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Enthalpy of NaOH(s) -> NaOH(g)

  1. Mar 7, 2013 #1
    Enthalpy of NaOH(s) --> NaOH(g)

    1. The problem statement, all variables and given/known data

    1. [itex]Na_{(s)} \rightarrow Na_{(g)}[/itex] [itex]\Delta H = 109 kJ[/itex]

    2. [itex]\Delta H^{\circ}_{diss} = 251\:kJ\:for\:O_{2}[/itex]

    3. [itex]\Delta H^{\circ}_{diss} = 435\:kJ\:for\:H_{2}[/itex]

    4. [itex]\Delta H^{\circ}_{diss} = 465\:kJ\:for\:O-H[/itex]

    5. [itex]\Delta H^{\circ}_{diss} = 255\:kJ\:for\:Na-O[/itex]

    6. [itex]\Delta H_{soln} = -46\:kJ\:for\:NaOH_{(s)}[/itex]

    7. [itex]\Delta H^{\circ}_{f} = -427\:kJ\:for\:NaOH_{(s)}[/itex]

    Predict [itex]\Delta H[/itex] for [itex]NaOH_{(s)} \rightarrow NaOH_{(g)}[/itex] ???

    2. Relevant equations
    see above

    3. The attempt at a solution
    I understand the standard states of all substances for #1-3 but I am unsure of #4,5,7. I don't understand what the standard states are for #4,5 (both reactants and products). I don't understand what the physical state of sodium is for #7 (is it solid or gas ?). Also how will the aqueous ions produced from dissociation of sodium hydroxide in #6 cancel out ? and how will covalent bonds of [itex]Na-O[/itex] or [itex]O-H[/itex] as reactants cancel out ?

    Thanks!! :)
  2. jcsd
  3. Mar 8, 2013 #2


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    Staff: Mentor

    You can assume bond dissociation enthalpy doesn't depend on the state. In #7 you start with elements in the standard state and end with NaOH in the standard state.
  4. Mar 8, 2013 #3
    How will I cancel aqueous hydroxide ion or sodium cation ? So the standard state of elemental sodium in #7 is solid ? Also, after O--H bond or Na--O bond dissociates do the Na or O or H assume a gaseous state ?
  5. Mar 29, 2013 #4
    Any new thoughts from my previous reply ?
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