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Enthalpy of neutralisation

by theincrediblea
Tags: enthalpy, neutralisation
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theincrediblea
#1
May24-14, 03:25 AM
P: 4
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
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Adithyan
#2
May24-14, 12:27 PM
Adithyan's Avatar
P: 106
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
#3
May24-14, 12:59 PM
Admin
Borek's Avatar
P: 23,599
Quote Quote by theincrediblea View Post
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
#4
May25-14, 07:06 AM
P: 4
Enthalpy of neutralisation

thanks adithyan for ur help


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