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Discussion on the enthalpy of hydrogen fluoride

2025-06-04View Original

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Today, I noticed that the enthalpy value of hydrogen fluoride given in the Matheson Gas Data Handbook is lower than that of liquid hydrogen fluoride at the same temperature. Could anyone explain why this is the case?
Reply #22025-06-04
The phenomenon in which the enthalpy of liquid HF is higher than that of gaseous HF is mainly related to the special intermolecular forces (hydrogen bonds) between HF molecules. When ordinary compounds evaporate from a liquid to a gas, heat is generally absorbed, causing the enthalpy of the gas to be higher than that of the liquid at the same temperature. However, there are extremely strong hydrogen bonds between HF molecules; these hydrogen bonds cause the liquid HF molecules to bind tightly together, releasing significant energy and thereby reducing the total energy of the system. Therefore, when liquid HF vaporizes into individually dispersed gaseous HF molecules, it is necessary to overcome strong hydrogen bonding forces, which requires a significant amount of energy. This results in the enthalpy of gaseous HF (especially the standard enthalpy) calculated under certain selected reference states and benchmarks being lower than that of liquid HF. Simply put, the strong hydrogen bond interactions between HF liquid molecules significantly reduce the energy within the system, resulting in the calculated enthalpy of liquid HF being higher than that of the gas state under the reference conditions used in the handbook. It is a special but reasonable phenomenon. .
Reply #32025-06-13
Brother on the second floor, hydrogen fluoride also has hydrogen bonds in its liquid state, just like liquid water.

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