Enthalpy of Fusion/Vaporization
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Enthalpy of Fusion/Vaporization
Is it possible for the enthalpy of vaporization to be less than the enthalpy of fusion? In what instances would knowing this be useful?
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Re: Enthalpy of Fusion/Vaporization
No, the enthalpy of vaporization will always be greater than the enthalpy of fusion. Gases have the greatest amount of intermolecular space and have little to no forces of attraction between them, so the heat needed to separate molecules when a liquid boils would be greater than when a solid melts, as the extent to which molecules are separated when a substance goes through fusion is much less than when it goes through vaporization.
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Re: Enthalpy of Fusion/Vaporization
The enthalpy of vaporization is always greater than the enthalpy of fusion. This is because it takes more energy to overcome the attractive forces between the molecules in a liquid and turn them into a gas than it does to overcome the attractive forces between the molecules in a solid and turn them into a liquid. The distance between molecules in gases is much greater than the distance between molecules in liquids, meaning much more energy is needed to separate the molecules to change into a gas than it is to change into a liquid.
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Re: Enthalpy of Fusion/Vaporization
Enthalpy of vaporization is always greater than enthalpy of fusion because vaporization is turning a liquid into a gas and fusion, aka melting, is turning a solid into a liquid. It requires more energy to overcome the intermolecular forces of a solid and transform into a liquid where the molecules just slide past each other. To overcome the intermolecular forces of molecules in liquid state and turn it into a gas, you would need a lot more energy because gas molecules travel fast and are very far apart. This can be useful to know when you're drawing a heating curve, knowing that the phase change line from solid to liquid (vaporization) should be higher than the liquid to gas line (fusion).
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