ΔΔHvap is usually greater than ΔHfus for a substance, yet the nature of interactions that must b

Name: Problem 39, Chapter 13: Chemistry
Uploaded: 2021-07-13T17:20:10-08:00
Duration: 1 min 42 s
Channel: Lottie Adams
Description: ΔΔHvap is usually greater than ΔHfus for a substance, yet the nature of interactions that must be overcome in the vaporization and fusion processes are similar. Why is ΔHvap greater?

step 1 Gases, liquids and solids are all made up of atoms, molecules and or ions. But the behaviour of

Question

$\Delta \Delta H_{\mathrm{vap}}$ is usually greater than $\Delta H_{\mathrm{fus}}$ for a substance, yet the nature of interactions that must be overcome in the vaporization and fusion processes are similar. Why is $\Delta H_{\text {vap }}$ greater?

    $\Delta \Delta H_{\mathrm{vap}}$ is usually greater than $\Delta H_{\mathrm{fus}}$ for a substance, yet the nature of interactions that must be overcome in the vaporization and fusion processes are similar. Why is $\Delta H_{\text {vap }}$ greater?

Chemistry

Kenneth W. Whitten,… 9th Edition

Chapter 13, Problem 39

Instant Answer

Solved by Expert Lottie Adams

07/13/2021

Step 1

Both vaporization and fusion processes involve breaking intermolecular forces between particles in a substance to change its phase. Show more…

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$\Delta \Delta H_{\mathrm{vap}}$ is usually greater than $\Delta H_{\mathrm{fus}}$ for a substance, yet the nature of interactions that must be overcome in the vaporization and fusion processes are similar. Why is $\Delta H_{\text {vap }}$ greater?

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Key Concepts

Intermolecular Forces

Both fusion and vaporization require energy to overcome intermolecular forces that hold the substance together. However, during vaporization, nearly all attractive forces between particles are overcome because the particles are separated far enough to behave as a gas. In fusion, although some interactions are weakened as a solid turns to a liquid, many of these forces remain present in the liquid state, making the energy requirement lower compared to vaporization.

Expansion Work and p?V Contribution

In vaporization, the process involves not only breaking intermolecular bonds but also doing work against external pressure to expand the substance into a gas. This p?V work—energy required to overcome the ambient pressure during expansion—adds to the total enthalpy change. In contrast, during fusion, the volume change is much less significant, so this additional work contribution is minimal, resulting in a lower enthalpy of fusion.

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