- #1
theincrediblea
- 4
- 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
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