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For a phase change \(\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \stackrel{{ }^{\circ} \mathrm{C}, 1 \mathrm{bar}}{\longrightarrow} \mathrm{H}_{2} \mathrm{O}(\mathrm{s})\) (a) \(\Delta \mathrm{G}=0\) (b) \(\Delta \mathrm{S}=0\) (c) \(\Delta \mathrm{H}=0\) (d) \(\Delta \mathrm{U}=0\)

Short Answer

Expert verified
Option (a) is correct: 9G = 0. All other options are incorrect.

Step by step solution

01

Identifying Phase Change Condition

The process described is the freezing of liquid water ( ext{H}_2 ext{O(l)}) to ice ( ext{H}_2 ext{O(s)}) at standard temperature and pressure (0^ ext{o} ext{C}, 1 bar). During a phase change, the system is in equilibrium.
02

Analyzing Gibbs Free Energy Change

At equilibrium, the Gibbs free energy change  ext{(9 G)} is zero. This means option (a) is correct:  ext{9G} = 0.
03

Assessing Entropy Change

For a phase change at a constant temperature and pressure, the entropy change  ext{(9 S)} is non-zero. It's calculated using  ext{9 S = 9 H/T}. Thus, option (b) is incorrect:  ext{9S} 2c 0.
04

Evaluating Enthalpy Change

In physical processes like freezing, the enthalpy change  ext{(9 H)} reflects latent heat and is non-zero, as energy is released during the transition from liquid to solid. Option (c) is incorrect:  ext{9H} 2c 0.
05

Understanding Internal Energy Change

Similarly, the internal energy change  ext{(9 U)} is related to enthalpy and non-zero in this context, contrary to some conditions where  ext{9 U} might be approximated as zero. Option (d) is incorrect:  ext{9U} 2c 0.

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Key Concepts

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

Phase Change
A phase change is a fascinating transformation that materials undergo when they change from one state of matter to another, such as from liquid to solid, or gas to liquid. This happens under specific conditions of temperature and pressure. For water turning to ice, this transformation occurs at 0°C and 1 bar pressure. During a phase change, the substance is at equilibrium—a state where the properties do not change over time, and the process can move in either direction without a net change in energy.
For instance, when water freezes, it is in thermal equilibrium between the water (liquid) and ice (solid) at the freezing point. At this point, while the observable properties like temperature remain constant, energy transformations within the material lead to important changes in thermodynamic properties.
Entropy Change
Entropy is a measure of disorder within a system, and during a phase change, entropy plays a crucial role. When a liquid like water freezes to form ice, the molecules become more ordered, leading to a decrease in entropy. However, since entropy also depends on the energy exchange, even though the molecules are more ordered, the entropy change \( \Delta S \) is not zero.
The formula \( \Delta S = \frac{\Delta H}{T} \) helps calculate the change in entropy during the process. Here, \( \Delta H \) represents the change in enthalpy, and \( T \) is the temperature in Kelvin. This equation shows that even as the system becomes more ordered (lowering entropy), there is still a non-zero change due to the energy exchange.
Enthalpy Change
Enthalpy, represented by the symbol \( H \), reflects the total heat content of a system. During a phase change, such as freezing, enthalpy changes because energy is either absorbed or released. In the case of freezing water, energy is released as the molecules arrange into a solid structure, illustrating a negative change in enthalpy \( \Delta H \).
This release of energy is often referred to as latent heat and is a non-zero factor during phase changes. It's important to note that this change is distinct from the energy changes that alter the temperature, as it specifically relates to the phase transition without changing temperature.
Internal Energy Change
The internal energy of a system, denoted as \( U \), comprises all the energy contained within the system's molecules. This includes kinetic and potential energy components. During phase changes, such as freezing, internal energy changes in tandem with enthalpy changes, driven by the same forces of temperature and pressure.
This change in internal energy \( \Delta U \) isn't zero, as the process involves a real exchange of energy. Even though under certain ideal conditions, \( \Delta U \) might be zero, real-world phase transitions like freezing involve energy release or absorption, leading to a measurable \( \Delta U \).
Equilibrium
Equilibrium in thermodynamics refers to a state where the macroscopic properties of the system are balanced and unchanging over time. During a phase change, the system as a whole remains at equilibrium, meaning the forces, energies, and reactions are balanced despite the transformations occurring at a molecular level.
For example, as water freezes, the transformation between the liquid and solid states happens concurrently without a net change in the system's energy. This dynamic equilibrium is what makes phase changes happen at a specific temperature and pressure consistently, and it is essential for calculating thermodynamic properties like entropy and enthalpy changes during these processes.

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Most popular questions from this chapter

A piston filled with \(0.04\) mol of an ideal gas expands reversibly from \(50.0 \mathrm{~mL}\) to \(375 \mathrm{~mL}\) at a constant temperature of \(37.0^{\circ} \mathrm{C}\). As it does so, it absorbs \(208 \mathrm{~J}\) of heat. The values of \(\mathrm{q}\) and \(\mathrm{w}\) for the process will be: \((\mathrm{R}=3.314 \mathrm{~J} / \mathrm{mol} \mathrm{K})(\operatorname{Ln} 7.5=2.01)\) (a) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (b) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=+208 \mathrm{~J}\) (c) \(\mathrm{q}=+208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\) (d) \(\mathrm{q}=-208 \mathrm{~J}, \mathrm{w}=-208 \mathrm{~J}\)

Match the following Column-I (a) \(\Delta \mathrm{G}<0\) (b) \(\Delta \mathrm{S}_{\text {Total }}<0\) (c) \(\Delta \mathrm{S}_{\text {total }}=0\) (d) \(\Delta \mathrm{G}=0\) Column-II (p) spontaneous (q) equili brium (r) \(\Delta \mathrm{H}>\mathrm{T} \Delta \mathrm{S}\) (s) \(\Delta \mathrm{H}<\mathrm{T} \Delta \mathrm{S}\) (t) \(\Delta \mathrm{H}=\Delta \mathrm{E}\)

For a reaction to occur spontaneously (a) \((\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S})\) must be negative (b) \((\Delta \mathrm{H}+\mathrm{T} \Delta \mathrm{S})\) must be negative (c) \(\Delta \mathrm{H}\) must be negative (d) \(\Delta \mathrm{S}\) must be negative

The bond energy of an \(\mathrm{O}-\mathrm{H}\) bond is \(109 \mathrm{kcal} \mathrm{mol}^{-1}\). When \(5 \times 10^{-3}\) mole of water is formed, the energy released in kcals is approximately

For the reaction, \(\mathrm{A}(\mathrm{g})+2 \mathrm{~B}(\mathrm{~g}) \longrightarrow 2 \mathrm{C}(\mathrm{g})+3 \mathrm{D}(\mathrm{g})\) The value of \(\Delta \mathrm{H}\) at \(27^{\circ} \mathrm{C}\) is \(19.0 \mathrm{kcal}\). The value of \(\Delta \mathrm{E}\) for the reaction would be (given \(\mathrm{R}=2.0 \mathrm{cal} \mathrm{K}^{-1} \mathrm{~mol}^{-1}\) ) (a) \(20.8 \mathrm{kcal}\) (b) \(19.8 \mathrm{kcal}\) (c) \(18.8 \mathrm{kcal}\) (d) \(17.8 \mathrm{kcal}\)

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