The enthalpies of neutralization of NaOH and NH₄OH by HCI are -57.3 kJ mol^-1 and -51.3 kJ mol⁺ respectively. The values of enthalpy of ionization ofNH₄OH is about

$\begin{array}{l}{\mathrm{H}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right)⟶{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\Delta }}_1{H}_{\mathrm{l}}^{\circ }=-57.3\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+}\left(\mathrm{aq}\right)⟶{\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\Delta }}_{\mathrm{r}}{H}_2^{\circ }=-51.3\mathrm{kJ}{\mathrm{mol}}^{-1}\\ {\mathrm{NH}}_4\mathrm{OH}⟶{\mathrm{NH}}_4^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right);\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{\Delta }}_1{H}^{\circ }=?\\ \text{ Obviously }\mathrm{\Delta }{H}^{\circ }=\mathrm{\Delta }{H}_2^{\circ }-\mathrm{\Delta }{H}_{\mathrm{l}}^{\circ }=6.0\mathrm{kJ}{\mathrm{mol}}^{-1}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\end{array}$