Question

Enthalpy of neutralization of `HCl` by `NaOH` is `-57.1 J//mol` and by `NH_(4)OH` is `-51.1 KJ//mol`. Calculate the enthalpy of dissociation of `NH_(4)OH`.

Answer 1

Given that, `H^(+)(aq)+NH_(4)OH(aq)rarrNH_(4)^(+)(aq)+H_(2)O(l)" "Delta H= -51.1 KJ//"mole"`
We may consider neutralization in two steps.
(i) Ionization `NH_(4)OH(aq)rarrNH_(4)^(+)(aq)+OH^(-)(aq)" "Delta H_(1)=?`
(ii) Neutralization `H^(+)(aq)+OH^(-)(aq)rarrH_(2)O(l)" "Delta H_(2)= -57.1KJ//"mole"`
Thus, `Delta H=Delta H_(1)+Delta H_(2)`
Therefore, `Delta H_(1)=Delta H- Dleta H_(2)= -51.1 KJ//mol +57.1 KJ mol^(-1)= 6.0 KJ//mol`