Heat liberated in the neutralization of 500ml of 1N HCl and 500ml of 

${\mathrm{NH}}_4\mathrm{OH} \mathrm{is} -1.36 \mathrm{K}.\mathrm{cals}$. the heat of ionization of NH₄OH is

$\therefore \text{ gm eq. }\cdot \mathrm{HCl}={\text{ gmeq }}_{{\mathrm{NH}}_4\mathrm{OH}}=1\times \frac{500}{1000}=0.5$

$Now, Heat liberated in the neutralisation of 0.5 gram equivalent of HCl and$${\mathrm{NH}}_4\mathrm{OH}=-1.36\mathrm{kcal}$

${\mathrm{NH}}_4\mathrm{OH}=\frac{-1.36}{0.5}=-2.72\mathrm{kcal}$

${\mathrm{H}}_2\mathrm{O}+{\mathrm{NH}}_4\mathrm{OH}+{\mathrm{H}}^{+}\to {\mathrm{H}}^{+}+{\mathrm{OH}}^{-}+{\mathrm{NH}}_4^{-}+{\mathrm{H}}_2\mathrm{O}\mathrm{\Delta H}=13.7+(-2.72)$