The entropy change for a phase transformation is : (a) \(\frac{\Delta U}{\gamma+d T}\) (b) \(\frac{\Delta T}{\Delta H}\) (c) \(\frac{\Delta H}{T}\) (d) \(\frac{\Delta H+\Delta G}{T}\)

Short Answer

Expert verified
The correct formula for entropy change during a phase transformation is (c) \(\frac{\Delta H}{T}\).

Step by step solution

01

Understanding Entropy Change in Phase Transformations

Entropy change, denoted as \( \Delta S \), during a phase transformation is defined as the heat absorbed or released by the system at a constant temperature divided by that temperature. This quantity can be found using the formula \( \Delta S = \frac{\Delta Q_{rev}}{T}\), where \( \Delta Q_{rev} \) is the reversible heat change and \(T\) is the absolute temperature.
02

Identify Correct Formulas for Heat Change

The heat change (\( \Delta Q_{rev}\)) during a phase transformation at constant pressure is equivalent to the change in enthalpy (\(\Delta H\)). Thus the correct formula for entropy change is \(\Delta S = \frac{\Delta H}{T}\). We need to pick the option that matches this correct formulation.
03

Choosing the Correct Answer

Option (c) \(\frac{\Delta H}{T}\) is the correct formula for entropy change during a phase transformation since it represents the enthalpy change (\(\Delta H\)) divided by the absolute temperature (\(T\)), aligning with the definition and formula for entropy change.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Entropy Change Formula
When we talk about entropy change in the context of phase transformations in physical chemistry, we're referring to the measure of a system's disorder or randomness increasing or decreasing as it changes state - for example, from solid to liquid or liquid to gas. This change is critical to understanding thermodynamic processes.

The entropy change formula, represented as \(\frac{\text{th{4}{\tiny H}}}{T}\) , where \(\text{th{4}{\tiny H}}\) is enthalpy change and T is absolute temperature, provides important insight into how much thermal energy per temperature unit is redistributed within a system or exchanged with the surroundings during a phase change. This formula is derived from the second law of thermodynamics, which tells us that for a reversible process, the entropy change is the heat absorbed or released divided by the temperature.

To effectively harness this understanding, it's crucial to think of entropy as a currency exchanged in every energy transaction. When a substance melts or vaporizes, it's like the molecules are 'paying' entropy to transition into a more disordered state. Keeping track of this 'payment', via the entropy change formula, helps us predict how a system will behave under varying conditions.
Enthalpy Change
Moving on to the concept of enthalpy change, denoted as \(\text{th{4}{\tiny H}}\), it represents the total heat content of a system and is a measure of the energy that is either absorbed or released during a process at constant pressure. In the scenario of phase changes, the enthalpy change is particularly insightful because it signifies the amount of energy needed to change a substance from one phase to another.

For instance, the enthalpy of fusion is the energy required for melting a solid into a liquid, while the enthalpy of vaporization concerns the energy needed to evaporate a liquid into gas. A key point to remember is that these processes are endothermic, meaning they consume heat, reflecting a positive enthalpy change. Conversely, when a gas condenses or a liquid freezes, energy is released, indicating a negative enthalpy change.

  • During melting: \(\text{th{4}{\tiny H}}\) > 0 (endothermic)
  • During freezing: \(\text{th{4}{\tiny H}}\) < 0 (exothermic)

To measure the enthalpy change in a laboratory, calorimetry can be used where the heat exchanged with the surroundings is calculated, providing practical insights into the energetic transformations during phase changes.
Phase Transitions in Physical Chemistry
Phase transitions are fascinating events in physical chemistry that involve a substance changing from one state of matter to another. This could be solid to liquid (melting), liquid to gas (vaporization), or even direct solid to gas (sublimation). Each transition involves a reconfiguration of the substance's molecules and changes in entropy and enthalpy.

These changes are governed by laws of thermodynamics and can be quantified by thermodynamic quantities such as the aforementioned entropy and enthalpy. The Clausius-Clapeyron equation is a key tool in describing the relationship between pressure and temperature during phase transitions, indicating the phase boundary on a phase diagram.

Additionally, each phase transition point has a characteristic energy associated with it—latent heat. This is the energy required to change the phase without changing the temperature. Understanding these concepts and their mathematical descriptions allows chemists and physicists to manipulate conditions to achieve desired phase states, vital in processes ranging from material science to culinary arts.
Reversible Heat Change
The last piece of the puzzle is understanding reversible heat change, which is central to the concept of entropy. For a phase change occurring reversibly, which is an idealized process that proceeds in such a way that the system and surroundings can be returned to their initial states without creating any changes to the universe, the heat change is an exact differential. This means that the path by which the process occurs doesn't affect the overall heat exchange.

In real-world applications, however, truly reversible processes are an abstraction, as actual processes are irreversible, involving dissipative factors like friction and turbulence. Nevertheless, considering processes as reversible provides a useful model to calculate the maximum work that can be extracted or the minimum work that must be input to drive a reaction or phase change, which is pivotal in the design of efficient engines and refrigeration systems.

Moreover, calculating the reversible heat change during a phase transition helps us to obtain an accurate measure of the entropy change. This is because, in reversible processes, the entropy can be quantified as the integral of the heat transfer divided by the temperature through which the transfer occurs, aiding the comprehension of thermodynamic efficiency and spontaneity of processes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One mole of a non-ideal gas undergoes a change of state from \((1.0 \mathrm{~atm}, 3.0 \mathrm{~L}, 200 \mathrm{~K})\) to \((4.0\) atm, \(5.0 \mathrm{~L}, 250 \mathrm{~K}\) ) with a change in internal energy \((\Delta U)=40 \mathrm{~L}\) -atm. The change in enthalpy of the process in L-atm : (a) 43 (b) 57 (c) 42 (d) None of these

\begin{aligned} &\text { Suppose that a reaction has } \Delta H=-40 \mathrm{~kJ} \text { and } \Delta S=-50 \mathrm{~J} / \mathrm{K} \text { . At what temperature range will it } \\ &\text { change from spontaneous to non-spontaneous? } \\ &\begin{array}{llll} \text { (a) } 0.8 \mathrm{~K} \text { to } 1 \mathrm{~K} & \text { (b) } 799 \mathrm{~K} \text { to } 800 \mathrm{~K} & \text { (c) } 800 \mathrm{~K} \text { to } 801 \mathrm{~K} & \text { (d) } 799 \mathrm{~K} \text { to } 801 \mathrm{~K} \end{array} \end{aligned}

At \(25^{\circ} \mathrm{C}, 1\) mole of \(\mathrm{MgSO}_{4}\) was dissolved in water, the heat evolved was found to be \(91.2 \mathrm{~kJ}\). One mole of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) on dissolution gives a solution of the same composition accompanied by an absorption of \(13.8 \mathrm{~kJ}\). The enthalpy of hydration, i.e., \(\Delta H\) for the reaction $$ \mathrm{MgSO}_{4}(s)+7 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}(s) \text { is : } $$ (a) \(-105 \mathrm{~kJ} / \mathrm{mol}\) (b) \(-77.4 \mathrm{~kJ} / \mathrm{mol}\) (c) \(105 \mathrm{~kJ} / \mathrm{mol}\) (d) None of these

What is the change in entropy when \(2.5\) mole of water is heated from \(27^{\circ} \mathrm{C}\) to \(87^{\circ} \mathrm{C} ?\) Assume that the heat capacity is constant. \(\left(C_{p, m}\left(\mathrm{H}_{2} \mathrm{O}\right)=4.2 \mathrm{~J} / \mathrm{g}-\mathrm{K} \ln (1.2)=0.18\right)\) (a) \(16.6 \mathrm{~J} / \mathrm{K}\) (b) \(9 \mathrm{~J} / \mathrm{K}\) (c) \(34.02 \mathrm{~J} / \mathrm{K}\) (d) \(1.89 \mathrm{~J} / \mathrm{K}\)

The heat of formation of \(\mathrm{NH}_{3}(g)\) is \(-46 \mathrm{~kJ} \mathrm{~mol}^{-1}\). The \(\Delta H\) (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of the reaction, \(2 \mathrm{NH}_{3}(g) \longrightarrow \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)\) is : (a) 46 (b) \(-46\) (c) 92 (d) \(-92\)

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free