Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Which of the following does not represent enthalpy change during phase transformation? (a) Standard enthalpy of fusion (b) Standard enthalpy of vaporisation (c) Standard enthalpy of sublimation (d) Standard enthalpy of formation

Short Answer

Expert verified
The enthalpy of formation (d) does not represent an enthalpy change during phase transformation.

Step by step solution

01

Understanding the Concepts of Enthalpy Changes

First, we need to understand what each option means. The enthalpy of fusion is the heat change that occurs when a substance goes from solid to liquid. Enthalpy of vaporization is the heat change when a substance goes from liquid to gas. The enthalpy of sublimation is the heat change for a solid going directly to gas. The enthalpy of formation is the change in heat when one mole of a compound is formed from its elements in their standard states.
02

Identifying Phase Transformations

Next, we determine which of the options does not involve a phase transformation. Phase changes involve a transition between solid, liquid, and gas states.
03

Eliminate Options Related to Phase Changes

Options (a), (b), and (c) all describe specific enthalpy changes that occur during transitions between different states of matter, which are phase changes. Therefore, they can be associated with phase transformations.
04

Identifying the Correct Answer

Since the enthalpy of formation describes the heat change when a substance is formed from its constituent elements and does not relate to a phase transition between solid, liquid, and gas, option (d) does not represent an enthalpy change during phase transformation.

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.

Standard Enthalpy of Fusion
Understanding the standard enthalpy of fusion is critical when studying phase transformations. This term refers to the heat absorbed by one mole of a substance as it changes from a solid to a liquid at constant pressure. It is important to note that this process occurs at the melting point of the substance.

For instance, when ice melts to form water, it absorbs heat from the surroundings. This absorption of heat without a change in temperature indicates that the energy is used to break the intermolecular forces that hold the solid structure together. A notable point is that the standard enthalpy of fusion is specific to each substance due to the unique intermolecular forces within different materials. In examinations or real-life applications, this concept helps in calculating the energy required to melt a given amount of substance.
Standard Enthalpy of Vaporisation
Moving on to the standard enthalpy of vaporisation, this is the amount of heat required to convert one mole of a liquid into gas at a constant pressure. This phase change is often observed when a liquid reaches its boiling point. During this process, the energy provided to the system is used to overcome the intermolecular attractions between the liquid molecules, allowing them to spread out and form a gas.

Different liquids have varying enthalpies of vaporisation, reflecting the strength of intermolecular forces within the liquid. For example, water requires more energy per mole to vaporise than ethanol because of hydrogen bonding. This concept is applied in processes such as distillation, where the separation of substances relies on their differing vaporisation enthalpies.
Standard Enthalpy of Sublimation
The standard enthalpy of sublimation is another fascinating concept in phase changes, describing the energy needed to transform one mole of a solid directly into a gas, bypassing the liquid phase. This endothermic process is observed in substances like dry ice (solid carbon dioxide), which sublimates at atmospheric pressure and room temperature.

Sublimation occurs when the molecules in the solid phase have enough energy to overcome not just the forces holding them in a solid structure but also the forces that would typically keep them in a liquid state. As such, the enthalpy of sublimation is approximately equal to the sum of the enthalpy of fusion and the enthalpy of vaporisation. Understanding this enthalpy allows scientists to predict and manipulate conditions under which materials will change their state.
Standard Enthalpy of Formation
The standard enthalpy of formation, in contrast to the previous concepts, is the change in enthalpy when one mole of a compound is formed from its elements in their standard states. It is a crucial concept in thermochemistry, as it helps determine the energy change during the synthesis of a compound.

This enthalpy value can be either positive or negative, depending on whether the formation of the compound absorbs or releases energy. For instance, the formation of water from hydrogen and oxygen gas releases a significant amount of energy, making it exothermic. Standard enthalpies of formation are fundamental in calculating the overall energy changes in chemical reactions using Hess's law. Additionally, these values are often referenced to determine the stability of compounds and predict reaction spontaneity.

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

A reaction attains equilibrium state under standard conditions, then what is incorrect for this? (a) Equilibrium constant \(K=0\) (b) Equilibrium constant \(K=1\) (c) \(\Delta G^{\circ}=0\) and \(\Delta H^{\circ}=T \Delta S^{\circ}\) (d) \(\Delta G=0\) and \(\Delta H=T \Delta S\)

At \(373 \mathrm{~K}\), steam and water are in equilibrium and \(\Delta H=40.98 \mathrm{~kJ} \mathrm{~mol}^{-1}\). What will be \(\Delta S\) for conversion of water into steam? \(\mathrm{H}_{2} \mathrm{O}_{(l)} \rightarrow \mathrm{H}_{2} \mathrm{O}_{(g)}\) (a) \(109.8 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (b) \(31^{-1} \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\) (c) \(21.93 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\) (d) \(326 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\)

The work done during the expansion of a gas from \(4 \mathrm{dm}^{3}\) to \(6 \mathrm{dm}^{3}\) against a constant external pressure of \(3 \mathrm{~atm}\) is \((1 \mathrm{~L} \mathrm{~atm}=101.32 \mathrm{~J})\) (a) \(-6]\) (b) \(-608 J\) (c) \(+304 \mathrm{~J}\) (d) \(-304 \mathrm{~J}\)

What is the entropy change when 1 mole oxygen gas expands isothermally and reversibly from an initial volume of \(10 \mathrm{~L}\) to \(100 \mathrm{~L}\) at \(300 \mathrm{~K} ?\) (a) \(19.14 \mathrm{JK}^{-1}\) (b) \(109.12] \mathrm{K}^{-1}\) (c) \(29.12 \mathrm{JK}^{-1}\) (d) \(10 \mathrm{~K}^{-1}\)

For the reaction: \(\mathrm{H}_{2(g)}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{HCl} ;\) \(\Delta H=-44 \mathrm{kcal}\) What is the enthalpy of decomposition of HCl? (a) \(+44 \mathrm{kcal} / \mathrm{mol}\) (b) - \(44 \mathrm{kcal} / \mathrm{mol}\) (c) \(-22 \mathrm{kcal} / \mathrm{mol}\) (d) \(+22 \mathrm{kcal} / \mathrm{mol}\)

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