Enthalpy of neutralisation

  1. why enthalpy of neutralisation of HF is greater than 68KJ.

    MY attempt

    Enthalpy of neutralisation= Enthalpy of Ionisation + Δ(H+ + OH-)

    Now,For very strong acid, enthalpy of Ionisation = 0,

    Hence enthalpy of neutralisation= (H+ + OH-)= -13.7KCal

    for weak acid, enthalpy of ionisation is always > 0

    ∴, enthalpy of neutralisation should always be less than 13.7 (57 KJ).

    Please explain
  2. jcsd
  3. HF + OH- = F- + H2O

    Enthalpy of this reaction = Enthalpy of neutralization of HF = Enthalpy of products - Enthalpy of reactants
    which comes out to be greater than 68kJ. So, here evidently the enthalpy of of ionization of HF is EXOTHERMIC.

    Hope this helps! :)
  4. Borek

    Staff: Mentor

    That's not true.

    Just because the acid dissociated long ago and the solution temperature got in equilibrium with the surroundings, doesn't mean enthalpy of ionization was zero.
  5. thanks adithyan for ur help
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