Question

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

Answer 1

Correct Answer - 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"`