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Enthalpy of neutralization of HCl by NaO...

Enthalpy of neutralization of HCl by NaOH is `-55.84` kJ/mol and by `NH_(4)OH` is `51.34` kJ/mol. The enthalpy of ionization of `NH_(4)OH` is :

A

107.18 kJ/mol

B

4.5 kJ/mol

C

`-4.5` kJ/mol

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
B
`DeltaH_("neut")`.
`=DeltaH_("ionization")+Delta_(r )H(H^(+)+OH^(-)rarrH_(2)O)`
`implies" "-51.34 = x - 55.84`
So, `" "DeltaH_("ionization") " or "x=4.5 " kJ"//"mol"`
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