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Standard enthalpy of formation

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    The standard enthalpy of formation for liquid water is -285.8 kJ / mole.

    Which of the following could be the standard enthalpy of formation for water vapor?

    A) -480.7 kJ / mole
    B) -285.8 kJ
    C) -241.8 kJ / mole
    D) +224.5 kJ / mole

    2. Relevant equations



    3. The attempt at a solution

    The correct answer in the solutions manual was C. The logic behind the answer is that the formation of water vapor must be exothermic, and it must be less exothermic than the formation of liquid water.

    Could anyone help me see the reasons behind those last 2 statements? Why is the formation of a gas less exothermic than formation of a liquid? Why must the process be exothermic? The formation of water vapor from liquid water seems like an endothermic process from my perspective, requiring the input of energy.

    Thanks
     
  2. jcsd
  3. Mar 23, 2010 #2

    danago

    User Avatar
    Gold Member

    True, but the question is talking about the enthalpy of formation, which by definition is the enthalpy change associated with the formation of the substance from its constitutent elements. i.e. The enthalpy change of the following reaction:

    [tex]2H_2+O_2\stackrel{}{\rightarrow}2H_2O_{(g)}[/tex]

    You are given that the enthalpy of formation of LIQUID water is -285.8 kJ / mole (i.e. when hydrogen and oxygen combine to form liquid water, -285.8 kJ is released). To form water VAPOR, energy is still released, but not as much because some of the excess energy is required to vaporize the water.
     
  4. Mar 23, 2010 #3
    Thanks for the explanation. It makes sense now.
     
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