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NaOH(s) -> NaOH(g)

  1. Dec 23, 2008 #1
    NaOH(s) --> NaOH(g)

    [tex]\Delta H^{\circ}_{diss}[/tex] (O-O) = 251 kJ
    [tex]\Delta H^{\circ}_{diss}[/tex] (O-H) = 465 kJ
    [tex]\Delta H^{\circ}_{diss}[/tex] (H-H) = 435 kJ
    [tex]\Delta H^{\circ}_{diss}[/tex] (Na-O) = 255 kJ

    [tex]\Delta H_{sol}[/tex] = -46 KJ for [tex]NaOH_{(s)}[/tex]

    [tex]NaOH_{(s)} \rightarrow Na^{+}_{(aq)} + OH^{-}_{(aq)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot[/tex][tex]\Delta H[/tex] = -46 KJ

    [tex]\Delta H^{\circ}_{f}[/tex] = -427 KJ for [tex]NaOH_{(s)}[/tex]

    [tex]Na_{(s)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(s)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot[/tex][tex]\Delta H[/tex] = -427 KJ

    I forgot one part:
    [tex]Na_{(s)} \rightarrow Na_{(g)} \cdot\cdot\cdot\cdot\cdot \Delta H[/tex] = 109 kJ

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


    Because im trying to find the enthalpy of one reaction, and i have other 'known' enthalpy reactions does this problem automatically turn into a Hess' Law type problem ? or is this still a bond energy/enthalpy reaction or just bond energy type problem ? Also i do not have a [tex]NaOH_{(g)}[/tex] in any of my reaction so how can i go about computing the targeted enthalpy ?
     
    Last edited: Dec 24, 2008
  2. jcsd
  3. Dec 24, 2008 #2
    Re: NaOH(s)-->NaOH(g)

    So does anyone know how to tackle this problem ?
     
  4. Dec 31, 2008 #3
    Re: NaOH(s)-->NaOH(g)

    Okay here is what i tried doing, I know there might be something wrong and such but please please please i beg you indicate my mistake this is my last question regarding Reaction heats. Next up Equilibrium lol .... anyways.. *ahem*....

    [tex]Na_{(s)} \rightarrow Na_{(g)} \cdot\cdot\cdot\cdot\cdot \Delta H[/tex]=109 KJ
    This means:

    [tex]Na_{(g)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(s)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot \Delta H^{\circ}_{f}[/tex] = -268 KJ

    My question regarding the above question is when i convert Na(s)-->Na(g) does my product NaOH(s) become ---> NaOH(g) so that my target equation ends up NaOH(g) as a product ? or does it still remain a solid and my calculation or method is wrong and should be discarded because a better method is out there....... so please help. I did another part it seems to make sense to change NaOH(s) ---> NaOH(g) because im using the heat of dissociation and do those values mean to dissociate a given compound to its gaseous atoms or what ??. I know that those values are the same for any compound so does that also mean its the same values for the same compound but in a different state like with NaOH(s) and NaOH(g) ?????

    I converted NaOH(s) into NaOH(g) to show my method but still i need a better explanation so i can understand and not just from theorizing "making things happen" which i dont like to do :( I also know this is sublimation lol thats my last point. I got -268 KJ for the second equation because im using bond energies or is this still not the correct method ? please help me.

    [tex] NaOH_{(s)} \rightarrow Na_{(s)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \Delta H = 427 KJ[/tex]
    [tex]Na_{(g)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(g)} \Delta H = -268 KJ[/tex]
    ==========================================
    [tex]NaOH_{(s)} \rightarrow NaOH_{(g)} \Delta H = 159 KJ[/tex]

    You can do the cancellation yourself: Na, O2, and H2 get cancelled. This answer is correct on the back of my paper but can anyone suggest a better method ? of if this is somewhat the correct method then can anyone please clean up my explanation to make things clear ? Thanks a bunch. Now i go sleep ..... ZZZzz
     
  5. Jan 1, 2009 #4
    Re: NaOH(s)-->NaOH(g)

    CAN somebody help me with this ?
     
  6. Jan 4, 2009 #5
    Re: NaOH(s)-->NaOH(g)

    Helooooooooooooooooooooooo ?????????????? Is there anyone in this forum ??????????
     
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