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The normal freezing point of 1 -propanol \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\right)\) is \(-127{ }^{\circ} \mathrm{C}\). (a) Is the freezing of 1-propanol an endothermic or exothermic process? (b) In what temperature range is the freezing of 1-propanol a spontaneous process? (c) In what temperature range is it a nonspontaneous process? (d) Is there any temperature at which liquid and solid 1-propanol are in equilibrium? Explain.

Short Answer

Expert verified
(a) The freezing of 1-propanol is an exothermic process. (b) Freezing is spontaneous at temperatures below -127 °C. (c) Freezing is nonspontaneous at temperatures above -127 °C. (d) Liquid and solid 1-propanol are in equilibrium at the freezing point of -127 °C.

Step by step solution

01

Determine the freezing process (endothermic or exothermic)

Freezing is a process in which a substance changes from a liquid phase to a solid phase. Recall that during freezing, the molecules of the substance lose energy which results in a decrease of their temperature. Thus, freezing is an exothermic process because heat/energy is released from the system to the surroundings.
02

Finding the temperature range for spontaneous freezing

For a process to be considered spontaneous, it must be energetically favorable. In the case of freezing, as it is an exothermic process, the system tends to release heat to the surroundings. The freezing point of 1-propanol is at -127 °C. Therefore, at temperatures below -127 °C, freezing of 1-propanol is a spontaneous process because heat release is favored under these conditions.
03

Finding the temperature range for nonspontaneous freezing

On the other hand, a nonspontaneous process would be one in which the system is absorbing heat or energy from the surroundings. In the case of freezing, this would occur when the temperature is above the freezing point. Thus, for 1-propanol, the nonspontaneous freezing process will occur at temperatures above -127 °C.
04

Existence of an equilibrium temperature

At the freezing point (-127 °C), liquid and solid 1-propanol can coexist in a state of equilibrium. This is because, at this temperature, the rate of freezing (liquid to solid) is equal to the rate of melting (solid to liquid) and thus, both phases are in dynamic equilibrium.

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Key Concepts

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

Phase Change
When a substance transitions from one state of matter to another, such as from liquid to solid or vice versa, it undergoes a phase change. In the context of the given exercise regarding the freezing point of 1-propanol, a phase change occurs when it goes from liquid to solid form when cooled below its freezing point of -127 °C.

Diving deeper into this phase transition, we referred to freezing as an exothermic process, which means it releases heat to the surroundings. The reason for this heat release is the loss of kinetic energy as the particles in the liquid slow down and form a rigid, structured solid. This explains why we feel a release of cold air when we observe substances like water freezing in an ice tray – it's the heat being expelled from the liquid as it becomes a solid.

Understanding both endothermic and exothermic phases changes are essential as they connect to thermal energy transfer and provide a basis for exploring many natural and industrial processes, from the making of ice cream to the crystallization of metals.
Thermodynamics
Thermodynamics is the branch of physics dealing with heat, work, and energy transformation. A central theme in thermodynamics is the concept of spontaneity, which refers to whether a process will occur without external influence. In the exercise, we differentiated between temperatures that result in the spontaneous and nonspontaneous freezing of 1-propanol.

Below its freezing point, the process of 1-propanol turning solid is energetically favorable as the system releases energy. This concurs with the Second Law of Thermodynamics, suggesting that for a spontaneous process, the overall entropy, or disorder, of the universe increases. In the case of freezing, the lowering of system temperature results in a decrease in entropy of the system, but the release of heat into the surroundings increases the universe's overall entropy.

It is important to recognize that temperature plays a crucial role in thermodynamics and in determining the spontaneity of a phase change. In the freezing of substances, the environment's temperature relative to the substance's freezing point is crucial in establishing whether the process will occur spontaneously.
Chemical Equilibrium
Chemical equilibrium refers to the state in which the forward and reverse reactions occur at the same rate, resulting in no net change in the concentrations of reactants and products. In terms of phase changes, we find this dynamic balance at the freezing point of a substance, in this case, 1-propanol.

At -127 °C, 1-propanol exists in both liquid and solid phases in equilibrium. It's dynamic because even though we see no net change, individual molecules are constantly shifting between the liquid and solid states – some liquid molecules are freezing, while an equal number of solid molecules are melting.

Understanding equilibrium is essential for grasping many scientific principles, such as the functioning of saturated solutions or the buffering capacity of blood. It also plays a key role in various applications, including industrial synthesis and the formulation of pharmaceuticals. The concept ensures that we can predict the conditions under which a reaction or phase change can be driven forward or pulled back, which is critical in process control and optimization.

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Most popular questions from this chapter

Predict the sign of the entropy change of the system for each of the following reactions: (a) \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) (b) \(\mathrm{Ba}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) (d) \(\mathrm{FeCl}_{2}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+2 \mathrm{HCl}(g)\)

(a) Using data in Appendix \(C\), estimate the temperature at which the free- energy change for the transformation from \(\mathrm{I}_{2}(s)\) to \(\mathrm{I}_{2}(g)\) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as WebElements (www.webelements.com), to find the experimental melting and boiling points of \(\mathrm{I}_{2}\). (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

How does the entropy of the system change when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes?

Suppose we vaporize a mole of liquid water at \(25^{\circ} \mathrm{C}\) and another mole of water at \(100{ }^{\circ} \mathrm{C}\). (a) Assuming that the enthalpy of vaporization of water does not change much between \(25^{\circ} \mathrm{C}\) and \(100^{\circ} \mathrm{C}\), which process involves the larger change in entropy? (b) Does the entropy change in either process depend on whether we carry out the process reversibly or not? 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?

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