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Calculating NaOH dissociation entalphy problem

  1. Aug 28, 2014 #1
    NaOH dissociation is exothermic so ΔH<0

    But when I have a reaction
    NaOH(s) => Na+(aq) + OH-(aq)

    and count ΔrH using standard entalphy change of formation:
    ΔfH (NaOH) = -425 kj/mol
    ΔfH (OH-(aq)) = -230 kj/mol
    ΔfH (Na+(aq)) = +242 kj/mol
    Data from Atkins phisical chemistry basics

    ΔrH= ΔfH (products) - ΔfH (substrates) = (-230 + 240,12) - (-425) = +435,12 kj/mol
    From this calculation it seems that reaction is pretty endothermic and that's surely not true.

    Can someone tell me where am I making mistake ?
     
  2. jcsd
  3. Aug 29, 2014 #2

    Borek

    User Avatar

    Staff: Mentor

    And not -242 kJ/mol?

    http://chemistry.about.com/od/chartstables/a/heatoformions.htm

    Generally speaking hydration of small ions is quite exothermic, so this positive value looks suspicious.

    BTW: you count fingers, but you calculate amount. Number of fingers - liczba palców, amount of water - ilość wody, palce są przeliczalne, ilość jest nieprzeliczalna.
     
  4. Aug 29, 2014 #3
    Thanks for so fast response.

    ΔfH (Na+(aq)) value in my book was given without sign (probably author just forgot about it) and I assumed that it must be positive since losing electron is usually positive (didn't take into consideration that is't (aq) ion) but now everything makes much more sense.

    Thank you
     
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