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6.3: Question 6.E.15 PASS - predict melting point

  • Page ID
    452264
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    Exercise \(\PageIndex{6.E.15}\)

    The melting point of H2O(s) is 0 °C. Would you expect the melting point of H2S(s) to be −85 °C, 0 °C, or 185 °C? Explain your answer.

    Answer

    −85 °C

    Water has hydrogen bonds which are a stronger intermolecular force. H2S must therefore melt at a lower temperature since it has diploe-dipole intermolecular attractions which are weaker than hydrogen bonds.

    See LibreText 6.1 Intermolecular Forces

    Strategy Map
    Step Hint
    1. Identify the difference between the two molecules. Recall Lewis Structures of both molecules (see LibreText section 4.4)
    2. Identify what intermolecular forces each molecule has.

    They do not have the same intermolecular forces.

    Solution

    Answer: −85 °C.

    Water has hydrogen bonds between the molecules which are a stronger intermolecular force, so it takes more energy to break some intermolecular attractions and change from a solid to a gas, so we expect H2S to melt at a lower temperature than H2O.

    Guided Solution

    Download Guided Solution as a pdf

    Guided Solution Hint
    This is a theory type problem where you must identify the intermolecular forces of the given molecules and decide if the melting point of one will be lower, the same, or higher than the other. See LibreText 6.1 Intermolecular Forces
    The melting point of H2O(s) is 0 °C. Would you expect the melting point of H2S(s) to be −85 °C, 0 °C, or 185 °C? Explain your answer.  
    Hydrogen bonding will occur when a hydrogen atom is bonded to what other atoms? Recall that hydrogen bonding occurs when a hydrogen atom is bonded to Fluorine, Oxygen, and Nitrogen.
    Since the molecules have different atoms you can expect that the molecules do not have the same melting point. The molecule with the stronger intermolecular forces will have the higher melting point.

    Complete Solution:

    Identify the intermolecular attractions in each molecule

    Although both H2O and H2S have hydrogen atoms, Sulfur is not one of the very electronegative atoms, therefore H2S does not contain hydrogen bonds, it contains dipole-dipole attractions.

    Compare the strength of the intermolecular attractions

    Dipole-Dipole attractions are not as strong as hydrogen bonds meaning the melting point of H2S must be lower than the melting point of H2O.

    Answer the question

    The melting point of H2S is expected to be −85 °C.

    Some intermolecular attractions must be broken to melt a substance in order to change it form a solid to a liquid, so molecules with stronger intermolecular attractions will melt a a higher temperature, molecules with weaker intermolecular attractions will melt at a lower temperature. Since H2S has dipole-dipole intermolecular attractions which are weaker than the stronger hydrogen bonds that H2O contains, H2S must have a lower melting point than H2O.

    Students often see a hydrogen atom and think of a hydrogen bond, but they must remember that hydrogen bonds only occur between hydrogen and specific atoms.

    Check your work!

    We expect our H2S melting point to be less than that for water, and −85 °C is less than the 0 oC melting point of water.

    Why does this answer make chemical sense?

    The two molecules do not have the same intermolecular forces. Hydrogen bonding is caused by very strong electronegativity difference between hydrogen and one of the highly electronegative atoms. The strong difference between the two causes the hydrogen atom to have a very strong partial positive charge. Although the bond between hydrogen and sulfur is strong, it is only a dipole-dipole attraction, meaning the forces between two H2O molecules is stronger than the forces between two H2S molecules. Since H2S has a weaker force, its melting point will require less energy and thus will be lower than H2O.

    (question source from page titled Liquids and Solids (Exercises) https://chem.libretexts.org/Bookshelves/General_Chemistry/Chemistry_1e_(OpenSTAX)/10%3A_Liquids_and_Solids/10.E%3A_Liquids_and_Solids_(Exercises), shared under a CC BY 4.0 license, authored, remixed, and/or curated by OpenStax, original source https://openstax.org/books/chemistry-2e/pages/10-exercises​​​​​​​, Access for free at https://openstax.org/books/chemistry/pages/1-introduction)


    6.3: Question 6.E.15 PASS - predict melting point is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.