1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar

    Staff: Mentor

    And not -242 kJ/mol?


    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted