Why is the Enthalpy of Neutralisation for HF Greater than 68KJ?

[theincrediblea]

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

 

[Adithyan]

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! :)
-Adithyan

 

[Borek]

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

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.

 

[theincrediblea]

thanks adithyan for ur help