranging into a fixed and orderly structure seen in solids.
  • The system loses latent heat and entropy drops significantly because molecular freedom is minimized.
  • After freezing into a solid, temperature decrease leads to a moderated drop in entropy as the molecules further slow their vibrational activity.
Understanding this transition helps appreciate how order increases as substances move from liquids to solids, reflecting entropy changes in thermodynamics.

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Most popular questions from this chapter

The crystalline hydrate \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s)\) loses water when placed in a large, closed, dry vessel: $$ \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2} \cdot 4 \mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}(s)+4 \mathrm{H}_{2} \mathrm{O}(g) $$ This process is spontaneous and \(\Delta H\) is positive. Is this process an exception to Bertholet's generalization that all spontaneous changes are exothermic? Explain.

(a) Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of \(\Delta S_{\text {surr }} ?\) (c) During a certain reversible process, the surroundings undergo an entropy change, \(\Delta S_{\text {surr }}=-78 \mathrm{~J} / \mathrm{K}\). What is the entropy change of the system for this process?

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) can be made by the controlled oxidation of methane: $$ \mathrm{CH}_{4}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g) $$ (a) Use data in Appendix C to calculate \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. (b) How is \(\Delta G^{\circ}\) for the reaction expected to vary with increasing temperature? (c) Calculate \(\Delta G^{\circ}\) at \(298 \mathrm{~K}\). Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a temperature at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

(a) What is the meaning of the standard free-energy change, \(\Delta G^{\circ},\) as compared with \(\Delta G\) ? (b) For any process that occurs at constant temperature and pressure, what is the significance of \(\Delta G=0 ?(c)\) For a certain process, \(\Delta G\) is large and negative. Does this mean that the process necessarily occurs rapidly?

The standard entropies at \(298 \mathrm{~K}\) for certain of the group \(4 \mathrm{~A}\) elements are as follows: \(\mathrm{C}(s,\) diamond \()=2.43 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) \(\mathrm{Si}(s)=18.81 \mathrm{~J} / \mathrm{mol}-\mathrm{K}, \mathrm{Ge}(s)=31.09 \mathrm{~J} / \mathrm{mol}-\mathrm{K},\) and \(\operatorname{Sn}(s)=\) \(51.818 \mathrm{~J} / \mathrm{mol}-\mathrm{K} .\) All but \(\mathrm{Sn}\) have the diamond structure. How do you account for the trend in the \(S^{\circ}\) values?

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